CH APTER 1 BUSINESS LOGISTICS/SUPPLY CHAIN VITAL SUBJECT A
12
(a) This problem introduces the student to the evaluation of alternate channels of production and distribution. To know whether domestic do mestic or foreign production is least expensive, the total of production and distribution costs must be computed from the source point to the marketplace. Two alternatives are suggested and they can be compared as follows. Production at Houston: Total cost = Production cost at Houston + Transportation and storage costs = $8/shirt×100,000 shirts + $5/cwt. ×1,000 cwt. = $805,000/year Production at Taiwan: Total cost = Production cost in Taiwan + Transportation and storage costs from Taiwan to Chicago Chica go + Import duty + Raw material transportation cost from Houston to Taiwan = $4/shirt×100,000 shirts + $6/cwt. ×1,000 cwt. + $0.5/shirt×100,000 shirts + $2/cwt. ×1,000 cwt. = $458,000/year Producing in Taiwan would appear to be the least expensive. (b) Other factors to consider before a final decision is made might be: (i) How reliable would international transportation be compared with domestic transportation? (ii) What is the business climate in Taiwan such that costs costs might change in favor of Houston as a production point? (iii) How likely is it that the needed transportation and storage will b e available? (iv) If the market were to expand, would there be adequate production capacity available to support the increased demand?
1
CH APTER 2 LOGISTICS/SUPPLY CHAIN STRATEGY AND PLANNING 13
The purpose of this exercise is to allow the student, in an elementary way, to examine the tradeoffs between transportation and inventory-related costs when an incentive transportation rate is offered. Whether the incentive rate should be implemented depends on the shipment size corresponding to the minimum of the sum of transportation, inventory, and order processing costs. These costs are are determined for various shipping quantities that might be selected to cover the range of shipment sizes implied in the problem. Table 2-1 gives a summary of the costs to Monarch for various shipment sizes. From Monarch's point of view, view, the incentive rate would be beneficial. Shipment sizes should be approximately doubled so that the 40,000 lb. minimum minimum is achieved. It is is important to note that the individual cost elements are not necessarily at a minimum at low shipment sizes, whereas order-processing costs are low at high shipment sizes. They are in cost conflict with each other. other. Transportation costs are low at high high shipment sizes, but exact costs depend on the minimum volume for which the rate is quoted. In preparation for a broader planning perspective to be considered later in the text, the student might be asked what the place of the supplier is in in this decision. How does he affect the decision, and how is he affected by it? This will focus the student's attention attention on the broader issues of the physical distribution channel.
2
TABLE 2-1 Company
Evaluation of Alternative Shipment Sizes for the Monarch Electric
57 motors or 10,000 lb. 9×8,750 = $78,750 0.25×200×57/2 = $1,425a
Current 114 motors or 20,000 lb. 5×8,750 = $43,750 0.25×200×114/2 = $2,850
171 motors or 30,000 lb. 5×8,750 = $43,750 0.25×200×171/2 = $4,275
Proposed 228 motors 285 motors or or 40,000 lb. 50,000 lb. 3×8,750 3×8,750 = $26,250a = $26,250 0.25×200×228/2 0.25×200×285/2 = $5,700 = $7,125
Type of cost Transportation R× D Inventory carrying b I ×C ×Q/2 c Order processing 5,000×15/57 5,000×15/114 5,000×15/171 5,000×15/228 = $1,316 = $658 = $439 = $329 D×S/Q Handling 0.30×8,750 0.30×8,750 0.30×8,750 0.30×8,750 = $2,625 = $2,625 = $2,625 = $2,625 H × D $34,904a Total $84,116 $49,883 $51,089 a Minimum values. b Students should be informed that average inventory can be approximated by one half the shipment size. c Demand D has been converted to units per year. LEGEND R = transportation rate, $/cwt. D = annual demand, cwt. I = = inventory carrying cost, %/year. C = = cost of a motor, $/motor. Q = shipment size in motors, where Q/2 represents the average number of motors maintained in inventory. = order processing costs, $/order. S = = handling costs, $/cwt. H =
3
5,000×15/285 = $263a 0.30×8,750 = $2,625 $36,263
CH APTER 3 THE LOGISTICS/SUPPLY CHAIN PRODUCT 3
The 80-20 principle applies to sales and items where 80 percent of the dollar volume is generated from 20 percent of the product items. items. While this ratio rarely holds exactly in practice, the concept does. We can apply it to these data by ranking the products by sales, and the percentage that the cumulative sales represent of the total. The following table shows the calculations. Cumulative sales as % of total
Cumulative items as % of total
Product code
Dollar sales
Cumulative sales
08776 12121
$71,000 63,000
$ 71,000 134,000
18.2 34.3
8.3 16.7
10732 11693 10614 12077 07071 10542 06692 09721 14217 11007 Total
56,000 51,000 46,000 27,000 22,000 18,000 14,000 10,000 9,000 4,000 $391,000
190,000 241,000 287,000 314,000 336,000 336,000 354,000 368,000 378,000 391,000
48.6 61.6 73.4 80.3 85.9 90.5 94.1 96.7 98.9 100.0
25.0 33.3 41.7 50.0 58.3 66.7 75.0 83.3 91.7 100.0
The 80-20 rule cannot be applied exactly, since the cumulative percent of items does not break at precisely 20 percent. However, we might decide that only products 08776 and 12121 should be ordered directly from vendors. The important principle derived from the 80-20 rule is that not every item is of equal importance to the firm, and that different channels of distribution distribution can be used to handle them. The 80-20 rule gives some rational basis for deciding which products should be shipped directly from vendors and which are more economically handled through a system of warehouses. 6
(a) Reading the ground transport rates for the appropriate zone as determined by zip code and the weight of 27 lb. (rounding upward of 26.5 lb.) gives the following total cost table for the four shipments. To zip code 11101 42117 74001 59615 a
Catalog price $99.95 99.95 99.95 99.95
UPS zone 2 5 6 8
Transport a cost $ 8.39 10.46 13.17 18.29
Use 27 lb.
4
Total cost $108.34 110.41 113.12 118.24
(b) The transport rate structure is reasonably fair, since ground rates generally follow distance and size of shipment. These are the factors most directly affecting transport costs. They are not fair in the sense that that customers within a zone are all charged the same rate, regardless of their distance from from the shipment origin point. However, all customers may benefit from lower overall rates due to this simplified zone-rate structure. 10
(a) This is a delivered pricing scheme where the seller includes the transport charges in the product price. price. The seller makes the transport arrangements. (b) The seller prices the product at the origin, but prepays any freight charges; however, the buyer owns the goods in transit. (c) This is a delivered pricing scheme where the freight charges are included in the product price, however the freight charges are then deducted from the invoice, and the seller owns the goods in transit. (d) The seller initially pays the freight charges, but they are then collected from the buyer by adding them to the invoice. The buyer owns o wns the goods in transit, since the pricing is f.o.b. origin. (e) The price is f.o.b. origin. The buyer pays the freight charges and owns the goods in transit. Regardless of the price price policy, the customer will ultimately ultimately pay all costs. If a firm firm does not consider outbound freight charges, the design of the distribution system will be different than if it does. Since pricing policy is an arbitrary arbitrary decision, it can be argued that transport charges should be considered in decision making, whether the supplying firm directly incurs them or not. 11
This shows how Pareto's law (80-20 principle) is useful in estimating inventory levels when a portion of the product line is to to be held in inventory. An empirical function that approximates the 80-20 curve is used to estimate the level of sales for each product to be held in inventory. According to Equation 3-2, the constant A is determined as follows: A =
X (1 − Y ) Y
−
X
=
0.25(1−.75 75) 0.75 − 0.25
=
0125 .
The 80-20 type curve according to Equation 3-1 is: Y =
(1 + A) A + X
=
(1 + 01 0.125) X 0125 . + X
This formula can be used to estimate the cumulative sales from the cumulative item proportion. For example, item 1 is 0.05 of the total number of items (20) so that:
5
Y =
(1 + 0.125)( 0.05) 0.125 + 0.05 05
=
0321 .
Of the $2,600,000 in total annual warehouse sales, item 1 should account for 0.321×2,600,000 = $835,714. By applying this formula to all items, the following inventory investment table can be developed which shows sales by item. The average inventory investment investment by item item is found by dividing the turnover ratio into the item item sales. The sum of the average inventory value for each item gives a total projected inventory of $380,000. Inventory Investment Table Product 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Cumulative item pro portion, X 0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.55 0.60 0.65 0.70 0.75 0.80 0.85 0.90 0.95 1.00
B
C
Cumulative sales, Y $ 835,714 1,300,000 1,595,454 1,800,000 1,950,000 2,064,705 2,155,263 2,228,571 2,289,130 2,340,000 2,383,333 2,420,689 2,453,226 2,481,818 2,507,142 2,529,719 2,550,000 2,568,293 2,584,884 2,600,000
Projected item sales $ 835,714 464,286 295,454 204,546 150,000 114,706 90,558 73,308 60,559 50,870 43,333 37,356 32,537 28,592 25,324 22,587 20,271 18,293 16,591 15,116
Turnover ratio 8 8 8 8 6 6 6 6 6 6 4 4 4 4 4 4 4 4 4 4 Total
Average inventory value $104,464 58,036 36,932 25,568 25,000 19,118 15,093 12,218 10,093 8,478 10,833 9,339 8,134 7,148 6,331 5,647 5,068 4,473 4,148 3,779 $380,000
12
This problem involves the application of Equations 3-1 and 3-2. We can develop an 8020 curve based on 30 percent of the items items accounting for 70 percent of sales. That is, A =
X (1 − Y ) Y
−
X
=
0.30(1 − 0.70) 0.70 − 0.30
=
0225 .
Therefore, the sales estimating equation is:
6
(1 + 0.225) X 0225 . + X
Y =
By applying this estimating curve, we can find the sales of A and B items. For example, 20 percent of the items, or 0.2×20 = 4 items, will be A items with a cumulative proportion of sales of: Y A
=
(1 + 0.225)( 0.20) 0.22 225 + 0.20 20
=
05765 .
and 3,000,000×0.5765 = 1,729,412. The A+ B item proportion will be: Y A
= +B
(1 + 0.225)( 0.50) 0.22 225 + 0.50
=
08448 .
and 3,000,000×0.8448 = 2,534,400. The product product group group B sales will be A+ B sales less A sales, or 2,534,400 − 1,729,412 = $804,988. The product group C will be the remaining sales, but these are not of particular interest in this problem. The average inventories for A and B products are found by dividing the estimated sales by the turnover ratio. That is, A: B:
1,729,412/9 804,988/5 Total inventory
= 192,157 = 160,988 353,155 cases
The total cubic footage required for this inventory would be 353,155×1.5 = 529,732 cu. ft. The total square footage for products A and B is divided by the stacking height. That is, 529,731/16 = 33,108 sq. ft. 13
This problem is an application of Equations 3-1 and 3-2. We first determine the constant A. That is, A =
X (1 − Y ) Y
−
X
=
0.20(1 − 0.65) 0.65 − 0.20
=
0156 .
and 0.75 =
(1 + 0.156) X 0156 . + X
Solving algebraically for X , we have:
7
X =
AxY
1 + A − Y
=
0.156 x 0.75 75 1 + 0.156 − 0.75
=
0288 .
That is, about 29 percent of the items (0.288×5,000 = 1,440 items) produce 75 percent of the sales. 14
The price would be the sum of all costs plus an increment for profit to place the automotive component in the hands of the customer. This would be 25+10+5+8+5+transportation cost, or 53+T . Based on the varying transportation cost, the following price schedule can be developed. Quantity 1 to 1,000 units 1,001 to 2,000 units >2,000 units
Price per unit 53+5=$58 53+4.00=57 53+3.00=56
Discount 0 1.7%a 3.5%
a
[(58 - 57)/58][100]=1.7%
8
CH APTER 4 LOGISTICS/SUPPLY CHAIN CUSTOMER SERVICE 6
(a) This company is fortunate to be able to estimate the sales level that can be achieved at various levels of distribution service. Because of this, this, the company should seek to maximize the difference between sales and costs. These differences are summarized as follows: Percent of orders delivered within one day Contribution to 50 60 70 80 90 95 100 profit -1.8 2.0 3.5 4.0 3.4 2.8 -2.0
The company should strive to make deliveries within one day, 80 percent of the time, for a maximum contribution to profit. (b) If a competing company sets its delivery time so that more than 80 percent of the orders are delivered in one day and all other factors that attract customers are the same, the company will lose customers to its competitor, as the sales curve will have shifted downward. Cleanco should adjust its service service level once again to to the point where the profit contribution is maximized. maximized. Of course, there is no guarantee that the previous level of profits p rofits can be achieved a chieved unless the costs of supplying the service can correspondingly be reduced. 7
(a)
The first task is to determine the increase in sales that can be attributed to the change in the service policy. To determine if there is is a significant significant change in the control group, we set up the following hypothesis test: z =
X 2 2 2
s
N 2
− +
X 1 2 1
s
N 1
=
224 − 1 85 2
2
61 79 + 102 102
=
39 36.48 48 + 61.18
=
3.94
Now, referring to the normal distribution table in Appendix A of the text, there is a significant difference at the 0.01 level in the sales associated with the control group. That is, some factors other than the service policy polic y alone are causing sales to increase. inc rease. Next, we analyze the test group in the same manner.
9
z =
2,295 − 1, 34 342 5762 56
3352 + 56
=
953 5,92 924 + 2,00 0 04
= 10.7
This change is also significant at the 0.01 level. The average increase in sales for the control group is 224/185 = 1.21, or 21 percent. The average sales increase in the test group is 2295/1342 = 1.71, or 71 percent. If we believe that 21percent of the 71percent increase in the test group is due to factors other than service policy, then 71 − 21 = 50 percent was the true service effect. Therefore, for for each sales unit, unit, an incremental increase in profit of (0.40×95)(0.50) = $19 can be realized. realized. Since the cost of the service improvement improvement is $2, the benefit exceeds the cost. The service improvement should be continued.
(b) The use of the before-after-with-control-group experimental design is a methodology that has been used for for some time, especially in marketing research studies. The outstanding feature of the design is that the use of the control group helps to isolate the effect of the single service service variable. On the other hand, there are a number of potential problems with the methodology: • The
sales distributions may not be normal. • The time that it takes for diffusing the information that a service change has taken place may distort the results. • The products in the control group may not be mutually exclusive from those in the test group. • The method only shows the effect of a single step change in service and does not develop a sales-service relationship. • It may not always be practical to introduce service changes into on-going operations to test the effect. 8
(a) The optimum service level is set at that point where the change in gross profit equals the change in cost. The change in gross profit: = ∆ P =
Trading margin × Sales response rate × Annual sales = 1.00×0.0015×100,000 = $150 per year per 1 percent change in the service level
The change in cost: ∆C =
Annual carrying cost × Standard product cost × ∆ z
10
× Demand
standard deviation for order cycle = 0.30×10.00×400×∆z Now, set ∆ P = = ∆C and and solve for ∆ z . 150 = 1200×∆z = 0.125 ∆ z = From the tabulated changes in service level with those changes in z , the service level should be set between 96-97 percent. (b) The weakest link in this analysis is estimating the effect that a change in service will have on revenue. This implies that a sales-service sales-service relationship is known. 9
The methodology is essentially the same as that in question 7, except that we are asked to find X instead instead of Y . That is, ∆ P
= 0.75×0.0015×80,000 = 90
and ∆C = =
0.25×1,000×500×∆z = 1250×∆z
Then, ∆ P = ∆C
90 = 1250×∆z ∆z = 0.072 From the normal distribution (see Appendix A), the z for for an area under the curve of 93 percent is 1.48, and for 92 percent, z is 1.41. Since the difference of 1.48 − 1.41 = 0.07, we can conclude that the the in-stock probability should be set at 92-93 percent. Of course, the change in z is found by taking the difference in z values for one percent differences in the area values under the normal distribution curve for a wide range of area percentages. 10
Apply Taguchi’s concept of the loss function. First, estimate the loss per item if the the target level of service is is not met. We know the profit per item as follows.
11
Sales price Cost of item Other costs Profit per item
$5.95 -4.25 -0.30 $1.40
Since one-half of the sales are lost, the opportunity loss per item would be: Profit per item
Sales lost
$1.40 × (1/2)(880) 880 Current sales
Opportunity loss =
=
$0.70 /item
Next, find k in in the loss function: L = k ( y − m) 2
out-of-stock % at point where ½ sales are lost
0.70 = k (10 − 5) 2 Target %
0.70 = k ( 25) k = 0.03
Finally, the point where the marginal supply cost equals the marginal sales loss is: ( y − 5) = y
B
2k
=
= 1.67 + 5 =
0.10 2(0.03)
= 1.67%
6.67%
The retailer should not allow the out-of-stock percentage to deviate more than 1.67 percent, and should not allow the out-of-stock level to fall below 1.67 + 5 = 6.67 percent.
12
CH APTER 7 TRANSPORT DECISIONS 1
Selecting a mode of transportation requires balancing the direct cost of transportation with the indirect costs of both vendor and buyer inventories plus the in-transit inventory costs. The differences in transport transport mode performance performance affect these inventory levels, and, therefore, the costs for maintaining them, as well as affect the time that the goods are in transit. We wish to compare these four four cost factors for each mode choice as shown in Table 7-1 of the manual. The symbols used are: R = transportation rate, $/unit D = annual demand, units = item value at buyer's inventory, $ C = C' = item value at vendor's inventory, $ T = time in transit, days Q = Shipping quantity, units
Rail has the lowest total cost. TABLE 7-1
An Evaluation of of the Transport Alternatives for the Wagner Company
Cost type Transport
Method R ×D
Rail 25×50,000 = $1,250,000 In-transit 0.25×475×50,000 inventorya I×C’×D×t/365 ×(16/365) = $260,274 Wager’s 0.25×475×(10,000/2) inventorya = $593,750 I×C’×Q/2 Electronic’s 0.25×500×(10,00/2) inventory = $625,000 I×C×Q/2 Total $2,729,024 a C’ refers to price less transport cost per unit.
Piggyback 44×50,000 = $2,200,000 0.25×456×50,000 ×(10/365) = $156,164 0.25×456×(7,000/2) = $399,000 0.25×500×(7,000/2) = $437,500 $3,192,664
Truck 88×50,000 = $4,400,000 0.25×412×50,000 ×(4/365) = $564,384 0.25×412×(5,000/2) = $257,500 0.25×500×(5,000/2) = $312,500 $5,534,384
2
As in question 1, this problem is one of balancing transport costs with the indirect costs associated with inventories. inventories. However, in this case we must account for the variability in transit time time as it affects the the warehouse inventories. We can develop the following decision table.
17
Service type Cost type
Method
Transport
R× D
In-transit inventory Plant inventory
/365 I ×C × D×t /365
Warehouse inventory Total
I ×C’ ×Q*/2 + I ×C ’×r
I ×C ×Q*/2
A
12×9,600 = $115,200 0.20×50×9,600 ×(4/365) = $1,052 0.30×50×(321.8/2) = $2,684 0.30×62×(321.3/2) + 0.30×62×50.5 = $3,927 $122,863
B 11.80×9,600 =$114,048 0.20×50×9,600 ×(5/365) = $1,315 0.30×50×(357.8/2) = $2,684 0.30×61.80×(321.8/2) + 0.30×61.80×60.6 = $4,107 $122,154
Recall that Q* = 2 DS / IC 1 00) / 0.3( 50) = 357.8 cwt. for the plant, assuming IC = 2(9,600)(10 the order cost is the same at plant plant and warehouse. However, for the warehouse, we must account for safety stock (r ) and for the transportation cost in the value of the product. Therefore, For A: Q*
=
2 DS / IC IC = 2(9, 600)(100) / 0.3( 62)
=
321. 3 cwt.
and for z = = 1.28 for an area under the normal no rmal distribution of 0.90, the safety stock is: r = zs LT (d ) = 1.28 × 1.5 × ( 9,600 / 365) = 50.5 cwt.
For B: Q*
=
2(9,60 6 00)(100) / 0.3( 61.80)
=
321. 8 cwt.
and r = 1.28 × 1.8 × ( 9,600 / 365) = 60.6 cwt.
Service B appears to be slightly less expensive. 3
The shortest route route method can be applied to this problem. The computational table is shown in Table Table 7-2. The shortest route is defined by tracing the links from from the destination node. They are shown in Table 7-2 as A D F G for a total distance of 980 miles.
18
TABLE 7-2
Tabulation of Computational Steps for the Shortest Route Method Applied to Transcontinental Trucking Company Problem
Solved nodes directly Its closest connected to connected unsolved unsolved Step nodes node 1 A B A D 2 A D B C 3 B C D C D F 4 C E C F D F 5 C F E G D F 6 E G F G a Asterisk indicates the shortest route
Total time involved 186 mi. 276 276 186+110= 296 186+110= 296 276+ 58= 334 276+300= 576 296+241= 537 296+350= 646 276+300= 576 296+350= 646 537+479=1016 276+300= 576 537+479=1016 576+404= 980
nth nearest node B
Its minimum time 186 mi.
Its last connection a AB
D
276
AD*
C
296
BC
E
537
CE
F
576
DF*
G
980
FG*
4
In this actual problem, the U.S. Army used the transportation method of linear programming to solve its allocation problem. The problem can be set up in matrix form as follows: Origin Destination
Letterkenny Fort Hood Fort Riley
Cleveland 150 150 325 50 275 100 375
South Charleston 100 150 350
San Jose 800
Demand
300 300 50
325
350 100
400
275
Fort Carson
100
300
100
250
100
450
Fort Benning
100
0
Supply
400
150
100 150
The cell values shown in bold represent the number of personnel carriers to be moved between origin and a nd destination points p oints for minimum transportation costs of $153,750. An alternative solution at the same cost would be:
19
Origin Cleveland S. Charleston Cleveland San Jose Cleveland San Jose Cleveland
Destination Letterkenny Letterkenny Fort Hood Fort Hood Fort Riley Fort Carson Fort Benning
Number of carriers 150 150 50 50 100 100 100
5
The general approach is to first find the route in ROUTER without regard to the rectilinear distances of the road network. Because this may produce an infeasible solution, specific travel distance are added to the database to represent actual distances traveled or to block infeasible paths from occurring. A reasonable routing plan is shown shown in Figure 7-1 and the ROUTER database that generates it is given in in Figure 7-2. The total distance for the route is 9.05 miles and at a speed of 20 miles per hour, the route time is approximately 30 minutes.
20
FIGURE 7-1 ROUTER Solution to the School Bus Routing Exercise 0.5
1.0
1.5
2.0
0 19
17
20
21
0
0.5 22
15
8 18
5 1
9
1.0 2
3
6
4
7
10
11
12
13 14
1.5 Denotes pickup point Depot location
FIGURE 7-2 Input Data for ROUTER for School Bus Routing Problem —PARAMETERS AND LABELS— Problem label – School Bus Routing Exercise Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NW DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate – 0.14 Vertical coordinate coordinate – 0.45 Earliest starting time (min) - 0 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 20 After how many clock hours will overtime begin - 9999 GENERAL DATA Percent of vehicle in use before allo wing pickups - 0 Horizontal scaling factor - 1 Vertical scaling factor - 1 Maximum TIME allowed on a route (hours) - 9999 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 0 To begin after - 9999
21
16
Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after - 9999 --STOP DATA— STOP LOAD VOL. LOAD NO DESCRIPTION TY WGHT CUBE HCRD VCRD ZN TIME BEG1 END1 1 Stop 1 D 1 0 0. 0.14 0.80 0 0 0 9999 2 Stop 2 D 1 0 0. 0.14 1.14 0 0 0 9999 3 Stop 3 D 1 0 0. 0.14 1.31 0 0 0 9999 4 Stop 4 D 1 0 0. 0.35 1.31 0 0 0 9999 5 Stop 5/22 D 1 0 0 .5 .52 0. 0.61 0 0 0 9999 6 Stop 6 D 1 0 0. 0.58 1.31 0 0 0 9999 7 Stop 7 D 1 0 0. 0.80 1.31 0 0 0 9999 8 Stop 8 D 1 0 1. 1.03 0.61 0 0 0 9999 9 Stop 9 D 1 0 1. 1.03 0.96 0 0 0 9999 10 Stop 10 D 1 0 1. 1.03 1.31 0 0 0 9999 11 Stop 11 D 1 0 1. 1.36 1.31 0 0 0 9999 12 Stop 12 D 1 0 1. 1.48 1.31 0 0 0 9999 13 Stop 13 D 1 0 1. 1.80 1.31 0 0 0 9999 14 Stop 14 D 1 0 1. 1.87 1.31 0 0 0 9999 15 Stop 15 D 1 0 1. 1.84 0.61 0 0 0 9999 16 Stop 16 D 1 0 1. 1.95 0.61 0 0 0 9999 17 Stop 17 D 1 0 1. 1.29 0.10 0 0 0 9999 18 Stop 18 D 1 0 1. 1.26 0.61 0 0 0 9999 19 Stop 19 D 1 0 1. 1.15 0.10 0 0 0 9999 20 Stop 20 D 1 0 0. 0.69 0.23 0 0 0 9999 21 Stop top 21 D 1 0 0. 0.14 0.2 0.26 0 0 0 999 9999
BEG2 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 9999 999
END2 9 99 9 9 99 9 9 99 9 9 99 9 9 9 99 9 99 9 9 99 9 9 99 9 9 99 9 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 99 999 9999
—VEHICLE DATA— --CAPACITY--
NO. 1
VEHICLE DESCRIPTION Bus
TP 1
NO 1
WGHT 9999
--VEHICLE-FIXED COST 0
CUBE 9999
PER MI COST 0
--DRIVER-FIXED COST 0
—SPECIFIED STOP-TO-STOP DISTANCES—
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
STOP 14 14 15 16 18 19 19 19 19 9 9 9 5/22 5/22 5/22 5/22 20
STOP Stop 14 Stop 14 Stop 15 Stop 16 Stop 18 Stop 19 Stop 19 Stop 19 Stop 19 Stop 9 Stop 9 Stop 9 Stops 5&22 Stops 5&22 Stops 5&22 Stops 5&22 Stop 20
STOP 16 15 17 17 9 8 20 5/22 18 20 19 21 1 21 20 9 21
STOP Stop 16 Stop 15 Stop 17 Stop 17 Stop 9 Stop 8 Stop 20 Stops5&22 Stop 18 Stop 20 Stop 19 Stop 21 Stop 1 Stop 21 Stop 20 Stop 9 Stop 21
DISTANCE 0.78 0.90 1.06 1.18 0.58 0.76 0.59 1.14 0.53 1.08 1.11 1.69 0.56 1.05 1.14 0.97 0.84
22
PER HR COST 0
OVER TIME COST 0
17 18 19 20 21 22
20 20 20 17 0 2
Stop 20 Stop 20 Stop 20 Stop 17 School Stop 2
21 0 5/22 0 5/22 5/22
Stop 21 School Stops 5&22 School Stops 5&22 Stops 5&22
0.84 1.03 0.55 2.43 1.37 1.03
6
Strategy 1 is to stay at motel M 2 and serve the two routes routes on separate days. Using the ROUTESEQ module in LOGWARE gives us the sequence of stops and the coordinate distance. The routes routes originating at M 2 would be: a
Route Stop sequence 1 8,6,1,4,2,3,5,7,9 2 10,13,14,17,18,16,12,15,11 a
Distance 95.55 mi. 86.45 182.00 mi.
Includes map scaling factor
The total cost of this strategy would be: Motel 3 nights @ 49.00 Travel 182 miles @ $.30/mi. Total
$147.00 54.60 $201.60
Strategy 2 is a mixed strategy involving staying at motels closest to the center of the stop clusters. The route sequences from from different motels are: Route Stop sequence Distance 1 4,2,3,5,7,9,8,6,2 98.50 mi. 2 18,17,13,14,10,11,15,12,16 80.30 178.80 mi.
The total cost of this strategy is: Motel M 1 1st night $ 40.00 nd 40.00 M 1 2 night rd 45.00 M 1 3 night a Travel 214.80 mi. @ 0.30/mi. 64.44 Total $189.44 a
178.80 + 36 = 214.80
Strategy 2 appears to be most economical. 7
(a) Since distances are asymmetrical, we cannot use the geographically based traveling salesman method in in LOGWARE. Rather, we use a similar module in STORM that that allows such asymmetrical matrices, or the problem is small enough to be solved by inspection. For this problem, the minimal minimal cost stop sequence sequence would be: 23
Bakery!Stop 5!Stop 3!Stop 4!Stop 2!Stop 1!Bakery with a tour time of 130 minutes. (b) Loading/unloading Loading/unloading times may be added to the travel times to a stop. The problem may then be solved as in part a. (c) The travel times between stop 3 and all other nodes are increased increased by 50 percent. The remaining times are left unchanged. Optimizing on this matrix shows no change in the stop sequence. sequence. However, the tour time increases to 147.50 minutes. 8
This may be solved by using the ROUTER module in LOGWARE. The screen set up for this is as follows:
24
FIGURE 7-3 Input Data for ROUTER for Sima Donuts --PARAMETERS AND LABELS— Problem label - Sima Donuts Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - NE DEPOT DATA Depot description - Atlanta Located in zone - 0 Horizontal coordinate - 2084 Vertical coordinate - 7260 Earliest starting time (min) - 180 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 45 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allo wing pickups - 0 Horizontal scaling factor - 0.363 Vertical scaling factor - 0.363 Maximum TIME allowed on a route (hours) - 40 Maximum DISTANCE allowed on a route (miles) - 1400 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 60 To begin after - 1200 Duration of 3rd break (minutes) - 60 To begin after - 2160 Duration of 4th break (minutes) - 60 To begin after - 2640 --STOP DATA— NO 1 2 3 4 5 6 7 8 9 10 11
STOP DESCRIPTION Tampa FL Clearwater FL Daytona Beach F Ft Lauderdale FL N Miami FL Oakland Park FL Orlando FL St Petersburg FL Tallahassee FL W Palm Beach F Puerto Rico
TY D P D D D P D P D D D
LOAD WGHT 20 14 18 3 5 4 3 3 3 3 4
VOL. CUBE 0 0 0 0 0 0 0 0 0 0 0
HCRD 1147 1206 1052 557 527 565 1031 1159 1716 607 527
VCRD ZN 8197 0 8203 0 7791 0 8282 0 8341 0 8273 0 7954 0 8224 0 7877 0 8166 0 8351 0
LOAD TIME 15 45 45 45 45 45 45 45 15 45 45
BEG1 360 360 360 180 360 180 180 180 600 360 360
END1 BEG2 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800 1440 1800
END2 2880 2880 2880 2880 2880 2880 2880 2880 2880 2880 2880
--VEHICLE DATA— -CAPACITY--
NO. 1 2 3
VEHICLE DESCRIPTION Truck #1-20 Truck #2-25 Truck #3-30
TP 1 2 3
NO 3 1 1
WGHT 20 25 30
CUBE 9999 9999 9999
--VEHICLE-FIXED COST 0 0 0
Making a run with ROUTER will give the route design.
25
PER MI COST 1.30 1.30 1.30
--DRIVER-FIXED COST 0 0 0
PER HR COST 0 0 0
OVER TIME COST 0 0 0
Pickup
FIGURE 7-4 Graphical Display of Route Design for Sima Donuts
The route design involves 3 routes for a total distance of 3,830 miles, a cost of $4,978.71, and a total time of 100.4 hours. The route details are as follows: Route #1 with 20-pallet truck Depot Start time − 3:00 a.m. of day 1 Daytona Beach Deliver 18 pallets Clearwater Pickup 14 pallets Depot Return time − 5:48 a.m. of day 2 Route #2 with 20-pallet truck Depot Start time − 3:00 a.m. of day 1 Orlando Deliver 3 pallets W Palm Beach Deliver 3 pallets Ft Lauderdale Deliver 3 pallets N Miami Deliver 5 pallets Miami-Puerto R. Deliver 4 pallets Depot Return time − 4:43 p.m. of day 2
26
Pickup
Route #3 with 30-pallet truck Depot Start time − 4:13 a.m. of day 1 Tallahassee Deliver 3 pallets Tampa Deliver 20 pallets St Petersburg Pickup 3 pallets Oakland Park Pickup 4 pallets Depot Return time − 4:03 p.m. of day 2 9
Given sailing times and dates when deliveries are to be made, loadings need to be accomplished no later than the following dates: To: A B C D From: 1 16 40 1 2 69 25 5
The problem can be expressed as a transportation transportation problem of linear linear programming. There will be 6 initial states [(1,1), (2,5), (1,16), (2,25), (1,40), and (2,69)] and 6 terminal states [( D ,15), ( A ,52), and ( A D,10), (C ,15), A,36), ( B B,39), (C ,52), A,86)]. The linear program is structured as shown in Figure 7-4. Using a transportation solution method, we determine one of the optimum solutions. There are several. The solution is read by starting with the slack on initial loading state 1. This tells us to next select the cell of terminal state 1. In turn, this defines defines initial state 3 and hence, terminal state 3. And so it goes until we reach the terminal state slack column. This procedure is repeated until all initial state slacks are exhausted. Our solution shows two routings. The first is is (1,1)!( D D,10)! (1,16)!( A A,36)!(2,69)!( A A,86). The second is (2,5)!(C ,15) ,15)!(2,25)!( B ,52). Two ships ships are needed. B,39)!(1,40)!(C ,52).
27
FIGURE 7-5 Transportation Matrix Setup and Solution for the Queens Lines Tanker Scheduling Problem
1 1
Load date Discharge date
2 5
100
XXa
100 C 15
100
1
A 36
100
2 69 Rim re striction
1
100
100
C 52 A 86
10
1
1
10
1
XX
10
XX 10
10
1
10
100
10
1
XX 100
XX
1 1
100
100
1
1
1
XX
XX 10
1
XX 100
100
1
1
100
XX
XX
1
XX
XX 100
100
10
XX 100
XX 100
100
XX 100
10
1
XX 100
1
1 1
100
XX 100
1
1
XX
XX
B 39
1
XX
XX 100
a
1 40
Slack
D 10
Slack Rim re striction
Loading points and dates 1 2 16 25
XX 10
XX 10
1
1
1
1
1
1
4
6
10
1
1
1
1
6
XX inadmissible cells given a high high cost
10
This is a problem of freight consolidation brought about by holding orders so they can be shipped with orders orders from subsequent periods. The penalty associated with holding the orders is a lost sales cost. (i) Orders shipped as received Weight × Rate a Hays 10,000 × 0.0519 Manhattan 14,000 × 0.0519 Salina 13,000 × 0.0408 Great Bend 10,000 × 0.0498 Transportation Lost sales Total a
= = = = =
Ship 8,000 lb. as if 10,000 lb.
Average period cost is $2,274.00
28
Cost $519.00 726.00 530.00 498.00 $2,274.00 .00 $2,274.00
(ii) Consolidate first period orders with second period orders = Weight × Rate Cost Hays 16,000 × 0.0519 = $830.40 a Manhattan 40,000 × 0.0222 = 888.00 Salina 26,000 × 0.0342 = 889.20 Great Bend 10,000 × 0.0498 = 498.00 Transportation $3,105.60 Lost sales 1,050.00 Total $4,155.60 a
Ship 28,000 lb. as if 40,000 lb.
The lost sales cost is 1,000 cases × $1.05 = $1,050.00 to hold one group of orders for 2 weeks. Average cost per period is $4,155.60/2 = $2,077.80. (iii) Hold all orders until the third period. Weight × Rate Hays 24,000 × 0.0426 Manhattan 42,000 × 0.0222 a Salina 40,000 × 0.0246 Great Bend 15,000 × 0.0498 Transportation Lost sales Total
= = = = =
Cost $1,022.40 932.40 984.00 747.00 $3,685.80 3,150.00 $6,835.80
a
Ship as if 40,000 lb.
Lost sales Hold 1st period orders for 2 periods 1,000×1.05.2 = $ 2,100 Hold 2nd period sales for 1 period 1,000×1.05 = 1,050 $ 3,150 Average period cost is $6,835.80/3 = $2,278.60 Summary Ship immediately $ 2,274.00 Hold orders 1 period 2,077.80 Hold orders 2 periods 2,278.60
Optimum
11
Routes are built by placing the trips end-to-end throughout the day from 4 a.m. until 11 p.m., respecting the times that a warehouse can receive a shipment. This is a 19-hour block of time per day, or there are 95 hours per week per truck in which a truck may operate. If there were no delivery time restrictions on warehouses and trips could be 29
placed end-to-end for a truck without any slack at the end of the day, the absolute minimum number of trucks can be found multiplying the number of trips by the route time, and then dividing the total by the 95 hours allowed per week. That is,
Warehouse location Flint Alpena Saginaw Lansing Mt. Pleasant W. Branch Pontiac Traverse City Petoskey
(1) Number of trips 43 5 8 21 12 5 43 6 5
(2) Total time per trip, hr. 1.25 10.50 2.25 3.75 5.50 6.00 2.75 10.50 11.75 Total
(3)=(1)(2) Total time, hr. 53.75 52.50 18.00 78.75 66.00 30.00 118.25 63.00 58.75 539.00
For 539 trip hours, 539/95 = 5.67 rounded to six trucks trucks needed per week. Now, it is necessary to adjust for the the problem constraints. constraints. A good schedule can be found by following a few few simple rules that can be developed by examining the data. First, begin the day with a trip where the driving time to a warehouse is just long enough for the truck to arrive at the warehouse just after it opens. One-half the driving time should exceed 6:30 − 4:00 = 2:30, or 2½ hr. Trips to Alpena, Traverse City, and Petoskey qualify. Second, use the short trips at the end of the day to avoid slack time. Third, allocate the trips to the days using the the longest ones first. Make sure that the total trip time for for a day does not exceed 19 hours. For a minimum of six trucks, the following feasible schedule can be developed by inspection. Truck 1
Truck 2
Truck 3
Truck 4
Truck 5
Truck 6
Day 1 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 Lansing 3.75 3 Flint 3.75 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr.
Day 2 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 3 Pontiac 8.25 2 Flint 2.50 Total =16.25 hr.
Day 3 Petoskey 11.75 W Branch 6.00 Flint 1.25 Total =19.00 T. City 10.50 2 Lansing 7.50 Total =18.00 Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00 Flint 1.25 Total =17.75 M Pleasant 5.50 6 Saginaw 13.50
Day 4 Petoskey 11.75 5 Flint 6.25
Day 5 Petoskey 11.75 5 Flint 6.25
Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 hr. M Pleasant 5.50 4 Pontiac 11.00
Total =18.00 hr. T. City 10.50 2 Lansing 7.50 Total =18.00 hr. Alpena 10.50 2 Lansing 7.50 Total =18.00 hr. M Pleasant 5.50 4 Pontiac 11.00 2 Flint 2.50 Total =19.00 M Pleasant 5.50 4 Pontiac 11.00
Total =16.50 hr. W Branch 6.00 2 Saginaw 4.50
Total = 16.50 W Branch 6.00 10 Flint 12.50
Total =19.00 hr.
Total =10.50 hr.
Total =18.50 hr.
30
Although this schedule meets the requirements of the problem, it might be improved by better balancing the workload across the trucks and the days. 12
(a) A sweep method solution is shown on the following figure. Five trucks are needed with a total route distance of (30+29+39+44+19.5)×10 = 1,615 miles.
20
2
18 Route #1 Load 19
Route #5 Load 9
3
1
4
3
16
3 2 5
4 3
3
14
4
Miles 12 x 10 10
3
4
2 3 Warehouse
8
5
5 4
6
6
Route #2 Load 20
4
Route #4 Load `8
5
Route #3 Load 17
3
2
2 4
0 0
2
4
6
8
10 12 14 Miles x 10
16
18
20
22 22
24 24
26
(b) The sweep method is a fast and relatively simple method for finding a solution to rather complex vehicle routing problems. Solutions can be found graphically without the aid of a computer. However, there are some limitations. Namely, •
• •
The method is heuristic heuristic and has an average error of about 10 to 15 percent. This error is likely to be low if the problem contains many points and the weight of each point is small relative to the capacity of the vehicle. The method does not handle h andle timing issues well, such as time windows. Too many trucks may be used in the route design.
13
This problem may be solved solved with the aid of ROUTER in LOGWARE. The model input data may be formatted as shown in Figure 7-6.
31
(a) The solution from ROUTER shows that four routes are needed with a minimum total distance of 492 miles. The route design is shown graphically in Figure 7-7. A summary for these routes is given in following partial output report. Route Route no 1 2 3 4 Total
Run Stop Brk Stem time, time, time, time, time, hr hr hr hr hr 1.2 1.0 .3 .0 .4 8.9 6.6 1.3 1.0 1.1 6.2 3.7 1.4 1.0 .9 7.5 5.1 1.5 1.0 1.4 23.8 16.4 4.4 3.0 3.8
Start time 08:59AM 08:32AM 08:42AM 08:30AM
Return No of Route time stops dist,Mi 10:12AM 3 29 05:25PM 19 199 02:54PM 14 112 04:02PM 12 152 48 492
Route cost,$ .00 .00 .00 .00 .00
(b) Note that route #1 is short and that a driver and a station wagon would be used for a route that takes 1.2 hours to complete. By “attaching” route #1 to route #3, the same driver and station wagon may be used, and the constraints of the problems are still met. The refilled station station wagon can leave the depot by 3:30-3:45 p.m. and still still meet the customers’ time windows and return to to the depot by 6 p.m.. Thus, only three three drivers and station wagons are actually needed for this problem.
FIGURE 7-6 Input Data for ROUTER for Medic Drugs --PARAMETERS AND LABELS— Problem label - Medic Drugs Grid corner with 0,0 coordinates (NW, SW, SE, or NE) - SW DEPOT DATA Depot description - Pharmacy Located in zone - 0 Horizontal coordinate - 13.7 Vertical coordinate - 21.2 Earliest starting time (min) - 480 Latest return time (min) - 9999 Default vehicle speed (miles per hour) - 30 After how many clock hours will overtime begin - 168 GENERAL DATA Percent of vehicle in use before allo wing pickups - 0 Horizontal scaling factor - 4.6 Vertical scaling factor - 4.6 Maximum TIME allowed on a route (hours) - 168 Maximum DISTANCE allowed on a route (miles) - 9999 LOAD/UNLOAD TIME FORMULA Fixed time per stop - 0 Variable time per stop by weight - 0 By cube - 0 BREAK TIMES Duration of 1st break (minutes) - 60 To begin after - 720 Duration of 2nd break (minutes) - 0 To begin after - 9999 Duration of 3rd break (minutes) - 0 To begin after - 9999 Duration of 4th break (minutes) - 0 To begin after – 9999
32
--STOP DATA— NO 1 2 3 46 47 48
STOP DESCRIPTION Covington House Cuyahoga Falls Elyria Westbay Westhaven Broadfiels Mnr
TY D D D D D D
LOAD WGHT 1 9 1 6 2 6
VOL. CUBE 0 0 0 0 0 0
HCRD VCRD ZN 23.40 12.90 0 13.40 13.40 0 6.30 16.80 0 8.40 18.00 0 8.50 18.10 0 18.20 22.90 0
LOAD TIME 2 18 5 10 5 2
BEG1 540 540 540 630 540 540
END1 BEG2 1020 9999 1020 9999 1020 9999 690 9999 1020 9999 1020 9999
END2 9999 9999 9999 9999 9999 9999
--VEHICLE DATA— -CAPACITY--
NO. 1
VEHICLE DESCRIPTION Station wagon
TP 1
NO 50
WGHT 63
CUBE 9999
--VEHICLE-FIXED COST 0
--DRIVER--
PER MI COST 0
FIXED COST 0
PER HR COST 0
OVER TIME COST 0
FIGURE 7-7 Graphical Solution for Medic Drugs
14
There is no exact answer answer to this problem nor is one intended. Several approaches might be taken to this problem. We could apply the savings method or the sweep method to solve the routing problem for each day of the week, given the current demand patterns. However, we can see that there is much overlap in the locations of the customers by delivery day of the week. We might encourage orders orders to be placed so that deliveries form tight clusters by working with the sales department and the customers. customers. Perhaps some incentives could be provided to help discipline discipline the order patterns. The orders should form a general pattern as shown below. Currently, the volume for Thursday exceeds the available truck capacity of 45 caskets. Maybe the farthest stops could be handled by a for-hire service rather than acquiring acq uiring another truck for such little usage.
33
Monday
Tuesday
Friday
Depot
Thursday
Wednesday
It appears that the truck capacity is about right, given that some slack capacity is likely to be needed. Once the pattern orders are established, either as currently given or as may be revised, apply principles numbers 1, 3, 4, 5, and 7.
34
CH APTER 8 FORECASTING SUPPLY CHAIN REQUIREMENTS 4
(a) The answer to this question is aided by b y using the FORECAST module in LOGWARE. A sample calculation calculation is shown as carried out by FORECAST. The results are then summarized from FORECAST output. An example calculation for an α = 0.1 is shown. Other α values would be used, ranging from 0.01 to 1.0. We first calculate a starting forecast by averaging the first four weekly requirements. That is, [2,056 + 2,349 + 1,895 + 1,514]/4 = 1,953.50 Now, we back cast this value and start the forecast at time 0. Thus, the forecasts and the associated errors would be:
F1 F2 F3 F4 F5 F6 F7 F8 F9 F10 F11
= = = = = = = = = = =
.1(2056) .1(2349) .1(1895) .1(1514) .1(1194) .1(2268) .1(2653) .1(2039) .1(2399) .1(2508)
+ + + + + + + + + +
.9(1953.50) .9(1963.75) .9(2002.20) .9(1991.48) .9(1943.73) .9(1868.76) .9(1908.00) .9(1982.50) .9(1988.15) .9(2029.24)
Forecast Error 1953.50 = 1963.75 = 2002.28 = 1991.48 = 1943.73 = 1868.76 -749.73 = 1908.00 399.24 = 1982.50 745.00 = 1988.15 56.50 = 2029.24 410.85 = 2077.12 478.76 Total squared error
Squared error
562,095.07 159,392.58 555,025.00 3,192.25 168,797.72 229,211.14 1,677,713.76
The standard error of the forecast is: S F =
Total sq squared er error = N
1,6 77 77, 713.76 76 . = 52879 6
Note: FORECAST does not use N-1 in the denominator.
Repeating this type of analysis, analysis, the following table can be developed. The results from FORECAST are shown.
73
α .01 .05 .1 .2 .5 1.0
S F 528.72 528.42 528.46 528.89 535.55 566.07
The α that minimizes S F is 0.05. (b) The forecast errors are computed in part a. (c) If we assume that the errors are normally distributed around the forecast, we can then construct a 95 percent confidence band on the forecast. That is, if Y is the actual volume in period 11, then the range of the forecast ( F 11 11 = 2,017.81 for α = 0.05) will be: = F 11 Y = 11 + z × S F = 2,017.81 + 1.96×528.42 Then, 982.11 ≤ Y ≤ 3,053.51 All values are in thousands. 5
(a) & (b) The solution to this problem was was aided by the use of the exponential smoothing module in FORECAST. Using the first first four week's data to initialize the the level/trend version of the exponential smoothing model and setting α and β equal to 0.2, the forecast for next week is F 11 11 = 2,024.47, with a standard error of the forecast of S F = 171.28. (c) Assuming that the forecast errors are normally distributed around F 1111, a 95 percent statistical confidence band can be constructed. The confidence band is: = F 1111 + z × S F Y = = 2,024.47 + 1.96×171.28 where z = = 1.96 for 2.5 percent of the area under the two tails of a normal distribution. The range of the actual weekly volume is expected to be: 1,688.76 ≤ Y ≤ 2,360.18
74
6
(a) The data may be restated as shown below.
Sales, S 27,000 70,000 41,000 13,000 30,000 73,000 48,000 15,000 34,000 82,000 51,000 16,000 500,000
t 1 2 3 4 5 6 7 8 9 10 11 12 78
S ×t 27,000 140,000 123,000 52,000 150,000 438,000 336,000 120,000 306,000 820,000 561,000 192,000 3,265,000
2
t
1 4 9 16 25 36 49 64 81 100 121 144 650
Trend value,a S t t 41,087 41,192 41,298 41,403 41,508 41,613 41,719 41,824 41,929 42,035 42,140 42,245
Seasonal index b 0.66 1.70 0.99 0.32 0.72 1.75 1.15 0.36 0.81 1.95 1.21 0.38
a
Computed from the linear trend line. For example, for period 1, S 1 = 40,981.6 + 105.3×1 = 41,087. b The ratio of the actual sales S to the trend line value S t t. For example, for period 1, the seasonal index is 27,000/41,087 = 0.66.
Given the values from the above table and that t = 78/12 = 6.5, N = = 12, and S = 500,000/12 = 41,666, the coefficients in the regression trend line would be: b=
S ×t−N ×S ×t t2 −
2
=
265,00 000 − 12 × 41, 666 × 6.5 3, 26
× t
650 − 12 × 6.52
. = 1053
and 666 − 105.3 × 6.5 = 40, 98 981.6 a = S − b × t = 41, 66 Therefore, the trend value S t t for any period t would be: S t t = 40,981.6 + 105.3×t
(b) The seasonal factors are determined by the ratio of the actual sales in a period to the trend value for that period. For example, the seasonal factor for period 12 (4th quarter of last year) year) would be 16,000/42,245 = 0.38. This and the seasonal factors factors for all past quarters are shown in the previous table.
75
(c) The forecasts using the seasonal factors from the last 4 quarters quarters are as follows:
t 13 14 15 16
Seasonal factors 0.81 1.95 1.21 0.38
S t t 42,351 42,456 42,561 42,666
Forecast 34,304 82,789 51,499 16,213
7
An exponential smoothing model is used to generate a forecast for period 13 (January of next year). The sales for January through through April are used to initialize the model, and an α = 0.2 is used as the smoothing constant. The FORECAST module is used used to generate the forecast. The results are are summarized as follows:
Forecast, F 13 13 Forecast error, S E
Region 1 219.73 26.89
Region 2 407.04 25.50
Region 3 303.30 17.54
Combined 938.26 61.41
Note that the sum of the forecasts by region nearly equals the forecast of the combined usage. However, whether a by-region by-region forecast is better better than an overall forecast that is disaggregated by region depends on the forecast error. The standard error of the forecast is the best indicator. A comparison of a bottoms-up forecast developed from regional forecasts to that of a forecast from combined data can be based on the law of variances. That is, if the usage rates within the regions are independent of each other, the estimate of the total error can be built from the individual regions and compared to that of the combined usage data. The total forecast error error (variance) from from the individual regions regions S C 2 might be estimated as the weighted average ave rage of the variances as follows: S C 2 =
F 1 F C
S E 21 +
F 2 F C
S E 22 +
F 3 F C
S E 23
where F i = forecasts of each region F C = forecast based on combined data S E 2 = variance of the forecast in each region i
2 T
forecast based on regional data S = total variance of the forecast Therefore, 21973 . 40704 . 30330 . 2689 . 2+ 2550 . 2+ 1754 . 2 93007 . 93007 . 93007 . 236 × 723.07 + 0. 43 438 × 650. 25 25 + 0.326 × 307. 65 65 = 0.23
S T 2 =
. = 55574
76
Then, S T = 555.74 74 = 23.57
Since S T T < S C C, it appears that a bottom-up, or regional, forecast will have a lower error than a top-down forecast. 9
(a) See the plot in Figure 8-1. It shows that there is a seasonal seasonal component with a very slight trend to the data as well as some random, or unexplained, variation. v ariation. FIGURE 8-1 Plot of time series data for Problem 9 300
250 s e c i r p 200 t i n u y l h t n 150 o m e g a r e 100 v A
50
0
r l r y t n r y t n r y t n r y t y t n p c n a a p l c a p l c a p l c a p l c J A J O J A J O J A J O J A J O J A J O
Time, months
(b) A time series model typically will involve only on ly two components: trend and seasonality. Using 2 years of data should be sufficient to establish establish an accurate trend line and the seasonal indices. We can develop the following following table for computing a regression line and seasonal indices.
77
Prices, P t t 211 210 214 208 276 269 265 253 244 202 221 210 215 225 230 214 276 261 250 248 229 221 209 214 5,575
Time, t 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 300
P ×t 211 420 642 832 1380 1614 1855 2024 2196 2020 2431 2520 2795 3150 3450 3424 4692 4698 4750 4960 4809 4862 4807 5136 69,678
2
t
1 4 9 16 25 36 49 64 81 100 121 144 169 196 225 256 289 324 361 400 441 484 529 576 4,900
Trend,a T t t 232.4 232.4 232.4 232.4 232.2 232.3 232.3 232.3 232.3 232.3 232.3 232.2 232.3 232.2 232.3 232.3 232.2 232.2 232.2 232.2 232.2 232.2 232.2 232.2
Seasonal index b 0.91 0.90 0.92 0.90 1.19 1.16 1.14 1.09 1.05 0.87 0.95 0.90 0.93 0.97 0.99 0.92 1.19 1.12 1.08 1.07 0.99 0.95 0.90 0.92
S t t
0.92 0.93 0.96 0.91 1.19 1.14 1.11 1.08 1.02 0.91 0.92 0.91
a
Computed from the trend regression regression line. For example, the period 1 trend is T 1 = 232.39 - 0.008×1 = 232.4. b The seasonal index is the ratio of the actual price to the trend for the same period. For example, the period 1 seasonal index is 211/232 = 0.91.
We also have N = = 24, t = 300/24 = 12.5, and P = 5575/24 = 232.29. Now, b=
6 78 − 24( 232.29 29)(12. 5) ! P × t − N × P × t = 69,67 . = −0008 2 2 2 4 , 90 9 0 0 2 4 ( 1 2 . 5 ) − ! t − N × t
and a = P t − b × t = 2 32.29 − ( −0.008)(12.5) = 232.39
Therefore, the trend equation is:
78
29 − 0.00 008 × t Tt = 232.29 Note that the trend is negative for the th e last two years of data, even though the 5-year trend would appear to be positive. Now, computing c omputing the trend value T t t for each value of t gives the results as shown in the previous table. The seasonal index is is a result result of dividing P t t by T t t for each period t . The indices are averaged for corresponding periods periods that are one year apart. Forecasting into the 5th year year shows the potential error in the method. That is, for th January of the 5 year, the forecast is F t t = T t t ×S t-12 t-12, or F 25 25 = [232.39 − 0.008×25][0.92] = 213.6. Repeating for each method, we have:
t 25 26 27 28 29 30 31 32 33 34 35 36
Actual price 210 223 204 244 274 246 237 267 212 211 188 188
Forecast Forecast price error 213.6 - 3.6 215.6 7.1 222.9 -18.9 211.3 32.7 276.3 - 2.3 264.6 -18.6 257.7 -20.7 250.7 16.3 236.8 -24.8 211.2 - 0.2 213.5 -25.5 211.2 -23.2 Total squared error
Squared error 13.0 50.4 357.2 1069.3 5.3 345.9 428.5 265.7 615.0 0.0 51.0 538.2 3,739.5
Revised seasonal a 0.91
a
The seasonal index for period 25 is .90. The average of the seasonal index for period 25 − 12 = 13, and this period is (0.92 + 0.90)/2 = 0.91.
The standard error of the forecast is S F = 3,739.5 / (12 − 2 ) = 19.34 . Now, the forecast for period 37 would be: F 37 = ( 232.39 − 0.008 × 37)( 0.91) = 211. 21
(c) Using the exponential smoothing module in the FORECAST software, the forecast for the coming period is F = = 201.26, with S F = 17.27. The smoothing constants given in the problem are the "best" that FORECAST could find. (d) Each model should be combined according to its its ability to forecast accurately. We can give each a weight in proportion to its forecast error, or standard error of the forecast (S F ). Hence, the following table can be developed:
79
(1) Model type Regression Exp. smooth. Total
Forecast error 19.34 17.27 36.61
(2) = (1)/36.61 Proportion of total error 0.528 0.472 1.000
(3)=1/(2) Inverse of error proportion 1.894 2.119 4.013
(4)=(3)/4.013 Model weights 0.472 0.528 1.000
Therefore, each of the model results is weighted according to the model weights. The weighted forecast for the upcoming January Januar y would be: (1) Model type Regression Exp. smooth.
(2)
Forecast Model weight 211.21 0.472 201.26 0.528 Weighted forecast
(3)=(1)× (2) (2) Weighted proportion 99.69 106.27 205.96
In a similar fashion, we can weight the forecast error variances to come up with a weighted forecast error standard deviation S Fw. That is, S Fw =
0.472 × 19.342 + 0.528 × 17.272 = 18. 28
A 95 percent confidence band using the combined results might be constructed as: Y = = 205.96 ± z ×18.28
where z is is 1.96 for 95 percent p ercent of the area under the normal distribution. Y = = 205.96 ± 1.96×18.28
Hence, we can be 95 percent sure that the actual price Y will will be within the following range: 170.13 ≤ Y ≤ 241.79 10
The plot of the sales data is shown in Figure 8-2. The plot reveals a high high degree of seasonality with a noticeable downward trend. A level-trend-seasonal level-trend-seasonal model seems reasonable. (b) Using the search capability within the FORECAST software, a Level-Trend-Seasonal form of the exponential smoothing model was found to give the lowest forecast error. A 14-period initialization initialization and 6 periods to to compute error statistics were were used. The respective smoothing constants were α = 0.01, β = 0.08, and γ = = 0.60. This produced
80
a forecast for the upcoming period (January 2004) of F = 6,327.60 and a standard error of the forecast of S F = 1,120.81. FIGURE 8-2 Plot of Time Series Data for Hudson Paper Company 30000
25000 s 0 0 0 2 0 0 0 0 n i s e l a s 1 5 0 0 0 e t a g e r g g 1 0 0 0 0 A
5000
0
n a J
r y p l A J
t r y c n a p l O J A J
t c n a O J
r y p l A J
t c n a O J
r y p l A J
t r y c n a p l O J A J
t c O
Tim e, m on ths
(c) Assuming that the forecast errors are normally distributed distributed around the forecast, a 95 percent confidence band on the forecast is given by: = F + + z ×S F Y = = 6,327.60 ± 1.96×1,120.81 Y = where z = 1.96 for 95 percent of the area under the normal distribution curve. Therefore, we can be 95 percent sure that the actual sales Y should fall within the following limits: 4,130.8 ≤ Y ≤ 8,524.4 11
(a) For A569, the BIAS = −165,698 and the RMSE = 126,567 when using the 3-month moving average. However, if a level only exponential smoothing smoothing model with an α = 0.10, the BIAS drops to –9,556 and the RMSE is 118,689. The model fits the data better and there is a slight improvement in the forecasting accuracy. For A366, the BIAS = 18,231 and the RMSE = 144,973 when using the 3-month moving average. A level-trend-seasonal model offers offers the best fit, but it is suspect since the data show a high degree of random variability rather than seasonality. Overall, a simple level-only level-only model is probably better in practice. The model has an α = 0.08, a BIAS = −3,227, and a RMSE = 136,256. This is an improvement over the 3-month moving average.
81
(b) Using the level-only models, the forecast for October for A569 = 193,230 and for A366 = 603,671. (c) The 3-sigma (99.7 percent) confidence confidence band on the forecasts would be: For A569, Y = = 193,230 ± 3(118,689), or 0 ≤ Y ≤ 549,297. For A366, Y = = 603,671 ± 3(136,256), or 194,903 ≤ Y ≤ 1,012,439. The actual October usage falls within the 3-sigma confidence bands for each of these products. The difference of the actual from the forecast for for each product is attributable to the substantial variability in the data, which is characteristic of purchasing in the steel processing industry.
82
WORLD OIL Teaching Note Strategy The purpose of this case study is to allow students to develop an appropriate forecasting model for some time series series data. Discussion may begin with the nature of this product one with which most students students should be very familiar. familiar. Based on the many one available forecasting approaches, students should be encouraged to select several for consideration. In this note, both exponential smoothing smoothing and time time series decomposition decomposition are evaluated. Both are appropriate here because (1) they can project from from historical time time series data, (2) they can handle seasonality, which appears to be present in the data, (3) there is enough data to construct and test the models, and (4) the forecast is for a short period into the future. Assistance with the computational aspects of this problem is available with the use of the FORECAST module in the LOGWARE software. Answers to Questions (1) Develop a forecasting procedure for this service station. Why did you select your method?
Both exponential smoothing and time series decomposition forecasting methods are tested using using the FORECAST module in LOGWARE. For exponential smoothing, an initialization period of one seasonal cycle (52 weeks) plus two weeks are used for a total of 54 weeks, a minimum minimum requirement in FORECAST. FORECAST. The last 30 weeks of data is used for computing the error statistics. statistics. This number of periods is arbitrary, arbitrary, but seems reasonably large so as to give stable stable statistical values. We wish to minimize the forecast forecast error over time, and FORECAST computes both MAD and RMSE statistics that can be used to to make comparisons among model types. Testing the the various exponential smoothing model types and the time series gives the following statistics. Smoothing constants Model type Level only.... Level-trend... Level-seasonal Level-trendseasonal...... TS decomp.....
β − .5 −
γ − − 1.0
MAD 37.82 45.85 38.97
BIAS -5.27 7.13 11.30
RMSE 67.61 67.80 45.71
.01 .2 − −
.4 −
30.27 59.46
-6.05 37.18
44.17 71.85
α .4 .2 .3
Forecast week 6 of this year 817.35 860.26 648.75
770.74⇐ 731.33
The MAD and RMSE statistics show how well the forecast has been able to track historical fuel usage rates. They are an indication of the accuracy of the forecasting process in the future on the average. We favor forecasting methods that can minimize these statistics. statistics. In this case, case, the Level-Trend-Seasonal version version of the exponential smoothing model seems to do this best. Both MAD and RMSE are the the lowest for this model type among the alternatives. 83
Further evidence of the performance of a forecasting method is obtained from a plot of the forecast against against the actual usage rates. This is shown in Figure 1. Note that the Level-Trend-Seasonal model tracks the usage rates quite well, especially in the more recent weeks. The modeling process has likely stabilized in the last 30 weeks of the data and is now tracking quite well. FIGURE 1 Fit of Level-Trend-Seasonal Exponential Smoothing Model to Fuel Usage Data on Mondays of the Week
(2) How should sh ould the periods of promotions, holidays, or o r other o ther periods p eriods where usage rates deviate from normal patterns, be handled in the forecast?
If the deviations occur at the same time within the seasonal cycle and with the same relative intensity, no special procedures are required. The adaptive characteristic characteristic of the the exponential smoothing process will automatically incorporate these deviations into the forecast. However, when the deviations are not regular, as promotions may be timed irregularly, they may best be handled as outliers in the time series and eliminated from the time series. series. The model may be fit fit without the outliers, and then the effect of them treated as modifications to to the forecast. These modifications can be handled manually.
84
(3) Forecast next Monday's fuel usage and indicate the probable accuracy of the forecast.
From the Level-Trend-Seasonal exponential smoothing model developed in question 1, where the smoothing constants are α = 0.01, β = 0.2, and γ = = 0.4, the forecast for Monday of week 6 would be 771 gallons. gallons. However, this forecast only represents the average fuel usage. Determining the accuracy of the forecast requires that the forecast track the mean of the actual usage, i.e., a bias of 0, and that the forecast errors be normally distributed. While the BIAS (sum of the forecast errors over the last 30 weeks) is not exactly 0, and will not likely ever be so, it is low (-6.05), such that we will assume good tracking by the forecast model. A histogram of the forecast forecast errors can reveal whether they follow follow the familiar bell-shaped pattern. Such a histogram is given given below. We can conclude that while the errors are not precisely normally distributed, we cannot reject the idea that they did not come from a normally distributed population. A goodness-of-fit test could be used to check this assumption. Although this test is not performed here, it is quite forgiving, such that the normal distribution of errors assumption is not likely to be rejected where the data show a reasonably normal distribution pattern. The distribution here qualifies. We can now proceed with developing a 95 percent confidence band around the forecast. The forecast of the actual fuel usage usage rate Y will be: F − z (σ F ) ≤ Y ≤ F + z (σ F ) !
!
where σ F is the standard error of the forecast. F is is the forecast, and z is is the number of standard deviations for for 95 percent of the area under a normal distribution. FORECAST RMSE ) as: computes the root mean squared error ( RMSE !
N
! ( At − F t ) RMSE =
2
t =1
N
85
HISTOGRAM FOR FORECAST ERROR OF LAST 30 WEEKS Class Width = 20.0000 Number of Classes = 10 0%
50%
100%
MID CLASS +----+----+---+----+----+----+----++----+----+------+----+----+-+----+----+-----+----+----+ +----+ < -80.0000 |
|
-70.0000 |******
|
-50.0000 |***
|
-30.0000 |********
|
-10.0000 |********
|
10.0000 |*****
|
30.0000 |********
|
50.0000 |******
|
70.0000 |*
|
90.0000 |*
|
110.0000 |
|
>= 120.0000 | | +----+----+----+----+----+----+----+----+----+----+
Since RMSE is uncorrected for degrees of freedom lost, we apply a correction factor (CF ) as a multiplier to RMSE to get the unbiased estimate of the standard error of the ˆ F ): forecast ( σ CF =
N N - n
where n is the number of degrees of freedom lost in the model building process. We estimate n to be the number of smoothing constants in the model, or three in this case. Hence,
σ F = RMSE × CF !
30 30 − 3 = 44.17 × 1. 05 054 . = 4417
= 4656 . Now, with z @95% = 1.96 from a normal distribution table, we can be 95 percent confident that the true 87-octane fuel usage Y on on Monday of week 6 will be: 771 − 1.96(46.56) < Y < < 771 + 1.96(46.56) 680 < Y < < 862 gallons
86
METRO HOSPITAL
You are the materials manager at Metro Hospital. Approximately one year year ago, the hospital began stocking a new drug (Ziloene) that helps the healing process for wounds and sutures. It is your responsibility responsibility to forecast and order the monthly supply supply of Ziloene. The goal is to minimize the combined cost of overstocking and understocking the drug. Orders are placed and received at the beginning of the month and demand occurs throughout the month. The following demand and cost data have been compiled. Costs . If more is ordered than is demanded, a monthly holding cost of $1.00 per case is incurred. If less less is ordered than is demanded, a $2.00 per case lost sales sales cost is incurred. The drug has a short short shelf life, and any overstocked product at the end of the month is worthless and no longer available to meet demand. Demand Demand . The demand for the twelve months of last year was: Month Cases
1 43
2 36
Last year's demand demand 3 4 5 6 7 8 24 69 34 75 90 67
9 59
10 51
11 77
12 50
You believe this demand to be representative of Metro's normal usage pattern. FIGURE 1 Plot of last year's monthly demand in cases
87
Decision Worksheet Month 1 (13) 2 (14) 3 (15) 4 (16) 5 (17) 6 (18) 7 (19) 8 (20) 9 (21) 10 (22) 11 (23) 12 (24) Total
Month 1 (25) 2 (26) 3 (27) 4 (28) 5 (29) 6 (30) 7 (31) 8 (32) 9 (33) 10 (34) 11 (35) 12 (36) Total
Cases ordered
Actual demand
Over @ $1/case
Short @ $2/case
Cost, $
Cases ordered
Actual demand
Over @ $1/case
Short @ $2/case
Cost, $
88
METRO HOSPITAL Exercise Note Purpose Metro Hospital is an in-class exercise designed to illustrate the relationship between good forecasting and the control of inventory related costs. It shows shows that accurate forecasting is a primary primary factor in minimizing minimizing inventory costs. Participants in this exercise use a variety of methods, often intuition, to forecast demand and to come up with a purchase quantity. Their performance performance is measured as over- or understock costs. costs. Using a simple exponential smoothing forecasting model and an understanding of the standard deviation of the forecast, an effective purchase plan can be constructed. This process results results in costs that are significantly lower than the majority of the participants are able to achieve using intuitive methods. Administration The descriptive material and the decision worksheet are to be distributed to the class at the time that the exercise is conducted. To hand out the material ahead of time may take away much of the drama from from the exercise. About one half hour should should be scheduled for running the exercise. The instructor asks the class to make a decision regarding the size of the order to be placed in the upcoming period and to record it on the worksheet. The participants are then informed of the demand for that period from Table 1 after the simulated time of one month has passed. Given that they now know the actual demand for for the period, the participants are asked to record their costs and then th en to place an order for the next nex t period. The pattern is repeated for at least twelve months, a full seasonal cycle. The participants are asked to sum their costs and to report report them to the exercise leader. They are displayed in a public place, such as a chalkboard, for all to see. Then, the exercise exercise leader announces his or her cost level that was achieved using a disciplined approach using a simple forecasting procedure and some basic statistics. TABLE 1 Actual Demand for Period 13 Through 36 Period 13 14 15 16 17 18 19 20 Demand 47 70 55 38 90 24 65 65
21 23
22 55
23 85
24 66
Period Demand
33 45
34 70
35 50
36 56
25 53
26 64
27 61
28 63
29 65
30 38
31 80
32 88
Quantitative Analysis The demand series was generated using a normal distribution with a reasonably high variance and a very slight upward trend. trend. To illustrate the use of a quantitative approach to forecasting, an exponential smoothing model was selected, although other methods such as time series decomposition would also be appropriate. The twelve historical data points were submitted to the FORECAST module in LOGWARE. A three-month initialization period and a three-month time period for computing error statistics were chosen. The smoothing constants for the level, level-trend, and level-trend-seasonal models were examined. Based on the root mean squared error RMSE ), the best model (RMSE ),
89
was the level-trend-seasonal ( RMSE RMSE = 13.59), but the level model with α = 0.19 and RMSE = 13.89 performed very well and is used here. The model is: F t +1 = 0.19 At + 0.81 F t
where F t +1 = forecast for next period t + 1 At = actual demand for current period t F t = forecast for current period t
LOGWARE gives a forecast value of 58.1 and this is used as the forecast value for period 13. Applying this simple, level only model to the second year demand as it is revealed in each period gives the following forecast values: TABLE 2 Simple Exponential Smoothing Forecast Values for the Next Year Actual Period demand Forecast 13 47 58.1 14 70 56.0 15 55 58.7 16 38 58.0 17 90 54.2 18 24 61.0 19 65 54.0 20 65 56.1 21 23 57.8 22 55 51.2 23 85 51.9 24 66 54.6
Now we must determine the order quantity. It can be calculated from + z(RMSE) Q = F t + 1
Recall the RMSE was 13.89 for this model. To be precise, precise, we calculate z by trial and error. The following order quantity and cost computations can be made for a z value value of 0.8 (Table 3).
90
TABLE 3 Purchase Order Quantity and Associated Inventory Costs Actual Order Units Units Period demand Forecast quantity over short * 13 47 58.1 69 22 14 70 56.0 67 3 15 55 58.7 70 15 16 38 58.0 69 31 17 90 54.2 65 25 18 24 61.0 72 48 19 65 54.0 65 20 65 56.1 67 2 21 23 57.8 69 46 22 55 51.2 62 7 23 85 51.9 63 22 24 66 54.6 66 *
Cost, $ 22 6 15 31 50 48 0 2 46 7 44 0 271
Q = 58.1 + 0.8(13.89) = 69.21, or 69
We see from the following graph (Figure 1) that z = = 0.8 is optimal. FIGURE 1 Plot of Total Annual Costs Against the Factor
z
310 305 300 295 $ 290 , t s o C 285
280 275 270 265 0
0.2
0.4
0.6
0.8 Z
91
1
1.2
1.4
Figure 2 graphically shows the good purchase pattern of Table 3. FIGURE 2 Plot of Forecast and Purchase Order Quantity on Product Demand
100
Order uantit
90 80 70 s e s a C
60 50
Demand
40
Forecast
30 20 10 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Time period
Summary The exercise leader should discuss that one of the problems with intuitively forecasting demand is overreacting to randomness in in the demand pattern. pattern. This has the the effect of causing extreme over and short short costs in inventories. A model for short short term forecasting that is integrated into the purchasing and inventory control process can help to avoid these extremes and give lower costs. Several forecasting models may perform well, such as exponential smoothing, a simple moving average, a regression model, or a times series decomposition model. One of the most practical for for inventory control control purposes is the exponential smoothing model. The results from a simple, level only model were illustrated above using the same information that was available to the participants. Recognizing that it is less costly to order too much than to order too little, the purchase quantity should exceed the forecast by some margin. The astute participant will likely approximate the standard deviation of demand from the range of the demand values, that is, σ = (Max - Min)/6. Then, one or two σ might might be used to add a margin of safety to the forecast and size of the purchase order. This simple approximation procedure can lead to reasonable results.
92
CH APTER 9 INVENTORY POLICY DECISIONS 1
The probability of finding all items in stock is the product of the individual probabilities. That is, (0.95)×(0.93) ×(0.87) ×(0.85) ×(0.94) ×(0.90) = 0.55 2
(a) The order fill rate rate is the weighted weighted average of filling the item mix on an order. We can setup the following table. (1)
(2) Frequency Order Item mix probabilities of order 1 0.20 .95×.95×.95×.90×.90 = .69 2 0.15 .95×.95×.95 = .86 3 0.05 .95×.95×.90×.90 = .73 4 0.15 .95×.95×.95×.95×.95×.90×.90 = .62 5 0.30 .95×.95×.90×.90×.90×.90 = .59 6 0.15 .95×.95×.95×.95×.95 = .77 Order fill rate
(3)=(1)× (2) (2) Marginal probability 0.139 0.129 0.037 0.094 0.178 0.116 0.693
Since 69.3 percent is less than 92 percent, the target order fill rate is not met. (b) The item service levels that will give an order fill rate of 92 percent must be found by trial and error. Although there are many combinations of item service levels that can achieve the desired service level, a service level of 99 percent for items A, B, C, D, E, and F , and 97 percent to 98 percent for the remaining items would be about right. The order fill rates can be found as follows. (1)
(2) Frequency of order 0.20 0.15 0.05 0.15 0.30 0.15 Order fill rate
Order Item mix probabilities 1 (.99)3×(.975)2 = .922 2 (.99)3 = .970 3 (.99)2×(.975)2 = .932 4 (.99)5×(.975)2 = .904 5 (.99)2×(.975)4 = .886 6 (.99)5 = .951
93
(3)=(1) × (2) (2) Marginal probability 0.184 0.146 0.047 0.136 0.266 0.143 0.922
3
This is a problem of push inventory control. The question is one of finding how many of 120,000 sets to allocate to each warehouse. We begin by estimating estimating the total requirements for each warehouse. That is, Total requirements = Forecast + z ×Forecast error From Appendix A, we can find the values for z corresponding corresponding to the service level at each warehouse. Therefore, we have:
Warehouse 1 2 3 4 Total
(1) Demand forecast, sets 10,000 15,000 35,000 25,000 85,000
(2) Forecast error, sets 1,000 1,200 2,000 3,000
(3) Values for z 1.28 1.04 1.18 1.41
(4)=(1)+(2)× (3) (3) Total requirements, sets 11,280 16,248 37,360 29,230 94,118
We can find the net requirements for each warehouse as the difference between the total requirements and the the quantity on hand. The following table can be constructed:
Warehouse 1 2 3 4
(1) Total requirements 11,280 16,248 37,360 29,230 94,118
(2)
(3)=(1) − (2) (2)
(4)
(5)=(3) + (4) (4)
On hand quantity 700 0 2,500 1,800
Net requirerequirements 10,580 16,248 34,860 27,430 89,118
Proration of excess 3,633 5,450 12,716 9,083 30,882
Allocation 14,213 21,698 47,576 36,513 120,000
There is 120,000 − 89,118 = 30,882 sets to be prorated. This is done by assuming that the demand rate is best expressed by the forecast and proportioning the excess in relation to each warehouse's forecast to the the total forecast quantity. That is, for warehouse 1, the proration is (10,000/85,000)×30,882 = 3,633 sets. Prorations to the other warehouses are are carried out in a similar similar manner. The allocation to each warehouse is the sum of its net requirements plus a proration of the excess, as shown in the above table. 4
(a) The reorder point system is defined by the order quantity and the reorder point quantity. Since the demand is known for sure, sure, the optimum order quantity is: Q*
=
2 DS / IC IC = 2(3, 200)( 35) / ( 0.15)( 55)
94
=
164. 78, or 165 cas cases
The reorder point quantity is: ROP = d
×
LT = (3, 20 200 / 52) × 1.5 = 92 units
(b) The total annual relevant cost of this design is: TC = D × S / Q + I × C × Q * / 2 =
(3,200)( 35) / 164.78 + ( 0.15)( 55)( 164. 78) / 2
=
679.69 6 9 + 679.97 97
=
$1,359 359.66
(c) The revised reorder point quantity would would be: ROP = 200 / 52) × 3 = 185 units . = ( 3, 20
The ROP is greater than Q*. It is possible possible under these circumstances the reorder quantity may not bring the stock level above the ROP quantity. In deciding whether the ROP has has been reached, we add any quantities on order or in transit to the quantity on hand as the effective quantity in inventory. Of course, we start with an adequate in-stock quantity that is at least equal to the ROP quantity. quantity. 5
(a) The economic order quantity formula can be used here. That is, Q*
=
IC = 2( 30 2 DS / IC 300)( 8, 500) / ( 0.10)( 8, 500)
=
77. 5, or 78 students
(b) The number of times that the course should be offered is: N *
*
= D / Q =
300 / 77.5 = 3.9, or about four times per year
6
This is a single-period inventory inventory control problem. We have: Revenue = $350/unit Profit = $350 − $250 = $100/unit Loss = 0.2×250 = $50/unit Therefore, CP n
=
100 1 00 + 5 0
=
0667 .
Developing a table of cumulative frequencies, we have:
95
Quantity 50 55 60 65 70 75
Frequency 0.10 0.20 0.20 0.30 0.15 0.05 1.00
Cumulative frequency frequency 0.10 0.30 0.50 * 0.80 ⇐Q 0.95 1.00
CP n lies between quantities of 60 and 65. We round up and select 65 as the optimal purchase order size. 7
This question can be treated as a single-order problem. problem. We have: Revenue = 1 + 0.01 = $1.01/$ Cost/Loss = 0.10(2/365) = $0.00055/$ which is the interest expense expe nse for two days Profit = 1.01 − 1.00055 = $0.00945/$ and CP n
=
0.00945 00945 . = 0945 0.00 0 0945 + 0.0 00 0055
For an area under the normal curve of 0.945 (see Appendix A), z = = 1.60. The planned number of withdrawals is: Q* = D + z × σ D = 120 + 1.60(20) = 152.00
The amount of money to stock in the teller machine over two days would be: Money = Q*×75 = 152.00×75 = $11,400 8
This is a single-period inventory control problem. (a) We have: Profit = 400 − 320 Loss = 320 − 300 Then,
96
CP n =
400 − 32 0 ( 400 − 320) + ( 320 − 300)
=
0.80
We now need to find the sales that correspond to a cumulative frequency frequency of 0.80. In the following table:
Sales 500 750 1,000 1,250 1,500
Cumulative frequency frequency 0.2 0.4 0.7 * 0.9 ⇐Q 1.0
Frequency 0.2 0.2 0.3 0.2 0.1 1.0
the cumulative frequency frequency table. Q* lies between 1,000 and 1,200 in the * roundup to Q = 1,250 units.
We choose to
(b) Carrying the excess inventory to next year, CP n =
80 80 + ( 0.2 × 320)
=
0556 .
where the loss is the cost of holding a unit until the next year. The Q* now lies between 750 and 1,000 units. We choose 1,000 units. Holding the excess units means a potential loss of 0.2×320 = $64/unit, whereas discounting the excess units represents a loss of only 320 − 300 = $20/unit. Therefore, Cabot will need fewer fewer units if they are held over in inventory. 9
(a) The optimum order quantity is: is: Q*
2 DS / IC =
=
2(1,250)(52)( 40) /(0.3)(56)
and the reorder point quantity is: ROP = d
×
LT
+
z × s d '
where sd'
=
sd LT = 475 2.5
=
7 51
and z P =0.80 = 0.84.
97
=
556 cases
Now, ROP = (1,250)( 2.5) + ( 0.84)( 751) = 3, 756 cases
Policy: When the amount of inventory inventory on hand plus any quantities on order or in transit falls below ROP , reorder an amount Q*.
(b) For the periodic review system, we first estimate the order review time: T*
=
250 = 0.44 44 weeks Q * / d = 556 / 1,25
The max level is: M *
=
d
× (T
*
+
LT ) + z × sd '
where sd ' now is: sd'
=
sd T *
+
LT = 475 0.44 44 + 2.5 = 814 cases
Hence, M *
44 + = 1,250( 0. 44
2.5) + 0.84( 814) = 4, 35 359 cases
Policy: Find the amount of stock on hand every 0.44 weeks and place a reorder for the amount equal to the difference between the quantity on hand and the max level * ( M M ) of 4,359 cases.
(c) The total annual relevant cost for these policies is: TC = DS / Q + ICQ / 2 + ICzsd ' + kDsd ' E ( z ) / Q
For the reorder point system: TC Q = 1250(52)(40)/556 + .3(56)(556)/2
+ .3(56)(.84)(751) + 10(1250)(52)(751)(.1120)/556 = 4,676.26 + 4,670.40 + 10,598.11 + 98,332.37 = $118,277.14 For the periodic review system: TC P = 1250(52)(40)/556 + .3(56)(556)/2
98
+ .3(56)(.84)(814) + 10(1250)(52)(814)(.1120)/556 = 4,676.26 + 4,670.40 + 11,487.17 + 106,581.29 = $127,415.12 (d) The actual service level achieved is given by: SL = 1 −
sd' E ( z ) Q
For the reorder point system: SLQ
= 1−
751 751( 011 01120 . 20) 556
= 1−
0.15
or demand is met 85 percent of the time. For the periodic review system: SL P = 1 −
814 814( 011 01120 . 20) 556
= 1−
0.16
or demand is met 84 percent of the time. (e) This requires an iterative iterative approach as follows: Compute Q =
2 DS / IC
Compute P = 1 − QIC / Dk , then z, then E(z)
Compute Q =
2 D( S
+
ks d' E( z ) ) / IC
Go back and stop when there is no change in either P or Q P or After the initial value of Q = 556.3, the process can be summarized in tabular form.
99
Step 1 2 3 4 5 6
Q 778.4 860.0 889.9 899.6 902.8 902.8
P 0.9856 0.9799 0.9778 0.9777 0.9767 0.9767
z 2.19 2.06 2.01 2.00 1.99 1.99
E (z) (z) 0.0050 0.0072 0.0083 0.0085 0.0087 0.0087
Now, for P = = 0.9767, z = = 1.99 ROP = = 1,250(2.5) + 1.99(751) = 4,620 cases
and the total relevant cost is: TCQ
=
DS / Q + ICQ / 2 + ICzsd'
=
65,000( 4) / 902.8 + 0.3( 56 56)( 902.8) / 2
+
kDsd' E( z ) / Q
+0.3( 56)(1.99)( 751)
000)( 751)( 0. 00 0087) / +10( 65, 00 =
902. 8
$40,275
This is considerably less than the $118,277.14 for the preset P at at 0.80. If you solve this problem using INPOL, INPOL, you will get a slightly slightly different answer. That * is, Q = 858. This simply simply is because z is is carried to two significant digits rather than the four significant digits used in the above calculations. 10
Refer to the solution of problem 10-9 for the general approach. (a) Q* = 556.3 cases and ROP = d
×
LT
+
z LT
= 1,250( 2.5) +
2
× sd +
d2
×
0.84 2.5 × 4752
=
3,12 125 + 0.84 84( 977.08 08)
=
3946 , cases
2 sLT 2
+ 1, 250 ×
(b) An approximation for T * = Q*/d , or *
T = 556/1,250 = 0.44 weeks
and approximating sd ' as 100
0.52
sd'
=
(T *
=
( 0.44 + 2.5)( 4752 ) + 1, 2502 ( 0.5) 2
= 1,027
+
LT )sd2
+
d2
×
s L2T
cases cases
So, Max
d (T *
=
+
LT ) + z × s d '
= 1,250( 0.44 +
2.5) + 0.84(1, 027)
4,537 537 cases cases
=
(c) According to INPOL, TC = 4,686 + 4,686 + 128,195 + 13,862 = $151,429 Q
TC = 4,686 + 4,686 + 134,751 + 14,571 = $158,694 P
(d) According to INPOL, SL = 80.28 percent Q
SL = 79.27 percent P
(e) According to INPOL, = 5,128 cases, Q* = 930 cases, ROP = TC = $49,532, SLQ = 99.22 percent Q
*
T = 0.76 weeks, MAX = 6,257 cases TC = $52,894, SL P = 99.18 percent P
11
(a) The production run quantity is: Q p*
=
2 DS p × = IC p − d
2(100)( 250)( 250) 0.25 2 5( 75)
(b) The production run cycle is: Q p*
0 00 / = 1,00
300 = 3.33 days
101
×
300 3 0 0 − 10
, = 1000
units
(c) The number of production runs is: D / Q p*
0 00 = = 100( 250) / 1,00
25 runs per year
12
(a) The order quantity is: Q*
=
IC = 2( 2,000)( 250)(1.00) / ( 0.30)( 35) 2 DS / IC
=
309 valves
and the reorder point quantity is: ROP = d
×
but sd
0 . Therefore, Therefore,
=
LT
+
z × sd LT
0 00 / 8)(1) = 250 valves ROP = = ( 2,00 (b) Boxes are set up that contain 309 valves - the optimum order order quantity. When an order arrives from a supplier, 250 valves are set aside in a separate box and are treated as the backup stock. The residual 309 − 250 = 59 valves are used on the production line. When the 59 valves at the production line are used up, the backup box containing 250 valves is brought to the production line and the empty box is sent to the supplier refilling. One hour later when the order arrives, there will be zero valves remaining at the production line. line. Then, 250 valves are set set aside and 59 are sent to the the production line. The cycle is then repeated. This problem approach is similar to that of the KANBAN system. system. Lead times are very short so that lead times times are virtually certain. Demand is certain, since it is fixed by the production p roduction schedule. Boxes or cards ca rds are used to assure movement of the most economic quantity. KANBAN is essentially classic economic reorder point inventory control under certainty. 13
(a) The economical quantity of cars to be called for at a time is found by the economic order quantity formula: Q*
=
IC = 2( 40 2 DS / IC 40)( 52)( 500) / ( 0. 25)( )( 90, 000)( 30) / 2, 000 = 78. 5, or 79 cars
(b) This is the reorder point quantity: ROP = d
×
LT
+
z × sd LT
where z = 1.28 from Appendix A for an area under the curve equal to 0.90. Therefore,
102
3.33) 1 = 40(1) + 1.28( 33 ROP = =
44.3 cars cars,, or 44. 44.3( 3(90 90,0 ,000 00 / 2,00 2,000) 0)
= 1,994 tons tons of soda ash 14
(a) This is a reorder point design under conditions of uncertainty for both demand and lead-time. We assume that the probability of an out of stock is given. Therefore, the order quantity is: Q*
2 DS / IC ) ( 365)( 50) / ( 0.30)( )( 45) IC = 2( 50)(
=
=
367. 7 units
and ROP = d
×
LT
+
z × s d '
where z = = 1.04 (see Appendix A) for the area under the curve equal to 0.85 and sd'
sd2 LT
=
+
d 2 sL2T
=
1 52 ( 7) + ( 502 )( 2 2 ) = 107. 6 units
Therefore, ROP = 107.6) = 461.9 units = 50( 7) + 1.04(10
(b) This is the periodic periodic review system system design under uncertainty. The complexity requires us to make some approximations here. The time interval for review review of the stock level is: T*
=
Q * / d = 367.7 / 50 = 7.35 35 days
The MAX level level is: MAX
=
d (T *
+
LT ) + z × sd '
where z = = 1.04 and sd ' is approximated as: sd'
=
(T *
=
( 7.35 + 7)(152 ) + 502 ( 22 )
= 115.0
+
2 LT )( sd2 ) + d 2 ( s LT LT )
units units
Therefore,
103
MAX = = 50(7.35 + 7) + 1.04(115.0)
= 837.1 units (c) Since the service level is specified, the probability is not set at the optimum level. Knowing the out-of-stock cost allows us to find the most appropriate service level. Since this is an iterative process, we use INPOL to carry out the calculations. The optimized service level yields a reorder point design of * = 571 units Q = 410 units and ROP =
and the total relevant cost drops from $12,642 in part a to $8,489. The demand in stock in part a was 97.74 percent, and it now increases to 99.81 percent. 15
(a) Find the common review time: T*
! s ) / I! C D
=
2( O +
=
2(100 + 0) / [( 0.3 / 52)( 2 × 25 × 2, 000 + 1× 90 × 500) ]
=
I
i
i
2.5 weeks weeks
Then, M A*
=
d A (T *
+
LT ) + z A
× sd A
T * + LT
where z = 1.282 for P = = 0.90 A
M A*
=
2,00 000( 2.5 + 1.5) + 1.282(10 100) 2.5 + 1.5 = 8, 25 256 units
=
500( 2.5 + 1.5) + 0.842( 70) 2.5 + 1.5 = 2,118 units
and M B*
where z = 0.842 for P = = 0.80. B
The control system works works as follows: the stock levels of both items are reviewed every 2.5 weeks. The reorder size for A is the difference between the amount on hand and 8,256 units. The reorder size for B is the difference between the amount on hand and 2,118 units.
(b) The average amount in inventory is expected to be: 104
AIL = d × T * / 2 + z × sd T * + LT
For A: AIL A
=
2,000(2.5) / 2 + 1.28(100) 2.5 + 1.5 = 2, 756 units
For B: AIL B
=
500( 2.5) / 2 + 0.842( 70 70) 2.5 + 1.5 = 743 units
(c) The service level level is given by: SL = 1 − sd'
×
E( z ) / d
×T
*
For A: SL A
= 1 − 1 00
2.5 + 1.5( 0.0475) / 2, 000( 2.5) = 0. 998
For B: SL B
= 1−
70 2.5 + 1.5( 0.1120) / 500( 2.5) = 0.987
(d) We set T * = 4 and cycle through the previous previous calculations. Thus, we have: M A*
= 11,301 units
M B*
=
2,888 units
AIL = 4,301
AIL = 1,138
SL = 0.999
SL =0 .991
A
A
B
B
16
This problem is one of comparing the combined cost of transportation and in-transit inventory. In tabular form, we have the following annual costs: costs: Cost type Transportation
Formula R× D
In-transit inventory
365 ICDT/ 365
Total
Rail 6(40,000)(1.25) = $300,000 0.25 25( 250)( 40,00 0 00)( 21)
Truck 11(40,000)(1.25) = $550,000 0.25( 250)( 40, 00 00)( 7)
365 = $143,836 $443,836
365 = $47,945 $597,945
You should select rail. 17
105
The two transport options from the consolidation point are diagrammed in Figure 9-1. Whether to choose one mode over the other depends more than transportation costs alone. Because the transport modes differ in the time in transit, the cost of the money tied up in the goods while in transit must be considered in the choice decision. This in-transit ICDt inventory cost is estimated from . The following design matrix can be developed. 365 Cost type Transportation In-transit inventory
Method R ×D ICDt/365 Total
*
Air $180,800 3,447* $184,247
Ocean $98,800 34,467 $133,267
ICDt/365 = 0.17(185)(20,000)(2)/365= 3,447
Ocean appears to be the lowest cost option even when a substantial in-transit inventory cost is included. The ocean option assumes that the trucking cost to to move the goods from the consolidation point to the Port of o f Baltimore is included in the ocean carrier rate. FIGURE 9-1 The Consolidation Operation for a Hydraulic Equipment Manufacturer
Multiple sourcing points
Consolidation point
Baltimore
20 days 2 days Sao Paolo
18
The demand pattern is definitely lumpy, since s = 327 > d = 169. To develop the minmax system of inventory control, we first find Q*. That is, d
Q*
=
IC = 2(169)(12)(10) / 0. 20( 09 2 DS / IC 0. 96 + 0. 048)
The ROP is is ROP = d
×
LT
+
z × s d ' + ED
where = 1.04 from Appendix A, z = 106
=
448. 5 units
ED = 8 units the the average daily demand rate,
and sd'
=
sd2 LT
=
327 2 ( 4) + 1692 ( 0.82 )
=
+
d 2 sL2T
6678 667 .8 units units
So, ROP = = 169(4) + 1.04(667.8) + 8
= 1,378.5 units The max level is: + Q* − ED M = ROP = ROP + *
= 1,378.5 + 448.5 − 8 = 1,819 units 19
(a) The basic relationship is: I T
=
Ii n
We know that I = $5,000,000. If there are 10 warehouses, the amount of inventory in a single one would be: T T
I1
=
I T / 10
=
5,00 0 00,00 0 00 / 3.162 = 1, 58 581,139
The inventory in all 10 warehouses would be $1,581,139×10 = $15,811,390. (b) The inventory in a single warehouse would be: I T
000,00 0 00 = 1, 00
9
=
3, 00 000,00 0 00
In each of three warehouses, we would have: I = 3,000,000 /
3 = $1, 732, 051
107
and in all three warehouses, we would have $1,732,051×3 = $5,196,152. 20
(a) The turnover ratio is the annual demand d emand (throughput) divided by b y the average inventory level. These ratios for each warehouse and for the total total system are shown shown in the table below.
Warehouse 21 24 20 13 2 11 4 1 23 9 18 12 15 14 6 7 22 8 17 16 10 19 3 5
Annual warehouse thruput 2,586,217 4,230,491 6,403,349 6,812,207 16,174,988 16,483,970 17,102,486 21,136,032 22,617,380 24,745,328 25,832,337 26,368,290 28,356,369 28,368,270 40,884,400 43,105,917 44,503,623 47,136,632 47,412,142 48,697,015 57,789,509 75,266,622 78,559,012 88,226,672 818,799,258
Average inventory level 504,355 796,669 1,009,402 1,241,921 2,196,364 1,991,016 2,085,246 2,217,790 3,001,390 2,641,138 3,599,421 2,719,330 4,166,288 3,473,799 5,293,539 6,542,079 2,580,183 5,722,640 5,412,573 5,449,058 6,403,076 7,523,846 9,510,027 11,443,489 97,524,639
Turnover ratio 5.13 5.31 6.34 5.49 7.36 8.28 8.20 9.53 7.54 9.37 7.18 9.70 6.81 8.17 7.72 6.59 17.25 8.24 8.76 8.94 9.03 10.00 8.26 7.71 8.40
Avg. = 5.59
Avg. = 8.66
The overall turnover turnover ratio is 8.40. Ranking the warehouses by throughput and averaging turnover ratios for the top three and the bottom three warehouses shows that the lowest volume warehouses have a lower turnover ratio (5.59) than the highest volume warehouses (8.66). There are several reasons why this this may be so: •
The larger warehouses warehouses contain the higher-volume items such as the A items in the line. These may carry less safety stock stock compared with the sales volume. Conversely, the low-volume warehouses may have more dead stock in them.
108
•
There may be start-up start-up (fixed) stock in the warehouses, needed to open them, that becomes less dominant with greater throughput.
(b) A plot of the the inventory-throughput data is shown in Figure 9-2. A linear regression line is also shown fitted to the data. The equation for this line is: is: Inventory = 200,168 + 0.1132×Throughput FIGURE 9-2
Plot of Inventory and Warehouse Thruput for California Fruit Growers’ Association
12
) s n o i l l i M ( $ , l e v e l y r o t n e v n i e g a r e v A
10
8 E s t im i m a t in i n g l in in e
6
4
2
0 0
20
40
60
80
100
A n n u a l w a r e h o u s e t h r u p u t , $ ( M il li o n s )
(c) The total throughput for the three warehouses is: Warehouse 1 12 23 Total
Throughput $21,136,032 26,368,290 22,617,380 $70,121,702
Using this total volume and reading the inventory level from Fig. 9-2 or using the regression equation, we have: Inventory = 200,168 + .01132(70,121,702) = $8,137,945
109
(d) Warehouse 5 has a throughput of $88,226,672. percent and 70 percent, we have:
Splitting this throughput by 30
0.30×88,226,672 = 26,468,002 0.70×88,226,672 = 61,758,670 88,226,672 Estimating the inventory for each of the new warehouses using the regression equation, we have: Inventory = 200,168 + 0.1132×26,468,002 = $3,196,346 and Inventory = 200,168 + 0 .1132×61,758,670 = $7,191,249 for at total inventory in the two warehouses of $10,387,595 $1 0,387,595 21
The order quantity for each item when there is no restriction on inventory investment is: Q*
=
IC 2 DS / IC
We first find the unrestricted order quantities. Q A*
=
2(51,000)(10) / 0.25( 17 1. 7)
Q B*
=
2( 25, 00 000)(10) / 0.25( 3. 25)
* = QC
2(9, 00 000)(10) / 0. 25( 2.50)
1, 527 units
=
=
=
784 units
537 units
The total inventory investment for these items is: IV
=
C A ( Q A / 2) + CB ( QB / 2) + CC ( QC / 2)
1,527 / = 1.75(15 =
2) + 3.25( 78 7 84 / 2) + 2.50( 53 5 37 / 2)
$3,28138 .
Since the total investment limit is exceeded, we need to revise the order quantities. For each product: Q*
=
2 DS / [C ( I
+ α )]
110
For product A: Q A*
=
2( 51,000)(10) / [1. 75( 0. 25 + α )]
For product B: Q B*
=
2( 25, 00 000)(10) / [ 3.25( 0. 25 + α )]
For product C: * QC =
2(9, 00 000)(10) / [ 2.50( 0.25 + α )]
Now, the investment limit must be respected so that: 3,000 = C A ( Q A / 2) + CB ( QB / 2) + CC ( QC / 2) Expanding we have: 3,000 = 1.75 2( 51, 000)(10) / [1. 75( 0. 25 + α )] +3.25
2( 25, 00 000)(10) / [ 3. 25( 0. 25 + α )]
+2.50
2(9,00 0 00)(10) / [ 2.50( 0. 25 + α )]
We now need to find an α value by trial and error that will satisfy this equation. We can set up a table of o f trial values. Investment Investment in
α 0.03 0.04 0.045 0.049 0.05 0.10
A 1,262.44 1,240.48 1,229.92 1,221.67 1,219.63 1,129.16
B 1,204.53 1,183.58 1,173.51 1,165.63 1,163.69 1,077.36
C 633.87 622.84 617.54 613.40 612.37 566.95
Total inventory value, $ 3,100.84 3,046.90 3,020.97 3,000.70 2,995.69 2,773.47
When the term I+α is the same for all products, as in this case, α may be found directly from Equation 10-30. We can substitute the value for α = 0.049 into the equation for Q* and solve. Hence, we have:
111
Q A*
=
2(51000)(10) / [1.75( 0.25 + 0. 049)]
Q B*
=
2(25 2 5,000)(10) / [ 3.25( 0. 25 + 0. 049)]
* = QC
2(9,000)(10) / [ 2.50( 0.25 + 0. 049)]
=
1, 396 units
=
=
717 units
491 units
Checking: 1.75(1,396)/2 + 3.25(717)/2 + 2.50(491)/2 = $3,000 22
We first check to see whether truck truck capacity will be exceeded. Since three items are to be placed on the truck at the same time, the items are jointly ordered. The interval for ordering follows Equation 9-23, or:
T *
2( O +
=
I
! S ) =
!C D i
0.25[ 50 50(10 100)( 52) + 30( 30 3 00)( 52) + 25( 20 2 00)( 52)]
i
120 025 . ( 988, 000)
=
2( 60 + 0)
i
=
0.022 years, years, or 1.144 1.144 weeks weeks
Now, from
! D T w *
i
i
Truck capacity capacity
≤
i
[100(70) + 300(60) + 200(25)][1.144] = 34,320 lb. The truck capacity of 30,000 lb. has been exceeded, and the order quantity or the order interval must be reduced. Given the revised Equation 9-31, the increment to add to I can can be found. That is,
α =
2O
" Truck capacity % $$ '' D w # ! i i &
− I
2
!C D i
i
2( 60)
=
2
" % 30,000 000 30)( 52) + 25( 20 20)( 52) ) $ ' (50(10)( 52) + 30( 30 # [100(52 5 2)( 70) + 300( 52 5 2)( 60) + 200( 52 5 2)( 10)] )] & 120 2
" 30,000 % 988,000 000) $ ' (988 # 2,340,000 & =
−
0.25
0.73 73895 − 0.25 25 = 0.48 48895
Revise T * , the order interval by:
112
−
0.25
T *
! S ) ( I + α )! C D 2( O +
=
i
i
i
120 0.738 73895( 988, 000)
=
2( 60 + 0) ( 0.25 + 0.48895)[50(10 100)( 52) + 30( 300)( 52) + 25( 200)( 52)] )]
=
=
0.012 01282 82 years, years, or 0.6667 weeks weeks
Once again, we check that the truck capacity has not been exceeded. [100(70) + 300(60) + 200(25)][0.66667] = 30,000 lb. Therefore, place an order every 4.7, or approximately five days. 23
The average inventory for each item is given by: AIL =
Q*
2
'
+ z × s d
2 DS . z @ 95% = 1.65 from the normal IC distribution in Appendix A. The results of these computations can be tabulated.
where sd'
s
' d *
Q
AIL
=
sd LT and Q* is found by Q*
A 7.75 188.38 106.98
B 15.49 238.28 144.70
C 19.36 421.23 242.56
=
D 11.62 361.98 200.16
E 27.11 565.14 327.30
Summing the AIL for each product gives a total inventory of 1,022 cases. 24
The peak quantity of an item to appear on a shelf can be approximated as the order quantity plus safety stock, or 250 boxes boxes Q + z × sd ' ≤ 250 ' where z @93% @93% = 1.48 from Appendix A and sd economic order quantity is
Q*
=
2 DS = IC
2(123 × 52)(1.25) 019 01 . 9(129 12 . 9)
=
=
sd LT = 19 1 = 19 boxes. The
255 255.42 boxe boxess
Checking to see if the shelf space limit will be exceeded by this order quantity
113
255.42 + 1.48(19) = 283.54 boxes The quantity is greater than the 250 allowed. Subtracting the safety stock from the the limit gives 250 − 28 = 222 boxes. The order quantity should be limited limited to this amount. 25
The plot of average inventory to period facility throughput (shipments) gives an overall indication of how the company is managing collectively its inventory for all stocked items. We can see see that the relationship is linear with a zero intercept. This suggests that that the company is establishing its inventory levels directly to the level of demand (throughput). An inventory policy, such as stocking to a number of weeks of demand, may be in effect. Overall, the inventory policy seems to be well executed in that the regression line fits the point for each warehouse quite well. The terminal with an inventory level of $6,000 seems to be an outlier and it should be investigated. If its its high turnover turnover ratio were brought in line with the other terminals, an inventory inven tory reduction from $6,000 $6,0 00 to $4,000 on the average could be achieved. The stock-to-demand inventory policy should be challenged. An appropriate inventory policy should show some economies of scale, i.e., the inventory turnover ratio should increase as terminal terminal throughput increases. Whereas the current policy is of the form I = 0.012 D , a better policy would be I = kD 0.7 , where D represents terminal throughput and I is the average inventory level. level. The coefficient 0.012 for for the current policy is found as the ratio of 6,000/500,000 = 0.0.12 for the last data point in the plot. The k value value for the improved policy needs to be estimated. From the cluster of the lowest throughput facilities, the average inventory level is approximately $2,000 with an average throughput of about $180,000. Therefore, from I = kD 0.7 2,000 = k (180,000) 0.7 2,000 = k ( 4,771.894) 2,000 k = 4,771.894 k = 0.419
Reading values from the plot, the following table can be developed showing the inventory reduction that might be expected from from revised inventory policy. Note (Note: If the inventory-throughput values cannot be adequately read from the plot, the values in the following table may be provided to the students.) Actual Terminal Inventory, Inventory, $ 1 2,000 2 1,950 3 2,000 4 2,050
Shipments, Shipments, $ 150,000 195,000 200,000 200,000
Estimated Estimated inventory, inventory, $ I = 0.012 D 1,800 2,340 2,400 2,400
114
Revised inventory, inventory, $ I = 0.419 D 0.7 1,760 2,115 2,152 2,152
5 6 7 8 9 Totals
3,900 6,000 4,500 4,300 5,500 32,200
320,000 330,000 390,000 410,000 500,000 2,695,000
3,840 3,960 4,680 4,920 6,000 32,340
2,991 3,056 3,435 3,558 4,088 25,307
Revising the inventory control policy has the potential of reducing inventory from the 32,340 − 25,307 linear policy by x100 = 21.7% . 32,340 26
We can use the decision curves cu rves of Figure 9-23 in the text tex t answer this question since it applies to a fill rate of 95 percent and an α = 0.7. First, determine K for for an inventory throughput curve for the item, which is K =
D1−α TO
=
(117 x12) 0.3 6
= 1.466
Next, X =
tD1−0.7
=
ICK
12(117 x12) 0.3 0.20( 400)(1.466)
=
0.90
and with z ≈1.96 from Appendix A Y =
zs LT KD
a
=
1.96(15) 2 (1.466)(117 x12) 0.7
=
0.18
The demand ratio r is is 42/177 = 0.36. The intersection of r and and X lies lies below the curve Y (use curve Y = = 0.25), so do not no t cross fill. 27
Regular stock For two warehouses, estimate the regular stock for the three p roducts.
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2dS RS =
Product A
Product B
Product C
RS A1
=
RS A2
=
Q
IC 2 2 2(3,000)( 25) =
0.02(15)
=
354 units
=
457 units
=
408 units
=
445 units
2 2(5,000)( 25) 0.02(15) 2
RS B1
=
RS B 2
=
2(8,000)( 25) 0.02(30) 2 2(9,500)( 25) 0.02(30) 2
RS C 1
=
RS C 2
=
2(12,500)( 25) 0.02(25) 2 2(15,000)( 25) 0.02(25) 2
=
559 units
=
612 units
Regular system inventory for two warehouses is RS 2W 2W = 354 + 457 + 408 + 445 + 559 + 612 = 2,835. Regular stock for a central warehouse
RS A
=
RS B
=
RS C =
2(8,000)( 25) 0.02(15)
=
2 2(17,500)( 25) 0.02(30) 2 2( 27,500)( 25) 0.02( 25) 2
577 units
=
604 units
=
829 units
Total central warehouse regular stock is RS 1W 1W =577 + 604 + 828 = 2,009 units.
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Safety Stock Product A
SS = zsd LT SS A1 = 1.65(500) 0.75 = 714 units SS A2 = 1.65( 700) 0.75 = 1,000 units
where z @0.95 @0.95 = 1.65 from Appendix A Product B SS B1 = 1.65( 250) 0.75 = 357 units SS B 2 = 1.65(335) 0.75 = 479 units
Product C SS = zsd LT SS C 1 = 1.65(3,500) 0.75 = 5,001 units SS C 2 = 1.65(2,500) 0.75 = 3,572 units
System safety stock is SS 2W 2W = 714 + 1,000 + 357 + 479 + 5,001 + 3,572 = 11,123 units For each product, the estimated standard deviation of demand on the central warehouse is: s A s B s B
= = =
s12 + s22 2
=
500 2
+
700 2
=
860 units
2
250 + 335 = 418 units 3,500 2 + 2,500 2 = 4,301 units
The safety stock is: SS = zs LT SS A = 1.65(860) .75 = 1,229 units SS B = 1.65( 418) .75 = 597 units SS C = 1.65( 4,301) .75 = 6,146 units
Total safety stock in the central warehouse SS 1W 1W = 1,229 + 597 + 6,146 = 7,972 units. Total inventory with two warehouses RS 2W 2W + SS 2W 2W = 2,835 + 11,123 = 13,958 units and for a central warehouse RS 1W units. Centralizing 1W + SS 1W 1W = 2,009 + 7,972 = 9,981 units. inventories reduces them by 13,958 – 9,981 = 3,977 units. 28
The solution to this multi-echelon inventory control problem is approached by using the base-stock control system method. The idea is that inventory at any an y echelon is to plan its inventory position plus the inventory from all downstream echelons.
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First, compute the average inventory levels for each customer. This requires finding Q and the safety stock. Q is found from the EOQ formula. For customer 1 Q1
2( 425 x12)(50) 0.2(35)
=
AIL1
=
Q1
2
+ zs d
LT 3
1
270 units
=
270
=
2
+ 1.65(65)
0.5
=
211 units
where
[email protected] =1.65 from Appendix A For customer 2 Q2
2(333 x12)(50) 0.2(35)
=
AIL2
=
Q2
LT 3
=
2( 276 x12)(50) 0.2(35)
=
2
+ zs d
239 units
=
2
239 2
+ 1.65(52)
0.5 = 180 units
For customer 3 Q3
=
AIL3
=
Q3
2
+ zsd
3
LT 3
=
218 units
218 + 1.65( 43) 0.5 2
= 159 units
Total customer echelon inventory is AILC = 211 + 180 + 159 = 550 units For the distributors echelon Q D
=
AIL D
2,000 units =
Q D
2
+ zsd
D
LT D
=
2,000 2
+ 1.28(94)
1.0
= 1,120 units
where
[email protected] =1.28 from Appendix A The expected inventory that the distributor will hold is the distributor echelon inventory less the combined inventory for the customers, or 1,120 - 550 = 570 units.
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COMPLETE HARDWARE SUPPLY, INC. Teaching Note Strategy Complete Hardware Supply is an exercise involving the control of inventoried items collectively. Data for a random sample of 30 items from the company's total of 500 items held in inventory are given. The objective is to manage the total dollar dollar value allowed to to be held as inventory. Several alternatives can be considered for changing inventory levels, some of which require an investment other than in inventory. The number of items that must be analyzed and the multiple scenarios that are to be examined can be computationally time consuming. It is strongly suggested that students use the INPOL module within LOGWARE LOGWARE to aid analysis. The current database has been prepared and is available in the LOGWARE software. The Base Case We begin with the the current data optimized as a reorder point design. The optimum order quantities and associated inventory levels are found. The base case costs are shown as follows:
Fixed order order quantity policy Purchase cost Transport costa
Carrying cost Order processing cost Out-of-stock cost Safety stock cost Total cost Total investment
$556,912 0 4,425 4,425 0 2,529 $568,291 $27,801
aIncluded in the purchase cost
We note that optimizing the current design shows that investment of $27,801 exceeds the allowed investment level of $18,000. Ways need to be explored to reduce this. Transmit Orders More Rapidly Instead of mailing orders to vendors, Tim O'Hare can buy a facsimile machine and transmit orders electronically. This scenario can be tested by reducing the lead times in the base case by two days, or (2/5) = 0.40 weeks and increasing order processing costs by two dollars, and then optimizing again. INPOL shows that there will be a slight increase in operating costs from $568,291 to $568,640, an incremental increase of $349. Projecting this to all 500 items, we have 349(500/30) = $5,817. Since both operating cost and inventory investment level increase, there is no economic incentive to implement this change. Faster Transportation Suggesting that vendors who are located some distance (>600 miles) from the warehouse use premium transportation is a possible way of reducing lead times, and therefore safety
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stock levels. Of course, the increase in transportation transportation cost for those those affected vendors is likely to lead to a price increase to cover these costs. This scenario is tested tested by reducing the lead-time in weeks to 2.2 for those vendors over 600 miles miles from the warehouse. For these same vendors, a five percent price increase is made. Compared with the base case, there is little change in the inventory investment ($27,801 vs. $27,746); however, operating costs increase. The total costs now are $585,490 compared with the base case of $568,291, an increase of $27,199. The major portion ($17,159) of this comes from the increase in price. We conclude that this is not a good option for Tim. Reduce Forecast Error Reducing the forecast error involves reducing the standard deviation of the forecast error. Testing this option requires taking 70 percent of the base-case forecast error standard deviations and optimizing the design once again. These changes have a positive impact on operating costs and inventory investment. Operating cost now is $567,529 and inventory investment is $24,739. This is a saving in operating costs of $762 per year. For all 500, we can project the savings savings to be 762(500/30) = $12,700. Based on a simple simple return on investment, we have:
ROI =
12,700 700 0.25, or or 25% / year 50,000 000
This would appear to be attractive since carrying costs are 25 percent per year and the company's return on investment probably makes up about 80 percent of this value. Reduce Customer Service At this point, we have only accepted the idea of reducing the forecast error. However, inventory investment remains too high. We can now try to reduce it by reducing the service levels. This is tested by dropping the the service index from its current 0.98 level level to a level where inventory investment approximates $18,000. This is done, assuming the forecast software will will be purchased and the forecast error error reduced by 30 percent. By trial and error, the service index is found to be 0.54, which gives an investment level of $18,028. The revised service level compared with with the base case is summarized below for the 30 items.
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Item 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Base case 99.88% 99.92 99.96 99.98 99.98 99.96 99.97 99.96 99.92 99.98 99.99 99.99 99.92 99.98 99.96
Revised 96.26% 98.02 98.54 99.15 99.45 98.60 98.84 98.61 97.29 99.26 99.70 99.43 97.30 99.14 98.84
Item 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
Base case 99.98% 99.90 99.95 99.89 99.97 99.69 99.97 99.97 99.96 99.92 99.97 99.93 99.89 99.97 99.91
Revised 99.56% 97.57 97.81 95.96 98.15 89.53 98.96 98.96 97.58 99.33 96.68 97.45 98.78 96.92 96.78
Notice how little the service level changes, even with a substantial reduction in the service index. Conclusions Tim can make a good economic argument for purchasing software that will reduce the forecast error. The only questions here are whether the software can truly produce produce at least the error reduction noted and whether a 25 percent return on investment is adequate for the risks involved. Arguing to accept a service reduction in order to lower the investment level is a little less obvious since since we do not know the the effect that service service levels have on sales. However, Tim may point out that the service levels need to be changed so little that it is unlikely that customers will detect the the change. He might also raise the question as to whether whether customer service levels were too high initially, and suggest that customers be surveyed as to the service levels that they do need. n eed.
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AMERICAN LIGHTING PRODUCTS Teaching Note Strategy American Lighting Products is a manufacturer of fluorescent lamps in various sizes for industrial and consumer use. As frequently happens in business, top management has requested that inventories be reduced across the board, but it does not want to sacrifice customer service. Sue Smith and Bryan White White have been asked to eliminate 20 percent of the finished goods inventory. inventory. Their plan is to reduce the number of stocking stocking locations and, thereby, eliminate the amount of inventory needed. Of course, they must recognize that with fewer stocking points, transportation costs are likely to increase and customer delivery times may increase as well. On the other hand, facility fixed cost may be reduced. The purpose of this case is to allow students to examine inventory policy and planning through aggregate inventory management procedures. They also can see the connection between location and inventory levels. Answers to Questions (1) Evaluate the company’s current inventory management procedures.
The company’s procedures for controlling inventory levels are at the heart of whether inventory reductions reductions are likely to be achieved through inventory consolidation. The company appears to be using some form of reorder point control for the entire system inventory, but it is modified by the need to produce in production lot sizes. It is not clear how the reorder point is established. established. If it is based on economic order quantity principles, then the effect of the principles becomes distorted by the need to produce to a lot size that is different from the economic order quantity. Therefore, average inventory levels in a warehouse will not be related to the squar squar e root of the warehouse’s throughput (demand), i.e., throughput raised to the 0.5 power.1 Rather, the throughput will be raised to a higher exponent between 0.5 and 1.0. The above ideas can be verified by plotting the data given in Table 1 of the case and then fitting a curve of the form I = α TP β . Note: The curve can be found from standard linear regression techniques when the equation is converted to a linear form through a logarithmic transformation, i.e., ln I = lnα + β ln lnTP . The results results are shown in Figure 1. 0.816 816 The inventory curve is I = 2.99TP with r = 0.86, where I and and TP are in lamps. The projected inventory reduction can be calculated by using this formula. From the plot of the inventory data, we can see that there is substantial variation about the fitted fitted inventory curve. There is not a consistent turnover ratio between the warehouses. This probably probably results results from the centralized centralized control policy. On the other other hand, improved control may be achieved by using a pull procedure at each MDC. MDC. The data available in the case do not let us explore ex plore this issue.
1
Based on the economic order quantity formula, the average inventory level (AIL) for an item held in inventory can be estimated as AIL = Q / 2 = 2 DS / IC / 2 . Collecting all constants into K , we have 0.5
AIL= K(D) , where D is demand, or throughput.
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FIGURE 1 Plot of MDC average inventory vs. annual throughput.
(2) Should establishing the LOC be pursued?
One of the ideas proposed in the case is to consolidate all Consumer product line items into one large large order center (LOC). Evaluating the impact impact of the LOC on inventory reduction requires that an assumption be made as to how much demand and associated inventory of the total belongs to Consumer products. Table 2 of the case gives gives the order and back order breakdown by sales channel. Using this data, total consumer consumer demand is 312,211 line items, or 33.4 percent of the total total line items. The assumption is that the same percentage applies to total demand. Hence, Consumer demand is 33.4%×169,023,000 = 56,453,682 lamps. lamps. From the inventory-throughput curve, we can estimate the amount of inventory inventory needed at the single single LOC. LOC. That is, is, I = 0.816 2.997(56,453,682) = 6,339,684 lamps. If Consumer products account for 33.4% of total inventory, then there are 33.4%×23,093,500 = 7,713,229 lamps in Consumer inventory. The reduction that can be projected is 7,713,229 − 6,339,684 = 1,373,545 lamps for a reduction of Reduction =
1,373,545 × 100 = 17.8% 7,713,229
in Consumer inventory levels, but only a 6 percent reduction in overall inventory levels. The 20 percent reduction goal is not achieved. Other alternatives need to be explored. (3) Does reducing the number of stocking locations have the potential for reducing system inventories inventories by 20 percent? Is there enough information information available to make a good inventory reduction decision?
The second alternative proposed in the case is to reduce the number of MDCs from eight to a smaller number. In order to evaluate evaluate this proposal, it needs to be determined which MDCs will be consolidated and the associated total demand flowing through the consolidated facilities. facilities. The inventory-throughput inventory-throughput relationship relationship can then be used to estimate the resulting inventory levels. For example, if the Seattle and Los Los Angeles MDCs are combined, the consolidated demand would be 4,922,000 + 21,470,000 = 26,392,000 lamps. The combined inventory is projected to be I = = 2.997(26,392,000)0.816 =
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3,408,852 lamps, compared with the inventory for the two locations of 4,626,333, as shown in Table 1. This yields a 26.3 percent reduction reduction from current levels. Table 1 shows other possible MDC consolidations and the resulting inventory reductions that can be projected. TABLE 1 Inventory Reduction for Selected MDC Combinations, Combination s, in Lamps Combined Combined Inventory Inventory MDC combination combination demand inventory reduction Seattle/Los Angeles 26,392,000 3,408,852 1,217,481 Kansas City/Dallas 29,194,000 3,701,403 50,181 Chicago/Ravenna 49,174,000 5,664,257 -557,590 Atlanta/Dallas 39,314,000 4,718,862 1,224,721 Kansas City/Chicago 39,271,000 4,714,650 -933,900 Ravenna/Hagerstown 64,046,000 7,027,231 1,715,607 K City/Dallas/Chicago 52,515,000 5,976,377 -36,377 Ravenna/H’town/Chicago 87,367,000 7,508,054 3,423,196 Atlanta/Dallas/K City 55,264,000 5,242,351 2,293,566
From the MDC combinations in Table 1, proximity to each other is a primary consideration in order to not increase transportation costs or jeopardize delivery service any more than necessary. Several options can be identified identified that yield yield a 20 percent inventory reduction. These are:
Option 1
2
3
Inventory Inventory reduction, lamps 1,217,481 3,423,196 4,640,677
MDC combinations LA/Seattle Ravenna/H’town/Chicago Total reduction
Total inventory reduction
20.1%
LA/Seattle Kansas City/Hagerstown Ravenna/Hagerstown Total reduction
1,217,481 1,224,721 1,715,602 4,157,804
18.0%
LA/Seattle Ravenna/Hagerstown Atlanta/Dallas/K City Total reduction
1,217,481 1,715,602 2,293,566 5,226,649
22.6%
Options 1 and 3 achieve the 20 percent reduction goal, although other MDC combinations not evaluated may also do so. The maximum reduction would be achieved with one MDC. The total inventory would be I = 2.997(169,023,000)0.816 = 15,512,812 lamps, for a system reduction of 32.8 percent. However, we must recognize that as the number of warehouses is decreased, decreased, outbound transportation costs will will increase. Inbound transportation costs to the combined MDC will remain about the same, since
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replenishment shipments are already already in truckload quantities. Some difference in cost will will result from differences differences in the length of the hauls to the warehouses. On the other hand, outbound costs may substantially increase, since the combined MDC locations are likely to be more removed from from customers then they are at present. Outbound transportation rates will be higher, as they are likely to be for shipments of less-than-truckload quantities. If the sum of the the inbound and outbound transportation transportation cost increases is is greater than the inventory carrying cost reduction, then the decision to reduce inventories must be questioned. Calculating all transportation cost changes is not possible, since the case study does not provide sufficient sufficient data on outbound transportation transportation rates. However, they should be determined before and after consolidation to assess the tradeoff between inventory reduction and transportation transportation costs increases. On the other hand, inbound transportation transportation costs can be found, as shown below for option 1, where the consolidation points are Los Angeles and Hagerstown. Annual TL rate, demand, Location $/TL lamps Seattle 1800 4,922,000 Los Angeles 1800 21,470,000 Ravenna 250 25,853,000 Hagerstown 475 38,193,000 Chicago 350 23,321,000 Total 113,759,000 a (4,922,000/35,000)×1800 = 253,131
Transport cost, $ 253,131a 1,104,171 184,664 518,334 233,210 2,293,510
Combined annual demand, lamps
Transport cost, $
26,392,000
1,357,302
87,367,000
1,185,695
113,759,000
2,542,997
There will be a net increase in inbound transportation costs of $2,542,997 − 2,293,510 = $249,487 for option 1. In addition, the annual fixed costs for the MDCs will be less, since the total space needed in the consolidated facilities should be less than that for the existing facilities. Again, the case study does not estimate the fixed costs for existing or potential locations. We do know that taking them into account would favor consolidation. In summary, the costs associated with option 1, that just meets the 20 percent inventory reduction goal, would be: Cost type Inventory carrying cost reduction Warehouse cost Warehouse fixed cost Outbound transportation cost Inbound transportation cost
Cost savings, $ 0.20×0.882×4,640,677 = 818,615 0.10×4,640,677 = 464,068 Unknown, but may be included in warehouse cost Unknown data data not given (249,487)
Although Sue and Bryan could report a substantial savings in inventory related costs, they should be encouraged to include fixed costs and transportation costs so as to report the true benefits of the inventory reduction redu ction plan.
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(4) How might customer service be affected by the proposed inventory reduction?
The general effect of inventory consolidation is to reduce the number of stocking points and make them more remote from customers. That is, the delivery distance will be increased if if inventory consolidation consolidation is implemented. Therefore, delivery customer customer service may be jeopardized and must be considered before deciding to consolidate inventories. From Table 3 of the case, it can be seen that customer lead times remain constant for a variety of locations with the the exception of Kansas City. Since consolidation points will be selected among the existing locations, outbound lead times will remain unaffected. Customer service due to location should be constant, at least for a moderate degree of consolidation. Customer service due to stock availability will be affected if safety stock levels are reduced after consolidation. Although the inventory-throughput relationship projects adequate safety stock to maintain the current first-time delivery levels, it does not account for any increase in lead times that may occur between the current system of MDCs and the consolidated ones. By comparing the weighted inbound lead times times for the existing distribution system and option 1, as shown in Table 2, the average inbound lead-time is slightly reduced through through consolidation. Lead-time variability is usually related related to average lead-time. This should have a favorable affect on inventory levels since since uncertainty is reduced. reduced. First-time deliveries should should not be adversely affected by consolidation, according to option 1. TABLE 2
A Comparison of Inbound Lead Times for the Existing Distribution System and a Consolidated Distribution System (Option 1) (a) Current Distribution System Inbound Weighted lead time, lead time, Master Distribution Distribution Center Shipments Shipments days days Atlanta 26,070,000 2 0.308 Chicago 23,321,000 1 0.138 Dallas 13,244,000 3 0.235 Hagerstown 38,193,000 1 0.226 Kansas City 15,950,000 2 0.094 Los Angeles 21,470,000 5 0.635 Ravenna 25,853,000 1 0.153 Seattle 4,922,000 6 0.175 Total 169,023,000 1.964
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(b) Consolidation Option 1
Master Distribution Distribution Center a Atlanta Dallas H’town/Ravenna/Chicago Kansas City Los Angeles/Seattle Total
Shipments 26,070,000 13,244,000 87,367,000 15,950,000 26,392,000 169,023,000
a
Inbound lead time, days 2 3 1 2 5
Weighted lead time, days 0.308 0.235 0.517 0.094 0.781 1.935
Consolidation is assumed to take place at the MDC with the largest number of current shipments.
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AMERICAN RED CROSS: BLOOD SERVICES Teaching Note Strategy The American Red Cross Blood Services has a mission to provide the highest quality blood components compo nents at the lowest possible cost. High quality qualit y blood products are provided to regional hospitals, but managing the inventory to meet demand as it occurs is a difficult problem. Blood is considered a precious product, especially by those who give it voluntarily. So, managing this perishable product carefully is a foremost foremost concern. Blood is a vital product to those in need of it for emergencies and a precious product to those requiring it for elective surgery and other treatments. The goal is to always always have what is needed but never so much that this this perishable product product has to outdated. Managing the blood inventory is quite difficult because (1) forecasting demand is not particularly accurate, (2) the planning horizon for collections can be up to a year long with uncertain yields, (3) the life of blood products ranges from 42 days to as short as five days, (4) once scheduled, blood donors are never turned away except for medical reasons, and (5) there is a limited opportunity to sell blood outside of the local region if too much is on hand. Overall, this situation has many characteristics of a “supply driven” inventory management problem, which requires inventory management techniques different from those for typical consumer products. The intended purpose of this case study is for students to examine an inventory situation where there is limited control over the amount of the product flowing into inventory. This supply-driven inventory situation situation is likely to be quite different from that discussed on the introductory level. Students are encouraged to to consider the various elements that affect inventory levels of individual individual products and how they interact. interact. These elements are (1) demand forecasting, (2) collections, (3) decision rules for creating blood derivatives, (4) product prices, and (5) inventory policy. It is expected that students students will be able to make general suggestions for improvement. Questions (1) Describ (1) Describee the inventory inventory management management problem facing blood services services at the American American Red Cross. Cross.
One of the major problems facing the American Red Cross (ARC) is that the availability of blood is supply-driven, meaning that quantities of blood received for processing to meet demand in the short term are unknown, yet they must be placed in inventory if demand is less than the collected quantities. Blood availability is is a function of number of factors that cannot be well-controlled by the regional blood center in the short run, causing wide variability in supply. The usage of blood at hospital blood banks, which creates the demand on ARC’s blood inventories, is also uncertain and varies from day to day and between hospital facilities. The yield of blood at the point of collection is random and does not necessarily give the product mix needed to meet demand. Different blood types can only be known by a probability distribution as to the percentage of the blood types that exist in the general population. In the short term, the demand for blood types may ma y differ from the collected c ollected
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quantities, resulting in a potential for under- and over-stocking, since blood is drawn from all qualified donors as they arrive at collection sites. Forecasting demand for blood products will likely be reasonably accurate for a base load. Surgery loads on hospitals are scheduled in advance so that blood blood needs will be known with a fair degree of certainty, although each operation will not typically use the full amount of blood allocated to it. However, emergency blood needs are not well predicted, and they can cause spikes in demand and unplanned draws on inventory. A problem is establishing how much accuracy is needed for good inventory management. Inventory policy for managing inventory levels is a mixed strategy of product pricing, derivative product selection for processing at the time of collection, conversion to other products later in the product life cycle, product sell off, emergency supply (call for blood), discount pricing, and stocking rules for hospitals. Although there are many avenues to controlling inventory levels, shortages and outdating cannot always be avoided. It is not clear that these procedures lead to an optimal optimal control of inventory levels. Competition from local independent blood banks that sell selected blood products at low prices makes it difficult for for ARC to cover costs. ARC provides a wider range of products, but it has difficulty-differentiating difficulty-differentiating price among derivative products so that it might compete effectively. Given pressures for hospitals to increase efficiency, they they will shop around for the the lowest-priced blood products. ARC is having difficulty maintaining maintaining its position as the dominant supplier of blood products in the region, which results in the greater uncertainty in managing inventory levels. In summary, blood is a precious product given by volunteers for the benefit of others. Donors have the right to expect that their contribution will be handled responsibly. To ARC, this means managing the blood supply so that recipients receive a high-quality product at the lowest possible price. To achieve this goal, ARC manages the blood supply through four inter-connected elements: (1) estimating the blood product needs over time, (2) planning the collection of whole blood, (3) deciding which derivative products and their amounts should be created from whole blood, and (4) controlling the inventory levels to to avoid outdating. The volunteer nature of the blood giving and donor attitudes surrounding it, long planning lead times and the associated uncertainties, rising competition among some products from local blood banks, and the uncertainties of blood needs all make blood supply management a unique inventory management problem. (2) Evaluate (2) Evaluate the current current inventory management management practices practices in light of ARC’s mission. mission.
Performance of blood management can be evaluated on two levels: customer service and cost. Tables 8 and 9 of the case show that in March standards were not quite met overall. Within specific product types, there was up to an eight percent deficit. Both order fill fill rate and item fill rate were less than 100 percent for most products. There would seem to be some room for improvement, especially in managing the variation among product types. From a cost standpoint, it is not known how efficiently the blood supply is managed since no costs are reported. reported. In addition, the revenue that the blood products generate generate is not known. We would like to know how prices of the various products are are set so that revenues might be maximized, considering competition among some of the product line.
129
We do expect that demand is price elastic, since hospitals do shop around for blood products that are available from local, commercial, and community blood banks. On the other hand, ARC is the sole regional supplier of certain products such as platelets. Setting product fill-rate standards at various levels levels can influence costs. We do not know this effect. Setting inventory levels by a “number of days of inventory” rule of thumb is simple but not as effective as planning inventory levels based on the uncertainties that occur in demand forecasts and supply lead times. The number-of-days of inventory rule rule does tend to lead to too much inventory or to too many out-of-stock situations. The plan for evaluation, if enough data were available, would be to establish a base case of cost and service. This, then, would provide a basis for evaluating the effect effect of change in the supply procedures. (3) Can you suggest any changes in ARC’s inventory planning and control practices that might lead to cost reduction or service improvement?
Suggestions for improvement in blood supply management stem from a basic understanding of the the nature of the demand-supply relationship. When supply is uncertain and all supply must be taken that is available, there is the possibility that significant excess inventory will will occur. The goal is to “manage” the demand in the short run to reduce inventory levels when overstocking occurs, rather than focusing on managing supply. Several approaches for doing this are: • Aggressively
price selected products that are in excess supply and are nearing their expiration dates, e.g. run a sale or offer price discounts. • Sell off excess supply to secondary demand sources or other regions of the ARC. Tempo rarily adjust return rules for hospitals. • Temporarily • Bring demand more in line with supply by converting products into derivative ones that have excess demand, e.g., reprocess whole blood into plasma. • Encourage hospitals to buy certain products in excess supply for a more favorable status in buying other products that are in short supply, such as phersis platelets and rare whole blood types. • Try to create excess demand for all products, especially those items that are available from local blood banks, through promotion of ARC’s distinct advantages, such as quality, high service levels, and a wide range of blood derivative products. • Offer “two-for-one” sales, such that if a hospital buys one blood product, it may receive another at a favorable price. u ncertain demand by maintaining a central inventory for all • Pool the risk of uncertain hospitals, or managing the inventories at all hospitals, as well at ARC, collectively. Provide quick deliveries or transfers among inventory inven tory locations. ARC should attempt to be the premier provider of blood products and leverage the advantage. This will allow it to maintain a degree of control over the demand for blood. Effectively controlling demand in turn allows it to control its costs and avoid product outdating.
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(4) Is pricing policy an appropriate mechanism to control inventory levels? If so, how should price price be determined? determined?
From the previous discussion, it can be seen that price plays a role in controlling demand. Since there appears a relationship between demand and price for some products, especially among those products offered by local blood banks that compete with ARC blood products, price may be an effective weapon to meet competition. Rather than setting price based on the cost of production, ARC might consider raising the price on products for which it is the sole provider, such as platelets, and then meeting the price of competitors on whole blood. Although ARC strives strives to be a nonprofit organization, organization, the increased volume that an effective pricing strategy promotes would allow more of the fixed costs to be covered. This may lead to lower lower overall average prices prices for ARC’s products. Blood could also be priced as a function of its freshness at two or more levels. Although blood that has been donated within 42 days legally can be utilized, the quality of blood does not remain the same for the the entire 42-day period. A chemical compound found in blood, called 2,3-DPG, decreases with the age of the stored blood, and is believed to be important in oxygen delivery. For this reason, certain procedures such as heart transplants and neonatal procedures require that blood be fresh, usually donated within 10 days or less. Thus, a simple pricing policy could be to charge a higher price for blood that is less than 10 days old, and a lower price for blood that is between 10 and 42 days old. Price differences here are based on product quality. quality.
131
CH APTE R 10 10 PURCHASING AND SUPPLY SCHEDULING DECISIONS 1
(a) The following requirements schedules will lead to the proper timing and quantities for the purchase orders. Desk style A Sales forecast Receipts Qty on hand 0 Releases to prod.
1 150 200 50 300
Week 2 3 4 5 6 7 8 150 200 200 150 200 200 150 300 300 300 30 0 20 0 0 10 100 25 250 50 15 150 0 300 300 30 0
1
2
Desk style B Sales forecast Receipts Qty on hand 80 Releases to prod.
60 20 100
3
60 100 60
60 0 100
Week 4 5 80 80 100 100 20 40 100 100
6 10 100 10 0 40 10 0
7
6
7
8
80 10 0 60
60 0
Desk style C Sales forecast Receipts Qty on hand 200 Releases to prod.
1 100 100 100
2 12 1 20 100 80 100
3 10 1 00 100 80
Week 4 5 80 80 10 0 0 20 100 100
60 10 0 60
8 60
0 10 0
80 10 0 60
Summing the releases for these three desk release schedules gives a production requirements schedule for desks in general and sheets of plywood in particular. That is, 1
2 Desk requirement 500 100 Plywood Ply wood sheets sheetsa 1500 30 300 a
Week 4 5 400 500 200 1200 15 1500 60 600
3
6
7 400 100 1200 30 300
8 0 0
Desk requirements times 3
Now, find the purchase order releases for the plywood sheets.
Sales forecast Receipts Qty on hand 2400 Rele Releas ases es to pro prod. d.
1 2 1500 300 60 0 900 1200 1000 1000 1000 1000
3 120 0 1 00 0 1 00 0 1000 1000
132
Week 4 5 6 7 1500 600 1200 300 1000 1000 1000 500 900 700 400 1000 1000
8 0 40 0
Therefore, purchase orders orders should be placed in weeks 1, 2, 3, and 4 for 1000 sheets each. (b) Using Equation 10-2 in the text, the probability of not having the plywood sheets at the time needed would be: P r = 1 −
P c Cc
+
=
P c
1−
5 01 . +5
=
0.02
From Appendix A, z @1-.02 @1-.02 = 2.05. Therefore, the lead-time should be: T*
=
LT
+
z × s LT
=
14 + 2. 05 05( 2) = 18.1 days
Another ½ week should be added to the current lead-time of 2 weeks. 2
(a) Using Equation 10-2, the probability of not having the item when needed for production is: P r = 1 −
P c Cc
+
=
P c
1−
150 ( 0.2 × 35 / 365) + 150
=
00001 .
The time to place an order ahead of need is: T * = LT + z × s LT = 14 + 3.6( 4) = 28 days where z @1-.0001 @1-.0001 = 3.6 from Appendix A. (b) Use part period cost balancing. The unit carrying cost is (0.2/52)×35 = 0.134. Then, (Q=250) Week 4 0.134×[500 + 200]/2 = 46.9 (Q=1350) Weeks 4 + 5 0.134×[(1350 + 1050)/2 + (1050 + 200)/2] = 244.6 The carrying cost closest to the order cost of $50 $ 50 is Q = 250. Order this amount. 3
Using the requirements planning procedure, we can develop a schedule of material flows through the network over the next 10 weeks. Whs e 1 Requi quiremen ments Sch d rec eipt s On-h On-han and d qty qty 1700 1700 Rel ease s
1 2 3 4 5 6 7 1200 12 120 00 12 1200 00 1 12 200 1 12 200 1 120 200 0 12 120 00 750 0 500 500 6800 6800 5600 5600 4400 4400 3200 3200 2000 2000 800 800 750 0 75 0 0
133
8 9 10 10 12 120 00 12 120 00 12 120 00 75 0 0 7100 7100 5900 5900 4700 4700
Whs e 1 1 Requ Requir irem emen ents ts 1200 1200 Sch d rec eipt s On-h On-han and d qty qty 1700 1700 500 500 Rel ease s 75 0 0
2 3 4 5 6 7 8 9 10 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 1200 7500 75 0 0 6800 6800 5600 5600 4400 4400 3200 3200 2000 2000 800 800 7100 7100 5900 5900 4700 4700 75 0 0
Whs e 2 Requi quiremen ments Sch d rec eipt s On-h On-han and d qty qty 3300 3300 Releases
1 2 3 4 2300 23 230 00 23 2300 00 2 23 300 7500 1000 1000 6200 6200 3900 3900 1600 1600 7500 750 0
5 6 7 2 23 300 2 230 300 0 23 230 00 750 0 6800 6800 4500 4500 2200 2200 75 00
8 9 10 10 23 230 00 23 230 00 23 230 00 75 00 7400 7400 5100 5100 2800 2800
Whs e 3 Requi quiremen ments Sch d rec eipt s On-h On-han and d qty qty 3400 3400 Releases
1 2 3 4 2700 27 270 00 27 2700 00 2 27 700 7500 700 700 5500 5500 2800 2800 100 100 7500 750 0
5 6 7 8 2 27 700 2 270 700 0 27 270 00 27 270 00 750 0 7500 4900 4900 2200 2200 7000 7000 4300 4300 7500
9 10 10 27 270 00 27 270 00 7 500 1600 1600 6400 6400 7 500
Reg n l wh s e A 1 2 3 4 5 6 7 8 9 10 Requirements 22 2 2500 0 0 15 0 0 0 0 15000 75 7 500 0 750 0 0 Sch d rec eipt s 1 500 0 15 0 0 0 On-hand On-hand qty 52300 29800 800 298 29800 298 29800 148 14800 1 14 4800 800 148 14800 7300 730 7300 1300 1300 1 13 300 Release Releases s to p lan t 1 50 0 0 15000 Whs e 4 Requi quiremen ments Sch d rec eipt s On-h On-han and d qty qty 5700 5700 Releases
1 2 3 4 5 6 4100 41 410 00 41 4100 00 4 41 100 4 41 100 4 410 100 0 7500 750 0 7500 1600 1600 5000 5000 900 900 4300 4300 200 200 3600 3600 7500 750 0 7500 7500
Whs e 5 Requi quiremen ments Sch d rec eipt s On-h On-han and d qty qty 2300 2300 Releases
1 2 3 4 5 1700 17 170 00 17 1700 00 1 17 700 1 17 700 7500 600 600 6400 6400 4700 4700 3000 3000 1300 1300 7500 750 0
Whs e 6 Requirements Sch d rec eipt s On-h On-han and d qty qty 1200 1200 Rel ease s
1
Reg n l wh s e B Requirements Sch d rec eipt s On-hand On-hand qty 3170 0 Release Releases s to p lan t
7 8 41 410 00 41 410 00 7500 7000 7000 2900 2900 75 00
9 10 10 41 410 00 41 410 00 75 0 0 6300 6300 2200 2200
6 7 8 9 10 10 1 170 700 0 17 170 00 17 170 00 17 170 00 17 170 00 75 00 7100 7100 5400 5400 3700 3700 2000 2000 300 300
2
3 4 5 6 7 8 9 10 10 900 900 900 900 900 900 900 900 900 750 0 7 500 300 300 6900 6900 6000 6000 5100 5100 4200 4200 3300 3300 2400 2400 1500 1500 600 600 7200 7200 750 0 7 500 900
1 2 22500 0 15000
3 7500
4 5 0 15 0 0 0
9200 24200 16700 16700 1 50 0 0
134
1700
6 7500 15000
7 8 0 7500
9200 1700 9200 1 5000
9 7 50 0 150 00
10 0
920 0 920 0
Pla n t Requirements Sch d rec eipt s On-hand qty 0 Releases-matls
1 0 0 0 40000
2 0 0
3 0
4 5 6 7 8 30000 0 0 30000 0 40000 2 0000 1 10 000 0 10000 10000 0 0 2 0 0 00
9 0 0
10 0 0
Summing the releases to the plant shows that the plant should place 30,000 cases into production in weeks 4 and 7. Because demand is shown to be constant , the average inventory must be one-half the order quantity. For the six field warehouses warehouses and a shipping quantity quantity of 7500, the average long run inventory would be (7500/2)×6 = 22,500 cases. For the regional warehouses, the average inventory would be (15,000/2)×2 = 15,000 cases. For the plant, the average inventory would be 20,000/2 = 10,000 cases. The total system system average inventory would would be 22,500 + 15,000 + 10,000 = 47,500 cases. 4
(a) The leverage principle shows the relative change that must be made in cost, price, or sales volume to affect a given given change in the profit level. Usually it is is used in reference to the cost of goods sold to show the impact that small changes in the cost of goods will have on profits and the important role that purchasing plays in the profitability of the firm. The following simple profit and loss statements will show how much change is needed nee ded in various activities to increase profits by 10 percent. p ercent.
Sales Cost of goods Labor & salaries Overhead P ro f i t
C u r re n t $55.0 27.5 15.0 8.0 $ 4.5
Sales (+4%) $57.2 28.6 15.6 8.0 $ 5.0
Pr Price (1%) $5 5 . 5 27. 5 1 5. 0 8. 0 $ 5. 0
L& L&S (-3%) $55.0 27.5 14.5 8.0 $ 5.0
OH ( -6 % ) $ 5 5. 0 27. 5 15 . 0 7. 5 $ 5. 0
CO G ( - 2% ) $5 5.0 2 7.0 1 5.0 8 .0 $ 5.0
Due to the magnitude of cost of goods sold, it requires less than a two percent change in COG to increase profits by 10 percent. (b) The current ROA as: Profit margin = (4.5/55)×100 = 8.2 percent Investment turnover = 55/20 = 2.75 ROA = 2.75×8.2 = 22.6 percent Reducing cost of goods by 7 percent will increase profits to 55 − 27.5×0.93 − 15 − 8 = $6.43 and the profit margin now is 6.43×100/55 = 11.7 percent. Inventory at 20 percent of total assets is $4 million. If the cost of goods is reduced by 7 percent, inventory value will decline to $4×0.93 = $3.72. Total assets will be 3.72 + 16 = $19.72 million. million. The investment investment turnover is 55/19.72 55/19.72 = 2.789. The ROA now will be 11.7×2.789 = 32.63 percent.
135
5
(a) A mixed purchasing strategy will generally be beneficial when prices show a definite seasonality, they are predictable, and inventory costs associated with forward buying are not excessive. In the problem, we should consider forward buying in the first first half of the year and hand-to-mouth hand-to-mouth buying in the last half. To test the various various strategies, compare (1) hand-to-mouth buying, (2) forward buying every 2 months, (3) forward buying every 3 months, and (4) forward buying for the first 6 months. The results are summarized in Table 10-1. The inventory for the hand-to-mouth buying strategy can be approximated as 50,000/2 = 25,000. The carrying cost would would be 0.30×4.98×25,000 = $37,350 per year. The carrying cost for the two month forward buying strategy is: 0.30×4.88×[(0.5×100,000/2) + (0.5×50,000/2)] = $54,900 For the 3-month forward buying strategy: 0.3×4.56×[(0.5×300,000/2) + (0.5×50,000/2)] = $119,700 From the total costs in Table 10-1, the best strategy is to forward buy the first sixmonth's requirements in January and hand-to-mouth buy bu y for the last six months. (b) Some possible disadvantages are: Prices may ma y fall rather than rise in the first six months • There may not be adequate storage space to accommodate such a large purchase. • The materials may be perishable and not easily stored. • Uncertainties in the requirements and carrying costs may void the strategy. •
136
TABLE 10-1 A Comparison of Various Forward Buying Strategies Strategies with Hand-to-Mouth Buying Hand-to-mouth buy buy
Jan Feb Mar Apr May Jun Jly Aug Sep Oct Nov Dec
Price, Quantity, $/unit units 4.00 50,000 4.30 50,000 4.70 50,000 5.00 50,000 5.25 50,000 5.75 50,000 6.00 50,000 5.60 50,000 5.40 50,000 5.00 50,000 50,000 4.50 50,000 4.25 50,000 Subtotals Inventory costs Totals
Average price/unit
Total $200,000 215,000 235,000 250,000 262,000 287,500 300,000 280,000 270,000 250,000 225,000 212,000 $2,987,500 37,350 $3,024,850
2-month forward buy
Price, $/unit 4.00
Quantity, units 100,000
3-month forward buy
Total $400,000
4.70
100,000
470,000
5.25
100,000
525,000
6.00 5.60 5.40 5.00 4.50 4.25
50,000 50,000 50,000 50,000 50,000 50,000 50,000
300,000 280,000 270,000 250,000 225,000 212,000 $2,932,500 54,900 $2,987,400
$4.98
$4.88
Price, $/unit 4.00
Quantity, units 150,000
Total $600,000
5.00
150,000
750,000
6.00 5.60 5.40 5.00 4.50 4.25
50,000 50,000 50,000 50,000 50,000 50,000
300,000 280,000 270,000 250,000 225,000 212,500 $2,887,500 72,150 $2,959,650
6-month forward buy
Price, $/unit 4.00
Quantity, units 300,000
6.00 5.60 5.40 5.00 4.50 4.25
50,000 50,000 50,000 50,000 50,000 50,000
Total $1,200,000
300,000 280,000 270,000 250,000 225,000 212,500 $2,737,500 119,700 $2,857,200
$4.81
137
6
(a) On the average, a total expenditure of 1.10×25,000 = $27,500 should be made for copper each month. (b) For the next 4 months, the dollar averaging purchases would be:
Month 1 2 3 4 a
(1 ) (2) Price, No. of $/lb. lb. 1.32 2 20 0, 8 3 3 1.05 2 26 6, 1 9 0 1.10 2 25 5, 0 0 0 0.95 2 28 8, 9 4 7 1 00,97 0
(3)=(1)× (2) (2) Tota l cost,$ 27,500 27,500 27,500 27,500 $110,000
(4)= (4)=(2 (2)/ )/2 2 A vera ge inventory, lb. 1 0 , 41 7 1 3 , 09 5 1 2 , 50 0 1 4 , 47 4 12,622a
50,486/4 = 12,622
The average per-lb. cost would be $110,000/100,970 = $1.089. The inventory carrying cost over 4 months would be 0.20×1.089×(4/12) ×12,622 = $916. If hand-to-mouth were used, we would have:
$4.56
6
(a) On the average, a total expenditure of 1.10×25,000 = $27,500 should be made for copper each month. (b) For the next 4 months, the dollar averaging purchases would be: (1 ) (2) Price, No. of $/lb. lb. 1.32 2 20 0, 8 3 3 1.05 2 26 6, 1 9 0 1.10 2 25 5, 0 0 0 0.95 2 28 8, 9 4 7 1 00,97 0
Month 1 2 3 4 a
(3)=(1)× (2) (2) Tota l cost,$ 27,500 27,500 27,500 27,500 $110,000
(4)= (4)=(2 (2)/ )/2 2 A vera ge inventory, lb. 1 0 , 41 7 1 3 , 09 5 1 2 , 50 0 1 4 , 47 4 12,622a
50,486/4 = 12,622
The average per-lb. cost would be $110,000/100,970 = $1.089. The inventory carrying cost over 4 months would be 0.20×1.089×(4/12) ×12,622 = $916. If hand-to-mouth were used, we would have: (1 ) (2) Price, No. of Month $/lb. lb. 1 1.32 2 25 5, 0 0 0 2 1.05 2 25 5, 0 0 0 3 1.10 2 25 5, 0 0 0 4 0.95 2 25 5, 0 0 0 1 00,00 0 a 50,00 50, 000/4 0/4 = 12,5 12,500 00
(3)=(1)× (2) (2) Tota l cost,$ 33,000 26,250 27,500 23,750 $110,500
(4)= (4)=(2 (2)/ )/2 2 A vera ge inventory, lb. 1 2 , 50 0 1 2 , 50 0 1 2 , 50 0 12,500a 1 2,50 0
The average per-lb. cost would be $110,500/100,000 = $1.105. The inventory carrying cost over 4 months would be 0.20×1.105×(4/12) ×12,500 = $921. If 100,000 lbs. of copper were purchased, the two strategies can be compared as follows. Purchase Inventory Total Strategy cost cost cost Dollar averaging $108,900 + 916 = $109,816 Hand-to-mouth 110,500 + 921 = 111,421
Dollar averaging buying would be preferred. 7
For an inclusive quantity discount price incentive plan, we first compute the economic order quantities for each range of price. Using Q*
=
2 DS / IC IC
we compute 138
Q1*
=
2(50 500)(15) / ( 0.20)( 49.95)
=
38. 75 case casess
Q2*
=
2(50 500)(15) / ( 0.20)( 44.95)
=
40. 85 case casess
Since Q2* is outside of the second price bracket, Q1* is is the only relevant quantity. Now we check the total cost at Q1* and at the minimum quantities within the price price break. We solve: TCi
=
Pi D + DS / Qi
+
ICi Qi / 2
At Q = 38.75 TC = = 49.95×500 + 500×15/38.75 + 0.2×49.95×38.75/2 = $25,362
At Q = 50 = 44.95×500 + 500×15/50 + 0.2×44.95×50/2 TC = = $22,850 At Q = 80 = 39.95×500 + 500×15/80 + 0.2×39.95×80/2 TC = = $20,388 Floor polish should be purchased in quantities of 80 cases. 8
This noninclusive price discount problem requires solving the following relevant total cost equation for various order quantities until the minimum cost is found . TCi
=
Pi D + DS / Qi
+
ICi Qi / 2
The computations can be shown in the table below given that D = 1,400, S = = 75, and I = = 0.25.
139
Q 20 50 100 200
P ri c e 795 795 795 795
P× D +D× S/Q +I× C × = Total cost ×Q /2 1,113,000.00 5,250.00 1,987.50 $1,120,237.50 1,113,000.00 2,100.00 4,968.75 1, 120, 068.7 5 1,113,000.00 1,050.00 9,937.50 1 , 1 2 3 , 9 8 7. 5 0 1,113,000.00 525.00 19,875.00 1 1, , 133, 400.0 0
300
200×795+100×750 300
1,092,0 2,000.00 .00
350.0 0.00
29,250. 50.00
1,12 ,121,600 600.00
400
200×795+200×750 400
1,081,5 1,500.00 .00
262.5 2.50
38,625. 25.00
1,12 ,120,387 387.50
500
200×795+200×750
1,068,2 8,200.00 .00
210.0 0.00
47,687. 87.50
1,11 ,116,097 097.50
550
+100×725 500 200×795+200×750
1,063,3 3,363.64 .64
190.9 0.91
52,218. 18.75 1,115,773.27⇐
1,059,3 9,333.33 .33
175.0 5.00
56,750. 50.00
+150×725 550 600
200×795+200×750
1,11 ,116,258 258.33
+200×725 600
The optimal purchase quantity is 550 motors. 9
(a) This problem is a good application of the transportation method of linear programming. We begin by determining the costs for the current sourcing arrangement. S ou r c e D ay t o n D ay t o n Kansas City Minneapolis
Destination Cincinnati Baltimore Dallas Lo Los Angeles
Pr P rice 3.40 3.40 3.45 3. 3.25
Tr T ransport 0.05 0.15 0.08 0.24
Volume Co C ost 5,000 $17,250 1,000 3,5 50 2,500 8,8 25 1,200 4,1 88 Total Total $33 $33,813 ,813
To optimize, we establish the following transportation cost matrix and solve it using any appropriate method, such as the TRANLP module in LOGWARE. Cincinnati 3.40
Dallas 3.44
Minneapolis Minneapolis
3.55
3.53
Los Angeles 3.49 1200 3.65
Baltimore 3.46
1200 3.63 4800
Kansas City
3.45 Dayton Requirements Requirements
Capacity
3.52 5000 5000
3.67 2500 2500
3.55 1200
1000 1000
9999
The total cost for this solution is $33,788, or a savings of $25 over the current cu rrent sourcing.
140
(b) Because Minneapolis is at capacity, this supplier supplier should be examined further. If unlimited capacity were available at Minneapolis, all requirements would be met by this supplier for a total cost of $33,248, or a savings of $565 for this material. (c) The above analysis does indicate indicate that too many suppliers are being used. Only two are needed if Minneapolis continues to supply at the current level. If Minneapolis can be expanded, it becomes the only supplier. Of course, whether the company would risk a single supplier for this material must be left unanswered. 10
(a) The deal-buying equation (Equation 10-5) can be applied to this problem. First, find find the optimal order quantity before the discount. Q*
=
2 DS = IC
2(120,000)( 40) 030 . (100)
=
566 units units
Next, find the adjusted order quantity after the discount has been applied. Q= !
dD
+
pQ *
( p − d ) I p − d
=
10(120 120,000 000) (100 − 5)( 0.30)
+
100( 566) (100 − 5)
=
42,700 700 units units
A large order size of 42,700 units should be placed. (b) The time that an order of this size will be held before it is depleted is given by: Q !
D
=
42,700 700 120 120,000 000
=
0.356 year years, s, or 18.5 18.5 weeks weeks
141
INDUSTRIAL DISTRIBUTORS, INC. Teaching Note Strategy The purpose of the Industrial Distributors case study is to illustrate the computation of purchase quantities under inclusive and noninclusive price discounts and transport rateweight breaks. The INPOL module of LOGWARE is helpful in conducting the analysis. analysis. As a teaching strategy, it may be worthwhile to begin any class discussion with the cost tradeoffs that are present in such such a problem as this. This will help to establish establish the nature of the total cost equation that needs to be solved in this problem. Answers to Questions (1) What size of replenishment orders, to the nearest 50 units, should Walter place, given the manufacturer's noninclusive price policy?
When price discounts are offered, purchase quantities are not simply determined by a single formula. Due to discontinuities in the total cost curve as a function of order quantity, the optimal order quantity is found by computing total costs for different quantity values. In this case of both price price and transport rate rate breaks plus warehousing warehousing costs that can be affected by the order size, the following annual total cost formula is to be solved. TC = PD + RD +
SD Q
+
ICQ
2
+ W ( Q - 3 00 )
where TC = total cost for quantity Q, $ PD = purchase cost for price P , $ RD = transport costs at rate R, $ SD/Q = ordering cost at quantity Q, $ ICQ/2 = carrying cost at quantity Q, $ W(Q-300) = public warehousing cost if Q is greater than 300 units, $ W = public warehousing rate, $ per unit per year D = annual demand, units P = price for orders of size Q, $ per unit R = transport per unit for shipments of size size Q, $ per unit processing cost, $ per order S = order processing I = annual carrying cost, % C = = product value, $ per unit Q = size of purchase order, units
Under noninclusive price discounts, price is an average, determined by the number of units in each break. For example, if 250 units units are to to be ordered, the average price price per unit would be computed as:
142
P 250 =
(100
×
$700 ) + (100
$68 0 ) + ( 50 250 ×
×
$67 0 )
=
$686.00
A table of annual costs can now be developed, as shown in Table 1. To the nearest 50 units, the optimal purchase quantity should be 250 units. (2) If (2) If the manufacturer's manufacturer's pricing policy policy were one where the prices in each quantity break break included all units purchased, should Walter change his replenishment order size?
The average price per unit is more easily determined in this case than the previous one. Since all units are included in the price break back to the first unit, the average price is simply the price associated with a given purchase quantity. Finding the optimal purchase quantity is simply a matter of determining the total cost for the quantities, found by the economic order quantity formula, assuming these quantities are feasible, and for the quantities at the transport rate-weight break. The comparison is made among the total costs of these alternatives. These costs are shown shown in Table 2. The order quantities, as determined by the economic order quantity formula for the base price of $700, would be: Q* =
2 DS = IC
2(1500)( 25 25) = 18.8, or 19 units 0.3( 70 700 + 7.2)
where C is the $700 price per unit at Baltimore plus the $45 transport cost from Baltimore, as determined by an LTL shipment (19 units × 250 lb. = 4,750 lb.) at $18 × 2.5 cwt. = $45 per unit. The Q values for the other prices in the schedule lie outside the feasible range of the price used to compute Q. The optimal strategy is to purchase 201 units per order, which is one unit into the last price break. Yes, Walter should alter his buying strategy. TABLE 1 Annual Costs by Quantity Purchased for Noninclusive Price Discounts Quantity
19 50 100 150 160 200
Average price $700.00
Purchase cost $1,050,000
Transport cost $67,500
700.00 700.00 693.33
1,050,000 1,050,000 1,039,995
67,500 67,500 67,500
750 371 250
5,588 11,287 16,613
0 0 0
1,123,838 1,129,158 1,124,363
692.50
1,038,750
45,000
234
17,340
0
1,101,324
690.00 686.00
1,035,000 1,029,000
45,000 45,000
187 150
21,707 26,850
0 0
1,101,818 1,101,000⇐Opt.
45,000 45,000
125 94
32,100 42,600
0 1,000
250 300 683.33 1,063,286 400 680.00 1,020,000 a EOQ at a price of ($700 + 45) per unit. b First price break. c Transport rate break. d Second price break.
Ordering cost $2,049
Carrying cost $2,045
Warehouse cost $0
Total cost $1,121,594
1,102,225 1,108,694
TABLE 2 Annual Costs by Quantity Purchased for Inclusive Price Discounts
143
19
Average price $700.00
19
680.00
Infeasible
19
670.00
Infeasible
101
680.00
1,020,000
67,500
371
10,984
0
1,098,855
160
680.00
1,020,000
45,000
234
17,040
0
1,082,274
670.00 1,005,000 45,000 201 201 a Feasible EOQ at a price of ($700 + 45) per unit. b Infeasible EOQ at a price of ($680 + 45) per unit. c Infeasible EOQ at a price of ($670 + 30) per unit. d First price break. e Transport rate break. f Second price break.
187
21,105
0
1,032,732⇐Opt.
Quantity
Purchase cost $1,050,000
Transport cost $67,500
Ordering cost $2,049
144
Carrying cost $2,045
Warehouse cost $0
Total cost $1,121,594
CH APTE R 12 12 STORAGE AND HANDLING DECISIONS 2
Various alternatives are evaluated in Tables 12-1 to 12-4. The annual costs of each alternative are plotted in Figure 12-1. The best economic choice is to use all public warehousing.
146
TABLE 12-1 Costs for a Pure Public Warehouse Strategy Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly a b Month lb. sq. ft. cation cost cation cost cost total cost Jan. 1,000,000 62,500 0% $0 $0 100% $30,000c $50,000d $80,000 Feb. 800,000 8 00,000 50,000 0 0 0 100 24,000 40,000 64,000 Mar. 600,000 37,500 0 0 0 100 18,000 30,000 48,000 Apr. 400,000 25,000 0 0 0 100 12,000 20,000 32,000 May 200,000 12,500 0 0 0 100 6,000 10,000 16,000 June 50,000 3,125 0 0 0 100 1,500 2,500 4,000 July 250,000 15,625 0 0 0 100 7,500 12,500 20,000 Aug. 450,000 28,125 0 0 0 100 13,500 22,500 36,000 Sept. 600,000 37,500 0 0 0 100 18,000 30,000 48,000 Oct. 700,000 43,750 0 0 0 100 21,000 35,000 56,000 Nov. 800,000 50,000 0 0 0 100 24,000 40,000 64,000 Dec. 900,000 56,250 0 0 0 100 27,000 45,000 72,000 Totals 6,750,000 421,875 $0 $0 $202,500 $337,500 $540,000 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c Given a turnover ratio of 2 and 100% of the demand through t hrough the rented warehouse, then 1,000,000 × 1.00/2 × 0.06 = $30,000 d 1,000,000 × 1.00 × 0.05 = $50,000
147
TABLE 12-2 Costs for a Mixed Warehouse Strategy Using a 10,000 Square Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 16% $9,792d $3,200e 84% $25,200f $42,000g $80,192 Feb. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 Mar. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Apr. 400,000 25,000 40 9,792 3,200 60 7,200 12,000 32,192 May 200,000 12,500 80 9,792 3,200 20 1,200 2,000 16,192 June 50,000 3,125 100 9,792 1,000 0 0 0 10,792 July 250,000 15,625 64 9,792 3,200 36 2,700 4,500 20,192 Aug. 450,000 28,125 36 9,792 3,200 64 8,640 14,400 36,032 Sept. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Oct. 700,000 43,750 23 9,792 3,200 77 16,170 26,950 56,112 Nov. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 Dec. 900,000 56,250 18 9,792 3,200 82 22,140 36,900 72,032 Totals 6,750,000 421,875 $117,504 $36,200 $147,930 $246,550 $548,184 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 10,000/62,500 = 0.16 d (35×10,000/20) + 10×10,000/12 = $9,792 per month e 1,000,000×0.16×0.02 = $3,200 f Given a turnover ratio of 2 and 84% of the demand through the rented warehouse, then 1,000,000 × 0.84/2 × 0.06 = $25,200 d 1,000,000 × 0.84 × 0.05 = $42,000
TABLE 12-2 Costs for a Mixed Warehouse Strategy Using a 10,000 Square Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 16% $9,792d $3,200e 84% $25,200f $42,000g $80,192 Feb. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 Mar. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Apr. 400,000 25,000 40 9,792 3,200 60 7,200 12,000 32,192 May 200,000 12,500 80 9,792 3,200 20 1,200 2,000 16,192 June 50,000 3,125 100 9,792 1,000 0 0 0 10,792 July 250,000 15,625 64 9,792 3,200 36 2,700 4,500 20,192 Aug. 450,000 28,125 36 9,792 3,200 64 8,640 14,400 36,032 Sept. 600,000 37,500 27 9,792 3,200 73 13,140 21,900 48,032 Oct. 700,000 43,750 23 9,792 3,200 77 16,170 26,950 56,112 Nov. 800,000 50,000 20 9,792 3,200 80 19,200 32,000 64,192 Dec. 900,000 56,250 18 9,792 3,200 82 22,140 36,900 72,032 Totals 6,750,000 421,875 $117,504 $36,200 $147,930 $246,550 $548,184 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 10,000/62,500 = 0.16 d (35×10,000/20) + 10×10,000/12 = $9,792 per month e 1,000,000×0.16×0.02 = $3,200 f Given a turnover ratio of 2 and 84% of the demand through the rented warehouse, then 1,000,000 × 0.84/2 × 0.06 = $25,200 d 1,000,000 × 0.84 × 0.05 = $42,000
148
TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 30,000 Square Foot Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 48%c $29,375d $9,600e 52% $15,600f $26,900g $80,575 Feb. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 Mar. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Apr. 400,000 25,000 100 29,375 8,000 0 0 0 37,375 May 200,000 12,500 100 29,375 4,000 0 0 0 33,375 June 50,000 3,125 100 29,375 1,000 0 0 0 30,375 July 250,000 15,625 100 29,375 5,000 0 0 0 34,375 Aug. 450,000 28,125 100 29,375 9,000 0 0 0 38,375 Sept. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Oct. 700,000 43,750 69 29,375 9,600 31 6,510 13,020 58,505 Nov. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 Dec. 900,000 56,250 53 29,375 9,600 47 12,690 21,150 21, 150 72,815 Totals 6,750,000 421,875 $352,500 $94,200 $61,200 $104,170 $612,070 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 30,000/62,500 = 0.48 d (35×30,000/20) + 10×30,000/12 = $29,375 per month e 1,000,000×0.48×0.02 = $9,600 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000 × 0.52/2 × 0.06 = $15,600 d 1,000,000 × 0.52 × 0.05 = $26,000
TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 30,000 Square Foot Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 48%c $29,375d $9,600e 52% $15,600f $26,900g $80,575 Feb. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 Mar. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Apr. 400,000 25,000 100 29,375 8,000 0 0 0 37,375 May 200,000 12,500 100 29,375 4,000 0 0 0 33,375 June 50,000 3,125 100 29,375 1,000 0 0 0 30,375 July 250,000 15,625 100 29,375 5,000 0 0 0 34,375 Aug. 450,000 28,125 100 29,375 9,000 0 0 0 38,375 Sept. 600,000 37,500 80 29,375 9,600 20 3,600 6,000 48,575 Oct. 700,000 43,750 69 29,375 9,600 31 6,510 13,020 58,505 Nov. 800,000 50,000 60 29,375 9,600 40 9,600 16,000 64,575 Dec. 900,000 56,250 53 29,375 9,600 47 12,690 21,150 21, 150 72,815 Totals 6,750,000 421,875 $352,500 $94,200 $61,200 $104,170 $612,070 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 30,000/62,500 = 0.48 d (35×30,000/20) + 10×30,000/12 = $29,375 per month e 1,000,000×0.48×0.02 = $9,600 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000 × 0.52/2 × 0.06 = $15,600 d 1,000,000 × 0.52 × 0.05 = $26,000
149
TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 40,000 Square Foot Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 64%c $39,167 $12,800e 36% $10,800f $18,000g $80,767 Feb. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 Mar. 600,000 37,500 100 39,167 12,800 0 0 0 51,167 Apr. 400,000 25,000 100 39,167 8,000 0 0 0 47,167 May 200,000 12,500 100 39,167 4,000 0 0 0 43,167 June 50,000 3,125 100 39,167 1,000 0 0 0 40,167 July 250,000 15,625 100 39,167 5,000 0 0 0 44,167 Aug. 450,000 28,125 100 39,167 9,000 0 0 0 48,167 Sept. 600,000 37,500 100 39,167 12,000 0 0 0 51,167 Oct. 700,000 43,750 91 39,167 12,800 0 1,890 3,150 57,007 Nov. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 Dec. 900,000 56,250 71 39,167 12,800 29 7,830 13,050 13, 050 72,847 Totals 6,750,000 421,875 $470,004 $115,000 $30,120 $50,200 $665,324 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 40,000/62,500 = 0.64 d (35×40,000/20) + 10×40,000/12 = $39,167 per month e 1,000,000×0.64×0.02 = $12,800 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000 × 0.36/2 × 0.06 = $10,800 d 1,000,000 × 0.36 × 0.05 = $18,000
TABLE 12-3 Costs for a Mixed Warehouse Size Strategy Using a 40,000 Square Foot Foot PrivatelyOperated Warehouse Privately-operated Rented WareSpace house require Private Monthly Monthly Rented Monthly Monthly thruput, ments, allo fixed cost variable allo storage handling Monthly Month lb.a sq. ft.b cation cost cation cost cost total cost Jan. 1,000,000 62,500 64%c $39,167 $12,800e 36% $10,800f $18,000g $80,767 Feb. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 Mar. 600,000 37,500 100 39,167 12,800 0 0 0 51,167 Apr. 400,000 25,000 100 39,167 8,000 0 0 0 47,167 May 200,000 12,500 100 39,167 4,000 0 0 0 43,167 June 50,000 3,125 100 39,167 1,000 0 0 0 40,167 July 250,000 15,625 100 39,167 5,000 0 0 0 44,167 Aug. 450,000 28,125 100 39,167 9,000 0 0 0 48,167 Sept. 600,000 37,500 100 39,167 12,000 0 0 0 51,167 Oct. 700,000 43,750 91 39,167 12,800 0 1,890 3,150 57,007 Nov. 800,000 50,000 80 39,167 12,800 20 4,800 8,000 64,767 Dec. 900,000 56,250 71 39,167 12,800 29 7,830 13,050 13, 050 72,847 Totals 6,750,000 421,875 $470,004 $115,000 $30,120 $50,200 $665,324 a Thruput (lb.) = Sales ($)/($5/lb.) b Space requirements (sq. ft.) = Thruput T hruput (lb.) × ½ × 1/0.40 × 0.1/10(5) = Thruput × 0.6250 c 40,000/62,500 = 0.64 d (35×40,000/20) + 10×40,000/12 = $39,167 per month e 1,000,000×0.64×0.02 = $12,800 f Given a turnover ratio of 2 and 52% of the demand through the rented warehouse, then 1,000,000 × 0.36/2 × 0.06 = $10,800 d 1,000,000 × 0.36 × 0.05 = $18,000
150
FIGURE 12-1 Total Annual Costs for a Combined Warehouse Size Using Private and Public Warehouse Space
670 650 s 0 0 0 $ , t s o c l a t o T
630 610 590 570 550 530 0
1 0 ,0 0 0
3 0 ,0 0 0
4 0 ,0 0 0
Private warehouse space, sq. ft. 3
The annual cost of public warehousing is:
FIGURE 12-1 Total Annual Costs for a Combined Warehouse Size Using Private and Public Warehouse Space
670 650 s 0 0 0 $ , t s o c l a t o T
630 610 590 570 550 530 0
1 0 ,0 0 0
3 0 ,0 0 0
4 0 ,0 0 0
Private warehouse space, sq. ft. 3
The annual cost of public warehousing is: Handling Storage Total
$ 600,000 300,000 $ 900,000
The costs of private warehousing are: Annual operating $ 250,000 Annual lease payment 3×150,000 = 450,000 Other fixed (one time) 400,000 The savings in operating costs of lease vs. v s. public warehousing is: Savings = $900,000 − 250,000 = $650,000/yr.
151
TABLE 12-5 Ten-Year Cash Flow Stream for Public vs. Leased Warehouse Comparison Savings DepreSavings less Savings: Pre-tax ciation less depre After-tax Lease vs. net cash schedepreTaxes ciation Savings net cash Year public flow dule ciation (35%) & tax less tax flow a 0 $0 (3,050) $0 0 0 0 0 ($3,050) b c 1 650 650 57 593 208 385 442 442 2 650 650 57 593 208 385 442 442 3 650 650 57 593 208 385 442 442 4 650 650 57 593 208 385 442 442 5 650 650 57 593 208 385 442 442 6 650 650 57 593 208 385 442 442 7 650 650 58 592 207 385 443 443 8 650 650 0 650 228 422 442 442 9 650 650 0 650 228 422 442 442 10 650 650 0 650 228 422 442 442 $6,500 $3,450 $400 $6,100 $2,139 $3,961 $4,361 $1,311 a Capitalization lease plus initial cash outlay, i.e., $2,650,154 + 400,000 = $3,050,154 b Depreciation charge for each of seven years is 1/7 = 0.1429 such that 400,000 ×0.1429 = $57,143 c Add back depreciation, i.e., 385 + 57 = $442
Discount factor 1/(1+i) j
0.9009 0.8116 0.7312 0.6587 0.5935 0.5346 0.4817 0.4339 0.3909 0.3522 NPV =
Discounted cash flow ($3,050) 398 359 323 291 262 236 213 183 165 149 ($471)
152
Capitalizing the lease over ten years, we have: h ave: (1 + 0.11)10 − 1 450,000 000 PV = 450 0.11(1 + 0.11)10
=
$2,650154 ,
The initial investment in $000s then is: Initial investment = $2,650 + 400 = $3,050 The ten-year cash flow stream stream is shown in Table 12-5. Since the savings are expressed to favor leasing and the net present value is negative, choose public warehousing. 4
Given: = $210/sq ft. k = = 100,000 sq. ft. S =
Capitalizing the lease over ten years, we have: h ave: (1 + 0.11)10 − 1 450,000 000 PV = 450 0.11(1 + 0.11)10
=
$2,650154 ,
The initial investment in $000s then is: Initial investment = $2,650 + 400 = $3,050 The ten-year cash flow stream stream is shown in Table 12-5. Since the savings are expressed to favor leasing and the net present value is negative, choose public warehousing. 4
Given: = $210/sq ft. k = = 100,000 sq. ft. S = = $0.01/ft.×10,000 = $100/ft. C = The width is: W *
=
C + 8 k
2C + 8k = =
S
100 + 8( 210) 100 100,000 000 2(10 100) + 8(210) 308 ft.
The length is: L*
=
S / W *
= 100,000 /
308 = 325 ft.
5
Space layout according to text Figure 12-4(a) can be determined by the application of Equations 12-8 and 12-9. These equations specify the best number of shelf spaces and the best number of double racks, respectively. Equations 12-10 and 12-11 give the the length and width of the building. The optimal number of shelf spaces would be:
153
m1*
=
1 ! dCh + 2aCs + 2C p $ ! K ( w + a ) L $ # & &% 2h L #" 2( dCh + C p ) &% #"
=
1 ! 400, 00 000( 0.001) + 2(10 10)( ) ( 0.50) + 2( 3. 00) $ ! 50, 00 000( 8 + 10)( 4) $ # & # & 4 " 2( 400,00 000)( 0. 00 001) + 3.00 2( 4) %" %
. , = 12048
or 121
The optimal number of double racks would be:
n1*
=
! 2( dCh + C p ) $ ! K ( w + a ) L $ # & &% 2h w + a #" dCh + 2aCs + 2C p &% #"
=
! $ ! 50, 00 1 2[( 400, 00 000)( 0.001) + 3.00] 000(8 + 10)( 4) $ # & # & 8 + 10 " 400, 00 000( 0.001) + 2(10 10)( ) ( 0.50) + 2( 3. 00) % " 2( 4) %
=
52
1
The warehouse length would be: u1
=
n1* ( w + a ) = 52( 8 + 10) = 936 ft.
and the width would be: v1
=
2a + m1* L = 2(10) + 121( 4) = 504 ft.
6
According to Equation 12-17, the number of truck doors can be estimated by: N =
DH CS
Therefore, N = (75×12,000) ×3/(3×12,000) ×8 = 9.37, or 10 doors 7
Summarizing the given information as follows:
Initial investment Useful life Salvage Salvage value value @15% of initial cost
Three Fi v e S ev e n type 1 typ e 2 type 3 units units u n it s $60,000 $50,000 $35,000 10 yr. 10 yr. 10 yr. $ 9,000
$ 7,500
154
$ 5,250
Annual operating operating expenses Return Return on investme investment nt before tax
$ 6,000
$12,500
20%
$21,000
20%
20%
An initial solution to this problem can be found through a discounted cash flow analysis. Three alternatives are to be evaluated.
! (1 + 0.2)10 − 1$ ! $ 1 0 00 + 6,00 0 00# 9 , 0 0 0 PV 1 = 60,00 − # (1 + 0.2)10 & 10 & " % " 0.2(1 + 0.2) %
PV 2
PV 3
=
60,00 000 + 6,00 000( 4.2) − 9, 00 000( 0.16)
=
$83,760
=
50,00 000 + 12,50 500( 4. 2) − 7, 50 500( 0.16)
=
50,00 0 00 + 52,50 500 − 1, 20 200
=
$101,300
=
35,00 000 + 21, 00 000( 4. 2) − 5, 25 250( 0.16)
=
35,00 000 + 88,20 2 00 − 8 40
=
$122,360
The low present value of the Type 1 truck indicates that from among these three alternatives, this would be the best buy. 8
Given: Initial cost of equipment = $4,000 Operating costs 500 + 40(t - 1)2 - 30(t - 1) Salvage value S n = I (1 (1 – t /7) /7) Rate of return on investment = 20% Replacement is expected to be with equipment of like kind The best replacement year can be found by comparing the equivalent annual cost of a sequence of similar equipment replaced every n years. The equivalent annual cost is: AC n
=
I +
'
n j=1
( C j /(1 + i ) j ) − ( S n /(1 + i ) n ) [i (1 + i ) n /(1 + i ) n
]
−1
Solving this equation for different years is facilitated if the equation is set up in tabular form, as shown in Table 12-6. The equipment should be replaced at the end of the third year of service although a 5year replacement cycle is also attractive.
155
TABLE 12-6 Equivalent Annual Cost Computations for Problem 8 (1) (2) (3) (4) (5)=(1+2-3)(4) Operating Salvage Factor Year, Initial costs, value, Equivalent annual i(1 + i ) n n j n S n / (1 + i) n investment, I ' C j / (1 + i) cost, AC n (1 + i ) − 1
1 2 3 4 5 6 7
$4,000 4,000 4,000 4,000 4,000 4,000 4,000
$416 770a 1,117 b 1,488 1,898 2,350 2,841
$2,857 1,984 1,323 827 459 191 0
a
$416 + [500 + 40(2-1) 2 – 30(2-1)]/(1+0.20) 2 = $770
b
$770 + [500 + 40(3-1) 2 – 30 (3-1)]/1+0.20) 3 = $1,117
1.20 0.65 0.47 0.39 0.33 0.30 0.28
$1,871 1,821 1,783 ⇐ 1,818 1,795 1,848 1,915
9
(a1) Layout by popularity involves locating the more frequently ordered items closest to the outbound dock. Based on the average number of daily orders on which the the item appears, the items closest to the outbound dock would be ranked as follows: B,I,E,A,F,H,J,C,G,D The storage space might then be: Inbound
D
J, C, G
H, F, A
H, J
A, E
E
B
B, I, E
Outbound (a2) Layout by cube places the smallest items nearest the outbound dock. Using the individual item size, the ranking would be b e as follows: A,E,I,C,J,H,G,B,F The layout of items in the storage bays b ays would be:
156
Inbound
F, D
B
H, D
H, G, B
D, J, C
E, A
E, I, C
E
Outbound (a3) The cube-per-order index is created by ratioing the average required cubic footage of a product to the average number of daily orders on which the item is requested. Hence, this index is found as follows:
Product A B C D E F G H I J
(1) Space required cu. ft. 5,000a 30,000 15,000 17,000 55,000 11,000 7,000 28,000 13,000 9,000
(2)
(3 ) =( 1 ) / ( 2 )
Da Daily orders 56 103 27 15 84 55 26 45 94 35
CPO index 89 291 556 1,133 655 200 269 622 138 257
a
500 sq. ft. stacked 10 ft. high
Locating the products with the lowest index values nearest to the outbound dock results in the following ranking and layout: A,I,F,J,G,B,C,H,E,D
157
Inbound
D
E
E, F
E
H, C
B
F, I, A
F, J, G, B
Outbound (b) All of the above methods assume (1) that the product is moved to the storage locations in large unit loads, but retrieved from the storage locations in relatively small quantities and (2) that only one product is retrieved during an out-and-back trip. Therefore, these methods do not truly apply to the situation of multiple picks on the same trip. However, they may be used with some degree degree of approximation if the products can be aggregated as one and grouped together or zoned in the same section of the warehouse. 10
This extra challenging problem requires some knowledge of linear programming. It may be formulated as follows: Let X ijij represent the amount per 1,000 units of product j stored in location i. Let C ijij be the handling time associated with storage bay i and product j. G j is the capacity of a bay for product j and R j is the number of units of product j required to be stored. The linear programming statement is: Objective function Zmin = .90 .90X 11 .75X 12 .90X 13 .80X 21 .65X 22 .95X 23 11 + .75 12 + .90 13 + .80 21 + .65 22 + .95 23 + .60 .60X 31 + .70 .70X 32 + .65 .65X 33 + .70 .70X 41 + .55X 42 + .45X 43 + .50 .50X 51 + .50 .50X 52 + .45 .45X 53 + .40 .40X 61 + .45X 62 + .35X 63
Subject to: Capacity restrictions on bays
158
20X 11 33.3X 12 16.7X 13 11 + 33.3 12 + 16.7 13
≤
100
20X 21 33.3X 22 16.7X 23 21 + 33.3 22 + 16.7 23
≤
100
20X 31 + 33.3 33.3X 32 + 16.7 16.7X 33
≤
100
20X 41 33.3X 42 16.7X 43 41 + 33.3 42 + 16.7 43
≤
100
20X 51 33.3X 52 16.7X 53 51 + 33.3 52 + 16.7 53
≤
100
20X 61 + 33.3 33.3X 62 + 16.7 16.7X 63
≤
100
and storage requirements restrictions on products X 11 11 + X 21 21 + X 31 31 + X 41 41 + X 51 51 + X 61 61
≥
11
X 12 + X 22 + X 32 + X 42 + X 52 + X 62
≥
4
X 13 + X 23 + X 33 + X 43 + X 53 + X 63
≥
12
Solving the linear programming problem by means of any standard transportation code of linear programming, such as LNPROG in LOGWARE, yields: The total minimum handling time is 138.68 hours
X 12 = 1.610 1.610 X 21 = 1.020 1.020 X 22 2.390 22 = 2.390 X 31 = 5.000 5.000 X 43 5.988 43 = 5.988 X 51 4.980 51 = 4.980 0.024 X 53 53 = 0.024
X 63 = 5.988 5.988
where X s are in thousands of units. That is, product 1 should be stored in bays 3, 4, and 5 in quantities of 1,020, 5,000, and 4,980, respectively. Product 2 should be stored in bays 1 and 2 in quantities of 1,610 and 2,390, respectively. Product 3 should be stored in bays 4, 5, and 6 in quantities of 5,988, 24, and 5,988, respectively. Graphically, this is:
Bay Product 1 2 3 % of bay capacity
1
2
3 5,000
1,610
1,020 2,390
53.7
100.0
100.0
159
4
5
6
4,980 5,988
24
5,988
100.0
100.0
100.0
Require -ments 11,000 4,000 12,000
CH APTE R 14 14 THE LOGISTICS PLANNING PROCESS 3
The MILES module within the LOGWARE software is used to solve this problem. It computes distance based on the great circle distance formula using longitude and latitude. (a) The estimated road distance is 1,380 miles. (b) The estimated road distance is 830 miles. (c) Since both latitudes are in the same hemisphere, no adjustments need to be made. The estimated distance is 244 miles, or 244×1.61 = 393 km. (d) In this case, one point is east and the other west of the Greenwich line. Therefore, we need to set a sign convention. Let's set west longitudes as + and east longitudes as −. Thus, 2.20o E longitude is entered into MILES as −2.20 o . The estimated distance is 250 miles, or 250×1.61 = 402.5 km. 4
Suppose that a certain linear grid coordinate system has been overlaid on a map of the United States. The grid numbers are calibrated in miles, miles, and there is a road road circuity factor of 1.21. Find the expected road distances between the following following pairs of points: Equation 14-1 in the text is used to approximate distances from linear coordinates. The K factor factor in the equation is set at 1.21. (a) Lansing, MI to Lubbock, TX L o c a t i on a. From To b. From To c. From To d. From To
D
= 1.21
X C oor di nat e Lansing, MI Lubbock, TX El Paso, TX Atlanta, GA Boston, MA Los Angeles, CA Seattle, WA Portland, OR
( 924.3 − 1, 488.6)2
Y Co o r din a t e 924.3 1488.6 1696.3 624.9 374.7 2365.4 2668.8 2674.2
+
(1, 675. 2 − 2, 579. 4) 2
+
( 2, 769.3 − 2, 318. 7)2
=
1 67 5 . 2 2 579.4 2 7 6 9. 3 2 318.7 1 3 2 6. 6 2 76 3 . 9 1 900.8 2 0 3 9. 7
1, 290 miles
(b) El Paso, TX to Atlanta, GA D = 1.21 (1, 696.3 − 624.9)2
(c) Boston, MA to Los Angeles, CA
202
=
1, 406 miles
763.9)2 D = 1.21 ( 374.7 − 2,365.4)2 (1, 326. 6 − 2, 76
=
2, 971 miles
(d) Seattle, WA to Portland, OR D = 1.21 ( 2,668.8 − 2, 674.2)2
+
(1, 900.8 − 2, 039. 7)2
=
168 miles
5
The plot of the truck class rates is shown in Figure Figure 14-1. The rates show a high degree of linearity. A linear regression was found with aid of the MULREG module in LOGWARE. The rate equation was determined to be: R = 5.1745 + 0.0041× D
The standard error of the estimate S E is 0.9766. The coefficient of determination r 2 is 0.928. The best single estimate of the rate at 500 miles is: R = 5.1745 + 0.0041×500 = $7.23/cwt.
Assuming the error around the regression line is normally distributed, a 95 percent confidence band would give a range for the actual rate. That is, Y = R ± 1.96×S E = 7.23 ± 1.914
where 1.96 is the normal deviate for the normal distribution representing 95 percent of the area in a two-tailed two-tailed distribution. The range of the estimate is: is: $5.32/cwt. ≤ Y ≤ $9.14/cwt. The r 2 value of 0.928 indicates that a linear rate equation explains about 93 percent of the variation in the data with distance. Such a simple relationship seems to represent the rates quite well.
203
FIGURE 14-1 Plot of Truck Class Rates 20 18 16 . 1 4 t w c 1 2 / $ , e 1 0 t a r 8 s s 6 a l C 4
Estimating line
2 0 0
500
1000
1500
2000
2500
3000
3500
Distance, miles 6
A plot of the average inventory level versus warehouse throughput is shown in Figure 142. The multiple regression software software in LOGWARE was used to test two equation forms. The first was of the form I
=
aTP b
and the other was of the form I
=
a + bTP
Both forms showed high r 2 values, with the exponential form being slightly higher at 0.9406. It was selected as the equation form to use. This equation was: I
=
0.83 0704 . × TP
where TP and and I are are both expressed in thousands of dollars. We can now estimate that for an annual warehouse throughput of $50,000,000, the average inventory would be: 704 × 50, 0000.83 I = 0.70 =
5,59393 593939 . 9, or $5,593,939 $5,593,939
This type of relationship is very useful in network planning, especially warehouse location, to estimate how inventory levels will change when sales are reallocated to a varying number of warehouses.
204
FIGURE 14-2 Plot of Inventory Levels and Warehouse Throughput for California Fruit Growers’ Association )
s 12 n oi ll i
M( $ e v el yr ot n e v ni e
10
l, 8
6 4
g 2 ar e v 0 A
Estimating line
0
20
40
60
Annual warehouse thruput, $(Millions)
205
80
100
USEMORE SOAP COMPANY Teaching Note
The purpose of this case study is to provide students with the opportunity to evaluate and design a large-scale production-distribution network using real data and cost relationships. To assist in the substantial amount of computational effort in this this problem, an interactive computer program (WARELOCA) is available in the LOGWARE collection of software modules. Major Issues The text of the case suggests a number of questions that are critical to productiondistribution network network design. These reduce to three major major issues, namely:
(1) Should plant capacity be added and, if so, when and where? (2) How many warehouses are optimal and where should they be located? (3) Should the current customer service level be retained? Although no change can be made in the network without potentially affecting other variables, the attempt here will be to treat these questions sequentially to converge on a good network design. Numerous computer runs were made to provide the basic information needed ne eded in the analysis. The more meaningful runs are summarized in Appendix A to this note. Tables 1 and 2 compare selected runs for both the current-year and the future-year time periods. This information is used throughout the analysis of the major issues. The Plant Expansion Issue An attempt to meet 5-year growth goals using current plant capacity will cause the system having a total capacity of 1,630,000 cwt. to be out of capacity in 1.7 years. That is,
5th-year demand Current demand Net increase
1,908,606 cwt. −1,477,026 431,580 cwt.
Therefore, the average annual growth rate rate is 431,580/5 = 86,316 cwt. So, in (1,630,000 − 1,477,026)/86,316 = 1.7 years all available capacity will be depleted. If no expansion of plant capacity occurs, then 1,908,606 − 1,630,000 = 278,606 cwt. will potentially be lost by the 5th year. Sales are $100 million on 1.477 million cwt. in volume for a product product value of $67.7/cwt. With a profit margin of 20 percent, the profit per cwt. would be 20%×$67.7/cwt, or $20/1.477, = $13. Thus, 278,606×13 = $3.16 million are lost in sales. The weighted profit loss over the five-year period would be: 2/5× (0) + ( 3/5)× ((0 + 3.6))/2) = $1.08m/yr.
206
TABLE 1 Current-Year Comparison of Network Alternatives ($000s)
C o s t t y pe
B e n c hm a rk
Production $30,762 Warehouse operations 1,578 Order processing 369 Inventory carrying 457 Transportation Inbound 2,050 Outbound 6,896 Total costs $42,112
I m p r o v ed b e n c hm a rk
O p t i m um n u m b er of whses
O p t i m um n u m b er of o f whses
R e l a x ed se s ervice (1)
R e l a x ed se s ervice (2)
Maxim um op p o r tu t unit y
$30,678 1,468 354 431
$3 $ 30,673 1,608 370 508
$3 $ 30,675 1,572 358 490
$3 $ 30,678 1,296 349 390
$30,673 1,420 354 445
$30 ,3 86 1 ,5 29 3 58 5 00
1,802 6,991 $41,725
1,976 6,310 $41,447
1,860 6,365 $41,321
1,249 7,238 $41,201
1,178 6,698 $41,043
1, 1 78 6, 4 58 $4 0, 4 09
Customer service: ≤
300 mi.
93 %
93 %
98%
92%
75%
88%
81 %
≤
600 mi.
98 %
98 %
100%
100%
98%
100%
9 4%
No. of stock stockin ing g poi n t s
22
21
31
30
19
26
40
N o . o f pl an t s
4
4
4
4
4
4
6
Savings Savings vs. benchmark
$0
$387
$665
$ 79 1
$9 1 1
$1,069
$1 ,7 0 3
$0
$278
$ 40 4
$5 2 4
$
$ 1, 31 6
Service to match benchma r k
600 mi constraint on current warehouses
600 mi constraint on opt no. of warehouses
Savings Savings vs. improve improved d benchmark $0 Comments:
682
U nli m it e d se rv i ce , wh ses , an d plant cap.
207
TABLE 2 Future-Year Comparison of Alternatives ($000s)
C o s t t y pe
Production Warehouse operations Order processing Inventory carrying Transportation Inbound Outbound Total costs
No p l a nt expa nsio n
A dd p l a nt @ M e m p h is
Add plant @ M e m p h is & C h i c a go
M e m p h is a n d o pt n o . of w h s e s
Mem p his a nd o p t n o . of w h s e s
$33,965 1,496 393 431
$39,517 1,842 462 505
$ 3 9,5 4 8 1, 84 7 45 4 49 7
$39,524 2,028 470 591
$ 39 , 52 2 1 , 97 6 4 60 57 3
1,647 7,230 $45,164
2,350 9,030 $53,705
2,0 0 0 9 , 03 6 $ 5 3, 38 2
2,614 8,117 $53,342
2 , 42 6 8 , 22 2 $ 53 , 17 9
Customer service: ≤
300 mi
98%
94 %
95%
98%
92%
≤
600 mi
99%
98 %
98%
100%
1 00 %
No. of stock stockin ing g poi n t s
20
21
20
31
30
N o . o f pl an t s
4
5
6
5
5
Comments:
Not all demand met
High service level
S er v ic e he l d a t be n ch ma r k
TABLE 2 Future-Year Comparison of Alternatives ($000s)
C o s t t y pe
Production Warehouse operations Order processing Inventory carrying Transportation Inbound Outbound Total costs
No p l a nt expa nsio n
A dd p l a nt @ M e m p h is
Add plant @ M e m p h is & C h i c a go
M e m p h is a n d o pt n o . of w h s e s
Mem p his a nd o p t n o . of w h s e s
$33,965 1,496 393 431
$39,517 1,842 462 505
$ 3 9,5 4 8 1, 84 7 45 4 49 7
$39,524 2,028 470 591
$ 39 , 52 2 1 , 97 6 4 60 57 3
1,647 7,230 $45,164
2,350 9,030 $53,705
2,0 0 0 9 , 03 6 $ 5 3, 38 2
2,614 8,117 $53,342
2 , 42 6 8 , 22 2 $ 53 , 17 9
Customer service: ≤
300 mi
98%
94 %
95%
98%
92%
≤
600 mi
99%
98 %
98%
100%
1 00 %
No. of stock stockin ing g poi n t s
20
21
20
31
30
N o . o f pl an t s
4
5
6
5
5
Comments:
Not all demand met
High service level
S er v ic e he l d a t be n ch ma r k
208
Based on a simple rate of return on investment, capturing this profit potential would yield 1.08/4 = 27 percent annually on a $4,000,000 investment investment for expansion. The return would increase to 90 percent per year with the full loss in the 5th year. year. The potential seems great enough to justify one unit of expansion (1,000,000 cwt.). Two units of expansion probably cannot be justified, since adequate capacity would be available from the first capacity unit to meet demand requirements. requirements. The only benefit would be from from the network design improvement. The savings would be about $323,000 per year in the fifth year (see Table 2) comparing one additional plant with two additional plants and keeping the current number of warehouses. The simple return on investment using fifth-year savings would only amount to about 8 percent (323,000×100/4,000,000 = 8.1%). The next question is: Where should the expansion take place at at an existing plant or at one of the two proposed locations? From a test of expanding any of the four existing plants or the two proposed plant locations (runs 10 through 16 in Appendix A of this note), it would appear that Memphis would be the lowest cost site in the 5th year with Chicago next at only an additional cost of $76,000 per year (compare runs 14 and 15 in Appendix A). Adding a plant at a new location rather than expanding an existing plant
Based on a simple rate of return on investment, capturing this profit potential would yield 1.08/4 = 27 percent annually on a $4,000,000 investment investment for expansion. The return would increase to 90 percent per year with the full loss in the 5th year. year. The potential seems great enough to justify one unit of expansion (1,000,000 cwt.). Two units of expansion probably cannot be justified, since adequate capacity would be available from the first capacity unit to meet demand requirements. requirements. The only benefit would be from from the network design improvement. The savings would be about $323,000 per year in the fifth year (see Table 2) comparing one additional plant with two additional plants and keeping the current number of warehouses. The simple return on investment using fifth-year savings would only amount to about 8 percent (323,000×100/4,000,000 = 8.1%). The next question is: Where should the expansion take place at at an existing plant or at one of the two proposed locations? From a test of expanding any of the four existing plants or the two proposed plant locations (runs 10 through 16 in Appendix A of this note), it would appear that Memphis would be the lowest cost site in the 5th year with Chicago next at only an additional cost of $76,000 per year (compare runs 14 and 15 in Appendix A). Adding a plant at a new location rather than expanding an existing plant site saves a minimum of $281,000 annually (compare runs 11 and 14 in Appendix A of this note), which results from placing plant capacity closer to warehouses. Selecting Warehouses A simple test on the number of warehouses in the network shows that transportation costs are dropping more rapidly than inventory related costs are increasing (see Figure 1). This means that 40 active warehouses will have the lowest total total cost. However, some of these warehouses will have low throughput. throughput. In order to maintain a minimum replenishment frequency and shipment size, size, a minimum throughput needs to be met. Approximately a truckload every two weeks, or 10,400 cwt. of throughput per year, is the minimum activity needed to open a warehouse. Therefore, any warehouse showing less than this throughput will be eliminated from consideration. Under various assumptions about plants and their capacities, demand growth, and service levels, 30 to 31 warehouses seem most economical with no deterioration on service over the benchmark network. The following table shows selected results.
209
T y p e of
P l a nt
Percent of d e m a n d
T o t al
N o. of
≤
co s t
w hs es
r un
Y e ar
c ap aci ti es
Ben chmar k Improved ben chmar k
Current
C urren t
93
$ 4 2, 11 2
22
Current
C urren t
93
4 1 , 72 5
21
5th yr.
C urren t + Memphis
94
53,70 5
31
92
41,32 1
3 30 0
5th year
C u rr e n t C u r re n t + Memphis
92
53,17 9
30
Improved ben chmar k Current yr. whses 5th yr. whses
3 00 m i .
Note that this conclusion about the number of warehouses depends on the previous conclusion that a Memphis plant should should be added by the fifth year. year. The number of warehouses should be increased from the present 22 in both the current year and the fifth year. 42
100 99
41.8
98 Service (right scale)
97
) s 0 0 0 , 41.6 0 0 0 ( $ , t s o c 41.4 l a t o T
96 95 94
Cost (left scale)
93
41.2
. i m 0 0 3 < d n a m e d f o %
92
Practical design
91 41
90 22
26
30
31
36
40
Number of warehouses
FIGURE 1 Cost and Customer Service Profiles for Alternative Network Designs
More detailed economic analysis shows that if the plants are held at current throughput levels, a savings realized from 30 warehouses would be $41,725,000 − 41,321,000 = $404,000 (see previous table). If current plant capacities are used and the Memphis plant is on-stream in year five, the savings of the added warehouse would be: $53,705,000 − 53,179,000 = $526,000
210
On the average, there can be savings of approximately ($404,000 + 526,000)/2 = $465,000 per year by increasing the number of warehouses to 30 from the current 22. Since these are public warehouses, little or no investment would be required to implement the change. Although the number of warehouses remains relatively unchanged from the current year to the 5th year, there is some shifting among the particular warehouses in the mix. The 30 warehouses in the current year should be numbers: 1,2,3,4,5,7,8,11,13,14,15,16,17,18,19,20,21,25,28,31,32,33,34,35,36,37,38,40,44,45 providing that the loading lo ading on the current plants is allowed up to the limits of their current capacity. When the Memphis plant is brought brought on-stream by the end of the second year, the warehouse mix should begin to evolve to numbers: 1,2,3,4,5,7,8,11,13,14,15,17,18,19,20,21,25,28,29,31,32,34,35,36,37,38,40,44,45,47 As the Memphis plant is bought on stream, the Memphis public warehouse is closed and the volume is is shifted to the Memphis plant as a warehouse. In addition, the Richmond, VA warehouse is closed and the Las Vegas, NV warehouse warehouse is opened. The number of warehouses remains at 30. Both in the base year and in the future year, the throughputs in the plants serving as warehouses are within acceptable limits as the following summary shows. P l a nt as a wa r e hou s e
Covin vington New York Arl ingto n Long Beach
T h r u p ut l imi t s
C u r r e n ty e ar s o l u t i on
Fu tu re y e ar so lut io n
450,000 000 cwt. wt. 380,000 140,000 180,000
254,47 ,471 cwt cwt. 302,043 66,592 95,943
306, 06,478 cwt. 38 0,523 6 6,161 11 7,288
Customer Service Currently, a high proportion of demand (93 percent) is located within 300 miles of a stocking point. Since the service distance may be up to 600 miles and still meet the company's service policy, should the service level be reduced somewhat to effect a cost saving? For example, using the improved benchmark as the the base case (run 2), 93 percent of the demand is within within 300 miles and 97.5 percent is within 600 miles. If a 600-mile 600-mile constraint is applied to the current network configuration (run 23), 75 percent of the demand is within 300 miles and 98 percent is still still within 600 miles. The total costs are reduced from $41,725,000 to $41,201,000, or a savings of $524,000 per year. In addition, if the number of warehouses in the network is optimized, the costs can be reduced by another $158,000 per year year (run 23 vs. run 22). However, $278,000 of the total $524,000 + 158,000 = $628,000 can be realized without a service change. This leaves approximately $404,000 that can be saved by a relaxed service restriction. The question now becomes one of whether the higher costs associated with the more restrictive service level level are justified. Since there is no sales-service sales-service relationship for this problem, we can only estimate the worth of the service. That is, can enough sales be
211
generated to cover the higher service level? If physical physical distribution costs for the com pany are 15 percent of sales, which is probably a conservative estimate, then 1/0.15 = $6.70 in sales must be generated for each dollar that is added to distribution costs. Therefore, to cover $404,000 in cost would require $404,000 × $6.70 , cwt. = 38124 $0.71 / lb l b.×100lb./cwt. increase in sales. In terms of overall demand, this would be 38,124×100/1,477,026 = 2.5 percent. But not all customers customers would experience a higher service level. Comparing the demand centers for 299,818 cwt. of demand shows a reduction in warehouse to customer miles. Thus, moving from a minimum cost network to one with with a high service level, where the percent of demand less than 300 miles increases from 75 percent to 93 percent, requires that the 38,124 cwt. increase in demand occur in the 299,818 cwt. of demand affected by the change. This would be a 13 percent increase. increase. The products are not highly differentiated from others in the marketplace so that service plays an important role role in selling selling these products. Whether a 93 − 75 = 18 percentage points increase in service can result in a 2.5 percent increase in overall sales s ales cannot be judged by the distribution department alone. The sales department must play an important part in indicating whether the additional sales are possible. possible. If they are not likely to be realized, there is no incentive for a network other than the minimum cost one. If this information is not available from sales, the conclusion is likely to be to maintain the status status quo as represented by the benchmark. That is, one-day one-day service is most likely to guide the design. Overall Analysis and Summary The recommended design would involve an immediate increase in the number of warehouses from 22 to 30. In addition, there should be an immediate reallocation reallocation of demand among the existing plants. No reduction in the customer service level seems justified at this time. Therefore, a total cost reduction of $42,112,000 − 41,321,000 = $791,000 per year seems immediately achievable (run (run 1 vs. run run 18). By the end of the 2nd year, the Memphis plant should be brought on stream and the network should begin to evolve from the current design design (run 24) to that for the fifth fifth year (run 25). The addition of a plant is justified from the high rate of return realized from the profit potential of being able to continue meeting the growth in demand. For the current year, a breakdown of the service and the cost changes show the following:
212
Cos t typ e Production Whse operations Order processing I n v e n t o r y c a r r y in g Transportation I nboun d Outbound Total costs ($000s)
Benchmark
Currentyear design
Change from be n ch m a r k
$3 0 ,7 6 2 1,57 8 36 9 45 7
$ 3 0 ,6 7 5 1,57 2 35 8 49 0
$ -87 - 6 -11 +33
- 0 . 3% -0 -0.4 - 3 .0 +7.2
2, 0 5 0 6,89 6 $ 4 2, 1 1 2
1,86 0 6 , 36 5 $ 4 1, 3 2 0
-190 -531 $-792
-9.3 -7.7 -1 . 9 %
By the fifth year, total distribution costs should be $53,179,000, or $53,179,000/1,908,606 = $27.86, compared with the current-year cost of 42,112,463/1,477,026 = $28.51 per cwt. If current year year costs are projected to the fifth year demand level, the 5th-year 5t h-year production/distribution costs might be 28.51×1,908,606 = $54,414,357, or a savings of $54,414,357 − 53,179,000 = $1,235,357 per year. Of course, these savings can only be realized through the addition of capacity at Memphis for $4,000,000. If this this capacity is useful for at least 15 years, the the amortization of $4,000,000/15 = $267,000 per year would yield a net savings of $532,000 per year. Overall, the design change appears app ears to be justified.
213
APPENDIX A Listing of Selected Computer Runs R u n . R un
No of
P l a nt
D e m a nd
S e r v i ce c o n-
No of
T o t al
P e r c e n t of d e m a nd w i t h i n
n o. d e s c r i p t i o n
plant s
c a p a c i ty
l e v el
s t r a i nt
w h s es
c osts
≤
1 Benchmark 2 Im I mproved benchmark 3 No No serv constraint 4 Ma M ax opportunity 5 Future yr-imp bmk 6 Test 27 whses 7 Test 32 whses 8 Test 37 whses 9 Test 42 whses 10 E Ex xp Covington 11 Ex Exp New York 12 E Ex xp Arlington 13 Exp Long Beach 14 Add Memphis 15 Add Chicago 16 Add Mem & Chi ea 17 No plant expansion served 18 Optimum whses capacity 19 Op Optimum whses 20 O Op ptimum whses 21 Test cust service 22 Test cust service 23 Te T est cust service 24 Optimum whses 25 Optimum whses
4 4 4 6 4 4 4 4 4 4 4 4 4 5 5 6
Cu rr en t C u rr en t Cu rre n t Cr nt +1 m C u rr en t C u rr en t C u rr en t C u rr en t C u rr en t C u rr en t C u rr en t C u rre n t Cu r r en t Cu r r en t C u rr en t
C u rr en t - mi Current 30 3 00 C ur re n t 9 00 0 Current 9000 5th yr. 300 Current 300 Current 300 Current 300 Current 300 5th yr. 3 30 00 5th yr. 30 300 5th yr. 3 30 00 5th yr. 30 3 00 5th yr. 300 5th yr. 300 5th yr. 300
22 21 18 40 21 26 31 36 40 21 21 21 21 21 20 20
$42,112 41,725 40,896 40,409 53,777 41,744 41,615 41,501 41,486 54,145 53,986 54,709 55,251 53,705 53,781 53,382
4
C u rr en t
5th yr.
30 0
20
4
C u rr en t
C u r ren t
300
4 5 4 4 4 4 5
S e e cm t C u rr en t C u rr en t C u rr en t C u rr en t C u rr en t S e e cm t
C ur re n t 5th yr. C ur re n t C ur re n t C ur re n t C u r ren t 5th yr.
30 300 3 30 00 60 0 60 0 60 6 00 375 375
300
≤
600 600
Comm Commen ents ts
93% 93 93 71 71 81 81 93 95 98 99 99 9 94 4 93 93 9 94 4 94 94 94 94 95
98% 98 89 94 98 1 00 1 00 1 00 1 00 98 98 98 98 98 98 98
Current network design No investment required
Covington cap + 1m cwt New York + 1m cwt Arlington cap + 1m cwt Long Beach cap + 1m cwt Add Memphis at 1m cwt Add Chicago at 1m cwt Add Chi & Mem at 1m cwt
45,164
98
100
Only 85.4% of demnd
31
41,447
98
100
Plants at current
30 31 31 26 19 30 30
41,563 53,342 40,996 41,043 41,201 41,321 53,179
97 97 9 98 8 80 88 75 75 92 92
100 100 100 100 98 100 100
Plants at current thruput Memphis at 1m cwt Whses at opt no = 31 Whses from opt no = 31 Whses from current 22 Service level at bmk Serv at bmk/Mem @ 1m cwt
Added plants at 1m cwt Plant cap + 1m cwt
214
ESSEN USA Teaching Note Strategy Essen USA is concerned with entire supply channel performance. The supply channel consists of four echelons ranging from factory to customers. The purpose of this case study is for the student to manipulate the supply channel variables using a channel simulator in in order to improve individual individual member and system-wide performance. The channel variables include forecasting methods, inventory policies, transportation services, production lot sizes, order processing costs, and stock availability levels. Students should seek to optimize channel performance, although it is not expected that the optimum actually can be found or verified. However, improving performance over existing levels is achievable. The SCSIM module of LOGWARE is used to simulate the demand and product flows throughout the multi-echelon multi-echelon supply chain. SCSIM is an ordinary Monte Carlo day-today type of simulator. simulator. Using a simulator for performance improvement requires thinking thinking of it in terms of as an experimental methodology. That is, a single run of the simulator is
ESSEN USA Teaching Note Strategy Essen USA is concerned with entire supply channel performance. The supply channel consists of four echelons ranging from factory to customers. The purpose of this case study is for the student to manipulate the supply channel variables using a channel simulator in in order to improve individual individual member and system-wide performance. The channel variables include forecasting methods, inventory policies, transportation services, production lot sizes, order processing costs, and stock availability levels. Students should seek to optimize channel performance, although it is not expected that the optimum actually can be found or verified. However, improving performance over existing levels is achievable. The SCSIM module of LOGWARE is used to simulate the demand and product flows throughout the multi-echelon multi-echelon supply chain. SCSIM is an ordinary Monte Carlo day-today type of simulator. simulator. Using a simulator for performance improvement requires thinking thinking of it in terms of as an experimental methodology. That is, a single run of the simulator is a particular event sequence generated from random numbers. Changing the seed number in the simulator causes a different set of random numbers to be generated and possibly another outcome from the same input data. A simulation run with a specified seed number should be viewed as a single statistical observation and multiple outcomes from various seed numbers should be treated as a statistical sample and analyzed accordingly, i.e., comparing means and standard deviations. Each simulation is run for a period of 11 years with results taken from years 2 through 11. The first year is not used since it can show unstable results due to to startup conditions. The results appear to reach steady steady state by the second year, and the results results for the 10 years thereafter are averaged to give a reasonable representation of channel performance for a given run. The database used to represent the current cu rrent performance of the channel, as derived from the case study, is summarized in the Appendix A of this note and a typical run report is shown in Appendix B. This case provides students with the opportunity to observe the operation of a multiechelon supply channel and to assess the impact of changing key operating variables on individual members as well as on channel-wide performance. The effect on cost and customer service as well as sales, inventory, and back order levels of demand patterns, demand forecasting methods, inventory control methods, transportation performance, production lot sizing, order processing procedures, procedu res, and item fill rates can be observed in both graphical and report forms. Most importantly, students can see the effects of supply chain decisions rather than project the results statistically. Questions
1. What can you say about the logistics performance performance throughout the supply channel for Essen and its customers?
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General observations It is recognized that Essen must deal with demand that has significant seasonal peaks at gift-giving times of the the year as shown in in Figure 1. Compared with a smooth demand pattern, this can cause increasing demand variability upstream from the customers, as illustrated in Figure 2. This “bull whip” effect is partly partly a result of the demand for an upstream member being derived from the order size and pattern of its immediate downstream channel member. Forecast accuracy, lead-time uncertainty, and inventory control method also affect demand variability and the resulting cost of that variability.
Figure 1 Typical Demand Pattern for Essen Over the Period of One Year
Retailer Distri warehouse
Essen Warewarehouse house
Factory Factory
-butor
Retailer Retailer
Figure 2 Increasing Demand Variability of Upstream Channel Members for a Four-Year Period
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Benchmark Running the simulator (SCSIM) with a seed number of 123456 and simulated period of 11 years with results taken from the last 10 years, the channel generates average annual sales of $109.5 million for a net average annual system profit contribution of $24.4 million, as shown in Table 1. The question arises as to whether channel performance can be improved and profits increased. At least two observations can be made that suggest there is room for improvement. improvement. First, the inventory levels for both the retailer’s warehouse and Essen’s warehouse are quite high compared with the Retailer (see Figure 3). It is possible that Retailer inventories are too low. low. However, the inventory turnover ratio is about seven for the Essen’s warehouse (see Table 1). This is not particularly high for a food product that might have a turnover at least in the range of 10 to 12. The turnover for the retailer’s warehouse appears more in line with industry norms of about 13 (see Table 1).
Retailer warehouse
Essen warehouse
Retailer
Figure 3 Inventory Levels for Four Years Using Benchmark Data
Second, the backorders at the Retailer level do not seem to recover well from the seasonal spike in demand. Correspondingly, the Retailer Retailer inventory turnover is 81 (see Table 1), which is quite high. The low percentage of demand filled filled on request (<50%) suggests that inadequate inventory is being maintained to meet reasonable fill rates. Third, customer service levels are also low for the retailer’s warehouse and Essen’s warehouse. Backorder occurrences are high for both channel members. Although inventory levels are adequate most of the time, seasonal demand rippling through the supply chain causes a significant number of back orders before inventory can be replenished. The observation is that there is an opportunity to improve channel performance, especially in terms of customer service. A major concern is how to mange the seasonal demand pattern that is causing the cyclical behavior throughout the echelons of the
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channel. Current performance of the the channel members is summarized summarized in Table 1 for four simulation runs using different seed numbers. Table 1 Average Annual Performance of Channel Members and the System at Benchmark Channel member
Run 1
Run 2
Run 3
Run 4
Average
$73,105,904 37,918 $1,928
$72,967,088 37,846 $1,928
$72,616,192 37,664 $1,928
$72,477,376 37,592 $1,928
$72,791,640 37,755 $1,928
6.52 <50% $5,578,291 $147.02
6.59 <50% $5,549,202 $146.50
6.61 <50% $5,521,236 $146.32
6.58 <50% $5,540,447 $146.65
6.58 <50% $5,547,294 $146.62
12.95 <50% $3,873,236 $101.94
13.01 <50% $3,895,406 $102.15
12.96 <50% $3,853,890 $102.14
12.89 <50% $3,912,699 $103.07
12.95 <50% $3,883,808 $102.33
81.02 <50% 38,017 $2,884,527 $76.19
80.38 53.02% 37,983 $2,768,697 $72.89
80.45 54.09% 37,774 $2,939,464 $77.82
80.60 <50% 37,804 $2,981,607 $78.87
80.61 <50% 37,895 $2,893,574 $76.44
$24,426,593 22.23% 123456
$24,591,053 22.40% 444444
$24,235,498 22.20% 555555
$24,340,563 22.28% 666666
$24,398,426 22.28%
Essen’s factory Total cost
Units produced Cost per unit Essen’s warehouse TO ratio Fill rate Cost
Cost per unit Retailer’s warehouse TO ratio Fill rate Cost
Cost per unit Retailer TO ratio Fill rate Units sold Cost
Cost per unit System Profit Profit as % of sales Seed Number
2. What steps steps would you suggest taking to improve logistics performance throughout the channel? Do any of the changes involve Essen? If so, so, does the company directly realize any cost and/or operating performance improvements?
A number of actions can be taken to lower costs and improve customer service. Improving the forecast, shortening the lead times, changing the inventory control policy, and changing production lot sizes are all variables that can be altered for possible performance improvement. The interactions among these variables and the large number of variable variable combinations preclude finding the optimal set. However, they can be explored in a systematic way to find improvement. The primary focus of this analysis will be to increase the fill rates at the the risk of increasing increasing costs. Ultimately, revenues, through improved customer service, may be preserved or increased to more than compensate for reduced profits.
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Retailer Level Start with the retailer because of the proximity to the customer. Fill rates need to be improved, probably in the 95-99% range as specified in the database. Inventory turns can be guided by the industry average of 12 turns per year. Where the two cannot be jointly met, service will prevail. Clearly, putting additional inventory at the retail point will improve customer service. Using the company’s current inventory policy of stocking to demand, the target level can be raised without changing the review time. Exploring different target levels shows 14 days to offer about 35 turns and a 99+% fill rate. Because of the high cost of a back order, total costs at the retail level drop significantly. Altering the forecasting method and the settings associated with the method yield little opportunity for improvement. Using an exponential smoothing smoothing model with a high smoothing constant to better follow the seasonal changes in demand results in increased costs. Lowering the the smoothing constant to 0.1 did not offer improvement either. Altering the number of periods in the moving average model did not improve costs and only degraded performance. Shortening the review time in the stock-to-demand reorder policy did have a positive effect on fill rate, but resulted in high costs and lower inventory turns. The tradeoff did not seem beneficial, given the fill rate and turnover targets. Retail Warehouse Level Determining an improved policy at this level is difficult because a 95% fill rate and 10 to 12 inventory turns is an illusive goal. Using service as the primary primary target, a stockto-demand control policy is used with a review period of 7 days and a target of 25 days of inventory. The forecasting forecasting method method is is moving average with a period period of 7 days. The performance achieved at this channel level is about nine inventory turns per year and a 97% fill rate. Essen’s Warehouse Level The performance at Essen’s warehouse level seems to mimic that at the retail warehouse level except that there there is more demand variability. Again, an inventory turnover ratio in the target range cannot be achieved while maintaining a high fill rate level. Trying to achieve high service levels with high levels of inventory is difficult, difficult, probably due to the extensive demand variability that filters back to this member of the channel. Multiple simulation simulation runs show that a high fill rate cannot consistently be achieved even when on the average inventory levels are high. However, average performance shows an 82% fill rate and 1.5 inventory turns per year based on a 7-day moving average forecasting model and a stock-to-demand inventory control policy with a review time of 7 days and an inventory target of 25 days. Essen’s Factory The concern with the factory level in the channel is whether product should be manufactured in a larger lot size, but with slightly higher production time variability. The reduced costs seem to out weigh the negative effects of increased variability. Producing in the larger lot size is favored.
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Overall Using the objective of improving customer service, it is not surprising that supply channel costs increase as shown from the reduced profit in Table 2 compared with Table 2. The average fill rate has increased for all members of the channel, but the cost effects effects are spread disproportionately disproportionately among the members. Even with a higher fill rate, rate, the retailer benefits from a substantial reduction reduction in the cost per unit sold. On the other hand, the cost for handling a unit of the product at Essen’s warehouse is substantially increased. Essen should take advantage of the cost reduction from producing in the larger batch size, but this does not offset the higher cost at the company’s warehouse. As an upstream channel member, Essen undoubtedly suffers from variability in demand, which it cannot entirely control. The retailer benefits from the action to increase fill rates across the channel. However, Essen is put at a disadvantage and may take a counter action to improve its cost position. Essen may simply lower its inventory level by reducing the reorder target quantity from 25 to 10 days. This reduces Essen’s per-unit warehouse cost, but it also increases the costs for for the retailer. The reduced inventory level at the Essen Essen warehouse causes lower fill fill rates for for the downstream retailer. Unless the retailer can find an incentive to reward Essen for its good service, it will be difficult for Essen to provide the level of service that the retailer would like and that is economically beneficial to Essen. Table 2 Average Annual Performance of Channel Members and the System as Revised Channel member
Run 1
Run 2
Run 3
Run 4
Average
$68,638,738 35,950 $1,909
$74,761,258 39,286 $1,903
$78,418,188 41,268 $1,900
$75,400,666 39,622 $1,903
$74,304,713 39,032 $1,904
1.47 90.65% $12,060,444 $317.51
1.42 100% $12,449,170 $326.28
1.51 63.16% $11,895,743 $311.87
1.46 72.92% $12,178,581 $319.61
1.47 81.68% $12,145,984 $318.82
9.24 96.64% $4,018,143 $105.70
9.02 96.29% $4,080,160 $107.09
9.14 97.60% $4,043,973 $106.55
9.28 98.09% $3,992,791 $105.51
9.17 97.16% $4,033,767 $106.21
35.44 99.52% 38,017 $727,378 $19.13
35.36 99.53% 37,936 $717,363 $18.91
34.99 99.70% 37,979 $709,772 $18.69
35.14 99.40% 37,730 $748,198 $19.75
35.23 99.54% 37,916 $725,677 $19.12
$24,423,849 22.23% 123456
$17,627,666 16.08% 111111
$14,690,765 13.38% 222222
$17,184,464 15.69% 333333
$18,481,686 16.85%
Essen’s factory Total cost
Units produced Cost per unit Essen’s warehouse TO ratio Fill rate Cost
Cost per unit Retailer’s warehouse TO ratio Fill rate Cost
Cost per unit Retailer TO ratio Fill rate Units sold Cost
Cost per unit System Profit Profit as % of sales Seed Number
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3. Would shipping by airfreight from Germany be a benefit to channel performance? To Essen?
No. Selling candies to end customers at $2,890 per thousand lb. using airfreight shipping results in obvious loss to Essen. The cost of shipping by air is is $1,833 per thousand lb., plus $1,000 material cost and $850 production cost results in a total cost much higher than selling price. There is no point to using airfreight. Running the simulation with the higher freight rate but lower variability confirms that channel profits would b e negative. 4. Is there a benefit to producing in the larger 20,000-pound batch size?
Yes. This was tested in question 2. From Tables 1 and 2, it can be seen that production costs drop from $1,928 per unit to $1,904 per unit. The overall channel cost reduction reduction overshadows the negative effects of greater length and variability in production time.
Appendix A Simulation Database for Essen USA Under Current Conditions Title: ESSEN USA Initialization 123456 11 2890
Seed value Length of simulation, years Annual price, $/unit
Customer demand pattern Generate daily demand 100 Average daily demand, units 15 Standard deviation of daily demand, units 1 Annual demand growth increment, % Monthly seasonal indices Month Index Month Index Month Index Month 1 0.25 4 0.75 7 0.75 10 2 1.25 5 0.75 8 0.75 11 3 1.25 6 0.75 9 0.75 12 Retailer/Level 1 Product item data 2220 1 35 25 1 0
Index 0.75 1.50 2.50
Item value in inventory, $/unit Customer order filling cost, $/unit Purchase order processing cost, $/order Inventory carrying cost, %/year Average customer order fill time, days Customer order fill time standard deviation, days
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98 In-stock probability, % 670 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 10 Target days of inventory 7 Review time in days Distributor/Level 2 Product item data 2220 Item value in inventory, $/unit 20 Retailer order filling cost, $/unit 75 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 2 Average retailer order fill time, days 0.2 Retailer order fill time standard deviation, days 95 In-stock probability, % 100 Back order cost, $/unit Forecasting method Moving average 30 Number of periods Reorder policy Stock-to-demand control method 45 Target days of inventory 30 Review time in days Warehouse/Level 3 Product item data 1710 Item value in inventory, $/unit 15 Distributor order filling cost, $/unit 75 Purchase order processing cost, $/order 20 Inventory carrying cost, %/year 3 Average distributor order filling time, days 0.3 Distributor order fill time, days 95 In-stock probability, % 25 Back order cost, $/unit Forecasting method Moving average 360 Number of periods Reorder policy Stock-to-demand control method 90 Target days of inventory 30 Review time in days
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Factory/Source Produ Product ct item item data data
850 10 10 8 2 1000
Production cost, $/unit Minimum production lot size, units Warehouse order filling cost, $/unit Average production time, days Production time standard deviation, days Purchase cost, $/unit
Transportation Transpo Transport rt between between Distrib Distributor utor and Retailer Retailer
25 1 0
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
Transpo Transport rt between between Warehou Warehouse se and Distrib Distributor utor
70 5 1
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
Transpo Transport rt between between Factory Factory and Warehous Warehouse e
78 9 3
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
Appendix B Benchmark Simulation Results with Seed Number 123456 and Simulation Length of 11 Years SUPPLY CHANNEL REPORT FOR SIMULATED YEARS 2 TO 11
Yearly average $109 $109,8 ,868 68,5 ,552 52 37,918,000 71,950,552
Sim ulat ed period $1,0 $1,098 98,6 ,685 85,5 ,520 20 379,180,000 719,505,520
32,230,300
322,303,000
949,025 2,656,010 2,957,604
9,490,250 26,560,100 29,576,040
38,017 759,890 569,145
380,168 7,598,900 5,691,450
1,638 683 675
16,380 6,825 6,750
FINANCIAL PERFORMANCE Reve Revenu nue e Cost of purchased goods Gross margin Production cost Transportation Transportation costs: Distributor to retailer Warehouse to distributor Factory to warehouse Sales orde Sales order r handl handlin ing g cost cost for: for: Customer orders Retailer orders Distributor orders Order Order pro process cessing ing cos cost t for: Orders to distributor Orders to warehouses Orders to factory Inventory Inventory costs
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260,414 1,627,909 1,988,984
2,604,145 16,279,092 19,889,837
R e t a i l er D i s t r i b u t or Ware house
2,584,458 535,730 363,478
25,844,580 5,357,300 3,634,775
Back ord Back order er cost costs s Reta iler D i s t r i bu t o r W a r e h o us e
$24, 24,426, 26,593
$24 $244,265 265,928 928
Net prof rofit contrib ribution ion
Appendix C Simulation Database for Essen USA as Revised for Service Improvement Title: ESSEN USA Initialization 123456 11 2890
Seed value Length of simulation, years Annual price, $/unit
Customer demand pattern Generate daily demand 100 Average daily demand, units 15 Standard deviation of daily demand, units 1 Annual demand growth increment, % Monthly seasonal indices Month Index Month Index Month Index Month 1 0.25 4 0.75 7 0.75 10 2 1.25 5 0.75 8 0.75 11 3 1.25 6 0.75 9 0.75 12
Index 0.75 1.50 2.50
Retailer/Level 1 Product item data 2220 Item value in inventory, $/unit 1 Customer order filling cost, $/unit 35 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 1 Average customer order fill time, days 0 Customer order fill time standard deviation, days 98 In-stock probability, % 670 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 14 Target days of inventory 7 Review time in days
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Distributor/Level 2 Product item data 2220 Item value in inventory, $/unit 20 Retailer order filling cost, $/unit 75 Purchase order processing cost, $/order 25 Inventory carrying cost, %/year 2 Average retailer order fill time, days 0.2 Retailer order fill time standard deviation, days 95 In-stock probability, % 100 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 35 Target days of inventory 7 Review time in days Warehouse/Level 3 Product item data 1710 Item value in inventory, $/unit 15 Distributor order filling cost, $/unit 75 Purchase order processing cost, $/order 20 Inventory carrying cost, %/year 3 Average distributor order filling time, days 0.3 Distributor order fill time standard deviation, days 95 In-stock probability, % 25 Back order cost, $/unit Forecasting method Moving average 7 Number of periods Reorder policy Stock-to-demand control method 25 Target days of inventory 7 Review time in days Factory/Source Produ Product ct item item data data
825 20 10 10 2.1 1000
Production cost, $/unit Minimum production lot size, units Warehouse order filling cost, $/unit Average production time, days Production time standard deviation, days Purchase cost, $/unit
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Transportation Transpo Transport rt between between Distrib Distributor utor and Retailer Retailer
25 1 0
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
Transpo Transport rt between between Warehou Warehouse se and Distrib Distributor utor
70 5 1
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
Transpo Transport rt between between Factory Factory and Warehous Warehouse e
78 9 3
Transport cost, $/unit Average time in-transit, days Transit time standard deviation, days
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