Ballou Logistics Solved Problems Chapter 13

CHAPTER 13 FACILITY LOCATION DECISIONS 1 (a) The center-of-gravity method involves finding the X,Y coordinates according to the formulas:

∑V R X X= ∑V R i

i

i

i

i

i

i

and

∑V R Y Y = ∑V R i

i i

i

i

i

i

These formulas can be solved in tabular form as follows. Point P1 P2 M1 M2 M3

X Y 3 8 8 2 2 5 6 4 8 8 Totals

Vi 5,000 7,000 3,500 3,000 5,500

Ri .040 .040 .095 .095 .095

ViRi ViRiXi ViRiYi 200.0 600.0 1600.0 280.0 2240.0 560.0 332.5 665.0 1662.5 285.0 1710.0 1140.0 522.5 4180.0 4180.0 1620.0 9395.0 9142.5

Now,

X=

9,395.0 = 5.8 1,620.0

and

Y =

9,142.5 = 5.64 1,620.0

This solution has a total cost for transportation of $53,614.91. This problem may also be solved using the COG module in LOGWARE. (b) Solving for the exact center-of-gravity method requires numerous computations. We now use the COG module of LOGWARE to assist us. A table of partial results is shown below.

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Iteration number 0 1 2 3 . . . 50

X coord 5.799383 5.901199 5.933341 5.941554 . . . 5.939314

Y coord 5.643518 5.518863 5.446919 5.402429 . . . 5.317043

Total cost 53,614.91 ⇐ COG 53,510.85 53,483.97 53,474.60 . . . 53,467.71

After 50 iterations, there is no further change in total cost. The revised coordinates are X = 5.94 and Y = 5.32 for a total cost of $53,467.71. (c) The center-of-gravity solution can be one that is close to optimum when there are many points in the problem and no one point has a dominant volume, that is, has a larger volume relative to the others. Otherwise, the best single location can be at a dominant location. The exact center-of-gravity approach has the capability to find the minimal cost location. Although the COG model only considers transportation costs that are constant per mile, the transportation cost can be the major consideration in single facility location. However, other costs such as labor, real estate, and taxes can also be important in selecting one location over another. These are not directly considered by the model. Although the COG model may seem of limited capability, it is a useful tool for locating facilities where transportation costs are dominant. Location of oil wells in the Gulf, truck terminals, and single warehouses are examples of application. It also can be quite useful to provide a starting solution to more complex location models. (d) Finding multiple locations by means of the center-of-gravity approach requires assigned supply and demand volumes to specific facilities and then solving for the center of gravity for each. In this problem, there are 3 market combinations that need to be considered. This creates three scenarios that need to be evaluated. They can be summarized as follows. Point volumes appear in the body of the table. Scenario

Whse 1

P1 1458a

P2 2042

2

3542

4958

1

2708

3792

2

2292

3208

1

3750

5250

M1 3500

M2

M3

I 3000 3500

5500

3000

II 5500 3500

5500

III 2 1250 1750 3000 Allocated as a proportion of the volume to be served through the warehouse. That is, 5000×3500/(3500 + 3000 + 5500) = 1458. The volumes associated with other supply points are computed similarly.

a

161

The COG module in LOGWARE was used to find the exact centers of gravity for each warehouse in each scenario. The computational results are: Scenario I II III

Warehouse 1 X Y 2.00 5.00 5.84 4.04 7.06 7.28

Warehouse 2 X Y 7.88 7.80 8.00 8.00 6.00 4.00

Total cost $39,050 35,699 ⇐ 46,568

Scenario II appears to be the best. 2 (a) The center-of-gravity formulas (Equations 13-5 and 13-6) can be solved using the COG module of LOGWARE, or they can be solved in tabular form as shown below. Point A B C D E F G H I J

X Y 50 0 10 10 30 15 40 20 10 25 40 30 0 35 5 45 40 45 20 50 Totals

Vi 9,000 1,600 3,000 700 2,000 400 500 8,000 1,500 4,000

Ri .75 .75 .75 .75 .75 .75 .75 .75 .75 .75

ViRi 6,750 1,200 2,250 525 1,500 300 375 6,000 1,125 3,000 23,025

ViRiXi 337,500 12,000 67,500 21,000 15,000 12,000 0 30,000 45,000 60,000 600,000

ViRiYi 0 12,000 33,750 10,500 37,500 9,000 13,125 270,000 50,625 150,000 586,500

Now,

X=

600,000 = 261 . 23,025

and

Y =

586,500 = 25.5 23,025

The total cost of this location is $609,765. The exact center-of-gravity coordinates are:

X = 2351 . , Y = 26.98 with a total cost of $608,478. (b) The number of points, even in this small problem, requires us to apply some heuristics to find which patient clusters should be assigned to which warehouses. We will use a clustering technique whereby patient clusters are grouped by proximity until two clusters are found. The procedure works as follows:

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• There are as many clusters as there are points, which are 10 in this case. • The closest points are found and replaced with a single point with the combined volume located at the center of gravity point. There is now one less cluster. • The next closest two points/clusters are found, and they are further combined and located at their center of gravity. • The process continues until only two clusters remain. The centers of gravity for these two clusters will be the desired clinic locations. Applying the clustering technique, we start by combining points D and F into cluster DF.

X=

40( 700)( 0.75) + 40( 400)( 0.75) = 40.00 700( 0.75) + 400( 0.75)

and

Y =

20( 700)( 0.75) + 30( 400)( 0.75) = 23.64 700( 0.75) + 400( 0.75)

Continuing this process, we would form two clusters containing A, C, D, and F, and B, E, G, H, I, and J. The centers of gravity would be: Cluster 1 - ACDF

X = 50.00, X = 0.00 Cluster 2 - BEGHIJ

X = 5.00, Y = 45.00 for a total cost of $241,828. These are the same results obtained from the MULTICOG module in LOGWARE. (c) The second clinic can save $608,278 − 241,828 = $366,458 in direct costs annually. This savings does not exceed the annual fixed costs of $500,000 required to maintain a second clinic. On economic grounds, it should not be built. 3 (a) The center-of-gravity location can be determined by forming the following table or by using the COG module in LOGWARE. The coordinates for each location must be approximated.

163

Point A B C D E F G H I

X Y 1.5 6.6 4.7 7.3 8.0 7.1 1.5 4.0 5.0 4.9 8.5 5.1 1.5 1.3 4.4 1.8 7.8 1.8 Totals

Vi 10,000 5,000 70,000 30,000 40,000 12,000 90,000 7,000 10,000

Ri .10 .10 .10 .10 .10 .10 .10 .10 .10

ViRi 1,000 500 7,000 3,000 4,000 1,200 9,000 700 1,000 27,400

ViRiXi 1,500 2,350 56,000 4,500 20,000 10,200 13,500 3,080 7,800 118,930

ViRiYi 6,600 3,650 49,700 12,000 19,600 6,120 11,700 1,260 1,800 112,430

The center-of-gravity coordinates are:

X=

118,930 = 4.34 27,400

and

Y =

112,430 = 410 . 27,400

with a total cost of $195,966. (b) After 100 iterations in the COG module, the exact center of gravity was found to be:

X = 4.75, Y = 4.62 with a total cost of $195,367. In this case, using the exact center of gravity coordinates as compared with the approximate ones reduced costs by only: 195,966 − 195,367 × 100 = 0.3% 195,966 (c) Additional costs can be included in the analysis, although not necessarily in the model. The COG model can evaluate the variable costs of location. Other costs are compared with these. 4 We begin by developing a 3-dimensional transportation problem. The cost matrix is developed in the same manner as that in text Figure 13-11. The initial throughput of W1 and W2 is found by assuming that an equal amount of the customer demand flows through each warehouse. The cell cost for W1-C1 would be: [100(200,000/2)0.7]/(200,000/2 )+ 2 + 4 + 0 = 3.2 + 2 + 4 = 9.2

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In fact, the cell costs are identical to those in text Figure 13-11, except that there is no fixed cost element. Using the transportation method of linear programming (e.g., the TRANLP module in LOGWARE), the cell cost and solution matrix for iteration 1 is shown in Figure 13-1. The solution shows that only W2 remains and the solution process can be terminated. A summary of the costs is shown in Table 13-1. The total cost is $2,213,714, and the product is produced in plant P2 and stocked in warehouse W2. No further iterations are needed since only one warehouse is used and no further dropping of warehouses is possible. TABLE 13-1 Summary Information for Solution to Problem 4 Whse 1 Whse throughput Costs: Transportation Inbound Outbound Inventory Warehousing Fixed Production Total

Whse 2

0

200,000

$0 0 0 0 0 0

$400,000 300,000 513,714 200,000 0 800,000 $2,213,714

Iteration 2 A repeat of the iteration 1 solution.

Stop iterating.

FIGURE 13-1 Cell Cost and Solution Matrix for Iteration 1 of Problem 4 Warehouses Customers W1 W2 C1 C2 C3 Capacity a 4 9 99 99 99 60,000 Plants P1 0 0 8 6 99 99 99 P2 999,999 0 200,000 a 0 99 9.2 8.2 10.2 Whses W1 60,000 60,000 0 0 0 99 0 6.2 5.2 6.2 W2 799,999 50,000 100,000 50,000 999,999 Capacity/ Req’mts 60,000 999,999 50,000 100,000 50,000 a

High rate of $99/unit for an inadmissible cell.

5 We begin by forming the cell cost matrix of a three-dimensional transportation problem, as shown in Figure 13-2. It is similar to the text Figure 13-1 except that the capacity for warehouse 1 is set at 75,000. Solving the problem by means of the transportation method shows the solution given in Figure 13-2. 165

FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 5 Warehouses Customers W1 W2 C1 C2 C3 Capacity 4 9 100 100 100 60,000 Plants P1 60,000 8 6 100 100 100 999,999 P2 40,000 100,000 0 100 9.7 8.7 10.7 999,999 Whses W1 899,999 100,000 100 0 8.2 7.2 8.2 W2 50,000 50,000 100,000 Capacity/ Req’mts 999,999 100,000 50,000 100,000 50,000

Given the solution from iteration 1, the per-unit inventory and fixed costs are revised.

Inventory W1

$100(100,000)0.7 = $3.16 / unit 100,000 units

W2

$100(100,000)0.7 = $3.16 / unit 100,000 units

Fixed

W1 W2

$100,000/100,000 = $1.00/unit $400,000/100,000 = $4.00/unit

Adding outbound transportation and warehouse handling to per-unit inventory and fixed costs gives the following cell costs.

W1 W2

C1 10.2 10.2

C2 9.2 9.2

C3 11.2 10.2

Revising the warehouse-customer cell costs and solving gives the same warehouse throughputs, so cell costs will no longer change. A stopping point is reached. The solution is the same as that in Figure 13-2. A summary of the costs is:

166

Cost type Production Inbound transportation Outbound transportation Fixed Inventory carrying Handling Subtotal Total

Warehouse 1 100,000 cwt. 60,000×4 = $240,000 40,000×4 = 160,000 60,000×0 = 0 40,000×4 = 160,000 100,000×3 = 300,000

Warehouse 2 100,000 cwt. 100,000×4 = $400,000 100,000×2 =

200,000

50,000×2 = 50,000×2 =

100,000 100,000 100,000 400,000 100(100,000)0.7 = 316,228 100(100,000)0.7 = 316,228 100,000×2 = 200,000 100,000×1 = 100,000 $1,476,228 $1,616,228 $3,092,456

Compared with the costs from the text example, the cost difference is $3,092,456 − 2,613,714 = $478,742. This is the penalty for restricting a warehouse with economic benefit to the network. 6 Prepare a matrix for a 3-dimensional transportation problem like that in text Figure 1311, except that the per-unit cell costs for warehouse 2 to customer are reduced by $1/unit to reflect the reduction in that warehouse’s fixed costs. That is, $200,000/200,000 = $1/unit instead of $400,000/200,000 = $2/unit. The matrix setup and first iteration solution are shown in Figure 13-3. FIGURE 13-2 Cell Cost and Solution Matrix for Iteration 1 of Problem 6 Warehouses

Plants

W1 4a

W2 9

C1 99b

C2 99

C3 99

8

6 200,000 99

99

99

99

9.7

0 799,999

999,999c

P1

W1

W2 Warehouse capacity & customer demand

Plant & warehouse capacities 60,000

P2 Warehouses

Customers

0 60,000 99b

60,000

999,999c d

8.7

10.7

7.2e 50,000

6.2 100,000

7.2 50,000

50,000

100,000

50,000

60,000

a

Production plus inbound transport rates, that is, 4 + 0 = 4. Used to represent an infinitely high cost. c Used to represent unlimited capacity. d Inventory carrying, warehousing, outbound transportation, and fixed rates, that is, 3.2 + 2 + 4 + 0.5 = 9.7. e 3.2 + 1 + 2 + 1 = 7.2. b

167

999,999c

The results show that one warehouse is to be used. Further computations are not needed, as further warehouse consolidation is not possible. The total network costs are the same as those in the text example minus the $200,000 reduction in fixed costs for a total coat of $2,413,714. 8 This problem requires us to rework the dynamic programming solution to the example problem given in the text. The only change is that the cost of moving from one location to another is now $300,000 instead of $100,000. We begin with the last year and determine the best action based on the highest net profits. The action will be to move (M) or to stay (S). For example, given the discounted moving cost of 300,000/(1 + 0.20)(4) = $144,676, we evaluate each course of action, assuming that we are in location alternative A at the end of year four. From the location profits of text Table 13-6, we generate the following table for location A. Alternative (x) A B P5(A) = max C D E

Location profit $1,336,000 1,398,200 1,457,600 1,486,600 1,526,000

Moving cost -

0 144,676 144,676 144,676 144,676

Net profit = $1,336,000 = 1,253,524 = 1,312,924 = 1,341,924 = 1,381,324

The best action in the beginning of the fifth year, if we are already in location A, is to move to location E. This is an entry in Table 13-2. Once each of the five alternatives is evaluated for the fifth year, then the fourth year alternatives are evaluated. The moving cost is 300,000/(1 + 0.2)(3) = $173,611. We now include the profits for the subsequent years in our calculations. After Table 13-2 is completed, we search the first column for the highest cumulative profit. This is initially to locate in location D and remain there throughout the subsequent years. 9 (a) Using PMED software in LOGWARE and the PMED02.DAT database, solve for the number of locations from one to nine. The best locations for each number of sites are given in the table below.

168

TABLE 13-2 Location-Relocation Strategies Over a Five-Year Planning Horizon with Cumulative Profits Shown from Year j to Year Five for Problem 8 Year from present date j Warehouse 5th 1st 4th 3rd 2nd location StraStraStraStraaltertegya P2(x) tegya P3(x) tegya P4(x) tegya P5(x) P1(x) natives (x) A $3,557,767 SA $3,363,767 SA $3,007,667 MD $2,268,289 MD $1,381,324 B 3,556,167 SB 3,379,667 SB 3,007,667 MD 2,268,289 MD 1,398,000 C 3,673,200 SC 3,500,900 SC 3,156,200 SC 2,319,800 SC 1,457,600 D b 3,720,300 SD 3,553,600 SD 3,216,000 SD 2,459,900 SD 1,486,600 3,374,700 SE 3,071,300 SE 2,355,800 SE 1,526,000 E 3,534,100 SE a Strategy symbol refers to “staying” (S) in the designated location or “moving” (M) to a new location as indicated. b Arrows indicate maximum profit location plan when warehouse is initially located at D.

169

Strategya ME SB SC SD SE

Number of sites 1 2 3 4 5 6 7 8 9

Total cost $95,035,220 49,311,400 37,152,360 30,163,340 24,739,040 24,230,760 23,780,600 25,018,250 27,624,060

Seven sites is optimal, and they are to be located at New York, Cincinnati, Chicago, Phoenix, Denver, and Seattle. (b) The optimal cost for four sites is $30,163,340 as found in part a. The company operates the same sites as found in the optimal solution for four sites. Therefore, the cost savings comes from a between assignment of customers to the sites. The savings of $35,000,000 − 30,163,340 = $4,836,660 without any major investment. The savings better the optimal four sites and the optimal seven sites is $30,163,340 − 23,780,600 = $6,382,740. For an additional three sites, the simple return on investment is: ROI =

Savings 6,382,740 = × 100 = 35.5% Investment 18,000,00

This is a benefit certainly worth considering. (c) Increase the annual volume for Los Angeles and Seattle markets by a factor of 10 and re-solve the problem as in part a. Selected results are as follows: Number of sites 6 7 8 9

Total cost 30,892,720 25,468,410 25,018,250 27,624,060

The optimal number of sites is eight. An additional site at Los Angeles is needed, compared with the seven locations found in part a. 10 (a) We can apply Huff's model of retail gravitation to this problem. The solution table (Table 13-3) can be developed. Summarizing, branch A can be expected to attract 11,735/(11,735 + 11,765) = 49.9 percent of the customers, and branch B should attract the remaining 50.1 percent.

170

TABLE 13-3 Estimate of the Number of Customers Attracted to Each Branch Bank Pij = S j / Tij2

Time to ja Customer i 1 2 3 4 5 6 7 8 9 a

Time =

A 0.28 0.10 0.28 0.40 0.20 0.50 0.45 0.57 0.67

B 0.72 0.50 0.45 0.20 0.40 0.50 0.28 0.20 0.54

Tij2 A 0.08 0.01 0.08 0.16 0.04 0.25 0.20 0.32 0.45

S j / Tij2 B 0.52 0.25 0.20 0.04 0.16 0.25 0.08 0.04 0.29

A 12.5 100.0 12.5 6.3 25.0 4.0 5.0 3.1 2.2

B 1.4 2.8 3.5 17.5 4.4 2.8 8.8 17.5 2.4

∑S j

A 0.90 0.97 0.78 0.26 0.85 0.59 0.36 0.15 0.48

j

/ Tij2 B 0.10 0.03 0.22 0.74 0.15 0.41 0.64 0.85 0.52

Eij = Pij Ci A 900 1,940 3,120 1,820 850 885 1,440 300 480 11,735

B 100 60 880 5,180 150 625 2,560 1,700 520 11,765

( X i − X )2 + (Yi − Y )2 / 50

(b) The economic analysis of the sites would be: Revenue (100×No.of customers) Operating expenses Profit

$1,173,500 300,000 $ 873,000

Return on investment 873,000/750,000 = 116.4% The ROI seems sufficiently high, so the branch should be constructed. (c) The size of a branch and its proximity to customers may be too simple to explain the market share of each. The nature of the services offered, the accessibility of the site, and the reputation of the bank may be just as important in estimating patronage. What will the countermoves be of the competing branch? Adding another branch and locating near branch A could substantially reduce its market share. How likely is this to happen? Are the customer numbers stable? Will a third or fourth bank be locating branches in the region? Can we expect that customers will drive such long distances to seek banking services? 11 This problem can be solved as an integer linear programming problem similar to the Ohio Trust Company example in the text. First, we create a table showing the counties that are adjacent to each county. That is,

171

Counties under consideration 1. Williams 2. Fulton 3. Lucas 4. Ottawa 5. Defiance 6. Henry 7. Wood 8. Sandusky 9. Paulding 10. Putnam 11. Hancock 12. Seneca 13. Van Wert 14. Allen 15. Hardin 16. Wyandot 17. Mercer 18. Auglaize 19. Marion 20. Shelby 21. Logan

Adjacent counties by number 2,5,6 1,3,6 2,4,6,7 3,7,8 1,6,9,10 1,2,3,5,7,10,11 3,4,6,8,10,11,12 4,7,12 5,10,13 5,6,7,9,11,13,14 6,7,10,12,14,15,16 7,8,11,16 9,10,14,17,18 10,11,13,15,18 11,14,16,18,19,21 11,12,15,19 13,18 13,14,15,17,20,21 15,16 18,21 15,18,20

Next, according to the problem formulation given in the Ohio Trust Company example, we can build the matrix as given in the prepared database called ILP03.DAT. The problem formulation is shown in Table 13-4. This matrix can be solved by the integerprogramming module (MIPROG) in LOGWARE for solution. Note that all coefficients are ones or zeros. A coefficient of one is given to each county and its adjacent counties. The sum of all constraints must be one or greater. The Xs take on the values of zero or one. An X of one means that the branch is located in the county. Solving this problem using the integer-programming module in LOGWARE shows that a minimum of five principal places of business are needed. They should be located in Henry, Wood, Putnam, Hardin, and Auglaize counties.

172

TABLE 13-4 Coefficient Matrix Setup for Problem 11 Obj Fun Constraints Williams Fulton Lucas Ottawa Defiance Henry Wood Sandusky Paulding Putnam Hancock Seneca Van Wert Allen Hardin Wyandot Mercer Auglaize Marion Shelby Logan

X1 1

X2 1

1 1

1 1 1

1 1

1

X3 1

1 1 1 1 1

X4 1

X5 1

X6 1

1

1 1 1

1 1 1 1 1 1 1 1

1 1

1 1 1

1 1 1 1

X7 1

1 1

X8 1

X9 1

1 1

1 1 1 1 1 1

X10 X11 X12 X13 X14 X15 X16 X17 X18 X19 X20 X21 1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1

1 1 1 1 1 1 ≥1 1

1 1

1 1 1 1 1 1

1 1 1 1 1 1 1

1 1

1 1

1 1 1

1

1

1 1 1 1 1 1

173

1 1 1 1

1 1 1 1

1 1

1 1

1

1

1

1

1 1

1 1

1 1 1

RHS ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥

1 1 1 1 1 1 1 1 1 1 1 1

≥ ≥ ≥ ≥ ≥ ≥ ≥ ≥

1 1 1 1 1 1 1 1

12 This problem can be solved with the aid of the PMED program in LOGWARE. A database has been prepared for it called PMED04.DAT. The database shows the fixed costs for a single site. When other numbers are to be evaluated, the FOC must be recalculated and entered into the database. The scaling factor is set at one for this problem. The fixed operating cost must be calculated for each possible number of locations. Using PMED in LOGWARE gives the following results based on recalculated FOC values, an estimation of vendor to laboratories transportation cost, and an enumerative search. No. of locations 1 2

Volume, lb. Sites Chicago 680,000 Cleveland 515,000 L. Angeles 165,000 Total 680,000 3 New York 235,000 Chicago 280,000 L. Angeles 165,000 Total 680,000 4 New York 200,000 Atlanta 145,000 Chicago 170,000 L. Angeles 165,000 Total 680,000 5 New York 200,000 Atlanta 100,000 Chicago 170,000 Dallas 60,000 L. Angeles 150,000 Total 680,000 6 New York 200,000 Atlanta 65,000 Miami 35,000 Chicago 170,000 Dallas 60,000 L. Angeles 150,000 Total 680,000 1 515,000x310x0.02=3,193,000 2

Outbound cost, $ 28,350,000

Inbound distance, mi. 0 310 1,750

15,083,500 713 0 1,750 10,641,999 713 585 0 1,750 7,515,250 713 585 0 790 1,750 6,079,249 713 585 1180 0 790 1,750 5,029,250

Inbound cost, $ 0 3,193,0001 5,775,000 8,968,000 3,351,100 0 5,775,000 9,126,100 2,852,000 1,696,500 0 5,775,000 10,313,500 2,852,000 1,170,000 0 948,000 5,250,000 10,220,000 2,852,000 760,500 826,000 0 948,000 5,250,000 10,636,500

FOC, $ 5,000,000 3,535,5342 3,535,534 7,071,068 2,886,751 2,886,751 2,886,751 8,660,253 2,500,000 2,500,000 2,500,000 2,500,000 10,000,000 2,236,067 2,236,067 2,236,067 2,236,067 2,236,067 11,180,335 2,041,241 2,041,241 2,041,241 2,041,241 2,041,241 2,041,241 12,247,446

Total cost, $ 33,350,000

31,122,568

28,428,352

27,828,750

27,479,584

27,913,196

5,000,000 2 / 2 = 3,535,534

The PMED program is used to find the best combination of sites for a particular number of sites to be found. The fixed cost must be adjusted for the number of sites being evaluated. It should be recognized that the model handles only the outbound leg of the network (sites to serve laboratories). The vendor to site transportation cost is included externally, as shown in the previous table. Calculating the distances between vendor and the selected sites easily can be done by using the MILES module in LOGWARE with a scaling factor of one. Then, inbound transport costs are a product of site volume, distance, and the inbound rate. 174

Searching from one to N sites shows that outbound transportation costs decrease while inbound and fixed costs increase with increasing numbers of sites. Initially, total cost declines until five sites are reached after which total cost increases. We select five sites as economically the best number. Their customer assignments are Location number 1 2 3 4 5

Assignments New York Atlanta Chicago Dallas Los Angeles

Volume 200,000 100,000 170,000 60,000 150,000

Customers 1, 2, 3, and 12 4 and 5 6, 7, 8, 9, 10, and 11 13, 14, and 16 15, 17, 18, 19, and 20

A map of the solution is as follows.

The total cost for 5 sites is $27,479,584. 13 After changing the fixed costs in the problem setup matrix, the following solution is found.

175

OPTIMAL SOLUTION Variable X(1) = X(2) = X(3) = X(4) = X(5) = X(6) = X(7) = X(8) = X(9) = X(10) = X(11) = X(12) = X(13) = X(14) = X(15) = X(16) = X(17) = X(18) = X(19) = X(20) = X(21) = X(22) = X(23) = X(24) = X(25) = X(26) = X(27) = X(28) = X(29) = X(30) = X(31) = X(32) =

Value .0000 10000.0000 50000.0000 .0000 .0000 .0000 50000.0000 90000.0000 .0000 .0000 .0000 .0000 .0000 30000.0000 20000.0000 .0000 .0000 .0000 20000.0000 .0000 40000.0000 .0000 .0000 .0000 1.0000 .0000 1.0000 1.0000 1.0000 .0000 .0000 .0000

Rate 8.0000 7.0000 9.0000 11.0000 10.0000 11.0000 12.0000 11.0000 13.0000 8.0000 7.0000 8.0000 6.0000 5.0000 7.0000 11.0000 10.0000 11.0000 9.0000 8.0000 10.0000 7.0000 6.0000 7.0000 100000.0000 500000.0000 140000.0000 260000.0000 220000.0000 70000.0000 130000.0000 110000.0000

Objective function value =

Cost .0000 70000.0000 450000.0000 .0000 .0000 .0000 600000.0000 990000.0000 .0000 .0000 .0000 .0000 .0000 150000.0000 140000.0000 .0000 .0000 .0000 180000.0000 .0000 400000.0000 .0000 .0000 .0000 100000.0000 .0000 140000.0000 260000.0000 220000.0000 .0000 .0000 .0000

Variable label P1S1W1C1 P1S1W1C2 P1S1W1C3 P1S1W2C1 P1S1W2C2 P1S1W2C3 P1S2W1C1 P1S2W1C2 P1S2W1C3 P1S2W2C1 P1S2W2C2 P1S2W2C3 P2S1W1C1 P2S1W1C2 P2S1W1C3 P2S1W2C1 P2S1W2C2 P2S1W2C3 P2S2W1C1 P2S2W1C2 P2S2W1C3 P2S2W2C1 P2S2W2C2 P2S2W2C3 zW1 zW2 yW1C1 yW1C2 yW1C3 yW2C1 yW2C2 yW2C3

3700000.00

Note that warehouse 2 is no longer used in favor of all products flowing through warehouse 1. 14 (a) The demand of customer 1 for product 1 increases to 100,000 cwt. In the problem matrix of ILP02.DAT the following cell values are changed.

Cell Dem P1W1C1, yW1C1 Dem P1W2C1, yW2C1 Cap-W1, yW1C1 Cap-W2, yW2C1 Obj. coef., yW2C1 Obj. coef., yW2C1

From -50000 -50000 70000 70000 140000 70000

To -100000 -100000 120000 120000 240000 120000

The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,500,000. 176

(b) Using the ILP02.DAT file in MIPROG of LOGWARE, the following changes are made to the following cells. Cell

Obj. Coef., P2S2W1C1 Obj. Coef., P2S2W1C2 Obj. Coef., P2S2W1C3 Obj. Coef., P2S2W2C1 Obj. Coef., P2S2W2C2 Obj. Coef., P2S2W2C1

From 9 8 10 7 6 7

To 12 11 13 10 9 10

The result shows that warehouse 2 is still the only warehouse used, and the products are sourced from plant 2. However, the costs have increased to $3,380,000. (c) Using the ILP02.DAT file in MIPROG of LOGWARE, the changes are made to the following cells. Cell

Obj. Coef., yW2C1 Obj. Coef., yW2C2 Obj. Coef., yW2C3

From 70,000a 130,000 110,000

To 280,000 520,000 440,000

a

The sum of demand for the same customer for all products multiplied by the handling rate, i.e., (50,000 + 20,000) × $4/cwt. = 280,000.

The solution for product 1 shows that 50,000 cwt. flows from plant 1 through warehouse 1 and on to customer 1. The remainder flows from plant 2 through warehouse 2 and on to customers 2 and 3. For product 2, plant 1 supplies warehouse 1 and customer 1 with 20,000 cwt. The remaining 90,000 cwt. flows from plant 2 through warehouse 2 to customers 2 and 3. The total cost is $3,920,000. (d) Making some slight revisions in file ILP02.DAT can adjust the capacities on plant 1. The cell changes to make are: Cell Cap-P1S1, RHS Cap-P1S2, RHS

From 60,000 999,999

To 150,000 90,000

Both warehouses are now used for both products. summarized as:

177

The solution can be

Product 1 1 1 1 2 2 2

Plant 1 – 50,000 cwt. 1 – 60,000 2 – 40,000 2 – 50,000 1 – 20,000 2 – 30,000 2 – 60,000

Warehouse 1 – 50,000 2 – 60,000 2 – 40,000 2 – 50,000 1 – 20,000 2 – 30,000 2 – 60,000

Customer 1 – 50,000 cwt. 2 – 60,000 2 – 40,000 3 – 50,000 1 – 20,000 2 – 30,000 3 – 60,000

The total cost is $3,270,000 (e) Again revising the ILP02.DAT file by changing cells Obj. coef., P2S1W2C3 and Obj. Coef., P2S2W2C3 to have a very high cost (999), these cells are locked out of consideration. The solution is the same as the text example except that both the customers products are served from warehouse 1. The total cost is $3,340,000. 15 This problem follows the form of the Ohio Trust Company example in the text. First, identify the zones that are within 30 minutes of any particular zone. That is, Zone no. 1 2 3 4 5 6 7 8 9 10

Zones within 30 minutes 1,2,4,7,8,9,10 1,2,3,4,7,9,10 2,3,4,5,8,9,10 1,2,3,4,5,6,10 3,4,5,6,7,8 4,5,6,7,8 1,2,5,6,7,9,10 1,3,5,6,8,9,10 1,2,3,7,8,9,10 1,2,3,4,7,8,9,10

Using the MIPROG module in LOGWARE, the following matrix can be defined.

178

The solution from MIPROG is: OPTIMAL SOLUTION Variable Value Rate X(1) = 1.0000 1.0000 X(2) = .0000 1.0000 X(3) = .0000 1.0000 X(4) = 1.0000 1.0000 X(5) = .0000 1.0000 X(6) = .0000 1.0000 X(7) = .0000 1.0000 X(8) = .0000 1.0000 X(9) = .0000 1.0000 X(10) = .0000 1.0000 Objective function value =

Cost 1.0000 .0000 .0000 1.0000 .0000 .0000 .0000 .0000 .0000 .0000

Variable label 1 2 3 4 5 6 7 8 9 10

2.00

The optimal solution is to place claims adjuster stations in zones 1 and 4. 16 This is a location problem where the dominant location factor is transportation cost is and this cost is determined from optimizing the multi-stop routes originating at the material yard. The ROUTER module in LOGWARE can be used to generate these routes for each yard location. A database file for this problem (RTR13.DAT) has been prepared. Solving this problem requires balancing the cost of transporting the merchandise to the customers with the operating cost of the material yards at various locations. Optimizing the routing from the current material yard gives the route design shown in Figure 13-3. A minimum of nine trucks are required to meet all constraints on the problem. The total daily cost for this location is the route cost + vehicle costs + yard operating cost, or P4,500.87 + 9 x P200 + P350 = P6,650.87. FIGURE 13-3 Optimized Routing from Current Material Yard Location

179

*** SUMMARY REPORT *** TIME/DISTANCE/COST INFORMATION

Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 9.9 6.4 2.5 1.0 5.0 08:00AM 05:56PM 4 205 3.4 1.6 1.8 .0 .7 08:00AM 11:25AM 3 52 10.0 6.5 2.5 1.0 3.1 08:00AM 05:58PM 4 207 7.4 4.1 2.3 1.0 2.2 08:00AM 03:24PM 3 130 9.6 6.4 2.3 1.0 5.3 08:00AM 05:37PM 3 204 9.4 5.8 2.6 1.0 4.3 08:00AM 05:22PM 4 185 9.6 6.7 1.9 1.0 6.2 08:00AM 05:34PM 2 214 9.6 5.9 2.7 1.0 2.8 08:00AM 05:35PM 4 189 5.9 2.8 2.2 1.0 1.5 08:00AM 01:56PM 3 89 74.8 46.1 20.7 8.0 31.1 30 1476

Route cost,$ 602.45 220.25 608.59 416.05 599.64 552.96 625.79 563.48 311.65 4500.87

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 925 0 92.5% 9999 0 0 2 1 1000 625 0 62.5% 9999 0 0 3 1 1000 900 0 90.0% 9999 0 0 4 1 1000 950 0 95.0% 9999 0 0 5 1 1000 900 0 90.0% 9999 0 0 6 1 1000 950 0 95.0% 9999 0 0 7 1 1000 825 0 82.5% 9999 0 0 8 1 1000 1000 0 100.0% 9999 0 0 9 1 1000 850 0 85.0% 9999 0 0 Total 9000 7925 0 88.1% 89991 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

Substituting yard location A for the current yard location and solving for the route design in ROUTER yields Figure 13-4. No routes can be put end-to-end so that one truck can be used instead of two, so the minimum number of trucks remains at nine. The total daily cost for this location is 3872.02 +9 x 200 + 480 = P6,152.092. FIGURE 13-4 Route Design for Yard Location A

180


Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 9.8 7.0 1.8 1.0 5.4 08:00AM 05:47PM 4 224 8.4 5.1 2.3 1.0 3.8 08:00AM 04:26PM 3 163 8.4 5.0 2.4 1.0 3.9 08:00AM 04:26PM 3 161 8.6 4.8 2.9 1.0 2.2 08:00AM 04:37PM 5 152 7.2 4.0 2.2 1.0 2.6 08:00AM 03:12PM 3 128 6.3 2.9 2.4 1.0 1.6 08:00AM 02:18PM 3 93 5.3 2.1 2.2 1.0 1.6 08:00AM 01:15PM 3 66 8.9 5.7 2.1 1.0 3.7 08:00AM 04:51PM 3 183 5.1 1.7 2.3 1.0 .6 08:00AM 01:03PM 3 55 68.0 38.3 20.7 9.0 25.5 30 1225

Route cost,$ 649.88 498.40 491.57 470.04 410.15 321.27 254.22 548.68 227.82 3872.02


Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%


Continuing with location B, nine trucks are required and the total daily cost for the route design in Figure 3 is 3370.42 + 9 x 200 + 450= P5,620.42. FIGURE 3 Design for Yard Location B

181


Route no 1 2 3 4 5 6 7 8 9 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 7.1 3.7 2.4 1.0 2.3 08:00AM 03:06PM 4 120 6.5 3.1 2.3 1.0 1.8 08:00AM 02:27PM 3 100 7.9 4.6 2.4 1.0 3.0 08:00AM 03:56PM 4 146 5.9 2.8 2.2 1.0 1.4 08:00AM 01:56PM 3 89 5.9 2.6 2.3 1.0 1.1 08:00AM 01:55PM 3 84 6.7 3.6 2.1 1.0 2.3 08:00AM 02:41PM 3 114 5.8 2.5 2.3 1.0 2.1 08:00AM 01:50PM 3 80 6.1 3.3 1.9 1.0 2.8 08:00AM 02:07PM 2 104 9.7 5.8 2.8 1.0 3.0 08:00AM 05:40PM 5 187 61.7 32.0 20.7 9.0 19.8 30 1024

Route cost,$ 389.30 340.53 454.97 311.84 300.41 375.39 290.22 350.46 557.31 3370.42


Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%


Finally, the route design from location C is shown in Figure 4. Although 10 routes are in the design, two of these can be dovetailed so that only nine trucks are needed. The total daily cost is 4479.99 + 9 x 200 + 420 = P6,699.99. FIGURE 4 Design for Yard Location C

182


Route no 1 2 3 4 5 6 7 8 9 10 Total

Route Run Stop Brk Stem time, time, time, time, time, Start Return No of Route hr hr hr hr hr time time stops dist,Mi 1.5 .6 .9 .0 .6 08:00AM 09:30AM 1 20 6.7 3.7 2.0 1.0 1.9 08:00AM 02:41PM 2 120 8.0 4.1 2.9 1.0 1.5 08:00AM 04:00PM 5 132 9.4 5.6 2.8 1.0 3.4 08:00AM 05:25PM 5 180 7.5 4.2 2.3 1.0 2.7 08:00AM 03:29PM 3 134 7.7 4.2 2.4 1.0 1.9 08:00AM 03:39PM 3 136 7.0 3.7 2.3 1.0 2.8 08:00AM 02:59PM 3 117 7.3 4.0 2.3 1.0 3.5 08:00AM 03:18PM 3 127 9.6 7.2 1.4 1.0 5.6 08:00AM 05:35PM 2 229 9.8 7.4 1.4 1.0 6.3 08:00AM 05:47PM 3 236 74.5 44.7 20.7 9.0 30.3 30 1432

Route cost,$ 140.91 389.21 420.20 540.48 425.69 429.10 383.04 408.53 663.40 679.42 4479.99

VEHICLE INFORMATION Route Veh Weight Delvry Pickup Weight Cube Delvry Pickup no typ capcty weight weight util capcty cube cube 1 1 1000 375 0 37.5% 9999 0 0 2 1 1000 875 0 87.5% 9999 0 0 3 1 1000 975 0 97.5% 9999 0 0 4 1 1000 925 0 92.5% 9999 0 0 5 1 1000 925 0 92.5% 9999 0 0 6 1 1000 1000 0 100.0% 9999 0 0 7 1 1000 950 0 95.0% 9999 0 0 8 1 1000 950 0 95.0% 9999 0 0 9 1 1000 550 0 55.0% 9999 0 0 10 1 1000 400 0 40.0% 9999 0 0 Total 10000 7925 0 79.3% 99990 0 0

Cube util .0% .0% .0% .0% .0% .0% .0% .0% .0% .0% .0%

Vehicle description Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1 Truck #1

From an economic analysis, it appears that yard location B is the best choice.

183

SUPERIOR MEDICAL EQUIPMENT COMPANY Teaching Note Strategy The purpose of the Superior Medical Equipment Company case study is to encourage students to apply the center-of-gravity methodology to a problem involving a single warehouse location. Although the methodology is somewhat elementary, it can be useful in providing some first approximations to good warehouse locations. It allows students to evaluate the financial implications of alternative network designs, and it is intended to be solved with the aid of the COG module provided in the LOGWARE software. Answers to Questions (1) Based on information for the current year, is Kansas City the best location for a warehouse? If not, what are the coordinates for a better location? What cost improvement can be expected from the new location?

When a single warehouse is to be located, the primary location costs are transportation, both inbound to the warehouse and outbound from it, and the warehouse lease, which varies with the location. The current location serves as a benchmark against which the costs for other locations can be compared. That is, based on information given in the case, the total relevant cost for the Kansas City location is: Inbound transportation Outbound transportation Lease $2.75/sq. ft. × 200,000 sq. ft. Total relevant cost

$

$

2,162,535 4,819,569 550,000 7,532,104

Using the COG module with inbound transport rates from Phoenix set at $16.73/1163 = $0.014/cwt./mile and from Monterrey set at $9.40/1188 = $0.008/cwt./mile, and the outbound transport rate from the unknown warehouse location set at $0.0235/cwt./mile, the coordinates for the best location are X = 7.61 and Y = 4.51, or approximately Oklahoma City. Total transportation cost for this location would be $6,754,082. The total relevant cost would be: Transportation Lease $3.25/sq. ft. × 200,000 sq. ft. Total

$ 6,754,082 650,000 $ 7,404,082

The annual cost savings can be projected as $7,532,104 − 7,404,082 = $128,022. (2) In five years, management expects that Seattle, Los Angeles, and Denver markets to grow by 5 percent, but the remaining markets to decline by 10 percent. Transport costs are expected to be unchanged. Phoenix output will increase by 5 percent, and Monterrey's output will decrease by 10 percent. Would your decision about the location of the warehouse change? If so, how?

184

A new benchmark for the fifth year can be computed from the data given in Tables 1 and 2. After adjusting the plant and market volumes according to the changes indicated, the fifth year benchmark costs can be computed as follows: Volume,

Rate,

Transport

Point cwt. 1 64,575 2 108,540 3 17,850 4 33,600 5 13,125 6 8,550 7 26,550 8 18,900 9 37,170 10 7,740 11 9,630 Totals 346,230

$/cwt. $16.73 9.40 33.69 30.43 25.75 18.32 25.24 19.66 26.52 26.17 27.98

cost, $ $1,080,340 1,020,276 601,367 1,022,448 337,969 156,636 670,122 371,574 985,748 202,556 269,447 $6,718,483

The total 5th-year benchmark relevant cost would be: Transportation Lease $2.75/sq. ft. × 200,000 sq. ft. Total

$ 6,718,483 550,000 $ 7,268,483

Optimizing the location with the fifth year data gives a location at X = 7.05 and Y = 4.52. The relevant costs for this location are: Transportation Lease $3.25/sq. ft. × 200,000 sq. ft. Total

$ 6,464,206 650,000 $ 7,114,206

The annualized cost savings would be $7,268,483 − 7,114,206 = $154,277. The average annual cost savings are (128,022 + 154,277)/2 = $141,150. The simple annual return on investment (moving cost) can be computed as: ROI =

$141,150 . × 100 = 471% $300,000

Management must now judge whether 47.1 percent annual return is worth the risk of changing warehouse locations. (3) If by year five increases are expected of 25 percent in warehouse outbound transport rates and 15 percent in warehouse inbound rates, would your decision change about the warehouse location?

185

It is assumed here that the fifth year demand level applies. A revised fifth year benchmark can be recomputed by applying the cost growth factors to the overall fifth year transport costs. That is, Inbound 1.15 × 2,100,616 = $2,415,708 Outbound 1.25×4,617,867 = Subtotal

$8,188,042 Lease

Total

5,772,334 550,000

$8,738,042

Running COG shows that the minimum transport cost location would be at coordinates X = 7.20 and Y = 4.62, which is near the previous location in question two. The shift in location is minimal. The cost for this location is: Transportation $ 7,939,545 Lease $3.25/sq. ft. × 200,000 sq. ft. 650,000 Total $ 8,589,545 The annualized cost savings would be $8,738,042 − 8,589,545 = $148,497. It can be concluded that: 1. Location is similar to the optimized fifth year location. 2. The increase in possible cost savings further encourages relocation from Kansas City and toward a site near Oklahoma City, OK. (4) If the center-of-gravity method is used to analyze the data, what are its benefits and limitations for locating a warehouse?

The center-of-gravity method locates a facility based on transportation costs alone. This is reasonable when only one facility is being located and the general location for it is being sought. Such costs as inventory carrying, production, and warehouse fixed are not included, but they are not particularly relevant to the problem. However, costs such as warehouse storage and handling, and other costs that vary by the particular site are not included but may be relevant in a given situation. Transportation costs are assumed linear with distance. This may not be strictly true, although distance may be nonlinear. The obvious benefits of the method are (1) it is a fast solution methodology; (2) it considers all possible locations (continuous); (3) it is simple to use; (4) its data are readily available; and (5) it gives precise locations through a coordinate system. Some potential limitations are (1) coordinates need to be linear; (2) transportation rates on a per-mile basis are constant; (3) volumes are known and constant for given demand and source points; and (4) locations may be suggested that are not feasible such as in lakes, central cities, or restricted lands.

186

Concluding Comments The analysis in the case seems to suggest a move from Kansas City to a region around Oklahoma City would be advantageous. A return on investment of 47 percent or higher is possible, however management must now seek a particular site in the area whose choice may add or detract from this savings potential. In any case, the COG method has assisted in the selection of good potential locations and testing their sensitivity to changes in costs and volumes. APPENDIX 1

LOGWARE COG Module Input Data for the Current Year

Title: SUPERIOR MEDICAL EQUIPMENT COMPANY Power factor: .5 Scale factor: 230 Point X-coordinate Y-coordinate Volume 1 3.60 3.90 61,500 2 6.90 1.00 120,600 3 0.90 9.10 17,000 4 1.95 4.20 32,000 5 5.60 6.10 12,500 6 7.80 3.60 9,500 7 10.20 6.90 29,500 8 11.30 3.95 21,000 9 14.00 6.55 41,300 10 12.70 7.80 8,600 11 14.30 8.25 10,700

187

Rate 0.0140 0.0080 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235 0.0235

OHIO AUTO & DRIVER'S LICENSE BUREAUS Teaching Note Strategy The purpose of this case study is to introduce students to a logistics problem in a service area. It also provides an application for the multiple center of gravity location methodology. The MULTICOG module in the LOGWARE software can effectively be applied. The module allows students to quickly evaluate alternatives as to the number of bureaus to use, the bureau locations, and the size of the territory that each bureau should serve. The case may be assigned as a homework problem, a short case study project, or as a case for class discussion. The later would be appropriate especially if adequate attention is given to the issues of how Dan should go about solving a problem such as this, what data he needs and where to obtain it, and what concerns he should have about changing the existing network design. This would encourage students to think beyond the computational aspects of the problem. Answers to Questions (1) Do you think there is any benefit to changing the network of license bureaus in the Cleveland area? If so, how should the network be configured?

The nature of the costs and the number of possible alternative network designs make it impractical to seek an optimal solution. Therefore, a possible approach to the analysis is outlined as follows. First, establish a benchmark against which changes to the network can be compared. Much of the data for this is given in the case write up. The costs can simply be applied to the size of each bureau and its associated staff. The cost for residents traveling to the bureaus is not known because the bureau territories are not known. However, an estimate can be made of travel costs by solving the problem in MULTICOG for eight bureaus. Since MULTICOG attempts to optimize bureau location, this travel cost is probably understated. The benchmark costs are summarized in Table 1. The location costs for the current operation are estimated to be $1,355,706. Second, what improvements can be made on the existing eight locations? Besides moving the locations of the bureaus, which results in resizing the facilities and adjusting the staff numbers, there are no obvious improvements to be made. Therefore, the benchmark remains the base for comparison. Third, it is now necessary to estimate the approximate number of bureaus that are needed to serve the area. Since the costs for a particular network design depend on the size of each bureau, which cannot be known until the problem is solved, an initial assumption must be made. It will be assumed that all bureaus are of the same size. Hence, for 5 bureaus, the average number of residents in each bureau’s territory would be the total number of residents divided by the number of bureaus, or 691,700/5 = 138,340. Rent, staff salaries, and utility expenses can be derived from this estimate. Table 2 is developed to show the bureau size and the number of staff for one to 10 bureaus. Table 3 extends the average costs from these estimates. A reduced number of bureaus, in the range of two, is about right.

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TABLE 1

Bureau 1 2 3 4 5 6 7 8 Totals

Benchmark Costs for the Current Network of License Bureaus Customer Size, Rent, Salaries, Utilities, Travel, Staff sq. ft. $ $ $ $ 1,700 4 37,400 84,000 6,800 1,200 4 26,400 84,000 4,800 2,000 5 44,000 105,000 8,000 1,800 4 39,600 84,000 7,200 1,500 4 33,000 84,000 6,000 2,200 5 48,400 105,000 8,800 2,700 5 59,400 105,000 10,800 1,500 5 33,000 105,000 6,000 14,600 36 321,200 756,000 58,400 220,106*

*Approximated by running MULTICOG at 8 bureaus.

TABLE 2 Average Size and Staff for Various Numbers of Bureaus (1) 691,700/(1) Avg. Avg. no. of bureau Staff per No. of residents per size, bureau bureaus bureau sq. ft. 1 691,700 4,500 10 2 345,850 3,000 7 3 230,567 2,500 6 4 172,925 2,000 5 5 138,340 2,000 5 6 115,283 2,000 5 7 98,814 1,500 4 8 86,463 1,500 4 9 78,855 1,500 4 10 69,170 1,500 4

189

TABLE 3 Average Costs by Number of Bureaus

No. of bureaus 1 2 3 4 5 6 7 8 9 10

Rent, $ 99,000 132,000 165,000 176,000 220,000 264,000 231,000 264,000 297,000 330,000

Staff salaries, $ 210,000 294,000 378,000 420,000 525,000 630,000 588,000 672,000 756,000 840,000

Utilities, $ 18,000 24,000 30,000 32,000 40,000 48,000 42,000 48,000 54,000 60,000

Resident travel, $ 662,319 430,922 354,239 298,000 278,181 249,287 237,635 220,106 206,496 198,600

Annual total cost, $ 989,319 880,922⇐ 927,239 926,965 1,063,181 1,191,287 1,098,636 1,204,106 1,313,496 1,428,600

The cost estimates can now be refined around two bureaus. A comparison with the benchmark costs and a return of the initial investment (costs related to changing the network design) are sought. A sample analysis for two bureaus, based on a design provided by MULTICOG, is shown below.

Bureau 1 2 Totals

Residents 290,200 401,500 691,700

Size, sq. ft. 2,500 3,500 6,000

Staff 6 8 14

Rent, $ 55,000 77,000 132,000

Staff salaries, Utilities, $ $ 126,000 10,000 168,000 14,000 294,000 24,000

Resident travel cost, $ 166,332 264,590 430,922

The total annual variable cost is $132,000 + 294,000 + 24,000 + 430,922 = $880,922. There are one-time costs due to staff separation and equipment moves. Compared with the benchmark, 36 − 14 = 22 staff members will be separated for a cost of 22 × $8,000 = $176,000. Equipment movement costs to two bureaus would be 2 × 10,000 = $20,000. Total movement costs would be $176,000 + 20,000 = $196,000. Annual variable cost savings compared with the benchmark would be $1,355,706 − 880,922 = $474,784. A simple return on investment would be: ROI =

$474,784 × 100 = 242% $196,000

Similar calculations are carried out for various numbers of bureaus. These results are tabulated in Table 4.

190

TABLE 4 Cost Savings and Return of Investment for Alternate Network Designs as Produced by MULTICOG Return on Total Annual Savings, investNo. of size, sq. Total variable cost, Moving $ ment, % bureaus ft. staff $ cost, $ 1 4,500 10 989,319 218,000 366,387 168 2 6,000 14 880,922 196,000 242 474,784⇐ ⇐ 3 7,500 18 927,239 176,000 428,467 243 4 8,500 21 960,965 160,000 394,741 246⇐ 5 9,500 24 1,029,181 146,000 326,525 223 6 11,500 29 1,157,287 106,000 198,419 187 7 12,000 31 1,200,635 110,000 155,071 141 8 13,000 34 1,272,106 96,000 83,600 87 Benchmark 14,600 36 1,355,706 ---------

The maximum annual savings occurs with a network containing two bureaus. However, the maximal return on investment occurs with four bureaus. ROI is selected as the appropriate measure on which to base this economic decision. The details for a design with four bureaus are given in Table 5. The design is shown pictorially in Figure 1 of this note. TABLE 5 Design Details for a Network with Four Bureaus Column grid Row grid coordicoordiBureau nate nate Residents Grid box number assignment 1 3.36 2.74 218,200 1,2,3,4,5,6,7,8,9,10,11,12,13, 14,15,16,17,18,19,20,21,22,23, 24,25,26,27,28,29,30,31,32,33 2 7.00 3.00 168,700 34,35,36,37,38,39,40,41,43,44, 45,46,47,50,51,52,53 3 9.74 5.66 195,000 42,48,49,54,55,56,60,61,62,63, 67,68,69,70,74,75,76,77,81,82, 83,84, 4 10.00 2.00 109,800 57,58,59,64,65,66,71,72,73,78, 79,80

(2) Do you think Dan Roger's study approach is sound?

Overall Dan can be praised for the simplicity of the methodology that he has chosen. The center-of-gravity approach is appropriate in this problem since there are no capacity limitations on the facilities, locations across a continuous space are desired, and transportation cost (except for some fixed costs) is the primary location variable. The data requirements are relatively straightforward, although they are not always easy to fulfill. This would likely be a problem with any other solution approach as well. The results of this methodology can only be used as a first approximation at best. Several criticisms of this particular approach can be offered as follows:

191

• The effect of bureau location on the resident's perception of service is not as well known as portrayed in the case. In addition, service may need to be represented by more than location. • Travel to the bureaus is assumed straight line. However, location in the area is likely to be influenced by a road network. Time may be more important than distance to residents. • The fixed costs associated with location are not handled directly by the center-ofgravity approach. • Residents are assumed to travel to the locations within their assigned territories. They may not strictly do this. • Good facilities may not be available at the indicated location coordinates. The analysis is particularly weak around the estimate of the resident travel cost. While an exact cost is not likely to be known, Dan should conduct a sensitivity analysis around this cost. He may find that the design does not change a great deal over a wide range of assumed values. If this is the case, he can feel comfortable that his recommendation is fundamentally sound. If not, he should seek to find a more precise value. (3) What concerns besides economic ones should Dan have before suggesting that any changes be made to the network?

A quantitative approach to location will rarely give the precise locations to be implemented. Rather, it provides a starting point for further analysis. There are a number of other factors to be considered before the revised network design can be implemented. First, there are site selection factors to be taken into account such as the availability of adequate space near the location coordinates, proximity to good highway linkages, and reasonable neighborhood reactions to this type of operation. Second, there are political concerns. Reducing the number of locations will result in a releasing some of the staff. Dan may experience some political resistance to this. Since staff is a large expense in the operation, currently about 2/3 of the costs, retention of a larger number of bureaus may be required. Of course, transferring staff to other governmental operations may be a way of dealing with this issue. This assumes their willingness to be relocated, although this is not likely to be a strong issue if relocation were to occur in the same area. Third, there may be difficulty in demonstrating the economics of network redesign. Although others may appreciate the costs of rent, salaries, and utilities, the cost of resident travel is subject to much interpretation. Those favoring many bureaus may argue the high cost while those wanting to reduce the number of bureaus may perceive it as not very significant.

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Grid row number

FIGURE 1 Four Bureau Locations and Their Territories

Current bureau locations

Grid column number

Revised bureau locations

SUPPLEMENT Sample Input Data File for MULTICOG in LOGWARE for the Ohio Auto & Driver's License Bureau Case Study Title: LICENSE BUREAU Number of sources: 4 Number of demand points: 84 Scaling factor: 2.5 POINT X-COORDINATE Y-COORDINATE 1 1 1 2 1 2 3 1 3 4 1 4 5 1 5 6 1 6 7 1 7 8 2 1 9 2 2 10 2 3 11 2 4 12 2 5 13 2 6 14 2 7 15 3 1 16 3 2 17 3 3 18 3 4 19 3 5 20 3 6 21 3 7

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VOLUME 4100 6200 7200 10300 200 0 0 7800 8700 9400 11800 100 0 0 8100 10500 15600 10500 200 0 0

RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12

SUPPLEMENT (Continued) POINT 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74

X-COORDINATE 4 4 4 4 4 4 4 5 5 5 5 5 5 5 6 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 9 9 9 9 9 9 9 10 10 10 10 10 10 10 11 11 11 11

Y-COORDINATE 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4 5 6 7 1 2 3 4

194

VOLUME 10700 12800 13800 15600 400 0 0 11500 13900 14500 13700 600 0 0 9300 14900 13700 10200 1200 0 0 10100 12600 16700 15800 12400 2600 0 8800 13700 15200 14100 10800 17200 500 5300 16700 13800 11900 13500 18600 12000 5100 17400 10300 9800 10300 15500 11700 7700 9200 7500 8500

RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12 .12

POINT 75 76 77 78 79 80 81 82 83 84

X-COORDINATE 11 11 11 12 12 12 12 12 12 12

Y-COORDINATE 5 6 7 1 2 3 4 5 6 7

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VOLUME 7800 9900 8700 4300 6700 5800 6800 5400 7100 6400

RATE .12 .12 .12 .12 .12 .12 .12 .12 .12 .12

SOUTHERN BREWERY Teaching Note Strategy The purpose of this case study is to provide students with the opportunity to design a distribution network where plant location is at issue. They first should identify the major costs and alternatives that are important to such a design problem. Second, they should be encouraged to apply the transportation method of linear programming to assist in the analysis of alternatives using the TRANLP module in LOGWARE. Finally, they should consider factors other than those in the analysis that might alter the course of their recommendation and be sensitive to the limitations and benefits of linear programming as a solution methodology. Answers to Questions (1) If you were Carolyn Carter, would you agree with the proposal to build the new brewery? If you do, what plan for distribution would you suggest?

If growth is uniform over the next five years, Southern can expect that demand for its products will exceed the currently available plant capacity. That is, demand is increasing at the rate of (0 + (595,000 − 403,000)) ÷ 5 = 38,400 barrels per year. Thus, the current annual capacity of 500,000 barrels will be used up in (500,000 − 403,000) ÷ 38,400 = 2.5 years. A major concern is whether it would be profitable to construct the new plant. Rough estimates of its profitability can be made, projecting profits with and without the new plant. We know that 595,000 − 500,000 = 95,000 barrels of beer would not be sold annually in the fifth year if additional capacity is not constructed. This represents a potential average lost revenue of $280/barrel × 0.20 × [(0 + 95,000) ÷ 2.5] = $2,128,000. (The figure of 2.5 years assumes that the new brewery can be brought on stream at approximately the time when capacity will be used up in the existing plants.) From the benchmark1 costs for the current system, as shown in Table 1, Southern is currently producing and distributing 403,000 barrels at a total cost of $60,015,000. This is an average cost per barrel of $60,015,000 ÷ 403,000 = $148.92. The overhead and sales expense is 27 percent, or $280 × 0.27 = $75.60 per barrel. Total costs per barrel are $148.92 + 75.60 = $224.52, which is about 80 percent of the sales dollar. The 20 percent profit margin seems valid. Therefore, the benefit of serving the potentially lost demand with a new brewery can be estimated using on a simple return on investment: ROI =

$2,128,000 = 0.21, or 21% $10,000,000

If management feels that this is an adequate return for such a project, Carolyn should proceed with her analysis. Let's assume that she has this approval. Next, she may wish to explore the opportunities available by improving upon the existing distribution system without the presence of the new plant. This is an improved 1

A benchmark refers to the costs of producing and distributing demand as currently allocated throughout the network.

196

benchmark2 and it can be found by solving a transportation-type linear programming problem of the type shown in Table 2. The results given in the TRANLP module of LOGWARE can be summarized in Table 3. This shows that with some slight reallocation of demand among the plants, production and distribution costs can be reduced by $60,015,000 − 59,804,000 = $211,000 annually. A comparison of the benchmark results in Table 1 and the improved benchmark results in Table 3 shows that Knoxville's demand should be shifted from Columbia to Montgomery and 28,000 barrels of Columbia's market should be shifted from the Columbia plant to the Montgomery plant. The Columbia plant is a high-cost producer and cost savings are achieved by trading production costs for transportation costs. That is, production costs will be reduced by $400,000 while transportation costs will be increased by $189,000. Note that this $211,000 savings can be realized without any capital investment. TABLE 1 Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Producport Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Columbia 22 3,190 282 3,472 4 Columbia Columbia 44 6,380 306 6,686 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Montgomery 13 1,781 179 1,960 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Montgomery 26 3,562 355 3,917 5,206 577 5,783 9 Jacksonville Montgomery 38 Total 403 $56,000 $4,015 $60,015 TABLE 2 TRANLP Problem Setup RICH-

RA-

KNOX-

COL-

AT-

SAVAN-

MONT-

TALL

JACK-

From\To

MOND

LEIGH

VILLE

UMBIA

LANTA

NAH

GOMRY

AHAS

SONVL

RICHMD

148.49

150.70

156.38

152.54

155.48

154.64

159.98

164.30

158.84

100

COLMBA

157.54

154.78

157.81

151.96

156.85

154.54

157.93

160.18

157.27

100

MONTGM

156.98

153.35

150.80

149.93

147.20

150.80

143.69

150.65

152.18

300

JACKVL

152.13

149.25

150.48

146.16

148.80

144.54

148.80

144.72

142.68

0

Demand

56

31

22

44

94

13

79

26

38

2

Supply

An improved benchmark refers to a reallocation of current demand in an optimal way, respecting plant capacity restrictions.

197

TABLE 3 Improved Benchmark of Production and Transportation Costs ($000s) for Current Demand Demand TransBrewery of in 000s Producport Market area origin barrels tion costs costs Total costs 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Montgomery 22 3,014 304 3,318 4 Columbia Columbia 16 2,320 111 2,431 4 Columbia Montgomery 28 3,836 362 4,198 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Montgomery 13 1,781 179 1,960 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Montgomery 26 3,562 355 3,917 5,206 577 5,783 9 Jacksonville Montgomery 38 Total 403 $55,600 $4,204 $59,804

Adding a plant at Jacksonville with a capacity of 100,000 barrels per year provides enough capacity to satisfy demand out to the fifth year. If the new plant were constructed and producing immediately, total costs could be reduced from the improved benchmark by $59,804,000 − 59,090,000 = $714,000 per year (compare the total costs in Tables 3 and 4.) The Columbia plant would not be needed if the lower-cost Jacksonville plant were on line. We do not know the savings for the fifth year demand level since not all demand can be served without the presence of the new plant. Therefore, a future-year benchmark cannot be determined. However, we do know how the new plant should be utilized within the system (see Table 5) and how demand allocation should be adjusted to accommodate it. Also, note that the Columbia plant is needed once again although its capacity is not required until the last one-half year of the fifth year planning horizon. This suggests that the Columbia plant should not be sold, but perhaps some alternate use could be made of the facility in the interim, such as subcontracting beer production to a non-competing company. The new plant is not likely to be brought on stream immediately nor is it needed for two and a half years, so Carolyn might suggest a distribution plan similar to that in Table 5. A careful inspection of this plan shows that only one barrel of demand in the Knoxville region is assigned to Richmond. Splitting demand to this extent is probably not practical and can be assigned to Columbia where there is excess capacity. Costs will rise only slightly. An interesting question is whether the Columbia plant, through modernization, could be made as efficient as the new brewery, and what the implications for distribution might be. We know that this could potentially save $145 − 135 = $10 per barrel in production costs. At a 100,000-barrel capacity, this is $1,000,000 in cost savings. If the modernization where to cost no more than $5,000,000, this option might be attractive. Of course, we would need to resolve the linear programming problem with Columbia's per barrel costs at $135 plus transportation costs. This would tell us how and to what extent demand would be allocated to Columbia and give a more accurate basis for determining

198

the cost savings. Similarly, it would be interesting to explore what it means to expand the capacity of an existing brewery at a lower investment cost per barrel than the construction of a new facility. TABLE 4 Production and Transportation Costs ($000s) for Current Demand with the Jacksonville Plant Demand Transin 000s Producport Brewery of Total costs origin barrels tion costs costs Market area 1 Richmond Richmond 56 $7,840 $475 $8,315 2 Raleigh Richmond 31 4,340 332 4,672 3 Knoxville Montgomery 22 3,014 304 3,318 4 Columbia Montgomery 21 2,877 272 3,149 4 Columbia Jacksonville 23 3,105 257 3,362 5 Atlanta Montgomery 94 12,878 959 13,837 6 Savannah Jacksonville 13 1,755 124 1,879 7 Montgomery Montgomery 79 10,823 550 11,373 8 Tallahassee Jacksonville 26 3,510 253 3,763 5,130 292 5,422 9 Jacksonville Jacksonville 38 Total 403 $55,272 $3,818 $59,090

(2) If the new brewery is not to be constructed, what distribution plan would you propose to top management?

Table 6 shows a linear programming solution where the new plant is not brought on stream and the demand in the markets is set at the five-year level. An interesting solution occurs when demand exceeds capacity. The most costly demand region to serve is not assigned to any plant. As can be seen in Table 6, portions of the demand in Tallahassee and Jacksonville should not be served, and essentially the entire Knoxville market should not be served at all. Top management may wish to adjust this plan for reasons other than economic ones.

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TABLE 5 Production and Transportation Costs ($000s) for Projected Fifth Year Demand With the Jacksonville Plant TransDemand port Producin 000s Brewery of Total costs costs barrels tion costs origin Market area 1 Richmond Richmond 64 $8,960 $543 $9,503 2 Raleigh Richmond 35 4,900 375 5,275 3 Knoxville Richmond 1 140 16 156 3 Knoxville Columbia 20 2,900 256 3,156 3 Knoxville Montgomery 12 1,644 166 1,810 4 Columbia Columbia 55 7,975 383 8,358 5 Atlanta Montgomery 141 19,317 1,438 20,755 6 Savannah Columbia 20 2,900 191 3,091 7 Montgomery Montgomery 119 16,303 828 17,131 8 Tallahassee Montgomery 28 3,836 382 4,218 8 Tallahassee Jacksonville 24 3,240 233 3,473 10,260 584 10,844 9 Jacksonville Jacksonville 76 Total 595 $82,375 $5,395 $87,770

TABLE 6 Production and Transportation Costs ($000s) for Projected Fifth Year Demand Without the Jacksonville Plant Expected Served demand in demand in Brewery of 000s 000s Market area origin barrels barrels Total costs 1 Richmond Richmond 64 64 $9,503 2 Raleigh Richmond 35 35 5,275 3 Knoxville Richmond 33 1* 156 4 Columbia Columbia 55 55 8,358 5 Atlanta Montgomery 141 141 20,755 6 Savannah Columbia 20 20 3,091 7 Montgomery Montgomery 119 119 17,131 8 Tallahassee Montgomery 52 40* 6,026 25* 3,931 9 Jacksonville Columbia 76 Total 595 500 $74,226 *Indicates market demand is not fully served due to inadequate plant capacity.

(3) What additional considerations should be taken into account before reaching a final decision?

A number of assumptions have been implied in the analysis shown above. For example, • Demand has been assumed to grow at a constant rate in the markets. • Production is assumed limited to exactly the values given without the possibility for expansion through overtime, additional shifts, or subcontracting.

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• Per-unit production and transportation costs are assumed to remain unchanged with the reallocation of demand throughout the network. • Customer service effects are not considered in reallocation of demand. • There is no change in per-unit costs throughout the five-year planning horizon. This case might end with a discussion of the appropriateness of using linear programming as a vehicle for analysis in a problem such as this. Mentioning that linear programming does not consider such factors as fixed costs, return on investment, or the many subjective factors (top management's intuition about location, vested interests, etc.) that are typically a part of such problems means that linear programming, at best, is a facilitating vehicle for analysis. It does not provide the final answer.

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Ballou Logistics Solved Problems Chapter 13

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