Chapter 8: Aggregate Planning in a Supply Chain
Exercise Solutions :
We define a comprehensive set of decision variables that are utilized in problems 8-1 to
8-3 depending on the problem context.
Decision Variables:
Ht = # of workers hired in month t (t = 1,..,12)
Lt = # of workers laid-off in month t (t = 1,..,12)
Wt = # of workers employed in month t (t = 1,..,12)
Ot = # of hours of overtime in month t (t = 1,..,12)
It = # of units (000s) held in inventory at the end of month t (t = 1,..,12)
Ct = # of units (000s) subcontracted in month t (t = 1,..,12)
Pt = # of units (000s) produced in month t (t = 1,..,12)
Parameters:
Dt = # of units (000s) demanded in time period t (t = 1,…12)
Problem 8-1:
12
12
12
12
t =1
i =1
i =1
i =1
Minimize 3200∑ Wt + 30∑ Ot + 3000∑ I t + 20000∑ Pt
Subject to:
Inventory constraints:
I t −1 + Pt − I t = Dt ,
t = 1,..,12
I 0 = I 12 = 50
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
Pt −
6Ot
960Wt
−
≤ 0, t = 1,...12
1000 1000
Workforce constraints:
Wt = 1250,
t = 0,...,12
(a) Worksheet 8.1 provides the solution to this problem and the corresponding aggregate
plan. The total cost of the plan is $360,400,000.
(b) If the number of overtime hours per employee were increased from 20 to 40 it would
result in decreasing the total cost to $356,450,000. So, it is advantageous to do it.
(c) If the number of employees is decreased to 1200 and the overtime hours per
employee are held at 20 and 40 then the total costs of the plan are $363,324,000 and
$357,422,000, respectively. If the number of employees is increased to 1300 and the
overtime hours per employee are held at 20 and 40 then the total costs of the plan are
$358,790,000 and $356,270,000, respectively. So, the value of additional overtime
increases as workforce size decreases.
(d) We add a new constraint: Pt ≤ 1291.667, t = 1,...12 . The cost will be $363,049,982.
Problem 8-2:
We now include the subcontract option in the model:
12
12
12
12
12
t =1
i =1
i =1
i =1
i =1
Minimize 3200∑ Wt + 30∑ Ot + 3000∑ I t + 20000∑ Pt + 26000∑ C t
Subject to:
Inventory constraints:
I t −1 + Pt + C t − I t = Dt ,
t = 1,..,12
I 0 = I 12 = 50
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
Pt −
6Ot
960Wt
−
≤ 0, t = 1,...12
1000 1000
Workforce constraints:
Wt = 1250,
t = 0,...,12
Worksheet 8-2 provides the solution to this problem.
(a) Without the subcontract option the total cost is $360,400,000 (from problem 8-1) and
the total number of units produced is 14,900,000. Thus, the cost per unit is $24.19.
(b) Third party production must be used in periods 5, 6, 7, and 10 for a total of 1,050,000
units. If the subcontract cost per unit decreases to $25 per unit then a total of 1,650,000
units must be acquired from this option.
(c) If the subcontract cost increases to $28 per unit then a total of 700,000 units must be
acquired from this option.
(d) The total number of units produced, including subcontract production, is 14,900. So,
using subcontract the total cost incurred is $357,450,000, which leads to an average cost
of $23.99 per unit. Without the subcontract option, the cost per unit is $24.18 (from a).
Therefore, it is still beneficial for Skycell to use subcontracting option even if the per unit
cost is higher. Without using subcontracting, Skycell would need to use overtime to
produce extra units to fulfill demand. The cost of using overtime is 1.5 times regular
labor cost. In addition, in the absence of subcontracting, holding costs will also increase
due to extra units being carried into future time periods.
Problem 8-3:
We define and include a new decision variable for this model called Tt, which represents
the number of temporary workers employed in time period t. Since hiring and layoff of
temporary workers is allowed, we include hiring and layoff costs in the model. Also,
note that there is no subcontract option available for this case. The LP model for this case
is shown below:
12
12
12
12
12
12
t =1
t =1
t =1
i =1
i =1
i =1
Minimize 800∑ H t + 1200∑ Lt + 3200∑ (Wt + Tt ) + 30∑ Ot + 3000∑ I t + 20000∑ Pt
Subject to:
Inventory constraints:
I t −1 + Pt − I t = Dt ,
t = 1,..,12
I 0 = I 12 = 50
Overtime constraints:
Ot − 20(Wt + Tt ) ≤ 0,
t = 1,...,12
Production constraints:
Pt −
6Ot
960(Wt + Tt )
−
≤ 0, t = 1,...12
1000
1000
Full-Time Workforce constraints:
Wt = 1250,
t = 0,...,12
Temporary Workforce constraints:
Tt − Tt −1 − H t + Lt = 0, t = 1,...,12
T0 = 0
Tt ≤ 50
(a)
Total cost = $
358,210,000
Ht
Lt
Tt
Period
# Hired
0
1
2
3
4
5
6
7
8
9
10
11
12
# Laid off
0
0
0
50
0
(0)
50
0
# Temp
0
0
50
-
0
50
50
50
50
50
50
Wt
Pt
#
Workforce Production
0
1,250
0
1,250
950
1,250
1184
1,250
1200
1,250
1404
1,250
1404
1,250
1404
1,250
1404
1,250
900
1,250
1100
1,250
1200
1,250
1346
1,250
1404
(b)
Total cost = $
356,658,667
Ht
Lt
Tt
Wt
Pt
#
Workforce Production
0
1,250
0
1,250
950
1,250
1100
1,250
1200
1,250
1326
1,250
1458
1,250
1458
1,250
1458
1,250
900
1,250
1100
1,250
1200
1,250
1292
1,250
1458
Period
# Hired
0
1
2
3
4
5
6
7
8
9
10
11
12
# Laid off
0
# Temp
0
0
(0)
0
100
0
(0)
(0)
96
4
-
100
100
100
100
96
100
0
100
-
(c) If Skycell only carries 1100 permanent employees but has 200 seasonal employees,
the total cost will be reduced to $356,984,667, a 0.34% cost saving compared to (a)
(d) Since there is no difference between regular employees and temporary employees in
terms of productivity and cost of hiring and layoffs, this problem is equivalent to using at
most 1300 employees and allowing hiring and layoffs. So, the formulation can be revised
as follows.
12
12
12
12
12
12
t =1
t =1
t =1
i =1
i =1
i =1
Minimize 800∑ H t + 1200∑ Lt + 3200∑ Wt + 30∑ Ot + 3000∑ I t + 20000∑ Pt
Subject to:
Inventory constraints:
I t −1 + Pt − I t = Dt ,
I 0 = I 12 = 50
t = 1,..,12
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
Pt −
6Ot
960Wt
−
≤ 0, t = 1,...12
1000 1000
Full-Time Workforce constraints:
Wt − Wt −1 − H t + Lt = 0,
W0 = 1250
Wt ≤ 1300,
t = 1,...,12
t = 0,...,12
Temporary Workforce constraints: (not needed)
Total cost = $
356,926,500
Ht
Period
Lt
# Laid off
# Hired
0
1
2
3
4
5
6
7
8
9
10
11
12
0
156
154
208
44
110
0
310
363
-
Wt
Pt
# Temp # Permanent # workforce Production
0
50
1,250
1,300
0
0
990
990
950
0
1,146
1,146
1100
50
1,250
1,300
1284
50
1,250
1,300
1404
50
1,250
1,300
1404
50
1,250
1,300
1404
50
1,250
1,300
1404
0
938
938
900
0
1,146
1,146
1100
0
1,190
1,190
1142
50
1,250
1,300
1404
50
1,250
1,300
1404
So, if the no layoff no hiring policy for the permanent employees can be relaxed, Skycell
will be able to further reduce the total cost. However, the reduction from case (c) is
marginal.
Problem 8-4:
We define a comprehensive set of decision variables that are utilized in problems 8-4 to
8-7 depending on the problem context.
Decision Variables:
Ht = # of workers hired in month t (t = 1,..,12)
Lt = # of workers laid-off in month t (t = 1,..,12)
Wt = # of workers employed in month t (t = 1,..,12)
Ot = # of hours of overtime in month t (t = 1,..,12)
IRt = # of routers (000s) held in inventory at the end of month t (t = 1,..,12)
ISt = # of switches (000s) held in inventory at the end of month t (t = 1,..,12)
CSt = # of switches (000s) subcontracted in month t (t = 1,..,12)
PRt = # of routers (000s) produced in month t (t = 1,..,12)
PSt = # of switches (000s) produced in month t (t = 1,..,12)
Parameters:
DRt = # of routers (000s) demanded in time period t (t = 1,…12)
DSt = # of switches (000s) demanded in time period t (t = 1,…12)
12
12
12
12
t =1
i =1
i =1
t =1
Minimize 1600∑ Wt + 15∑ Ot + 2000∑ IRt + 1000∑ IS t
Subject to:
Inventory constraints:
IRt −1 + PRt − IRt = DRt ,
IS t −1 + PS t − IS t = DS t ,
IR0 = IR12 = 100
IS 0 = IS12 = 50
t = 1,..,12
t = 1,..,12
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
333PRt + 166.7 PS t − Ot − 160Wt ≤ 0,
Workforce constraints:
Wt = 6300,
t = 0,...,12
t = 1,...12
(a)
Total cost = $
135,429,000
Router
Switch
Productio
n
Production
Period
0
1
2
3
4
5
6
7
8
9
10
11
12
1700
1600
2600
2500
800
1800
1200
1400
2500
2800
1000
1100
2648
2848
1150
1804
800
1800
2400
3248
1048
1204
1000
1050
(b) If the overtime allowed per employee per month is increased to 40, the total cost
becomes $134,552,000. So, cost reduction is incurred as a result of such an action.
Comparing the output of (a) and (b) it can be observed that the production schedule of
switches (PSt) is impacted, as is ISt. On the other hand, the production plan for routers
remains the same.
(c) From the table below it is evident that as the number of employees increases, the
value of increasing total number of overtime hours allowed per employee per month
decreases. When the number of employees is 6700, there is no cost difference between
using 20 and 40 hours of overtime per employee.
Total cost
# employees Overtime 20
Overtime 40 Cost difference % change
5900 135,429,000 132,135,000
3,294,000 2.4323%
6300 135,429,000 134,552,000
877,000 0.6476%
6700 138,302,000 138,302,000
0
0
Problem 8-5:
In this case we add hiring and layoff options.
12
12
12
12
12
12
t =1
t =1
t =1
i =1
i =1
t =1
Minimize 700∑ H t + 1000∑ Lt + 1600∑ Wt + 15∑ Ot + 2000∑ IRt + 1000∑ IS t
Subject to:
Inventory constraints:
IRt −1 + PRt − IRt = DRt , t = 1,..,12
IS t −1 + PS t − IS t = DS t , t = 1,..,12
IR0 = IR12 = 100
IS 0 = IS12 = 50
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
333PRt + 166.7 PS t − Ot − 160(Wt − 0.5( H t −1 + H t )
) ≤ 0,
t = 1,...12
Workforce constraints:
Wt − Wt −1 − H t + Lt = 0,
W0 = 6300
t = 1,...,12
The solution to this problem is shown in Worksheet 8-5. Part (b) can be solved by
replacing the production constraints in the above formulation by:
333PRt + 166.7 PS t − Ot − 160Wt ≤ 0,
t = 1,...12
(a)
Total cost = $
Ht
137,118,107
Lt
Wt
Ot
Router
Switch
#
Productio
Period
# Hired # Laid off Workforce Overtime n
Production
0
0
0
15
0
1
0
0
6,300
0.0
1700
2645
2
0
0
6,300
0.0
1600
2845
3
0
0
6,300 86500.0
2600
1362
4
0
0
6,300 126000.0
2600
1599
5
0
671
5,629
0.0
800
800
6
0
0
5,629
0.0
1800
1800
7
241
0
5,869
0.0
1200
3117
8
0
0
5,869
0.0
1400
2716
9
0
0
5,869 117385.9
2500
1334
10
0
0
5,869 117385.9
2800
733
11
0
0
5,869
0.0
1000
1000
12
431
0
6,300
0.0
1000
1050
(b) The total cost will be $131,256,258, a reduction of 4.275% from (a). Comparing the
hiring and layoff results of (a) and (b), we find more hiring and layoff happens in (b). As
can be seen, when a new employee is 100% productive when they are hired, FlexMan’s
cost will reduce. So, it is reasonable that FlexMan will use the option of more new hires
and lay them off when appropriate.
(a)
Ht
# Hired
Period
0
1
2
3
4
5
6
7
8
9
10
11
12
(b)
Lt
Ht
Lt
# Laid off # Hired # Laid off
0
0
0
0
0
0
0
616
0
0
0
0
0
0
1451
0
0
0
0
0
0
671
0
3069
0
0
0
0
241
0
0
0
0
0
0
0
0
0
3718
0
0
0
0
0
0
0
0
1483
431
0
0
0
Problem 8-6:
In this case we add the subcontract option
Minimize
12
12
12
12
12
12
12
12
t =1
t =1
t =1
i =1
i =1
t =1
t =1
t =1
700∑ H t + 1000∑ Lt + 1600∑ Wt + 15∑ Ot + 2000∑ IRt + 1000∑ IS t + 6000∑ CRt + 4000∑ CS t
Subject to:
Inventory constraints:
IRt −1 + PRt + CRt − IRt = DRt ,
t = 1,..,12
IS t −1 + PS t + CS t − IS t = DS t , t = 1,..,12
IR0 = IR12 = 100
IS 0 = IS12 = 50
Overtime constraints:
Ot − 20Wt ≤ 0,
t = 1,...,12
Production constraints:
333PRt + 166.7 PS t − Ot − 160(Wt − 0.5( H t −1 + H t )
) ≤ 0,
t = 1,...12
Workforce constraints:
Wt − Wt −1 − H t + Lt = 0,
W0 = 6300
t = 1,...,12
(a) FlexMan should use the subcontractor to produce 446,000 units of routers in period 4
and 1,446,000 units of routers in period 10. FlexMan should not use the subcontractor
for switch
(b) FlexMan should not use the subcontractor at all.
(c) When the productivity of new employees is low, FlexMan suffers from loss of
capacity. When the new employees have the full productivity, FlexMan can use the
same amount of cost (of salary) in acquiring a higher output rate, thus reducing the
average unit cost. Therefore, the choice of whether or not to use a subcontractor will
vary with the productivity of new employees.
Problem 8-7:
We utilize the same formulation as in Problem 8-4, but impose constraints on the ending
inventory to allow for safety stock as shown below:
IRt ≥ 0.15 DRt +1
IS t ≥ 0.15 DS t +1
(a)
Total cost = $
142,960,650
Router
Switch
Productio
n
Production
Period
0
1
2
3
4
5
6
7
8
9
10
11
12
1770
1750
2585
2245
950
1710
1230
1565
2545
2530
1000
950
2504
2545
1037
2310
770
1815
2308
2915
953
1739
865
1050
(b) Comparing cost in (a) to cost in 8-4 (a), we observe an increase of $7,531,650 in total
cost, an 5.56% of cost increase by providing this service contract.
(c) From the table below we can see that keeping a safety stock of 5% of the following
month’s demand for routers will be the best policy for FlexMan.
Total Cost
Router 15%, Switch 15%
Router 5%, Switch 15%
Router 15%, Switch 5%
Savings
142,960,650
138,902,350
142,125,650
2.84%
0.58%
Chapter 10: Managing Economies of Scale in the Supply Chain: Cycle Inventory
Exercise Solutions
1. The economic order quantity is given by
2 DS
. In this problem:
hC
D = 109,500 (i.e., 300 units/day multiplied by 365 days/year)
S = $1000/order
H = hC = (0.2)(500) = $100/unit/year
So, the EOQ value is 1480 units and the total yearly cost is $147,986
The cycle inventory value is EOQ/2 = 1480/2 =740
Worksheet 10.1 provides the solution to this problem.
2.
(a) If the order quantity is 100 then the number of orders placed in a year are: D/Q =
109500/100 = 1095. So, 1095 orders are placed each year at a cost of $1000/order. Thus, the
total order cost is $1,095,000.
Cycle inventory = Q/2 = 100/2 = 50 and the annual inventory cost is (50)(0.2)(500) = $5,000
(b) If a load of 100 units has to be optimal then corresponding order cost can be computed by
using the following expression:
2 DS
hC
Q
100 
S 
( 2)(109500) S
(0.2)(500)
(100) 2 (0.2)(500)
 $ 4.57 per order
( 2)(109500)
This analysis is shown in worksheet 10-2.
3.
(a) We first consider the case of ordering separately:
For supplier A:
2( 20000)(400  100)
= 4,472 units/order
(0.2)(5)
Total cost = order cost + holding cost = (20000/4472)(500) + (4472/2)(0.2)(5) = $4,472
Order quantity (Q) =
Similarly, for suppliers B and C the order quantities are 1768 and 949 and the associated total
costs are $1,414 and $949, respectively.
So, the total cost is $6,835
1
(b) In using complete aggregation, we evaluate the order frequency (n*) as follows:
So, n* of the case is =
D A hC A  DB hC B  DC hCC
2S *
S* = 400 + 3(100) = $700
So, n* =
20000(0.2)(5)  2500(0.2)( 4)  9000(0.2)(5)
= 4 orders/year
2(700)
For supplier A:
Q = D/n = 20000/4 = 5000 units/order
Total cost = order cost + holding cost = 4(500) + (5000/2)(0.2)(5) = $4,500
Similarly, for suppliers B and C the order quantities are 625 and 225 and the associated total
costs are $650 and $513, respectively.
So, the total cost is $5,663.
Worksheet 10-3 provides the solution to this problem
4.
(a) This is a quantity discount model and the decision is to identify the optimal order quantity
in the presence of discounts. We evaluate the order quantities at different unit prices using the
economic order quantity equation as shown below:
For, price = $1.00 per unit
2( 20000)(400)(12)
 30,984
(0.2)(1)
Since Q > 19,999
Q= EOQ =
We select Q = 20,000 (break point) and evaluate the corresponding total cost, which includes
purchase cost + holding cost + order cost
(20000)(12)
 20000 
 400
 0.2  (0.98) 
Total Cost = ( 20000)(12)(0.98)  
2
20000


241,960
= $
Similarly we evaluate the EOQs at prices of p = 0.98 (Q = 31298) and p = 0.96 (Q = 31623,
which is not in the range so use Q = 40001). The corresponding total costs are $241,334 and
$236,640.
So, the optimal value of Q = 40001 and the total cost is $236,640
2
The cycle inventory is Q/2 = 40001/2 = 2000.5
(b) If the manufacturer did not offer a quantity discount but sold all plywood at $0.96 per
square foot then Q = 31,623 and the total cost is $ 233,436
This analysis is shown in worksheet 10 -4
5. We solve this problem using a similar approach as in the previous case except the equation
used for computing the order quantity at a particular price level in the presence of marginal
unit quantity discounts is as shown below:
Q at for a price level Ci =
2 D ( S  Vi  qi C i )
hCi
For price = $1.00 per unit
2(20000)(12)(400  0  (0)(1))
= 30,984
(0.2)(1)
Since Q > 19,999, we adjust Q = 20,000 and the corresponding total cost is $ 246,800
Q=
The same procedure is followed for the other unit prices and the optimal quantity is 63,246 at
a total cost of $242,663.
Worksheet 10 -5 shows the analysis and problem solution
6. In the case of no promotion, we can use the EOQ expression to compute the order quantity.
So, Q =
2(1000)(52)(200)
 6450 units/order
0.25( 2)
In the presence of discount,
Qd =
dD
CQ *

C  d h C  d
Qd =
0.2(1000)(52) 2(6450)

= 30,277 units/order
( 2  0.2)0.25
( 2  0.2)
Dominick’s order given the short-term price reduction must be 30,277.
Worksheet 10-6 shows the solution to this problem
7. In this problem, the goal is to obtain an annual demand for which TL costs are equal to LTL
costs. As the annual demand increases, the optimal batch size grows making TL more
economical. Above the threshold obtained, Flanger should use TL. Below the threshold they
should use LTL.
Thus, we equate the two cost functions as shown below:
3
TL Costs:
Optimal order quantity QTL =
2( D )(500)
(0.2)(50)
D
(100)
QTL
D
( 400)
Annual trucking cost =
QTL
Annual order cost =
Annual holding cost =
Total Cost for TL =
QTL
(10)
2
D
D
Q
(100) +
( 400) + TL (10)
QTL
QTL
2
LTL Costs:
Optimal order quantity QLTL =
2( D )(100)
(0.2)(50)
D
(100)
Q LTL
Annual trucking cost = D (1)
Annual order cost =
Annual holding cost =
Total Cost for TL =
Q LTL
(10)
2
D
Q
(100) + D (1) + LTL (10)
Q LTL
2
Equating the TL and LTL costs results in a demand value of 3056. If the demand goes
beyond this value then the TL option will prove economical and if the demand is below this
value then LTL is the optimal choice.
Worksheet 10-7 solves this problem in EXCEL by using the solver option.
(b) If the unit cost is increased to $100 then the new threshold is 6112. Thus, as unit cost
increases the LTL option becomes preferable.
(c) If the LTL cost decreases to $0.8 per unit then the new threshold value becomes 4775.
8.
(a) LTL costs with one supplier per truck:
Optimal order quantity QTL =
2(3000)(100)
= 245 units
(0.2)(50)
4
 245 
12 = 0.98 months
Time between orders = 
 3000 
3000
(100) = $1225
245
Annual trucking cost = 3000(1) = $3000
245
(10) = $1225
Annual holding cost =
2
Annual order cost =
Total Cost for TL = $5449
(b) TL costs with one supplier per truck:
Optimal order quantity QTL =
2(3000)(1000)
= 775 units
(0.2)(50)
 775 
12 = 3.1 months
Time between orders = 
 3000 
3000
(100) = $387
775
3000
(900) = $3486
Annual trucking cost =
775
775
(10) = $3873
Annual holding cost =
2
Annual order cost =
Total Cost for TL = $7746
(c) TL costs with two suppliers per truck:
In the presence of aggregation we solve for optimal order frequency n*
So, n* of the case of 2 suppliers is =
D1 hC1  D2 hC 2
2S *
S* = 800 + 2(100) + 2(100) = $ 1200
Thus, n* =
(3000)(10)  (3000)(10)
= 5 orders/year
2(1200)
Optimal order quantity (Q) per supplier = D/n = 600 units
Order cost per product =
3000
(100) = $500
600
5
3000
(800  100(2)) / 2 = $2500
600
600
(10) = $3000
Annual holding cost per product =
2
Annual trucking cost per product =
Total Cost for TL = $6000
(d) The optimal number of suppliers that need to be grouped is 4 with an order quantity of
490 units and total cost of $4,899. The truck capacity of 2000 units would not be sufficient if
more than 4 suppliers are aggregated.
(e) When demand is 3000 the aggregated TL option with four suppliers is optimal, and when
the demand decreases to 1500 the LTL option is optimal. As demand increases to 1800, the
aggregated TL option with four suppliers is optimal.
Worksheet 10-8 shows the results and analysis for this problem
9. We compute the total cost for the fast moving product and a similar approach can be
utilized to evaluate the total costs for medium and slow moving products.
6
(a)
Fast moving products:
2(30000)(200)
= 3464 units/batch
5(0.5)
3464
(365) = 42
Days of demand =
30000
EOQ = Q =
Annual setup cost =
30000
( 200) = $1732
3464
Annual holding cost =
3464
(0.5)(5) = $1732
2
Total cost per product = $3464
Total cost for all fast moving products (5 products) = $17320
Similar analysis for the medium and slow products results in batch sizes of 2191 and 980,
respectively.
(b)
The total costs for three product groups are:
Fast moving = $17,320
Medium moving = $21,908
Slow moving = $34,292
So, the total cost across all products is $73,522.
(c)
For the fast moving products the total time required is:
30000 30000

(0.5) =304.3 hours
100
3464
Similarly, for the medium and slow moving products the number of hours needed is 122.7
and 25.2, respectively.
Worksheet 10-9 demonstrates these computations.
7
10.
(a)
In situations where full truckloads are used the number of deliveries for large, medium, and
small customers in a given year is 5, 2, and 0.7, respectively, which is obtained by dividing
annual demand by truck capacity in each case.
For the Large customer:
Order quantity = Q = 12 units/order (truck capacity)
The transportation cost for large customer is given by:
nL(S+sL) = 5(800+250) = $5250
The holding cost is given by:
(12/2)(10000)(0.25) = $15,000
So, the total cost is $20,250
The days of inventory carried at the large customer are:
(12/2)(365)/60 = 37 days of inventory
For the medium and small customers the total costs are $17,100 and $15,700, respectively,
and the inventory carried by these customers is 91 and 274 units, respectively.
Thus, the overall cost of this plan for the three customers is $53,050
Worksheet 10-10 shows these evaluations.
(b) In this case, we evaluate separate EOQs for each of three cases.
For the Large customer:
Order quantity = Q =
2 D( S  s L )
=
hC L
2(60)(800  250)
= 7.1 units/order
0.25(10000)
Number of orders (nL) = D/Q = 60/7.1 = 8.5 orders/year
The transportation cost for large customer is given by:
nL(S+SL) = 8.5(800+250) = $8874
The holding cost is given by:
(7.1/2)(10000)(0.25) = $8,874
So, the total cost is $17,748
The days of inventory carried at the large customer are:
8
(7.1/2)(365)/60 = 22 days of inventory
For the medium and small customers the total costs are $11,225 and $6,481, respectively, and
the inventory carried by these customers is 34 and 59 units, respectively.
Thus, the overall cost of this plan for the three customers is $35,454
(c) In this case we utilize complete aggregation, i.e., each truck has products that are shipped
to all customers.
In the presence of aggregation we solve for optimal order frequency n*
So, n* of the case is =
DL hC L  DM hC M  DS hC S
2S *
S* = 800 + 3(250) = $1550
So, n* =
60(0.25)(10000)  24(0.25)(10000)  8(0.25)(10000)
= 8.6 orders/year
2(1550)
For the Large customer:
Order quantity = Q = D/n* = 60/8.6 = 6.97 units/order
Transportation cost:
nL(S+SL) = 8.6(800+250) = $9,044
The holding cost is given by:
(6.97/2)(10000)(0.25) = $8,707
So, the total cost is $17,751
The days of inventory carried at the large customer are:
(6.97/2)(365)/60 = 21.2 days of inventory
For the medium and small customers the total costs are $5,636 and $3,314, respectively, and
the inventory carried by these customers is 21.2 and 21.2 units, respectively.
Thus, the overall cost of this plan for the three customers is $26,702
(d) In the case of partial aggregation we evaluate relative delivery frequency. In this case not
every customer is supplied with the product in every order.
Step 1: we identify most frequently ordered product assuming each product is ordered
independently.
For the large customer:
9
nL =
hC L D L
=
2( S  s L )
0.25(10000)(60)
= 8.5 orders/year
2(800  250)
For the medium and small customers the order frequency is 5.3 and 3.1, respectively.
Thus, the most frequent ordering of the product comes from the large customer.
Step 2: We identify the frequency with which other customer orders are included into the
most frequently ordered.
We evaluate n M and n L
Since we are already accounting for the fixed cost for the large customer, we only consider
the product specific costs for medium and small customers. Thus:
nM
=
hC M DM
=
2s M
0.25(10000)(24)
= 11
2(250)
and similarly, n L = 6.3
We now evaluate the frequency with which medium and small customers order relative to the
large customer.
m M = n L n M = 8.5/11 = 0.77 => we round up to the closest integer, i.e., 1
Similarly, m S = 2
Step 3: Having decided the order frequency for each customer, we recalculate the order
frequency for the most frequently ordering customer, i.e., the large customer:
n=
D L hC L  DM hC M  DS hC S
=
2( S  s M m M  s L m L )
60(0.25)(10000)  24(0.25)(10000)  8(0.25)(10000)
2(800  ( 250 / 1)  ( 250 / 2))
= 9.37 orders/year
Step 4: For medium and small customers, we evaluate the order frequency:
nM = n/mM = 9.37/1 = 9.37
nS = n/mS = 9.37/2 = 4.68
The total costs are evaluated as in the previous problem except for the fact that the order costs
for medium and small customers only includes the product specific costs.
The total cost for tailored aggregation is $ 26,693
These evaluations are shown in different worksheets in 10-10
11.
(a) From the retailer’s standpoint, the optimal order quantity is:
10
Q=
2(240000)(200)
= 9798 units/order
0.2(5)
Retailer costs:
Order costs = (240000/9798)(200) = $4,899
Holding costs = (9798/2)(0.2)(5) = $4,899
Retailer total cost = $9,798
Crunchy’s costs:
Order costs = (240000/9798)(1000) = $24,495
Holding costs = (9798/2)(0.2)(3) = $2,939
Crunchy total cost = $27,434
Total cost = $37,232
(b) In jointly optimizing the order quantity is:
Q=
2(240000)(200  1000)
= 18974 units/order.
0.2(5)  0.2(3)
Retailer costs:
Order costs = $2,530
Holding costs = $9,487
Retailer total cost = $12,017
Crunchy’s costs:
Order costs = $12,649
Holding costs = $5,692
Retailer total cost = $18,341
Total cost = $30,358
11
(c) In this case, we equate the total costs associated with ordering at the EOQ and the
breakpoint levels for the retailer in determining the discount level. The goal seek option is
utilized to obtain the discount per unit at break point, which is equal to $0.00917. Worksheet
10-11 provides details of the analysis.
12.
(a) Given that Demand is estimated to be equal to 2,000,000 – 2,000p and the production
costs for Orange is $100 per unit, we get the optimal price by setting P equal to
(2,000,000 + 2,000(100))/4000 giving Orange a wholesale price equal to $550.
At this wholesale price Good Buy would set a retail price equal to (2,000,000 +
2,000(550))/4000 or $775.
Profits for Orange at this price would be $202,500,000 and Good Buy would have a
profit of $101,250,000.
(b) If Orange offers a $40 discount to Good Buy, then the new price would be (2,000,000
+ 2,000(510))/4000 or $755. Good Buy would pass along $20 or 50% of the discount
offered by Orange.
Worksheet 10-12 provides details of the analysis.
13.
(a) Good Buy should purchase is lots equal to SQRT[(2DS)/hC] =
SQRT{(2x450000x10000)/(.2x550)] = 9,045
(b) Given the $40 discount by Orange for the next two weeks, Good buy should
adjust its lot size to (40)(450000)/(550-40)(.2) + (550x9045)/(550-40) = 16,814.
Equation 10.15
The lot size increase about 86%.
Worksheet 10-13 provides details of the analysis.
12
13
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Chapter 11: Managing Economies of ScaleUncertainty in athe Supply Chain: Safety
Inventory
Exercise Solutions
1.
DL = LD = (2)(300) = 600
 L = L D =
2 ( 200) = 283
ss = FS-1 (CSL)  L = FS-1 (0.95)  283 = 465 (where, FS-1 (0.95) = NORMSINV (0.95))
ROP = DL + ss = 600 + 465 = 1065
Excel Worksheet 11-1 illustrates these computations
2.
DL = (T+L) D = (2+3)(300) = 1500
L =
T  L D
=
2  3 ( 200) = 447
ss = FS-1 (CSL)  L = FS-1 (0.95)  447 = 736 (where, FS-1 (0.95) = NORMSINV (0.95))
OUL = D(T+L) + ss = 1500 + 736 = 2236
Excel Worksheet 11-2 illustrates these computations
3.
DL = LD = (2)(300) = 600
 L = L D =
2 ( 200) = 283
ESC = (1 – fr)Q = (1 – 0.99)500 = 5
We use the following expression to determine the safety stock (ss):
ESC   ss[1  F S (
ss
L
)]   L f S (
ss
L
)
1
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
ESC = –ss[1 – NORMDIST(ss/L, 0, 1, 1)] + L NORMDIST(ss/L, 0, 1, 0).
We utilize the GOALSEEK function in EXCEL in determining safety stock (ss) by using ss as the
changing value that results in an ESC value of 5.
Excel Worksheet 11-3 illustrates these computations.
Goal Seek set-up:
SET CELL: C29
TO VALUE: 5
BY CHANGING CELL: C27
This results in an ss value of 477 and the reorder point of:
ROP = DL + ss = 600 + 477 = 1,077
4.
DL = LD = (2)(250) = 500
 L = L D =
2 (150) = 212
ss = ROP – ss = 600 – 500 = 100
CSL = F(DL + ss, DL, L) = F(600, 500, 212) = NORMDIST (600, 500, 212, 1) = 0.68
ESC   ss[1  F S (
ss
L
)]   L f S (
ss
L
)
ESC = –ss[1 – NORMDIST(ss/L, 0, 1, 1)] + L NORMDIST(ss/L, 0, 1, 0) = 43.86
Fill rate (fr) = 1- (ESC/Q) = 1- (43.86/1000) = 0.96
Excel Worksheet 11-4 illustrates these computations
5.
DL = LD = (2)(250) = 500
L =
L  2D  D 2 s 2L
=
2
2
2
2 150  250 1.5
= 431
ss = FS-1 (CSL)  L = FS-1 (0.95)  431 = 709 (where, FS-1 (0.95) = NORMSINV (0.95))
2
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
ESC   ss[1  F S (
ss
L
)]   L f S (
ss
L
)
ESC = –ss[1 – NORMDIST(ss/L, 0, 1, 1)] + L NORMDIST(ss/L, 0, 1, 0) = 9
Fill rate (fr) = 1- (ESC/Q) = 1- (9/1000) = 0.991
Standard deviation of lead
time
Required safety inventory
1.50
1.00
0.50
0.00
709
539
405
349
Excel Worksheet 11-5 illustrates these computations
6.
Following are the evaluations for the Khaki pants:
Disaggregated Option:
DL = LD = (4)(800) = 3200
 L = L D =
4 (100) = 200
Coefficient of variation =  /  = 100/800 = 0.13
ss per store = FS-1 (CSL)  L = FS-1 (0.95)  200 = 329 (where, FS-1 (0.95) = NORMSINV (0.95))
Total safety inventory = (329)(900) = 296,074
Total value of safety inventory = (296,074)(30) = $8,882,210
Total annual safety inventory holding cost = (8,882,210)(0.25) = $2,220,552
Holding cost per unit sold = 2220552/(800)(900) = $3.08
Aggregated Option:
DC  kD
= (900)(800) = 720000
3
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl

C
D
 k =
900 (100) = 3000
DL = LDC = (4)(800)(900) = 2,880,000
L =
L DC =
4 (3000) = 6000
ss = FS-1 (CSL)  L = FS-1 (0.95)  6000 = 9869 (where, FS-1 (0.95) = NORMSINV (0.95))
Total safety inventory = 9869
Total value of safety inventory = (9869)(30) = $296,070
Total annual safety inventory holding cost = (296070 )(0.25) = $74,018
Holding cost per unit sold = 74018/(800)(900) = $0.1
Savings in the holding cost per unit sold from aggregation = $3.08 - $0.1 = $2.98
Following are the evaluations for the Cashmere Sweaters:
Disaggregated Option:
DL = LD = (4)(50) = 200
 L = L D =
4 (50) = 100
Coefficient of variation =  /  = 50/50 = 1
ss per store = FS-1 (CSL)  L = FS-1 (0.95)  100 = 164 (where, FS-1 (0.95) = NORMSINV (0.95))
Total safety inventory = (164)(900) = 147,600
Total value of safety inventory = (147,600 )(100) = $14,760,000
Total annual safety inventory holding cost = $14,760,000 )(0.25) = $3,690,000
Holding cost per unit sold = $3,690,000 1/(50)(900) = $82
Note: the above are also incorrect on the worksheet.
Aggregated Option:
DC  kD
= (900)(50) = 45000
4
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl

C
D
 k =
900 (50) = 1500
DL = LDC = (4)(50)(900) = 180000
L =
L DC =
4 (1500) = 3000
ss = FS-1 (CSL)  L = FS-1 (0.95)  3000 = 4935 (where, FS-1 (0.95) = NORMSINV (0.95))
Total safety inventory = 4935
Total value of safety inventory = (4935)(100) = $493,456
Total annual safety inventory holding cost = (493456)(0.25) = $123,364
Holding cost per unit sold = 123364/(50)(900) = $2.74
Savings in the holding cost per unit sold from aggregation = $82.84 - $2.74 = $79.50
Centralization results in savings for both products, but it is evident that savings in holding cost
per unit sold from aggregating Cashmere Sweaters is higher than Khaki pants. So, Cashmere
Sweaters are better for centralization.
Excel Worksheet 11-6 illustrates these computations.
7.
Disaggregated Option:
France:
DL = LD = (8)(3000) = 24000
 L = L D =
8 2000 = 5657
ss at France = FS-1 (CSL)  L = FS-1 (0.95)  5657 = 9305
(where, FS-1 (0.95) = NORMSINV (0.95))
The ss at the other five countries is evaluated in a similar manner, which results in a total ss for
Europe of 48,384
Aggregated Option:
5
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
D
C
6
  Di = 3000 + 4000 + 2000 + 2500+ 1000 + 4000 = 16500
C
D
i 1
6

2
i
=
2000 2  2200 2  1400 2  1600 2  800 2  2400 2 = 4445.22
i 1
DL = LDC = (8)(16500) = 132,000
L =
L DC =
8 ( 4445.22) = 12573
ss = FS-1 (CSL)  L = FS-1 (0.95)  12573 = 20,681 (where, FS-1 (0.95) = NORMSINV (0.95))
Inventory savings from aggregation = 48,384 – 20,681= 27,704
Excel Worksheet 11-7 illustrates these computations.
8.
(a)
Disaggregated Option:
From the previous problem, we know that the total ss for Europe is 48,384
Holding cost = (200)(0.25)(48384) = $2,419,200
Aggregated Option:
ss = 20,681
Holding cost = (200)(0.25)(20681) = $1,034,036
Savings from aggregation = $2,419,200 -$1,034,036 = $1,385,164
(b) If the $5/unit additional cost of assembly from centralization then the total additional costs =
(132000)(52)(5) =
Savings = $1,385,164 - So, it is not economical to aggregate
(c) If the lead time changes to 4 weeks, we evaluate the safety stocks and associated costs in a
similar manner.
The holding cost from the disaggregated option = (200)(0.25)(34213) = $1,710,650
The holding cost from the aggregated option = (200)(0.25)(14623) = $731,150
Savings from aggregation = $1,710,650 - $731,150 = $979,500
Excel Worksheet 11-8 illustrates these computations.
6
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
9.
Since the demand at various locations is not independent, we utilize the following expressions for
the aggregated option:
DC =
ik1 Di
k
var(D C )   σi2  2   ij  i  j
i 1
i j
 CD  var(DC )
For  = 0.2
 CD  var(DC ) =
L =
L DC =
2
2
2000 .  2400  2(0.2){(2000)(2200) .  (800)(2400)
20002  . .  24002  2(0.2){(2000)(2200)  . .  (800)(2400)}
8
= 17307
ss = FS-1 (CSL)  L = FS-1 (0.95)  17307 = 28467 (where, FS-1 (0.95) = NORMSINV (0.95))
Inventory savings from aggregation = 48,384 – 28,467= 19,918
Holding cost savings from aggregation = (200)(0.25)(19918) = $995,900
The following table shows the savings as the correlation coefficient increases from 0 to 1 with
increments of 0.2
Corr.Coeff.
995876.3
0
0.2
0.4
0.6
0.8
1
4
$979,474
$704,191
$489,462
$307,211
$146,047
$0
Replenishment Lead Time (Weeks)
6
8
10
12
$1,199,606 $1,385,185 $1,548,684 $1,696,498
$862,454
$995,876 $1,113,424 $1,219,694
$599,466
$692,203
$773,907
$847,772
$376,255
$434,462
$485,743
$532,105
$178,870
$206,542
$230,921
$252,961
$0
$0
$0
$0
7
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
Excel Worksheet 11-9 illustrates these computations.
10.
Using Sea Transportation:
Average batch size = DT = (5000)(20) = 100,000
 L = L D =
36 ( 4000) = 24,000
ss = FS-1 (CSL)  L = FS-1 (0.99)  24000 = 55832 (where, FS-1 (0.99) = NORMSINV (0.99))
Days of safety inventory = 55832/5000 = 11.17 days
Average cycle inventory = batch size/2 = 100000/2 = 50,000
Days of cycle inventory = 50000/5000 = 10 days
Total inventory cost (cycle + safety) = (50000 + 55832)(100)(0.2) = $2,116,640
Transportation cost per year = (0.5)(5000)(365) = $912,500
Annual Holding Cost + Transportation Cost = $2,116,640 + $912,500 = $3,029,140
In-Transit Inventory = DL = (5000)(36) = 180,000
Cost of Holding In-Transit Inventory = (180000)(100)(0.2) = $3,600,000
Total Costs (including in-transit inventory) = $3,029,140 + $3,600,000 = $6,629,140
Using Air Transportation:
Average batch size = DT = (5000)(1) = 5,000
 L = L D =
4 ( 4000) = 8,000
ss = FS-1 (CSL)  L = FS-1 (0.99)  8000 = 18,611 (where, FS-1 (0.99) = NORMSINV (0.99))
Days of safety inventory = 18611/5000 = 3.72 days
Average cycle inventory = batch size/2 = 5000/2 = 2,500
Days of cycle inventory = 2500/5000 = 0.5 days
Total inventory cost (cycle + safety) = (2500 + 18611)(100)(0.2) = $422,220
8
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Transportation cost per year = (1.5)(5000)(365) = $2,737,500
Annual Holding Cost + Transportation Cost = $422,220 + $2,737,500 = $3,159,720
In-Transit Inventory = DL = (5000)(4) = 20000
Cost of Holding In-Transit Inventory = (20000)(100)(0.2) = $400,000
Total Costs (including in-transit inventory) = $3,159,720 + $400,000 = $3,559,720
Based on the results air transportation would be the optimal choice, but if Motorola does not have
the ownership of in-transit inventory then sea transportation is the optimal choice.
Excel Worksheet 11-10 illustrates these computations.
9
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
11.
Using Sea Transportation:
Average batch size = DT = (5000)(20) = 100,000
L =
L T  D
=
36  20 ( 4000) = 29,933
ss = FS-1 (CSL)  L = FS-1 (0.99)  29933 = 69,635 (where, FS-1 (0.99) = NORMSINV (0.99))
Days of safety inventory = 69635/5000 = 13.93 days
OUL = D(T+L) + ss = 5000(36+20) + 69635 = 349,635
Average cycle inventory = batch size/2 = 100000/2 = 50,000
Days of cycle inventory = 50000/5000 = 10 days
Total inventory cost (cycle + safety) = (50000 + 69635)(100)(0.2) = $2,392,700
Transportation cost per year = (0.5)(5000)(365) = $912,500
Annual Holding Cost + Transportation Cost = $2,392,700 + $912,500 = $3,305,200
In-Transit Inventory = DL = (5000)(36) = 180,000
Cost of Holding In-Transit Inventory = (180000)(100)(0.2) = $3,600,000
Total Costs (including in-transit inventory) = $3,029,140 + $3,600,000 = $6,905,200
Using Air Transportation:
Average batch size = DT = (5000)(1) = 5,000
L =
L T  D
=
4  1( 4000) = 8,944
ss = FS-1 (CSL)  L = FS-1 (0.99)  8944 = 20,807 (where, FS-1 (0.99) = NORMSINV (0.99))
Days of safety inventory = 20807/5000 = 4.16 days
OUL = D(T+L) + ss = 5000(1+4) + 20807 = 45,807
Average cycle inventory = batch size/2 = 5000/2 = 2,500
10
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
Days of cycle inventory = 2500/5000 = 0.5 days
Total inventory cost (cycle + safety) = (2500 + 20807)(100)(0.2) = $466,150
Transportation cost per year = (1.5)(5000)(365) = $2,737,500
Annual Holding Cost + Transportation Cost = $466,150 + $2,737,500 = $3,203,650
In-Transit Inventory = DL = (5000)(4) = 20000
Cost of Holding In-Transit Inventory = (20000)(100)(0.2) = $400,000
Total Costs (including in-transit inventory) = $3,203,650 + $400,000 = $3,603,650
Based on the results air transportation would be the optimal choice. Even if Motorola does not
have the ownership of in-transit inventory, air transportation is the optimal choice.
Excel Worksheet 11-11 illustrates these computations.
12.
ss = ROP – DL = 750 – 300(2) = 750-600 = 150
 L = L D =
2 (100) = 141.42
CSL = F(DL + ss, DL, L) = F(750, 600, 141.42) = NORMDIST (ss/L, 0,1,1) = 85.56%
ESC   ss[1  F S (
ss
L
)]   L f S (
ss
L
)
ESC = –ss[1 – NORMDIST(ss/L, 0, 1, 1)] + L NORMDIST(ss/L, 0, 1, 0) = 10
Fill rate (fr) = 1- (ESC/Q) = 1- (10/1500) = 0.993
If the ROP increased from 750 to 800 the fill rate will increase to 0.996
Excel Worksheet 11-12 illustrates these computations.
13.
Fill rate (fr) = 1- (ESC/1500) = 0.999
So, ESC = 1.5
11
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
ESC   ss[1  F S (
ss
L
)]   L f S (
ss
L
)]
ESC = –ss[1 – NORMDIST(ss/L, 0, 1, 1)] + L NORMDIST(ss/L, 0, 1, 0) = 1.5
We use the GOALSEEK function in determining the safety stock (ss) by using ss as the changing
value that results in an ESC value of 1.5.
Goal Seek set-up:
SET CELL: A15
TO VALUE: 1.5
BY CHANGING CELL: D12
This results in an ss value of 271 and a reorder point of = 300(2) + 271 = 871
Excel Worksheet 11-13 illustrates these computations.
14.
(a)
Disaggregated Option:

LT
=
L T  D
=
3  7 (50) = 158
ss per store = FS-1 (CSL)  L = FS-1 (0.99)  158 = 367.83
(where, FS-1 (0.99) = NORMSINV (0.99))
Total safety inventory = (367.83)(25) = 9195.7
Aggregated Option:
DC  kD
= (25)(300) = 7500
k
var( D C )   σi2  2   ij  i  j
i 1
i j
 CD  var(DC )
12
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl

C
D
 k =
25 (50) = 250 (we are assuming that
 = 0. If  is not 0 then the covariance
terms have to be included)

LT
=
L  T  DC =
3  7 ( 250) = 791
ss = FS-1 (CSL)  L = FS-1 (0.99)  791 = 1839.14 (where, FS-1 (0.99) = NORMSINV (0.99))
Units savings from aggregation = 9195.7 – 1839.14 = 7356.56
Inventory savings = (7356.56) (10) = $73,566
Annual holding cost savings = (73,566 )(0.2) = $14,712
Increase in delivery cost = (300)(25)(365)(0.02) = $54,750
Since the increase in transportation costs outweighs the savings received from aggregation, we do
not recommend aggregation for this case.
(b)
We utilize the same approach as in (a) by changing the daily demand mean and standard deviation
to 5 and 4, respectively
Units savings from aggregation = 735.66 – 147.13 = 588.52
Inventory savings = (588.52) (10) = $ 5885.2
Annual holding cost savings = ($5885.2 )(0.2) = $1177
Increase in delivery cost = (5)(25)(365)(0.02) = $913
Since the increase in transportation costs does not outweigh the savings received from
aggregation, we recommend aggregation for this case.
(c) Yes. The benefit from aggregation decreases as  increases. When  = 0.5, we do not
recommend aggregation in both cases.
Excel Worksheet 11-14 illustrates these computations.
15.
(a)
Popular Variant at Large Dealer:
Decentralized:
ss (at each large dealer) = FS-1 (CSL) 
L D = F
S
-1
(0.95) 
4 (15) = 49.35
13
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
ss (across all large dealers) = (5)(49.35) = 246.73
Popular Variant at Small Dealer:
Decentralized:
ss (at each small dealer) = FS-1 (CSL) 
L D = F
-1
S
(0.95) 
4 (5) = 16.45
ss (across all small dealers) = (30)(16.45) = 493.46
(b )
Popular Variant all Inventories Centralized:
Demand per period = demand at large dealers + demand at small dealers
= (50)(5) + (10)(30) = 550
Standard deviation of demand per period = 5(15) 2  30(5) 2 = 43.30
ss (at regional warehouse) = FS-1 (CSL) 
L D = F
-1
S
(0.95) 
4 ( 43.30) = 142.45
reduction in safety inventory from complete aggregation = 246.73 + 493.46 – 142.45 = 597.74
holding cost savings per year = (597.74)(20000)(0.2) = $2,390,942.52
production + transportation cost increase per year = (550)(100)(52) = $2,860,000
(c)
Popular Variant only Small Dealer Inventories Centralized:
Demand per period = demand at small dealers
= (10)(30) = 300
Standard deviation of demand per period = 30(5) 2 = 27.39
ss (at regional warehouse) = FS-1 (CSL) 
L D = F
-1
S
(0.95) 
4 ( 27.39) = 90.09
reduction in safety inventory from small dealer centralization = 493.46 – 90.09 = 403.36
holding cost savings per year = (403.36)(20000)(0.2) = $1,613,440
production + transportation cost increase per year = (300)(100)(52) = $1,560,000
(d) Centralizing inventories from small dealers and decentralizing at large dealers is the optimal
strategy
(e) Similar analysis can be performed for the uncommon variant (See EXCEL worksheet 11-15
for more details)
14
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
(f) For the popular variant, centralize inventories from small dealers and decentralize at large
dealers. For the uncommon variant, centralize all inventories.
Excel Worksheet 11-15 illustrates these computations.
16.
High volume variant without component commonality:
ss (for the variant) = FS-1 (CSL) 
L D = F
-1
S
(0.95) 
4 ( 200) = 657.94
ss (across all high volume variants) = (1)(657.94) = 657.94
Low volume variant without component commonality:
ss (for the variant) = FS-1 (CSL) 
L D = F
-1
S
(0.95) 
4 ( 20) = 65.79
ss (across all low volume variants) = (9)(65.79) = 592.15
(b & c)
With complete commonality:
Demand per period = demand for high volume variant + demand for low volume variant
= (1000)(1) + (28)(9) = 1252
Standard deviation of demand per period = 1( 200) 2  9( 20) 2 = 208.81
ss = FS-1 (CSL) 
L D = F
S
-1
(0.95)  8686.91
reduction in safety inventory from complete commonality = 657.94 + 592.15 – 686.91 = 563.18
holding cost savings per year = (563.18)(1000)(0.2) = $112,635.54
component cost increase per period = (1252)(25)(52) = $1,627,600
additional cost at which complete commonality is justified = 112635.54/(1252)(52) = $1.73
Commonality is not justified across all variants because of increased costs.
(d & e)
Only low volume variant uses commonality
Demand per period = demand for low volume variant
= (28)(9) = 252
Standard deviation of demand per period = 9( 20) 2 = 60
15
Full file at http://testbanksolutions.org/Solution-Manual-for-Supply-Chain-Management,-4-E-Sunil-Chopra,-PeterMeindl
ss = FS-1 (CSL) 
L D = F
S
-1
(0.95) 
4 (60) = 197.38
reduction in safety inventory from low volume variant using commonality
= 592.15 – 197.38 = 394.76
holding cost savings per year = (394.76)(1000)(0.2) = $78,952.97
component cost increase per period = (252)(25)(52) = $327,600
additional cost at which complete commonality is justified = 78952.97/(252)(52) = $6.03
Commonality is not justified across low volume variants because of increased costs.
Excel Worksheet 11-16 illustrates these computations.
16
Chopra/Meindl 5/e
Instant download and all chapters Solutions Manual Supply Chain Management 5th
Edition Sunil Chopra, Peter Meindl
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CHAPTER 5
Case Questions (see corresponding Chapter 5 case Excel spreadsheet)
1. What is the cost SportStuff.com incurs if all warehouses leased are in St. Louis?
Demand in 2007 is as shown in Table 5-15
Demand 2007
Northwest
320,000
Southwest
200,000
Upper Midwest
160,000
Lower Midwest
220,000
Northeast
350,000
Southeast
175,000
Total Demand
1,425,000
The capacity of the current warehouse in St. Louis is 2,000,000 units per year,
more than enough to accommodate 2007’s demand.
Costs are calculated as:
Fixed Warehouse Cost = $220,000
Variable Warehouse Cost = ($0.20/unit)(1,425,000 units) = $285,000
Holding Cost = $475,000+0.165*1,425,000 = $710,125
Shipping Cost = $1,068,750
Shipping Recouped from Customer = (1,425,000/4)*$3 = $1,068,750
Total Network Costs = $1,254,500
Note that the spreadsheet used for these calculations employed the linear holding
cost function of $475,000+0.165F, which results in a holding cost of $710,125
and a total plan cost of $1,254,500.
Subsequent holding costs will use the single linear function to determine holding
costs.
If demand increases by 80% per year for 2008, 2009, and 2010 and
SportStuff.com wishes to use St. Louis as their only warehouse center, the
following demands and costs are realized. The optimal solution for 2008 calls for
one large warehouse rather than two small ones.
Demand 2008
Northwest
576000
Southwest
360000
Upper Midwest
288000
Chopra/Meindl 5/e
Lower Midwest
Northeast
Southeast
Total Demand
396000
630000
315000
2,565,000
Total Shipping Cost
Total Holding Cost
Warehouse Cost
Total Shipping Recoup
$1,994,625
$898,225
$888,000
$1,923,750
TOTAL COST
$1,857,100
For 2009, one small and one large warehouse is optimal.
Demand 2009
Northwest
1036800
Southwest
648000
Upper Midwest
518400
Lower Midwest
712800
Northeast
1134000
Southeast
567000
Total Demand
4,617,000
Total Shipping Cost
Total Holding Cost
Warehouse Cost
Total Shipping Recoup
$3,590,325
$2,473,610
$1,518,400
$3,462,750
TOTAL COST
$4,119,585
For 2010, one small and two large warehouses are optimal
Demand 2010
Northwest
1866240
Southwest
1166400
Upper Midwest
933120
Lower Midwest
1283040
Northeast
2041200
Southeast
1020600
Total Demand
8,310,600
Total Shipping Cost
Total Holding Cost
Warehouse Cost
Total Shipping Recoup
$6,462,585
$5,538,747
$2,632,120
$6,232,950
TOTAL COST
$8,400,502
Chopra/Meindl 5/e
Chopra/Meindl 5/e
2. What supply chain network configuration do you recommend for SportStuff.com?
The progression if location is not restricted to St. Louis is as follows:
For 2008: Small warehouses in Seattle and St. Louis
Northwe Southwe Upper
Lower Northea
st
st
Midwes Midwes
st
t
t
Seattle
576,000
360,000
0
0
0
Denver
0
0
0
0
0
St. Louis
0
0
288,00 396,00 630,000
0
0
Atlanta
0
0
0
0
0
Philadelphi
0
0
0
0
0
a
Total
576,000
360,000 288,00 396,00 630,000
Demand
0
0
Total Shipping Cost
Total Holding Cost
Warehouse Cost
Total Shipping Chg
$1,688,625
$1,373,225
$1,033,000
$1,923,750
TOTAL COST
$2,171,100
Southea
st
Total
Supply
0
0
315,000
936,000
0
1,629,00
0
0
0
0
0
315,000
A lower total system cost is achievable if SportStuff abandons their small St. Louis
warehouse and opts for a single large warehouse in Atlanta.
Total Shipping Cost $2,068,875
Total Holding Cost
$898,225
Warehouse Cost
$888,000
Total Shipping Chg $1,923,750
TOTAL COST
$1,931,350
Chopra/Meindl 5/e
For 2009: Small warehouses in Seattle and St. Louis plus a small warehouse in Atlanta
Northwest Southwest
Upper
Lower
Northeast Southeast
Midwest Midwest
Seattle
1,036,800
648,000
0
0
0
0
Denver
0
0
0
0
0
0
St. Louis
0
0
518,400
497,096
984,504
0
Atlanta
0
0
0
215,704
149,496
567,000
Philadelphia
0
0
0
0
0
0
Total
1,036,800
648,000
518,400
712,800
1,134,000
567,000
Demand
Total Shipping Cost
Total Holding Cost
Warehouse Cost
Total Shipping Chg
$2,897,775
$2,186,805
$1,663,400
$3,462,750
TOTAL COST
$3,285,230
Total
Supply
1,684,800
0
2,000,000
932,200
0
For 2010, Small warehouses in all cities results in
Total Shipping Cost
$4,965,995
Total Holding Cost
$3,746,249
Warehouse Cost
$2,892,120
Total Shipping Chg
$6,232,950
TOTAL COST
$5,371,414
Northwest Southwest
Seattle
1,866,240
Denver
0
St. Louis
0
Atlanta
0
Philadelphia
0
Total
1,866,240
Demand
116,640
1,049,760
0
0
0
1,166,400
Upper
Midwest
0
933,120
0
0
0
933,120
Lower
Northeast
Midwest
0
0
0
0
1,283,040
41,200
0
0
0
2,000,000
1,283,040 2,041,200
Southeast
0
0
0
1,020,600
0
1,020,600
A lower cost solution of small warehouses in Seattle and Atlanta and large warehouses in
Denver and Philadelphia results in a savings of $248,018.
Total Shipping Cost
$5,082,976
Total Holding Cost
$3,271,249
Warehouse Cost
$3,002,120
Total Shipping Chg
$6,232,950
TOTAL COST
$5,123,395
Total
Supply
1,982,880
1,982,880
1,324,240
1,020,600
2,000,000
Chopra/Meindl 5/e