Last Name _________________________ First Name _________________________ ID _______________________
Operations Management I 73-331-91 (Distance)
Winter 2003 Final Exam Solution
Tuesday, April 15, 8:30 – 11:30 a.m.
Instructor: Mohammed Fazle Baki
Aids Permitted: Calculator, straightedge, and 3 one-sided formula sheets.
Time available: 3 hours
Instructions:
 This exam has 32 pages including this cover page, 1 blank page and 8 pages of Table
 It’s not necessary to return the tables and unused blank pages
 Please be sure to put your name and student ID number on each odd numbered page
 Show your work
Grading:
Question
Score
Question
Score
1
/15
2
/8
3
/15
4
/10
5
/12
6
/6
7
/6
8
/8
9
/8
10
/4
11
/8
Total
/100
Name:_________________________________________________
ID:_________________________
Question 1: (15 points) Circle the most appropriate answer
1.1 Jidoka
a. prevents defects
b. lights signal quality problem
c. is the authority to stop production line
d. makes problems visible
1.2 Which of the following is false about lot sizing heuristic procedures?
a. Lot for lot minimizes carrying cost
b. Lot for lot maximizes ordering cost
c. Silver-Meal heuristic and least unit cost perform best if the costs change over time
d. Silver-Meal heuristic and least unit cost perform best if the costs do not change over
time
1.3 Material requirements planning is used for
a. dependent demand and for assembly
b. dependent demand and for both assembly and manufacturing
c. independent demand and for assembly
d. independent demand and for both assembly and manufacturing
1.4 We find an optimal Q, R  policy subject to some service constraint because it’s difficult to
a. estimate a penalty cost, p
b. estimate lead time
c. find an optimal Q without any service constraint
d. find an optimal R without any service constraint
1.5 Backorder is charged if
a. production is initiated by customer order
b. production is initiated by demand forecast
c. the excess demand is lost because the customer goes elsewhere
d. the excess demand is backlogged and fulfilled in a future period
1.6 A rotation cycle policy is used
a. to solve the lot sizing problem with capacity constraint
b. if multiple products are manufactured using the same production facility
c. to minimize hiring and firing when seasonal products are manufactured
d. if the unit production time decreases due to learning and experience
1.7 Consider the all unit discount schedule. If a larger quantity is ordered,
a. both the unit cost and total cost may decrease
b. unit cost may decrease but the total cost increases
c. both the unit cost and total cost may increase
d. total cost may decrease but the unit cost increases
1.8 The following assumption of the EOQ model is not used in the finite production rate model:
a. Demand and cost parameters are known and fixed.
b. Shortages are not permitted.
c. The inventory increases instantaneously at one point of time when an order is received.
d. There is no price discount or resource constraint
2
Name:_________________________________________________
ID:_________________________
1.9 The following are some characteristics of the EOQ model cost curve
a. Total annual holding cost is the same as total annual ordering cost
b. The total annual cost does not change much from the optimal value if the order quantity is
near the EOQ value
c. Both of the above
d. None of the above
1.10 Consider a seasonal demand series with an increasing trend. Compare multiplicative and
additive forecasting series.
a. There will be more fluctuation in the multiplicative series in the later years.
b. There will be more fluctuation in the additive series in the later years.
c. Both series will demonstrate the same fluctuation every year.
d. Any of a, b, or c can be true.
1.11
a.
b.
c.
d.
Which of the following is not a part of the firing cost?
Severance pay
The costs of a decline in worker morale
The potential for decreasing the size of the labor pool in the future
The time and cost to advertise positions
1.12
a.
b.
c.
d.
The coefficient of correlation, r
varies from -100 to 100
varies from -1 to +1
demonstrates a weak or no relationship if r is too small
demonstrates a weak or no relationship if r is too large
1.13
a.
b.
c.
d.
Which of the following is false?
Forecasts are usually wrong
Long-term forecasts are less accurate than short-term forecasts
Aggregate forecasts are less accurate than disaggregate forecasts
Aggregate forecasts are more accurate than disaggregate forecasts
Consider learning curve. Consider the straight line that best fits ln u , ln Y u  points. Which of the
following is true?
1.14
a.
b.
c.
d.
1.15
a.
b.
c.
d.
The intercept, c is an estimate of the time required by the first unit.
The log of intercept, ln c  is an estimate of the time required by the first unit.
If the slope is less, then the learning is faster.
If the slope is more, then the learning is faster.
Economies of scale can be represented as f  y   kya where, k is a constant and
a 1
a 1
a 1
a0
3
Name:_________________________________________________
ID:_________________________
Question 2: (8 points)
A start-up firm has kept careful records of the time required to manufacture its product, a shutoff
valve used in gasoline pipelines.
Cumulative Number of Units Produced
Number of Hours Required for Next Units
5
26.2
15
14.9
25
11.5
75
6.5
a. (2 points) Compute the logarithms of the numbers in each column.
Cumulative Number of
Number of Hours Required
Ln(u)
Units Produced
For the Next Unit
u
Y(u)
5
15
25
75
26.2
14.9
11.5
6.5
1.60943791
2.7080502
3.21887582
4.31748811
Ln(Y(u))
3.265759411
2.701361213
2.442347035
1.871802177
b. (4 points) Estimate the time required to produce the first unit and the appropriate percentage
learning curve that fits these data. Use an exact least squares fit of the logarithms computed in
part a.
i
x
y
xy
x^2
ln(u)
ln(y(u)
1
1.609437912
3.26575941
5.256037009
2.590290394
2
2.708050201
2.70136121
7.315421776
7.333535892
3
3.218875825
2.44234704
7.861611828
10.36116158
4
4.317488114
1.87180218
8.08148365
18.64070361
Sum
11.85385205
10.2812698
28.51455426
38.92569147
Average
2.963463013
2.57031746
n
 n  n

n xi y i    xi   y i 
 i 1  i 1   4(28.5145)  (11.8539)(10.2813)  0.5145
Slope  i 1
2
n
4(38.9257)  (11.8539) 2
 n 
n xi2    xi 
i 1
 i 1 
Intercept  y  slope( x)  2.5703  (0.5145)( 2.9635)  4.0950
Time required to produce the first unit, a  e intercept  e 4.0950  60.0372 hours
Rate of learning, L  e slope.ln2   e 0..5145ln2   0.700045  70%
c. (2 points) Estimate the time required to produce 130 th unit.
b = -slope = 0.5145
Y (u)  au b  60.0372(130) 0.5145  4.9072 hours
4
Name:_________________________________________________
ID:_________________________
Question 3: (15 points)
A popular brand of tennis shoe has had the following demand history by quarters over a two-year
period.
Quarter
Demand
Quarter
2001
Demand
2002
1
35
1
47
2
50
2
58
3
55
3
64
4
45
4
56
a. (6 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=
Period
Demand
MA(4)
Centered
(B/D)
MA
Ratio
A
B
C
D
E
1
35
49
0.714285714
2
50
49
1.020408163
3
55
47.75
1.151832461
4
45
46.25
50.25
0.895522388
5
47
49.25
52.375
0.897374702
6
58
51.25
54.875
1.056947608
7
64
53.5
53.625
1.193473193
8
56
56.25
53.625
1.044289044
Period
Seasonal
Factors
1
2
3
4
Total
0.80583021
1.03867789
1.17265283
0.96990572
3.98706664
Final
Seasonal
Factors
0.8084
1.0420
1.1765
0.9731
4.0000
(Continued…)
5
Name:_________________________________________________
ID:_________________________
b. (6 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
y
Sum
Average
1
2
3
4
5
6
7
8
36
4.5
43.29303212
47.98247238
46.75055139
46.24624736
58.13635741
55.65966796
54.40064162
57.5508856
410.0198558
51.25248198
43.29303212
95.96494476
140.2516542
184.9849894
290.6817871
333.9580078
380.8044913
460.4070848
1930.345991
1
4
9
16
25
36
49
64
204
n
 n  n

n xi y i    xi   y i 
 i 1  i 1   8(1930.3460)  36(410.0199)  2.0299
Slope  i 1
2
n
8(204)  (36) 2
 n 
2
n xi    xi 
i 1
 i 1 
Intercept  y  slope( x)  51.2525  (2.0299)( 4.5)  42.1178
c. (3 points) Predict the demand for the third quarter of 2003
Deseasonalized demand, y  42.1178  2.0299 x
For the third quarter of 2002, x  11.
So, the deseasonalized demand, y  42.1178  2.0299  11  64.4470
To get the demand, reseasonalize, y  64.4470  1.1765  75.8191
Question 4: (10 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
1
(in thousands of gallons)
600
2
700
3
650
4
200
6
Name:_________________________________________________
ID:_________________________
Assume that there are currently 250 employees with the company. Employees are hired for at least
one full quarter. Hiring costs amount to $500 per employee and firing costs are $1,000 per
employee. Inventory costs are $0.20 per gallon per quarter. It is estimated that one worker produces
1,750 gallons of paint each quarter. Assume that Paris currently has 300,000 gallons of paint in
inventory and would like to end the year with an inventory of at least 100,000 gallons.
a. (5 points) Determine the minimum constant workforce plan (i.e., level strategy) for Paris Paint.
Assume that stock-outs are not allowed.
Quarter Forecast Beg/ End Production
(000
Inventory Requirement
gallons)
(000
(000 gallons)
gallons)
A
1
2
3
4
B
600
700
650
200
C
300
100
Cumulative
Production
Requirement
(000 gallons)
Cumulative
units
produced
per worker
(000 gallons)
D
E
F
600-300=300
300
1.75 (given)
700
300+700=1000 1.75+1.75 = 3.5
650
1000+650=1650 3.50+1.75 = 5.25
200+100=300 1650+300=1950 5.25+1.75 = 7
E/F
Workers
Required
= Round
up (G)
G
171.4
285.7
314.3
278.6
H
172
286
315
279
Since the maximum workers required is 315, the minimum constant workforce plan must use 315
workers. So, the number of workers to hire = 315 – 250 = 65 workers.
b. (5 points) Determine the hiring, firing, and inventory holding cost of the plan derived in part a.
Demand
= Forecast
(000
gallons)
Beginning
Inventory
1
600
300 (given)
315(1.75)=551.25
551.25 + 300.00 – 600 = 251.25
2
700
251.25
551.25
551.25 + 251.25 – 700 = 102.50
3
650
102.50
551.25
551.25 + 102.50 – 650 = 003.75
4
200
003.75
551.25
551.25 + 003.75 – 200 = 355.00
Quarter
Production
(000 gallons)
(000
gallons)
Ending Inventory = Production +
Beginning Inventory – Demand
(000 gallons)
Total ending inventory = (251.25+102.50+3.75+355.00) = 712.50 thousand gallons
Inventory holding cost = 712,500  0.20 = $142,500
Hiring cost = 65(500) = $32,500
Total cost = 142,500+32,500 = $175,000
7
Name:_________________________________________________
ID:_________________________
Question 5: (12 points)
Suppose that Item A has a production rate of 720 items per year, unit cost of $10.00, a setup cost of
$80, and a monthly demand of 30 units. It is estimated that cost of capital is approximately 15
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.
2 K
2 K
2 K
28030  12
EPQ=



h'
 
 
0.15  0.03  0.02  101  30  12 
h1  
Ic1  
720 
 P
 P

280360
280360
57,600


 240 units
20.5
1
 360 
0.20  101  
 720 
b. (3 points) What are the uptime, downtime and cycle time of Item A?
Q * EPQ
240
Cycle time, T 


 0.6667 years


12  30
Q * 240
Uptime, T1 

 0.3333 years
P 720
Downtime, T2  T  T1  0.6667  0.3333  0.3334 years

Item B has a production rate of 1200 items per year, a unit cost of $20.00, an ordering cost of $77.5,
and a monthly demand of 25 units. Recall that the cost of capital is approximately 15 percent per
year. Storage cost amounts to 3 percent and breakage to 2 percent of the value of each item.
c. (3 points) What is the cycle time if both Items A and B are produced in a single facility?
T* 

2 K j
 h'
j
j

2K1  K 2 

h'1 1  h' 2 2
280  77.5
 
 360 
21 
360  Ic 2 1  2
 720 
 P2

2


280  77.5
  
 
h1 1  1 1  h2 1  2
 P1 
 P2

2

2157.5
 25  12 
360  0.20201 
25  12
1,200 

315
315
315


 0.25  0.50 years
360  4  0.75  300
360  900
1,260
d. (4 points) What are the optimal order quantity and uptime of items A and B? Assume both items
A and B are produced using the same facility.
Item A

QA*   AT *  30  120.5  360  0.50  180 units
Uptime = Q A* / P  180 / 720  0.25 year
Item B
QB*  BT *  25  120.5  300  0.50  150 units
Uptime = QA* / P  150 / 1200  0.125 year
8
Name:_________________________________________________
ID:_________________________
Question 6: (6 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 $225. The demand is normally distributed with a
mean of 1500 and a standard deviation of 100. Find the optimal order quantity if
a. (3 points) any lawn mower unsold at the end of summer season are marked down to $100 and
sold in a special fall sale.
C o  Purchase price – salvage value = $150-100=$50
Cu  Selling price – Purchase price = $225-150=$75
p
Cu
75

 0.60
Cu  Co 75  50




Find z * such that P    z  z *  0.60 Or, P   z  0  P 0  z  z *  0.60


Or, P 0  z  z *  0.60  P   z  0  0.60  0.5  0.10
Hence, from Table A-1 z *  0.25 (the z -value for which area = 0.10)
Q *    z *  1,500  0.25  100  1,525 units
b. (3 points) any lawn mower unsold at the end of summer season are marked down to $25 and
sold in a special fall sale.
C o  Purchase price – salvage value = $150-25=$125
Cu  Selling price – Purchase price = $225-150=$75
p
Cu
75

 0.375
Cu  Co 75  125




Find z * such that P    z  z *  0.375 Or, P   z  0  P 0  z  z *  0.375


Or, P 0  z  z *  P   z  0  0.375  0.5  0.375  0.125
Hence, from Table A-1 z *  0.32 (the z -value for which area = 0.125)
Q *    z *  1,500  0.32  100  1,468 units
Question 7: (6 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 240 radios each year. The annual demand follows a normal distribution with a standard
deviation of 50. The store pays $60 for each radio. The holding cost is 25 percent per year. Fixed
costs of replenishment amount to $200. If a customer demands the radio when it is out of stock, the
customer will generally go elsewhere. Replenishment lead-time is one month.
9
Name:_________________________________________________
ID:_________________________
a. (3 points) Find an optimal (Q,R) policy with probability(no stockout)=0.94.
Step 1: Q  EOQ 
2 K

h
2 K

Ic
2200240
96,000

 6,400  80 units
0.2560
15
Step 2: Find z for which area on the left = F z    =probability(no stockout) = 0.94
From Table A-1, z  1.555 for area = 0.94-0.50 = 0.44
From Table A-4, z  1.555 for F z   0.94
Step 3: Compute reorder point, R    z
    2401 / 12  20
   y   50 1 / 12  14.43
R    z  20  1.555  14.43  42.44 units
Hence an optimal policy is Q  80 units, R  42.44 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  60  80 15  80



 $600
2
2
2
2
Safety stock = R    42.44  2401/ 12  42.44  20  22.44 units
Annual holding cost, safety stock = hR     1522.44  $336.60
Annual holding cost =
hQ
 hR    = 600 + 336.60 = $936.60
2
c. (1 point) Compute the annual ordering cost resulting from the (Q,R) policy obtained in part a.
Annual ordering cost =
K 200  240

 $600
Q
80
Question 8: (8 points)
Consider Question 7 again. Find an optimal (Q,R) policy with fill rate = 0.98. Use the iterative method
and show 2 iterations. Show your computation and summarize your results in the table below:
h  Ic  0.25  60  15,  1 / 12  0.0833,     240  0.0833  20,    y   50 0.0833  14.43
Iteration 1
2k
2(200)(240)

 80 units
h
15
Step 2: n  Q(1   )  80(1  0.98)  1.60
n 1.60
L( z )  
 0.1108
 14.43
z  0.85 (Table A-4)
R    z  20  0.85  14.43  32.27
Step 3: 1  F ( z )  0.1977 (Table A-4)
Step 1: Q  EOQ 
(Continued…)
10
Name:_________________________________________________
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
ID:_________________________
200 Note: K and h
15
are input data
240 input data
0.083333333 input data
20 <--- computed
14.43375673 input data
0.98 input data
Iteration 1
Iteration 2
Q=
80
EOQ
n=
1.6
Q(1   )
L(z)=
0.1108513
n /
Table A1/A4, pp. 835 - 41
z=
0.85
  z
R=
32.268693
Table A1/A4, pp. 835 - 41
Area on the right=1-F(z)
0.1976625 0.214764
2
Modified Q= n /(1  F ( z ))  2 K / h  ( n /(1  F ( z )))
88.50308 88.66533
n=
1.7700616
1.773307
Q(1   )
L(z)=
0.1226335 0.122858
n /
Table
A1/A4,
pp.
835
41
z=
0.79
0.79
  z
R=
31.402668 31.40267
2
2
2
2
1.60
2(200)( 240)  1.60 
n
2 K 
n 


 
Step 4: Q 

 
  88.5014
0.1977
15
1  F( z)
h
 0.1977 
 1  F( z) 
(not near 80, more iterations are necessary)
Step 5: n  Q(1   )  88.5014(1  0.98)  1.7700
n 1.7700
L( z )  
 0.1226
 14.43
z  0.79 (Table A-4)
R    z  20  0.79  14.43  31.4027
Iteration 2
Step 3: 1  F ( z )  0.2148

1.7700
2(200)( 240)  1.60 
n
2 K 
n
 


Step 4: Q 

 
  88.6635
0.2148
15
1  F ( z)
h
 0.2148 
 1  F ( z) 
(same as before, stop the process after finding R )
Step 5: n  Q(1   )  88.6635(1  0.98)  1.7733
n 1.7733
L( z )  
 0.1229
 14.43
z  0.79 (Table A-4)
R    z  20  0.79  14.43  31.4027
Q and R converge. An optimal policy is Q=89, R=31 (rounded to the nearest integer)
11
Name:_________________________________________________
ID:_________________________
Question 9: (8 points)
Each unit of A is composed of two units of B and one unit of C. Items A, B and C have on-hand
inventories of 20, 30 and 10 units respectively. Item B has a scheduled receipt of 50 units in period
1, and C has a scheduled receipt of 100 units in Period 1. Lot-for-lot (L4L) is used for Item A. Item B
requires a minimum lot size of 50 units. Item C is required to be purchased in multiples of 100. Lead
times are two periods for Item A, and one period for each Item B and C. The gross requirements for
A are 30 in Period 3, 50 in Period 6, and 90 in Period 9. 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.
A
B(2)
C
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
50
90
Requirements
A
Scheduled
receipts
On hand from
20
20
20
0
0
0
0
0
0
0
LT=2 prior period
Net
10
50
90
requirements
Time-phased Net
10
50
90
Q=
Requirements
L4L
Planned order
10
50
90
releases
Planned order
10
50
90
delivery
(Continued…)
12
Name:_________________________________________________
ID:_________________________
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
20
100
180
Requirements
B
Scheduled
50
receipts
On hand from
30
60
60
60
10
10
10
0
0
0
LT=1 prior period
Net
40
170
Requirements
Time-phased Net
40
170
Q >= Requirements
50
Planned order
50
170
releases
Planned order
50
170
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
50
90
Requirements
C
Scheduled
100
receipts
On hand from
10
100 100 100
50
50
50
60
60
60
LT=
prior period
1
Net
40
requirements
Time-phased Net
40
Requirements
Q=
Planned order
100
100
releases
Planned order
100
delivery
13
Name:_________________________________________________
ID:_________________________
Question 10: (4 points)
A single inventory item is ordered from an outside supplier. The anticipated demand for this item
over the next 7 months is 10, 13, 11, 15, 14, 9, 6. Current inventory of this item is 2, and the ending
inventory should be 1. Assume a holding cost of $3 per unit per month and a setup cost of $100.
Assume a zero lead time. Determine the order policy for this item over the next 7 months.
Use the Silver-Meal heuristic.
Net requirements: r1  10  2  8, r2  13, r3  11, r4  15, r5  14, r6  9, r7  6  1  7
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
Q
8
21
32
47
15
29
38
45
I1
I2
I3
13
24
39
11
26
15
14
23
30
9
16
7
I4
I5
I6
I7
Holding
Cost
0
39
105
240
0
42
96
159
Ordering
Cost
100
100
100
100
100
100
100
100
100
69.5
68.33
85 stop
100
71
65.33
64.75
a. (3 points) State your order policy:
Month
a.
b.
Lot size to order
32
45
b. (1 point) 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
10
13
11
15
14
9
6
Beginning Inventory
2
24
11
0
30
16
7
Net Requirements
8
15
Time-phased Net Requirements
8
15
Planned order Release
32
45
Planned Deliveries
32
45
Ending Inventory
24
11
0
30
16
7
1
Question 11: (8 points)
Consider Question 10 again.
a 
(2 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?
14
Name:_________________________________________________
Month
Production
Requirement
Capacity
1
2
3
4
5
6
7
10-2=8
13
11
15
14
9
6+1=7
12
12
12
12
12
12
12
ID:_________________________
Cumulative
Production
Requirement
8
8+13=21
21+11=32
32+15=47
47+14=61
61+9=70
70+7=77
Cumulative capacity
< 12
< 12+12=24
< 24+12=36
< 36+12=48
> 48+12=60 Shortage
Since the cumulative capacity is less than the cumulative production requirement in period 5, there is
no feasible solution. There will be a shortage in Month 5.
(b) (3 points) Suppose that the maximum order size is 13 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.
Month
1
2
3
4
5
6
7
Production
Requirement
10-2=8
13
11
15
14
9
6+1=7
Actual Production
Production
Capacity
13
13
13
13
13
13
13
89
13
11 13
15 13
14 13
9
7
Excess Capacity
54
20
4
6
First, observe that production requirement in Month 4 is 2 units more than the capacity. So, backshift 2 units to Month 3 Again, production requirement in Month 5 is 1 unit more than the capacity.
So, back-shift 1 unit to Month 1. Final production schedule is as follows:
A feasible solution obtained by lot-shifting technique:
Month
1
2
3
4
Actual Production
9
13
13
13
5
13
6
9
7
7
(c ) (3 points) Improve the solution obtained in Part (b) . Assuming a maximum order size of 13 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 7 units of Month 7?
Check if it’s better to backshift 4 units to Month 6 and 3 units to Month 1
Additional holding cost = 4(1)(3)+3(6)(3) = 66 < 100 = savings in ordering cost
Back-shift
Improved solution:
Month
Actual Production
1
12
2
13
3
13
4
13
15
5
13
6
13
7
--