Solution Manual—Chapter 2 Study Questions
1. Setup costs. (a) No, if anything it is harder because plants today generally generally produce more products and are more more complex than those in 1913. (b) labor cost (e.g., worker doing setup and any workers workers who become less productive productive during setup), materials cost (e.g., grease, gaskets, etc.), t ool wear (e.g., of tools used to t o perform setup), lost product (e.g., useless pink paint as process is switched from red to white), lost capacity (i.e., product that was needed but not produced due to setup). (c) Capacity, because the fewer fewer setups that that are performed, the greater the system system capacity. However, note that the value of capacity depends on the current status of the plant (i.e., how heavily loaded that process center is) and is therefore very difficult to specify. 2. Inventory carrying costs. • Carrying cost can be a surrogate for WIP, since the higher the carrying cost, the less WIP the model will be inclined to carry. • Another term for interest rate is “discount “discount rate.” This charge is supposed to represent the opportunity cost of money (i.e., the rate of return that could be attained if the money were invested in the best alternative use. 3. Harris’ “higher” “higher” mathematics mathematics was: • Calculus (differential calculus, to be specific). • Newton and Leibnitz invented it in the 17th century. • Most Americans study calculus in high school or freshman year of college. • Any engineer, in 1913 or now, has more than enough background in calculus to understand the derivation of the EOQ formula. 4. Appropriate (A) and less appropriate (L) applications of EOQ: • automobile manufacturer ordering screws from a vendor: A cars to paint per bat ch of a particular color: L • automobile manufacturer deciding on how many cars • a job shop ordering bar stock: A • office ordering copier paper: A • a steel company deciding how many slabs to move at once between the casting furnace and the rolling mill: L 5. Ways to live with with the assumption of “constant demand.” • One could average the demand as far out as the forecast is available, pretend it is constant, and use EOQ. segments (e.g., quarters) and use in EOQ to compute lot sizes for each • Average demand over time segments quarter. • Use Wagner-Whitin to compute dynamic lot sizes. 6. The key difference difference between Wagner-Whitin Wagner-Whitin and the EOQ assumptions is that Wagner-Whitin Wagner-Whitin assumes non-constant (although still deterministic) demand. 7. Wagner-Whitin “insights:” never • The WW Property insight is more for the model than the plant. It says that we should never order/produce if we have inventory to cover the demand in a period. • The problem with this property is that i t depends critically on the deterministic assumptions of the model. If everything were were deterministic, we would never never plan to both carry inventory and incur a setup in a period. But, given randomness in demand, demand, production, yield, etc., it is conceivable that we might
5
do exactly this in practice. 8. Four criticisms of the validity of the Wagner-Whitin model are: • Demands are assumed fixed and known. • Setup costs are very difficult to specify, since as in EOQ, t hey may really be a proxy for capacity, which is context sensitive. • Wagner-Whitin property implies that production should be for an integer number of periods of demand. This completely ignores capacity and yield issues, which may make it attractive to produce different quantities than permitted under this property. • Variable lot sizes may present process problems where there are physical reasons for maintaining lot sizes that are multiples of some number (e.g., tote sizes are set, negatives in an expose operation can produce a specific number of items before wearing out, a casting furnace produces 250 tons of steel per load (heat) and cannot be run with partial loads, etc.). 9. The key difference between EOQ and (Q,r ) is that demand is stochastic in (Q,r ) but deterministic in EOQ. The base stock model also has stochastic demand, but unlike the (Q,r ) model, it assumes that replenishment lot sizes are always equal to one. 10. The statement “the reorder point, r , affects customer service, while the replenishment quantity, Q, affects replenishment frequency” is true in rough terms but is not precisely true because: • Q certainly does govern order frequency (i.e., if annual demand is D, then the number of orders per year is F = D/Q). • Once Q is fixed, then r determines the customer service (i.e., the probability of an order being filled out of stock). • However, if we change Q for a given r , customer service changes. The reason is that if Q is made large, then the number of times per year that inventory falls to a level where stockouts occur is reduced. Thus, for a fixed r , service level increases with Q. This interaction between Q and r in determining service level is the reason for introducing Type I Service, which by definition is governed only b y r . This greatly simplifies solution of the model and for cases where Type I Service is a good approximation of the actual service level, provides useful heuristics. 11. Increasing the variability of the demand process tends to require a higher level of safety stock (i.e., a higher reorder point) in all of the following conditions: • Demand is deterministic. Set Q equal to the mean demand during a replenishment lead time. Thus, setting r = Q provides perfect service (i.e., we would just stock out as the replenishment order arrived). • Demand during replenishment lead time is more variable. Suppose it ranges between Q/2 to 2Q, still with a mean of Q. Now, to provide perfect service, we would have to set $r = 2Q. In general, the longer the right hand tail of the distribution of demand during a replenishment lead time, the higher r will have to be in order to achieve a given service level. Increasing variability (i.e., the standard deviation) of demand during a replenishment lead time tends to increase the right hand tail (note that the left hand tail is bounded by zero) and hence increases the safety stock required to attain a given service level. 12. When stocking parts purchased from vendors in a warehouse you could use a (Q,r ) model to determine whether a vendor of a part with a higher price but a shorter lead time is offering a good deal by solving the model with both price/leadtime pairs and see which resulted in the l owest overall cost. Other factors you should consider in deciding to change vendors include issues related to vendor reliability, quality, cost of certifying the new vendor (if you have a vendor certification program), etc. 13. In a multi-product reorder point problem subject to an aggregate service constraint, the effect of increasing the cost of one of the part will be to cause the fill rate for that part to go down while the fill rate for other parts to goes up. The reason is that it becomes more costly to hold inventory of the altered part necessary to attain the original fill rate and i s therefore economical to hold less of this p art, at the expense of
6
poorer service, and make up the overall service by holding more of the other parts. 14. One would expect the event that this particular man brings a bomb on the plane (denote it as Event A) and the event that another person brings a bomb on the plane (denote as Event B) should be independent (unless there is a conspiricy). Now the probability of two independent events occuring is P{ A ∩ B} = P{ A} • P{ B} but the probability of Event B occuring given Event A has occurred is P{ A ∩ B} P{ A} P{ B} P{B| A} = = = P{B} P{ A} P{ A} so this guy will not alter the probability that another person brings a bomb on the plane. Even worse, the authorities won’t buy his logic and he will wind up in jail .
Problems
1. (a) Suppose we have the following outcomes: ( ( ( ( ( ( ( ( ( (
H ,H ) H ,H ) T ,H ) H ,H ) H ,T ) H ,T ) H ,H ) T ,H ) H ,H ) T ,H )
( T ,H ) ( H ,T ) ( H ,H ) ( T ,T ) ( T ,T ) ( T ,T ) ( H ,T ) ( H ,T ) ( T ,H ) ( T ,T ) ( H ,H ) ( T ,H ) ( H ,T ) ( H ,T ) ( T ,T ) ( H ,H ) ( H ,T ) ( H ,H ) ( H ,H ) ( T ,H ) ( T ,T ) ( H ,H ) ( T ,T ) ( H ,H ) ( T ,H ) ( H ,H ) ( T ,H ) ( T ,T ) ( H ,H ) ( T ,H ) ( H ,T ) ( H ,H ) ( H ,T ) ( H ,H ) ( T ,T ) ( T ,T ) ( H ,T ) ( T ,T ) ( H ,T ) ( H ,T )
Of these, 38 have at least one head. Of the 38, 16 have two heads. Thus our estimate is 16/38 or 0.42. Although this appears quite high, it is not unlikely given our small sample size. (b) We now estimate the probability of two heads given that the first penny is a head. There are 29 instances of the first penny being a head. As before there are 16 instances of having two heads. Thus our estimate is 16/26 or 0.62. 2. We examine the probability of winning the fabulous prize first with the decision not to switch and then with the decision to switch. The only way to win if we do not switch is for the fabulous prize to be behind the door we initially choose. Since there are three doors, this has a probability of 1/3. To compute the probability of winning given that we switch, we condition on whether the prize is behind the first door. The probability of switching and winning given the prize was behind the first door is, obviously, zero. Now consider the probability of winning given the prize was not behind the firsr door. Since the prize is not behind the first choice of door and since the host will not reveal the prize when opening up another door, it must be behind the remaining door. Thus, the probability of winning when switching and when the prize is not behind the first door is one. Then, P{win} = P{win1st door correct} P{1st door correct}+ P{win|1st door incorrect} P{1st door incorrect} P{win} = 0 × 1 / 3 + 1 So it is best to switch.
×2 /3
= 2 /3
Note that this requires that the host always play the game the same way. He will always make the offer to choose another door regardless of the contestant’s first choice.
7
3. D = 3100/yr, c = $5, h = ic = 0.2(5) = $1/yr (a) F = 20, Q = D/F = 3100/20 = 155 2 AD
(b) 100 = Q =
h
A = ( 100 / 78.74 ) 2 2 AD
(c) Q =
h
=
2 A(3100) 1
A
= $1.61
2(10)(3100)
=
= 78.74
= 249
1
4. This satisfies all the assumptions of the EOQ model. The only tricky thing is to keep the units consistent. (a) D = 60 units / wk × 52 wk / yr = 3120units / yr h = ic = 0.25 / yr × $0.02 = $0.005 / yr
A = $12 *
=
2 AD
=
2 × 12 × 3120
= 3869.88 ≅ 3870 h 0005 . The time between orders is given by Q * 3870 = = 124 T * = . yr = 14.88 mo D 3120 (b) Setup cost is 3120units / yr D = $12 = $9.67 / yr A Q 3870units Q
Holding cost is
Q
h=
3870units
$0.005 / yr = $9.675/ yr . 2 2 The costs are essentially the same. This is always true in the case of the EOQ model. (c) The problem could be where to store all the items. If we were to order 1.24 years worth of styrofoam ice chests at a time it could take up a lot of room.
5. (a) Total (holding plus setup) cost would be TC = hQ / 2 + DA / Q = ($0.005 / yr )(3870units) / 2 + (6240units / yr )($12) / (3870 units) = $29.02 / yr (b) The optimum cost would be
2 ADh
=
2(12)(6240)( 0.005)
= $27.36 / yr .
(c) Using the wrong value for the demand (100% forecast error) in the EOQ formula results in an increase in cost of only 6%. EOQ is quite robust with respect to parameter values. 6. (a) The EOQ with 60 per week was computed to be 3,870 and the optimal reorder period was 1.24 years or 14.88 months. The closest power of two is 16 months or 1.33 years with a cost of . yr)(3120/ yr)($0.005/ yr) / 2 +$12 /1.33yr = $19 .37 / yr TC(1.33) = TDh / 2 + A / T = (133 The power of two on the other side of 14.88 mo is 8 mo or 0.67 yr with a cost of TC( 0.67) = TDh / 2 + A / T = (0.67 yr)(3120 / yr)($0.005/ yr) / 2 +$12 / 0.67yr = $23.20/ yr (b) The minimum cost without the power of two restriction is
2 ADh
=
2(12)(3120)( 0.005)
= $19.35 / yr
so 16 months has a cost that is only 0.1% over the optimal while the 8 month solution is around 20% over optimal. The total cost in the EOQ model is relatively insensitive to order quantity used. Since the order period is directly proportional to the order quantity the cost is not very sensitive to the period used as well.
8
(c) The robustness of the EOQ to errors in parameter estimates and the effectiveness of a power-of-2 policy are basically the same phenomenon. 7. Db = 1000/yr, cb=$200, hb = (0.2)(200) = $40; D s = 500/yr, c s=$150, hb = (0.2)(150) = $30 (a) Qb
=
Q s
=
2 ADb hb
2 AD s h s
=
2(50)(1000)
=
2(50)(500)
40
30
= 50;
T b
= 40.82;
=
T s
Qb Db
=
=
Q s D s
50 1000
=
= 0.05 yrs × 365 = 18.25 days
40.82 1000
= 0.082 yrs × 365 = 29.8 days
Note that there will be virtually no chance to share trucks. COST b
=
2 ADh
=
2(50)(1000)(40)
= $2,000;
COST s
=
2 ADh
=
2(50)(500)(30)
= $1,224.74
TOTAL COST = $3,224.74
Of this, the fixed trucking part is:
D 1000 = 40 b = 40 = $800; 50 Qb TOTAL = $1,289.96
FIXEDb
FIXED s
D 500 = 40 s = 40 = $489.96 40.82 Q s
Note that trucking is 1,289.96/3,224.74 = 40% of annual cost. (b) Round T b = 14 days, T s = 28 days, so Qb
=
T b Db
365
=
14(1000) 365
= 38.35 ≈ 38;
Q s
=
T s D s
365
=
28(500) 365
= 38.35 ≈ 38
Cost, not including fixed trucking (i.e., using A = $10) is: COST b
=
h b Qb
2
+
ADb Qb
=
40(38) 2
+
10(1000) 38
= $1,023.15;
COST s
=
h s Q s
2
+
AD s Q s
=
30(38) 2
+
10(500) 38
= $701.58
TOTAL = $1,724.73
Note that sheet stock orders always overlap with bar stock orders and so never incur a trucking cost. Total trucking cost is cost of bar stock orders: Trucking Cost b
= 40
Db Qb
= 40
1000 38
= $1,052.63
Hence, total cost is $1,052.63 + 1,724.73 = $2,777.36, which is 9.1% less than original cost of $3,224.74. (c) Suppose we leave Qb = 38, but decrease T s = 14 days, so Q s = T s D s/365 = 14(500)/365 = 19.17 ≈ 19 days. Then COST b is the same, as is fixed trucking cost, but COST s
=
hb Qb
2
+
ADb Qb
=
30(19) 2
+
10(500) 19
= $548.16
9
which is even lower. The reason is that since we are getting fixed trucking for free for the sheet stock, the true fixed cost is lower than the $50 we originally used. Hence, it makes sense to order more frequently to reduce the holding costs. (d) Note that had we known sheet stock would require no additional trucking cost, then we should have used A = $10 for it, which would have led to
Q s
=
2 AD s h s
=
2(10)(500) 30
= 18.26
which would round to 14 days on a power-of-two schedule using weeks as the base. Under a power-of-two schedule, we know that an it em with a larger order interval T will always overlap with a part with a smaller T after rounding. So, an approach to this type of problem would be: i) ii) iii) iv)
compute EOQ’s and hence T values for all parts round smallest T to nearest power-of-two (call it T o* ) recompute Qi* and T i* using the reduced value of A (i.e., w/o trucking cost) for all other parts round remaining parts to nearest power-of-two (but not below T o*)
8. (a) The schedule is developed in reverse. From the table we see that demand for period 12 is made in period 11. Thus, the quantity made in period 11 contains demand for both period 11 and 12, Q11 = D11 + D12 = 79 + 56 = 135 We are now left with a 10 period problem. From the table, demand for period 10 is made in period 10, or Q10 = 67 Again, demand for period 9 is made in period 8 so that Q8 = 67 + 45 = 112 If we continue in this fashion we obtain, Q5 = 121, Q4 = 97, Q1 = 98 . (b) If we start at period 6, a different schedule emerges. Q6 = D 4 + D 5 + D 6 = 148 Demand for period 3 is produced in period 1, so Q1 = 69 + 29 + 36 = 134 9. (a) Planning Horizon Last Period of Production 1 2 3 4 5 6 Zt* jt*
1 2000
2000 1
2 3200 4000
3200 1
3 4200 4500 5200
4200 1
4 4800 4900 5400 6200
4800 1
5 8000 7300 7000 7000 6800 6800 5
6
7800 8800 7800 5
Optimal solution is to produce 2,900 units in month 1 (for months 1-4) and 1800 units in month 5 (for months (5 and 6).
10
(b) Monthly planning periods are only appropriate if we make “monthly runs”. If we set up and run a product many times in a month, then a month is too long and maybe we should use weeks or something shorter. Factors affecting the choice of planning period include rate of production (slower enables longer periods), setup cost/time (higher/longer enables longer periods), and maybe number of products in the system (where more means fewer runs per year). (c) While a fixed setup cost will limit the number of setups, and hence capacity, it is not sensitive to varying load. When there are relatively few products being produced, capacity is not stretched and so setups are not so costly. But when a lot of products are being produced and capacity is tight, then they are very costly. The Wagner Whitin approach does not recognize this and hence can lead to inappropriate solutions.
10. This is a single period stochastic inventory model—the “news vendor problem.” Since the shirts sell for $20 and cost $5 any unit we are short will “cost” us $15. Since shirts that are not sold at the event can be discounted and sold for $4, the excess cost is $1. Thus, co = 1
= 15
c s
At this point it is clear that printing too few shirts is worse than printing up too many so their policy is not a good one. If G ( x ) represents the distribution function for the demand, from the news vendor model we have G (Q * )
=
c s c s
+ co
=
15 15 + 1
= 15 / 16 = 0.9375 = s
where s is the “service rate.” The question now is what distribution function to use. We estimate the mean to be 12,000 and know there is a “significant amount of uncertainty.” Since 12,000 is a large number, the normal distribution should be a reasonable approximation. If X is a random variable representing the demand during the event we can write and expression for Q * in terms of the mean and standard deviation of the demand. P{ X
≤ Q*} =
c s c s
= s
+ co
or P
X − 12000 σ
≤
Q*
− 12000 σ
= s
so that Q*
− 12000 σ
. = z s = 153
or Q*
= 12000 + 1.53σ
If demand were Poisson, the standard deviation would be the square root of the mean or 12000 = 109.54 so that Q = 12,167.6 ≅ 12 ,168 .
σ
=
*
11. c s = 500, co = 200 (a)
G(Q*) = cs/(cs + co) = 500/(500+200) = 0.714. Q* = µ + z0.714σ = 10,000 + (0.57)2500 = 11,425
11
(b) News vendor model does consider lost sales vs. materials cost, and does address uncertain demand forecast (how reasonable is it, though?). But it does not consider: • longer term inpacts, such as market share considerations, corporate image (e.g., if they can’t deliver), etc. • risk attitude – is expected value appropriate as decision criterion? • variants on contracts, which may allow purchase quantity within a range
12. The chairs are made in-house and so we are attempting to determine the appropriate parameters for a base-stock system. We assume that the wholesalers order once per month. (a) The holding cost is h = $5 while the backorder cost is b = $20 . The distribution of demand during a month is well approximated by a normal distribution with a mean of 1,000 chairs and a standard deviation of 200 chairs. Then, if X represents the demand during one month, b 20 G( R * ) = = = 0.8 h + b 5 + 20 and so G ( R * ) = P{ X
≤
R*} = P
X − 1000
200
≤
R*
− 1000 200
= 0.8
The value of the standard normal with 0.8 probabi lity is obtained from a standard normal table or using the Excel function NORMSINV(0.8) and yields 0.84. Then the order up to point is computed as R * = 1000 + 0.84 × 200 = 1168 . (b) If the sale is lost (as opposed to backordered) then the shortage cost must be the profit that would have been made which is $100. The computation is then similar, 100 = 0.9524 G( R * ) = 5 + 100 z 0.9524 = 1.67 R *
. × 200 = 1334 = 1000 + 167
(c) Since the cost of being short is higher in the second case, we want to carry more inventory to avoid that possibility.
9. This is a classic (Q,r) problem with service and order frequency constraints. The data is as follows: D = 30 units / yr
= 21 day × 1 yr / 365 day = 0.0575 yr θ = D = 1726 . units
(a) Jill wants to order twice per year. Therefore, Q must be Q = D / 2 = 15 units The desired service level (fill rate) is 0.98. Table 2.6 is for a value of θ = 1726 . which is exactly the value
12
of theta in this case. Then 1 S (Q, r ) = 1 − [ B( r ) − B ( r + Q )] ≥ 0.98 Q B (r ) − B (r + Q ) ≤ 0.02Q = 0.3 B (2) − B ( 2 + 15) = 0.389 − 0.000 = 0.389 S (15,2) = 1 −
1
S (15,3) = 1 −
1
[0.389 − 0.000] = 0.9741 15 B (3) − B (3 + 15) = 0.140 − 0.000 = 0.140 15
[0.140 − 0.000] = 0.9907
The smallest value of r that satisfies this is r = 3 , which yields a fill rate of 0.9907. A value of r = 2 is almost close enough but yields a fill rate of 0. 9741 just slightly below the desired 0.98. (b) If Jill orders 6 times per year, the order quantity becomes 5. Then 1 1 S (5,4) = 1 − [ B ( 4) − B (4 + 5)] = 1 − [0.042 − 0.000] = 0.9916 5 5 so r=4 is the smallest value that yields a fill rate of 0.98.. The reorder point (and therefore the safety stock) must be increased in order to obtain t he same fill rate because there are more opportunities to run out of stock in a year. In the second case, there are 5 opportunities while in the first there were only 2. | (c) We need to compare total cost, TC = Purchase Cost + Holding Cost. To represent holding cost, we make use of a carrying cost percentage, i. The total cost without the discount and Q=5 and r=4 (we know this is the r needed to achieve S=0.98 when Q=5 from part (b)) is TC 1 = $100 D+$100i(Q/ 2 + r − θ + B(Q,r)) = $3000 /yr + $100i( 5 / 2 + 4 − 1.726 + 0.015 ) = $3000 / yr + 478.9i (Note that we computed B(5,4)=B(5)+⋅ ⋅ ⋅ +B(9) using Table 2.6.) The total cost with the discount and Q=15, r=3 is TC 2 = $90 D+$90i(Q/ 2 + r − θ + B(Q,r)) = $ 2700 /yr + $90i( 15 / 2 + 3 − 1.726 + 0.056 ) = $2700 / yr + 794.7i We should take the discount if TC 2 < TC 1. Simple algebra shows that this occurs when i < 0.95. So, if the carrying cost percentage is less than 95% it is worthwhile to buy in the larger batches to take advantage of the discount.
13
14. Constraints
Costs
F (orders/year)
20
S (service level)
(a)
ci I
Fixed setup cost (A)
98%
Di
thetai
l i
$5
Backorder cost (b)
$15
Holding rate (h)
3%
sigmai
Qi
r i
Fi
150
7
1
7.0
2.6
1.0
13.0
7.0
B
15
30
0.5
15.0
3.9
1.0
23.0
30.0
37
ci I
Bi
Ii
Holding Order
Total
($/unit) (units/ (m os) (uni ts) (unit s) (unit s) (unit s) (order (fill mo) freq) rate)
A
(b)
Si
Di
(backor (inventory ($/yr) ($/yr) ($/yr) der invest) level) 0.987 0.010 $1,051.45 $378.52 35.00 413.52 0.981
0.024
18.5 98.18%
l
thetai
i
sigmai
Qi
r i
Fi
Si
$135.36
$48.73 150.00 198.73
$1,186.81 $427.25 185.00 612.25
Bi
Ii
Holding Order
Total
($/unit) (units/ (m os) (uni ts) (unit s) (unit s) (unit s) (order (fill mo) freq) rate)
A
150
7
1
7.0
2.6
4.0
12.0
1.8
(backor (inventory ($/yr) ($/yr) ($/yr) der invest) level) 0.988 0.009 $1,126.41 $405.51 8.75 414.26
B
15
30
0.5
15.0
3.9
26.0
18.0
1.2
0.980
37
0.033
1.5 98.16%
$247.99 $1,374.40
$89.28 $494.78
5.77
14.52 509.30
The higher values of Q make it possible to achieve the same service with lower r values. But inventory is higher due to increased cycle stock caused by bulk ordering. Total cost (inventory plus order cost ) is reduced considerably, however, by ordering in bulk. (c)
ci I
Di
l
thetai
i
sigmai
Qi
r i
Fi
Si
Bi
($/unit) (units/ (m os) (uni ts) (unit s) (unit s) (unit s) (order (fill mo) freq) rate)
A
150
7
1
7.0
2.6
4.0
9.0
1.8
(backor der level) 0.913 0.094
B
15
30
0.5
15.0
3.9
26.0
22.0
1.2
0.997
37
0.004
1.5 98.11%
Ii (inventory invest)
$689.10 $307.55 $996.65
This lowers service for part A (expensive one) and raises it for part B, so the same average service is achieved with lower total inventory. Note that it's even below the base stock inventory level where Q=1! (d)
ci I
A B
Di
l
i
($/unit) (units/ (days) mo) 150 7 30 15
30
15
thetai sigmaL sigmai (units) 7.0 15.0
Qi
r i
Fi
Si
Bi
Ii
(units) (units) (units) (order freq) 7 3.1 4.0 9.0 1.8 15
15.5
26.0
37
44.0
(fill (backorder (inventory invest) rate) level) 0.913 0.094 $ 689.10 1.2 1.000 0.000 $ 637.50 1.5 98.35% 0.094 $1,326.6
The variability in the lead times inflates the reorder points - in this case for part B (rounding can result in no change). Note that predictions of service, backorder level, and inventory level are no longer exact, since these are for the model with fixed lead times. In general, they're all too low.
14
95.05
15. Constraints
Costs
F (orders/year) S (service level)
a,b)
ci
Di
20 90%
l
i
Fixed setup cost (A) Backorder cost (b) Holding rate (h) thetai
sigmai
Qi
r i
$5.00 $7.00 2.0% Fi
Type 1 S
Si
Bi
Ii
(order (approx (backorder (inventory i ($/unit)(units/mo) (mos) (units) (units) (units) (units) freq) fill rate) (fill rate) level) invest) approx 1 12 15 0.5 7.5 2.7 25.0 11.0 0.6 0.92 0.994 0.006 $198.07 exact 1 12 15 0.5 7.5 2.7 25.0 6.0 0.6 0.38 0.922 0.133 $139.60 Note that Type 1 service underestimates true service by a lot, leading to a much larger r and higher inventory.
c)
ci
Di
l
i
thetai
sigmai
Qi
r i
Fi
Type 2 S
Si
Bi
Ii
(order (approx (backorder (inventory i ($/unit)(units/mo) (mos) (units) (units) (units) (units) freq) fill rate) (fill rate) level) invest) approx 1 12 15 0.5 7.5 2.7 25.0 6.0 0.6 0.922 0.922 0.133 $139.60 exact 2 12 15 0.5 7.5 2.7 25.0 6.0 0.6 0.922 0.922 0.133 $139.60 Type 2 service is very accurate because when Q is this large, the term B(r+Q) is negligible.
d)
exact
ci
Di
l
i
thetai
sigmai
Qi
r i
Fi
Si
Bi
Ii
(order (backorder (inventory i ($/unit)(units/mo) (mos) (units) (units) (units) (units) freq) (fill rate) level) invest) 2 12 15 0.5 7.5 2.7 13.0 8.0 1.2 0.934 0.087 $ 91.05
Note that when Q is reduced, we get slightly higher service at a much smaller inventory investment. But of course, we order twice as often. If we neglect the cost or capacity considerations of placing orders, we can always minimize inventory costs by choosing Q=1. But if we consider either order frequency (capacity) or fixed order cost, then EOQ may give a perfectly reasonable Q.
15
16
