Last Name _________________________ First Name _________________________ ID _______________________
Operations Management I 73-331 Fall 2003
Odette School of Business
University of Windsor
Final Exam Solution
Monday, December 15, 8:30 – 11:30 a.m.
Chrysler Hall North G 133
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and 3 one-sided formula sheets.
Time available: 3 hours
Instructions:
 This exam has 34 pages including this cover page, 1 blank page and 8 pages of Table
 It’s not necessary to return tables and formula sheets
 Please be sure to put your name and student ID number on each odd numbered pages
 Show your results up to four decimal places
 Show your work
Grading:
Question
Score
Question
Score
1
/15
2
/10
3
/6
4
/12
5
/4
6
/12
7
/9
8
/8
9
/10
10
/5
11
/9
Total
/100
Name:_________________________________________________
ID:_________________________
Question 1: (15 points) Circle the most appropriate answer
1.1 Consider estimating learning curve parameter values using regression on ln u  and ln Y u .
What is a best interpretation of intercept, c ?
a. The slope is ln c 
b. The slope is e c
c. An estimate of the time required by the first unit is ln c 
d. An estimate of the time required by the first unit is e c
1.2 Consider the EOQ model with price breaks. The optimal solution is
a. the cheapest realizable EOQ
b. one of the EOQs
c. one of the breakpoints
d. the cheapest realizable EOQ or a cheaper breakpoint
1.3 Forecasting error is described by
a. weighted moving average
b. mean absolute deviation
c. both
d. none
1.4 The dynamic capacity addition model assumes all of the following except
a. a constant rate of increase of demand over a finite planning horizon
b. the same capacity addition at an equal interval of time
c. economies of scale
d. continuous compounding
1.5 Which of the following are advantages of the Exponential Smoothing model?
a. Less computation
b. Less memory requirement
c. Both
d. None
1.6 Smoothing cost includes
a. cost to advertise and interview candidates
b. severance pay
c. both
d. none
1.7 Chase strategy attempts to produce
a. a constant amount each period
b. as much as needed
c. both
d. none
1.8 Annual holding cost and is the same as annual ordering/setup cost at
a. EOQ and EPQ
b. EOQ but not EPQ
c. EPQ but not EOQ
d. none of the above
2
Name:_________________________________________________
ID:_________________________
1.9 Which of the following is not a characteristic of the rotation cycle policy?
a. There is only one setup for each product in each cycle
b. The products are produced in the same sequence in each cycle
c. Only one product is produced at any time
d. For each product, the economic production quantity (EPQ) is produced in each cycle
1.10
a.
b.
c.
d.
To find an optimal Q, R  policy with Type II service, the penalty cost is
not estimated
estimated from the cost of backorder
estimated from the cost of lost sales
computed using the standardized loss function
1.11
a.
b.
c.
d.
Which of the following is not an input to the MRP system?
The production schedule of the finished products
The production schedule of the components/subassemblies
Bill of Materials
Inventory records
1.12
a.
b.
c.
d.
Which of the following is not an assumption in the space constraint model?
A known, fixed and uniform demand rate
Known and fixed cost parameters
A single product carried in the inventory
The same order size every cycle
1.13
a.
b.
c.
d.
Least Unit Cost (LUC) method performs best if
1.14
a.
b.
c.
d.
Andon
is the authority to stop a production line
makes problems visible
lights signal quality problems
prevents defects
1.15
a.
b.
c.
d.
The Q, R  policy is used for
multi-period component/subassemblies
multi-period finished products
single-period discrete demand
single-period continuous demand
the production environment is make-to-order or assemble-to-order
the production environment is make-to-stock or assemble-to-stock
inventory costs do not change over time
inventory costs change over time
3
Name:_________________________________________________
ID:_________________________
Question 2: (10 points)
A Japanese steel manufacturer is considering expanding operations. From experience, it estimates
that new capacity additions obey the law f  y   ky 0.65 , where the cost, f  y  , is measured in millions
of dollars and y is measured in tons of steel produced. If the demand for steel is assumed to grow at
the constant rate of 3,000 tons per year and future costs are discounted using a 16 percent discount
rate
a. (2 points) Determine the optimal timing of
plant additions.
Figure 1-14
1.00
rx  0.80 (from Figure 1 - 14)
rx 0.80
x

 5 years
r 0.16
Function = a
0.90
a  0.65 (given)
0.80
0.70
0.60
0.50
0.40
0.30
0
1
u = rx
b. (2 points) Determine the optimal size of each addition.
Optimal size = xD  3,000  5  15,000 tons per year
c. (2 points) If the size of the refinery is doubled, what is the percentage increase in the construction
costs?
f 2 y  k 2 y 

 2 0.65  1.57  57% increase
0.65
f y
ky
0.65
d. (2 points) If a plant size of 30,000 tons per year costs 18 million dollars, find k .
f  y   ky a  k 
f y
18

 0.02214
a
y
30,0000.65
e. (2 points) Continue from parts a, b and d. What is the present cost of the next 2 additions? The
first one is added today and the second one after the number of years obtained in part a.
For y  15,000 tons, the cost f  y   ky a  0.0221415,000
0.65
 11.4719 million dollar
Present cost
 11.4719 
11.4719
1  0.165
 11.4719  11.4719  0.4761  11.4719  5.4619  16.9338 million dollar
4
2
Name:_________________________________________________
ID:_________________________
Question 3: (6 points)
The Paris Paint Company is in the process of planning labor force requirements and production
levels for the next four quarters. The marketing department has provided production with the
following forecasts of demand for Paris Paint over the next year:
Quarter
Demand Forecast
(in thousands of gallons)
1
450
2
800
3
750
4
200
Assume that there are currently 275 employees with the company. Employees are hired for at least
one full quarter. Hiring costs amount to $400 per employee and firing costs are $800 per employee.
Inventory costs are $0.25 per gallon per quarter. It is estimated that one worker produces 1,500
gallons of paint each quarter. Assume that Paris currently has 200,000 gallons of paint in inventory
and would like to end the year with an inventory of at least 300,000 gallons.
a. (3 points) Determine the minimum constant workforce plan (i.e., level strategy) for Paris Paint.
Assume that stock-outs are not allowed.
Quarter
Production
Requirement
(000 gallons)
Cumulative
Production
Requirement
Units
Produced Per
Worker
(000 gallons)
(000 gallons)
Cumulative
Units
Produced Per
Worker
Workers
Required
(000 gallons)
1
450-200=250
250
1.5 (given)
1.5
250 / 1.5  167
2
800
250+800=1,050
1.5
3
1050 / 3  350 *
3
750
1,050+750=1,800
1.5
4.5
1800 / 4.5  400
4
200+300=500
1,800+500=2,300
1.5
6
2300 / 6  384
Since the maximum workers required is 400, the minimum constant workforce plan must use 400
workers. So, the number of workers to hire = 400 – 275 = 125 workers.
b. (3 points) Determine the hiring, firing, and inventory holding cost of the plan derived in part a.
Quarter
Beginning
Inventory
Production
(000 gallons)
Ending Inventory = Production + Beginning
Inventory – Demand
(000 gallons)
(000 gallons)
1
200
400(1.5)=600
600+200-450=350
2
350
600
600+350-800=150
3
150
600
600+150-750=0
4
0
600
600+0-200=400
Total ending inventory = (350+150+0+400) = 900 thousand gallons
5
Name:_________________________________________________
ID:_________________________
Inventory holding cost = 900,000  0.25 = $225,000
Hiring cost = 125(400) = $50,000
Total cost = 225,000+50,000 = $275,000
Question 4: (12 points)
A popular brand of tennis shoe has had the following demand history by quarters over a two-year
period.
Quarter
Demand
Quarter
2002
Demand
2003
1
25
1
33
2
35
2
47
3
45
3
55
4
40
4
50
a. (4 points) Determine the seasonal factors for each quarter by the method of centered moving
averages
4
The demand is quarterly, there are 4 quarters in each year.
N=
Centered
(B/D)
Period
Demand
MA(4)
MA
Ratio
A
B
C
D
E
1
25
38.5
0.649350649
2
35
38.5
0.909090909
3
45
37.25
1.208053691
4
40
36.25
39.75
1.006289308
5
33
38.25
42.5
0.776470588
6
47
41.25
45
1.044444444
7
55
43.75
43.75
1.257142857
8
50
46.25
43.75
1.142857143
Period
1
2
3
4
Total
Seasonal
Factors
0.71291062
0.97676768
1.23259827
1.07457323
3.9968498
Final
Seasonal
Factors
0.7135
0.9775
1.2336
1.0754
4.0000
6
Name:_________________________________________________
ID:_________________________
b. (4 points) Compute the deseasonalized demand series. Using the method of linear regression,
determine the slope and intercept of the straight line that best fits the deseasonalized series.
xy
x
Deseasonalized
x2
Demand
Sum
Average
1
2
3
4
5
6
7
8
36
4.5
35.03989219
35.80425166
36.47949307
37.1947644
46.25265769
48.07999508
44.58604708
46.4934555
329.9305567
41.24131958
35.03989219
71.60850331
109.4384792
148.7790576
231.2632884
288.4799705
312.1023296
371.947644
1568.659165
1
4
9
16
25
36
49
64
204
n
 n  n

n xi y i    xi   y i 
 i 1  i 1   8(1568 .6592 )  36(329.9306 )  1.9993
Slope  i 1
2
n
8(204 )  (36) 2
 n 
2
n xi    xi 
i 1
 i 1 
Intercept  y  slope(x)  41.2413  (1.9993 )( 4.5)  32.2443
c.
(4 points) Predict the demand of all quarters of 2004. Plot the original demand of 2002-2003 and
predicted demand of 2004.
Deseasonalized demand, y  32.2443  1.9993 x
First quarter of 2004: x  9 , y  32.2443  1.9993  9  50.2383
To get the predicted demand, reseasonalize, y  50.2383  0.7135  35.8436
Second quarter of 2004: x  10 , y  32.2443  1.9993  10  52.2376
To get the predicted demand, reseasonalize, y  52.2376  0.9775  51.0642
Third quarter of 2004: x  11, y  32.2443  1.9993  11  54.2369
To get the predicted demand, reseasonalize, y  54.2369  1.2336  66.9050
Fourth quarter of 2004: x  12 , y  32.2443  1.9993  12  56.2363
To get the predicted demand, reseasonalize, y  56.2363  1.0754  60.4776
7
Name:_________________________________________________
ID:_________________________
Demand
Demand
80
70
60
50
40
30
20
10
0
0
1
2
3
4
5
6
7
8
9
10
11
12
Period
Question 5: (4 points)
Green City sells a particular model of lawn mower, with most of the sales being made in the summer
months. Green city makes a one-time purchase of the lawn mowers prior to each summer season at
a cost of $150 each and sells each lawn mower for $210. The demand is normally distributed with a
mean of 1200 and a standard deviation of 80. Find the optimal order quantity if
a. (2 points) any lawn mower unsold at the end of summer season are marked down to $75 and
sold in a special fall sale.
cu  Selling price – purchase price = 210-150 = $60/unit
co  Purchase price – salvage value = 150-75 = $75/unit
For the optimal order quantity Q , Probability(demand  Q ), p 
cu
60

 0.4444
cu  co 60  75
Find the standard normal z -value for which cumulative area on the left, p  0.4444 .
Using Table A-1
Table A-1 gives the area between z  0 and positive z -values.
Since p  0.4444  0.50, the z -value is negative and corresponds to area = 0.50-0.4444 = 0.0556
Hence, z  0.14.
Using Table A-4
Since p  0.4444  0.50, the z -value is negative and corresponds to 1  F z   0.4444
Hence, z  0.14.
Q *    z  1,200   0.14  80  1,188.8  1,189 units
b. (2 points) any lawn mower unsold at the end of summer season are marked down to $120 and
sold in a special fall sale.
cu  Selling price – purchase price = 210-150 = $60/unit
co  Purchase price – salvage value = 150-120 = $30/unit
8
Name:_________________________________________________
For the optimal order quantity Q , Probability(demand  Q ), p 
ID:_________________________
cu
60

 0.6666
cu  co 60  30
Find the standard normal z -value for which cumulative area on the left, p  0.6666 .
Using Table A-1
Table A-1 gives the area between z  0 and positive z -values.
Since p  0.6666  0.50, the z -value is positive and corresponds to area = 0.6666-0.50 = 0.0166
Hence, z  0.43.
Using Table A-4
Since p  0.6666  0.50, the z -value is positive and corresponds to F z   0.6666
Hence, z  0.43
Q *    z  1,200  0.43  80  1,234.4  1,234 units
Question 6: (12 points)
Suppose that Item A has a production rate of 1,152 items per year, unit cost of $16.00, a setup cost
of $144, and a monthly demand of 48 units. It is estimated that cost of capital is approximately 20
percent per year. Storage cost amounts to 3 percent and breakage to 2 percent of the value of each
item.
a. (2 points) Compute EPQ of Item A.
K  $144
  1248  576 units per year
I  0.20  0.03  0.02  0.25
h  Ic  0.2516  $4 per unit per year
P  1,152 units per year
EPQ, Q * 
2 K

h'
2 K

 
h1  
 P
2144 576 

576 

4.001 

 1,152 
2144 576 
 288 units
2
b. (3 points) What are the maximum inventory and cycle time of Item A? What is the percentage of
idle time of the facility if the facility is dedicated to produce Item A only?
576 

 
Maximum inventory, H  Q * 1    2881 
  144 units
 P
 1,152 
Cycle time, T 
*
Q*

288
 0.50 year
576

Q
288
Uptime, T1 

 0.25 year, Downtime, T *  T1  0.50  0.25  0.25 year
P 1,152
T
0.25
Percentage of downtime in each cycle = 2* 
 0.50  50%
0.50
T
*
9
Name:_________________________________________________
ID:_________________________
Item B has a production rate of 1500 items per year, a unit cost of $32.00, an ordering cost of
$90.90, and a monthly demand of 30 units. Recall that the cost of capital is approximately 20 percent
per year. Storage cost amounts to 3 percent and breakage to 2 percent of the value of each item.
c. (2 points) What is the cycle time if both Items A and B are produced in a single facility? Consider
negligible setup times for both Items A and B.
T* 


2 K j
 h'
j
j

2K A  K B 

h' A  A  h' B  B
2144  90.90
 
576 

4.001 
576  Ic B 1  B
 1,152 
 PB

 B

2144  90.90
  
 
h A 1  A  A  hB 1  B
 PA 
 PB


 B

2234.90
 30  12 
2  576  0.25321 
30  12
1,500 

469.8
469.8

=0.375 years
1,152  6.08  360
3,340.8
d. (5 points) Continue from part c. What are the maximum inventory of Items A and B? What is the
percentage of idle time of the facility if the facility is used to produce only Items A and B?
QA   AT *  576  0.375  216 units
QB  BT *  30  12  0.375  135 units
Q
216
Uptime of A, TA  A 
 0.1875 years
PA 1,152
Q
135
Uptime of B, TB  B 
 0.09 years
PB 1,500
  
576 

Maximum inventory, H A  Q A* 1  A   2161 
  108 units
 1,152 
 PA 
  
360 

Maximum inventory, H B  QB* 1  B   1351 
  102.6 units
P
1
,
500


B 

Idle time = T *  TA  TB  0.375  0.1875  0.09  0.09 years in each cycle
IdleTime 0.0975

 0.26  26%
Hence, percentage of idle time =
0.375
T*
Question 7: (9 points)
The home appliance department of a large department store is planning to use a lot size-reorder
point system to control the replenishment of a particular model of FM table radio. The store sells an
average of 360 radios each year. The annual demand follows a normal distribution with a standard
deviation of 60. The store pays $80 for each radio. The holding cost is 25 percent per year. Fixed
costs of replenishment amount to $100. If a customer demands the radio when it is out of stock, the
customer will generally go elsewhere. Replenishment lead-time is three weeks. Assume 48 weeks in
a year.
10
Name:_________________________________________________
ID:_________________________
a. (3 points) Find an optimal (Q,R) policy with probability(stockout)=0.35.
Step 1: Q  EOQ 
2 K

h
2 K

Ic
2100 360

0.2580
72,000
 3,600  60 units
20
Step 2: Probability (stockout) = 0.35
F z  = Probability (no stockout) = 1-Probability(stockout) = 1-0.35 = 0.65
Find z for which area on the left = F z   0.65
From Table A-1, z  0.385 for area = 0.65-0.50 = 0.15
From Table A-4, z  0.385 for F z   0.65
Step 3: Compute reorder point, R    z
    3603 / 48  22.5
   y   60 3 / 48  15
R    z  22.5  0.385  15  28.275 units
Hence an optimal policy is Q  60 units, R  28 units
b. (2 points) Compute the annual holding cost resulting from the (Q,R) policy obtained in part a.
Annual holding cost, regular =
hQ IcQ 0.25  80  60 20  60



 $600
2
2
2
2
Safety stock = R    28.275  22.5  5.775 units
Annual holding cost, safety stock = hR     0.25  805.775  $115.5
Annual holding cost =
hQ
 hR    = 600 + 115.5 = $715.50
2
c. (2 points) Compute the annual ordering cost resulting from the (Q,R) policy obtained in part a.
Annual ordering cost =
K 100  360

 $600
Q
60
d. (2 points) Compute the expected annual number of units stockout resulting from the (Q,R) policy
obtained in part a.
L z  
0.2374  0.2339
 0.23565 for z  0.385 from Table A-4
2
n  Lz   150.23565  3.53475 units/cycle
Annual number of units stockout =
n 2.53475  360

 21.2085 units
Q
60
11
Name:_________________________________________________
ID:_________________________
Question 8: (8 points)
Consider Question 7 again. The question is re-written below:
The home appliance department of a large department store is planning to use a lot size-reorder
point system to control the replenishment of a particular model of FM table radio. The store sells an
average of 360 radios each year. The annual demand follows a normal distribution with a standard
deviation of 60. The store pays $80 for each radio. The holding cost is 25 percent per year. Fixed
costs of replenishment amount to $100. If a customer demands the radio when it is out of stock, the
customer will generally go elsewhere. Replenishment lead-time is three weeks. Assume 48 weeks in
a year.
Find an optimal (Q,R) policy with fill rate = 0.95. Use the iterative method and show 2 iterations.
Show your computation on the next page and summarize your results in the table below:
Summary of results:
Fixed cost (K)
Holding cost (h)
Mean annual demand (lambda)
Lead time (tau) in years
Lead time demand parameters:
mu
sigma
Type 2 service, fill rate, beta
Step 1
Step 2
Step 3
Step 4
Step 5
100 Note: K and h
20
are input data
360 input data
0.0625 input data
22.5 <--- computed
15.00 input data
0.95 input data
Iteration 1 Iteration 2
Q=
60
EOQ
Q(1   )
n=
3
L(z)=
0.2
n /
Table A1/A4, pp. 835 - 41
z=
0.49


z

R=
29.85
Table A1/A4, pp. 835 - 41
Area on the right=1-F(z)
0.3121
0.3483
2
Modified Q= n /(1  F ( z ))  2 K / h  ( n /(1  F ( z )))
70.3774
70.9477
n=
3.5189
3.5474
Q(1   )
L(z)=
0.2346
0.2365
n /
Table A1/A4, pp. 835 - 41
z=
0.39
0.385
R=
  z
28.35
28.275
h  Ic  0.25  80  20,  3 / 48  0.0625,     360  0.0625  22.5,    y   60 0.0625  15
Iteration 1
2k
2(100)(3600)

 60 units
h
20
Step 2: n  Q(1   )  60(1  0.95)  3
n
3
L( z )  
 0.20
 15
z  0.49 (Table A-4)
R    z  22.5  0.49  15  29.85
Step 3: 1  F ( z )  0.3121 (Table A-4)
Step 1: Q  EOQ 
12
Name:_________________________________________________
ID:_________________________
2
2
2
2

3
2(100)(360)  3 
n
2 K 
n
 


Step 4: Q 

 
  70.3774
0.3121
20
1  F ( z)
h
 0.3121 
 1  F ( z) 
(not near 60, more iterations are necessary)
Step 5: n  Q(1   )  70.3774(1  0.95)  3.5189
n 3.5189
L( z )  
 0.2346

15
z  0.39 (Table A-4)
R    z  22.5  0.39  15  28.35
Iteration 2
Step 3: 1  F ( z )  0.3483 (Table A-4)

3
2(100)(360)  3 
n
2 K 
n
 



 
Step 4: Q 
  70.9476 (same as
0.3483
20
1  F ( z)
h
 0.3483 
 1  F ( z) 
before, stop the process after finding R )
Step 5: n  Q(1   )  70.9476(1  0.95)  3.5474
n 3.5474
L( z )  
 0.2365

15
z  0.385 (Table A-4)
R    z  22.5  0.385  15  28.275
Q and R converge. An optimal policy is Q=71, R=28 (rounded to the nearest integer)
Question 9: (10 points)
Each unit of A is composed of three units of B and two units of C. Each unit of B is composed of four
units of C and five units of D. Items A, B and C have on-hand inventories of 50, 100 and 200 units
respectively. Item B has a scheduled receipt of 40 units in period 1, and D has a scheduled receipt of
150 units in Period 1. Lot-for-lot (L4L) is used for Item A. Item B requires a minimum lot size of 50
units. Each of the Items C and D is required to be purchased in multiples of 100. Lead times are one
period for each Items A, B and C, and two periods for Item D. The gross requirements for A are 30 in
Period 2, 25 in Period 5, and 80 in Period 8. Find the planned order releases for all items to meet the
requirements over the next 10 periods.
a. (2 points) Construct a product structure tree.
Level 0
A
Level 1
B(3)
C(4)
D(5)
C(2)
13
Level 2
Name:_________________________________________________
ID:_________________________
b. (2 points) Consider Item A. Find the planned order releases and on-hand units in period 10
Period
1
2
3
4
5
6
7
8
9
10
Item Gross
30
25
80
Requirements
A
Scheduled
receipts
On hand from
50
50
20
20
20
0
0
80
0
0
LT=
prior period
1
Net
5
80
requirements
Time-phased Net
5
80
Q=
Requirements
L4L
Planned order
5
80
releases
Planned order
5
80
delivery
c. (2 points) Consider Item B. Find the planned order releases and on-hand units in period 10.
Period
1
2
3
4
5
6
7
8
9
10
Item Gross
15
240
Requirements
B
Scheduled
40
receipts
On hand from
100 140 140 140 125 125 125
0
0
0
LT=
prior period
1
Net
115
Requirements
Time-phased Net
115
Q >= Requirements
50
Planned order
115
releases
Planned order
115
delivery
d. (2 points) Consider Item C. Find the planned order releases and on-hand units in period 10.
Period
1
2
3
4
5
6
7
8
9
10
Item Gross
10
460 160
Requirements
C
Scheduled
receipts
On hand from
200 200 200 200 190 190
30
70
70
70
LT=
prior period
1
Net
270 130
requirements
Time-phased Net
270 200
Q=
Requirements
100
Planned order
300 200
releases
Planned order
300 200
delivery
14
Name:_________________________________________________
ID:_________________________
e. (2 points) Consider Item D. Find the planned order releases and on-hand units in period 10.
Period
1
2
3
4
5
6
7
8
9
10
Item Gross
575
Requirements
D
Scheduled
150
receipts
On hand from
150 150 150 150 150
75
75
75
75
LT=
prior period
2
Net
425
requirements
Time-phased Net
425
Q=
Requirements
100
Planned order
500
releases
Planned order
500
delivery
Question 10: (5 points)
A single inventory item is ordered from an outside supplier. The anticipated demand for this item
over the next 7 months is 13, 11, 14, 13, 7, 8, 5. Current inventory of this item is 3, and the ending
inventory should be 4. Assume a holding cost of $2 per unit per month and a setup cost of $75.
Assume a zero lead time. Determine the order policy for this item over the next 7 months.
Use the Least Unit Cost heuristic.
Net requirements: r1  13  3  10, r2  11, r3  14, r4  13, r5  7, r6  8, r7  5  4  9
Months
1 to 1
1 to 2
1 to 3
1 to 4
4 to 4
4 to 5
4 to 6
4 to 7
7 to 7
Q
10
21
35
48
13
20
28
37
9
I1
I2
11
25
38
14
27
7
15
24
8
17
I3
I4
I5
I7
Holding
Cost
13
22
78
156
9
14
46
100
a. (3 points) State your order policy:
Month
1
4
7
I6
Lot size to order
35
28
9
15
Ordering
Cost
75
75
75
75
75
75
75
75
75
7.5
4.62
4.37
4.81 stop
5.77
4.45
4.32
4.73 stop
8.3333
Name:_________________________________________________
ID:_________________________
b. (2 points) Using the table below, show the ending inventory that results from your order policy at
the end of each month:
Month
1
2
3
4
5
6
7
Gross Requirements
13
11
14
13
7
8
5
Beginning Inventory
3
25
14
0
15
8
0
Net Requirements
10
13
5
Time-phased Net Requirements
10
13
5
Planned order Release
35
28
9
Planned Deliveries
35
28
9
Ending Inventory
25
14
0
15
8
0
4
Question 11: (9 points)
Consider Question 10 again. The question is re-written below:
A single inventory item is ordered from an outside supplier. The anticipated demand for this item
over the next 7 months is 13, 11, 14, 13, 7, 8, 5. Current inventory of this item is 3, and the ending
inventory should be 4. Assume a holding cost of $2 per unit per month and a setup cost of $75.
Assume a zero lead time. Determine the order policy for this item over the next 7 months.
a 
(3 points) Suppose that the maximum order size is 12 per month. Does there exist a feasible
solution? If there does not exist a feasible solution, what is first month when there will be a shortage?
Month
Production
Requirement
Capacity
1
2
3
4
5
6
7
13-3=10
11
14
13
7
8
5+4=9
12
12
12
12
12
12
12
Cumulative
Production
Requirement
10
21
35
48
55
63
72
Cumulative capacity
<= 12
<= 24
<= 36
<= 48
<= 60
<= 72
<= 84
Yes, there is a feasible solution.
Consider Question 10 again. The question is re-written below:
A single inventory item is ordered from an outside supplier. The anticipated demand for this item
over the next 7 months is 13, 11, 14, 13, 7, 8, 5. Current inventory of this item is 3, and the ending
inventory should be 4. Assume a holding cost of $2 per unit per month and a setup cost of $75.
Assume a zero lead time. Determine the order policy for this item over the next 7 months.
(b) (3 points) Suppose that the maximum order size is 12 per month. Use lot-shifting technique to
obtain a feasible solution (without using holding and setup cost). Show your final solution in the table
given below.
16
Name:_________________________________________________
Month
1
2
3
4
5
6
7
Production
Requirement
13-3=10
11
14
13
7
8
5+4=9
Actual Production
ID:_________________________
Production
Capacity
Excess Capacity
10 11 12
11 12
12
12
7
8
9
A feasible solution obtained by lot-shifting technique:
Month
1
2
3
4
Actual Production
12
12
12
12
5
7
6
8
7
9
Consider Question 10 again. The question is re-written below:
A single inventory item is ordered from an outside supplier. The anticipated demand for this item
over the next 7 months is 13, 11, 14, 13, 7, 8, 5. Current inventory of this item is 3, and the ending
inventory should be 4. Assume a holding cost of $2 per unit per month and a setup cost of $75.
Assume a zero lead time. Determine the order policy for this item over the next 7 months.
(c ) (3 points) Improve the solution obtained in Part (b) . Assuming a maximum order size of 12 units
per month and using the back-shifting technique, find another solution that has less total holding and
setup cost than the solution obtained in Part (b) . Show your final solution in the table given below.
Back-shift 9 units of Month 7?
Check if it’s better to backshift 4 units to Month 6 and 5 units to Month 5
Additional holding cost = 4(1)(2)+5(2)(2) = 28 < 75 = savings in ordering cost
Back-shift
Improved solution:
Month
Actual Production
1
12
2
12
3
12
4
12
17
5
12
6
12
7
--