Sample Final Exam
The breakdown of the final exam will be:
Chapter
11. Supply Chain Management
12. Inventory Management
13. Aggregate Planning
14. Material Requirements Planning
15. Short Term Scheduling
16. Project Management
17. Maintenance and Reliability
B. Linear Programming
C. Transportation Problem
D. Waiting Lines
Multiple Choice
10
Long Answer
10
10
10
10
10
10
10
10
10
Formulae
M/M/1 Queue
Average number of customers in the system LS =

 
Average time a customer spends in the system WS =
1
 
2
    

Average time a customer spends waiting in the queue Wq =
    

Utilization factor for the system  =


Probability of 0 customers in the system P0 = 1 
Average number of customers waiting in the queue Lq =

Probability of more than k customers in the system Pk =  

k 1
M/M/S Queue
Probability of 0 customers in the system P0 =
1
M 1 1   
   
 n 0 n!   
Average number of customers in the system LS =
n
 1    M M
 

 M!    M  
   
 
for M > 
M
M  1! M   
2
P0 


   
M
Average time a customer spends waiting in the system WS =
Average number of customers waiting in the queue Lq = LS -
M  1! M   


Average time a customer spends waiting in the queue Wq = WS -
1 Lq
=
 
M/D/1 Queue
2
2    

Average time a customer spends waiting in the queue Wq =
2    

Average number of customers in the system LS = Lq +

Average number of customers waiting in the queue Lq =
Average time a customer spends waiting in the system WS = Wq +
1

2
P0 
1

=
LS

Limited Population Queue
T = average service time
U = average time between unit service requirements
M = number of service channels
X = service factor – look up in table D.7 page 818
D = probability unit will have to wait in queue
F = efficiency factor
H = average number of units being served
J = average number of units not in the queue or in service ie units working
L= average number of units waiting for service
Service Factor X =
T
T U
Average number waiting L = N(1 – F)
Average waiting time W =
LT  U  T 1  F 
=
N L
XF
Average number running J = NF(1 – X)
Average number being serviced H= FNX
Economic Order Quantity
Q* =
2DS
H
Expected Number of Orders N = D / Q*
Expected time between orders T = Days per year / N
Total Annual Cost TC = (D / Q) S + (Q / 2) H
Demand per day d = D / Days per year
Reorder Point ROP = dL
Economic Order Quantity With Production
Setup Cost = (D / Q) S
Holding Cost = ½ HQ[ 1 – d/p ]
Q* =
2DS
H 1  d / p 
Economic Order Quantity with Quantity Discount
Q* =
2DS
IP
Multiple Choice
1. In a retail environment, the purchasing function is usually managed by
a) an expediter.
b) a buyer.
c) a purchasing agent.
d) the keiretsu.
2. Which of the following is not a purchasing strategy?
a) negotiation with many suppliers.
b) short-term relationships with few suppliers.
c) vertical integration.
d) keiretsu.
3. Which of the following best describes how short-term schedules are prepared?
a. directly from the aggregate plans.
b. directly from the capacity plans.
c. from master schedules which are derived from aggregate plans.
d. from inventory records for items that have been used up.
4. Orders are processed in the sequence in which they arrive if the _______ rule sequences the jobs.
a. earliest due date
b. first come, first served
c. slack time remaining
d. critical ratio
5. Which of the following statements regarding Gantt charts is true?
a. Gantt charts are visual devices that show the duration of activities in the project.
b. Gantt charts give a timeline and precedence relationships for each activity of a project.
c. Gantt charts use the four standard spines of Methods, Materials, Manpower and Machinery.
d. Gantt charts are expensive.
6. Which of these statements regarding time-cost tradeoffs in CPM networks is true?
a. Crashing is not possible unless there are multiple critical paths.
b. Crashing a project often reduces the length of long-duration, but non-critical activities.
c. Activities not on the critical path can never be on the critical path, even after crashing.
d. Crashing shortens the project duration by assigning more resources to one or more of the critical tasks.
7. The objective of maintenance is to
a. ensure that no breakdowns occur.
b. ensure that preventative maintenance costs are kept as low as possible.
c. maintain the capability of the system while controlling costs.
d. ensure that maintenance employees are fully utilized.
8. As the non-redundant components in a system decreases, all other things being equal, the reliability of the
system usually
a. increases.
b. stays the same.
c. decreases.
d. increases, then decreases.
Long Answer / Calculation
1. A craftsman builds two kinds of birdhouses, one for wrens and one for bluebirds. Each wren birdhouse takes
four hours of labour and four units of lumber. Each bluebird house requires two hours of labour and twelve units
of lumber. The craftsman has available 60 hours of labour and 120 units of lumber. Wren houses profit $6 each
and bluebird houses profit $15 each. Write out the objectives and constraints. Solve by corner points.
2. The Great Canadian Piano company manufactures grand pianos at five factories across Canada, and they
operate two distribution centers. The tables below show quantities supplied in one month by each factory,
quantities demanded in one month by each distribution center, and the transportation costs between each
facility.
a) Use the Northwest Corner Rule to find an initial solution.
b) Find the optimal transportation solution.
c) What is the cost of your solution?
Quantities Supplied
by Factories
Factory 1: 20 pianos
Factory 2: 30 pianos
Factory 3: 15 pianos
Factory 4: 20 pianos
Factory 5: 25 pianos
Quantities Required by
Distribution Centers
Center 1: 70 pianos
Center 2: 40 pianos
Transportation Costs
DC 1
Factory 1
5
Factory 2
8
Factory 3
4
Factory 4
12
Factory 5
7
DC 2
9
2
5
11
7
3. A crew of mechanics at the Highway Department garage repair vehicles that break down at an average of  =
5 vehicles per day (approximately Poisson in nature). The mechanic crew can service an average of  = 10
vehicles per day with a repair time distribution that approximates an exponential distribution.
a) What is the utilization rate for this service system?
b) What is the average time before the facility can return a breakdown to service?
c) How much of that time is spent waiting for service?
d) How many vehicles are likely to be in the queue at any one time?
4. Refer to question 3 above. How would your answers to parts b, c, and d change if there were two crews?
5. At the order fulfillment center of a major mail-order firm, customer orders, already packaged for shipment,
arrive at the sorting machines to be sorted for loading onto the appropriate truck for the parcel’s address. The
arrival rate at the sorting machines is at the rate of 150 per hour following a Poisson distribution. The machine
sorts at the constant rate of 160 per hour.
a) What is the average number of packages waiting to be sorted?
b) What is the average number of packages in the sorting system?
c) How long must the average package wait until it gets sorted?
6. A printing company estimates that it will require 750 reams of a certain type of paper in a given period. The
cost of carrying on e unit in inventory for that period is 40 cents. The company buys from a wholesaler in town,
sending its own truck to pick up the orders at a fixed cost of $16.00 per trip. Treating this cost as the order cost,
what is the optimum number of reams to buy at one time? How many times should lots of this size be bought
during this period? What is the minimum cost of maintaining inventory of this item for the period? Of this total
cost, how much is carrying cost and how much is ordering cost?
7. Kite Fabrication has the following aggregate demand requirements and other data for the upcoming four
quarters:
Quarter
Demand
1
1800
2
1100
Previous quarter’s output:
Beginning inventory:
Stockout cost:
Inventory holding cost:
Hiring workers:
Firing workers:
Subcontracting cost:
Unit cost:
Overtime:
3
1600
4
900
1300 units
0 units
$150 per unit
$40 per unit at end of quarter
$40 per unit
$80 per unit
$60 per unit
$30 per unit
$15 extra per unit
Which of the following production plans is better: Plan A-chase demand by hiring and firing; Plan B-pure level
strategy; or plan C-1200 level with the remainder by subcontracting? Explain how the relative costs of storage,
shortage, overtime, subcontracting, layoffs, etc., influence the choice of strategy.
8. Consider the bill of materials for product J and the data given in the following table. The gross requirements
for J are 100 units in week 6 and 200 units in week 8. Develop the MRP tables for each item for an 8-week
planning period. Use the lot-for-lot lot sizing rule.
Item
Lead Time
J
K
L
M
1
2
2
1
J
K (1)
L (4)
M (2)
Quantity
on Hand
0
10
0
20
Scheduled
Receipts
30 in week 3
10 in week 4
Item J
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
Item K
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
Item M
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
Item L
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
Answers
Multiple Choice
1. b
2. b
3. c
4. b
5. a
6. d
7. c
8.
Long Answer / Calculation
1.
Your answer should include and identify all
of the following:
Let
The graph of this system shows four corner points:
X1 = Quantity of wren houses
X2 = Quantity of bluebird houses
Objective Function:
Maximize Profit = $6(X1) + $15(X2)
Such That
Constraint 1: 4(X1) + 2(X2)  60
(labour)
Constraint 2: 4(X1) + 12(X2)  120
(lumber)
Constraint 3: X1  0
(non-negativity)
Constraint 4: X2  0
(non-negativity)
Point 1:
Point 2:
Point 3:
Point 4:
X1 = 0
4(X1) + 12(X2) = 120
4(0) + 12(X2) = 120
X2 = 120 / 12
X2 = 10
-4(X1) + -2(X2) = -60
4(X1) + 12(X2) = 120
0(X1) + 10(X2) = 60
X2 = 60 / 10
X2 = 6
4(X1) + 12(X2) = 120
4(X1) + 12(6) = 120
X1 = (120 – 72) / 4
X1 = 12
X2 = 0
4(X1) + 2(X2) = 60
4(X1) + 2(0) = 60
X1 = 60 / 4
X1 = 15
X1 = 0
X2 = 0
Profit = $6(X1) + $15(X2)
Profit = $6(0) + $15(10)
Profit = $150
Profit = $6(X1) + $15(X2)
Profit = $6(12) + $15(6)
Profit = $162
Profit = $6(X1) + $15(X2)
Profit = $6(15) + $15(0)
Profit = $90
Profit = $6(X1) + $15(X2)
Profit = $6(0) + $15(0)
Profit = $0
 The craftsman should produce 12 wren houses and 6 bluebird houses.
2. Quantity Demanded = Quantity Supplied, so we do not need a dummy factory or distribution center. We use
the Northwest Corner Rule to determine our initial solution:
DC 1
Supply
DC 2
Factory 1
5
9
8
2
4
5
12
11
20
20
Factory 2
30
30
Factory 3
15
15
Factory 4
5
Factory 5
Demand
20
15
7
7
25
25
40
70
Let’s evaluate the stepping stone method for the Factory 2 to DC 2 route:
DC 1
Supply
DC 2
Factory 1
5
9
8
2
4
5
12
11
20
20
Factory 2
30
30
Factory 3
15
15
Factory 4
5
Factory 5
Demand
20
15
7
7
25
25
40
70
Improvement Index = +2 – 8 + 12 – 11 = -5
This is not greater than or equal to zero, so we could improve our costs by using this route. We come up with
our next solution:
DC 1
Supply
DC 2
Factory 1
5
9
Factory 2
8
2
Factory 3
4
5
12
11
20
20
30
30
15
15
Factory 4
10
Factory 5
Demand
25
70
20
10
7
7
25
40
If we calculate the Improvement Index for each empty cell left in the solution, we find that all are greater than or
equal to zero: hence we now have the optimal solution.
Total cost of the solution = 20 ($5) + 30 ($2) + 15 ($4) + 10 ($12) + 10 ($11) + 25 ($7) = $625
3. This is a single channel system (1 crew) with Poisson arrivals and exponential service times: hence, we use
the M/M/1 model.
a) Utilization factor for the system  =


= 5 / 10 = 50%
b) Average time a vehicle spends in the system WS =
1
= 1 / (10 – 5) = 0.2 days.
 
c) Average time a vehicle spends waiting in the queue Wq =

d) Average number of customers in the system LS =
 

    
= 5 / 10(10 – 5) = 0.1 days.
= 5 / (10 – 5) = 1
4. This becomes a multiple channel system (2 crews): hence, we use the M/M/S model.
We first have to calculate P0:
P0 =
1
M 1 1    n  1    M M
 
     
 n 0 n!     M!    M  
   
M
b) WS =
M  1! M   
c) Wq = WS -
2
P0 
1

=
1
 5  5  2   5  2 2  10
    
10  10    10  2  10  5
10510 
2
=
1! 2  10  5
2
 0.92 
= 0.92
1
= 0.11 days.
10
1
= 0.11 – 1 / 10 = 0.01 days.

M
 
2
   
5
5

10


5


10
P0  =

0
.
92

d) LS =
= 0.55
2
 1! 2  10  52
10
M  1! M   
5. There is a constant service rate, and arrivals follow a Poisson distribution: Hence, we use the M/D/1 model.
 = 150,  = 160
a) Average number of packages waiting in the queue Lq =
b) Average number of packages in the system LS = Lq +


2
= 1502 / 2(160)(160-150) = 7.03
2    
= 7.03 + 150 / 160 = 7.97
c) Average time a package spends waiting in the queue Wq =
hours

= 150 / 2(160)(160 – 150) = 0.047
2    
6. Optimum number to buy at one time Q* =
2DS
=
H
2  750  16
= 245
0.40
Expected Number of Orders N = D / Q* = 750 / 245 = 3.06
Total Annual Cost TC = (D / Q) S + (Q / 2) H = (750 / 245) 16 + (245 / 2) 0.40 = $97.98
Portion that is ordering cost = (D / Q) S = (750 / 245) 16 = $48.98
Portion that is carrying cost = (Q / 2) H = (245 / 2) 0.40 = $49
7. Total Expected Demand: 1800 + 1100 + 1600 + 900 = 5400
Plan A – Chase – Notice how we use the previous quarter’s output of 1,300.
Quarter Forecast
0
1,300
1
1,800
Production Cost
Hiring Cost
1,800 x $30 = $54,000
500 x $40 =
$20,000
2
1,100
1,100 x $30 = $33,000
3
1,600
1,600 x $30 = $48,000
4
900
900 x $30 = $27,000
Layoff Cost
Total Cost
$74,000
700 x $80 =
$56,000
$89,000
500 x $40 =
$20,000
$68,000
700 x $80 =
$56,000
 = $112,000
 = $40,000
$83,000
 = $314,000
Total costs for Plan A are $314,000.
Plan B – Level – Notice how a backorder affects the next period.
Average Demand = Total Expected Demand / # Periods = 5,400 / 4 = 1,350
Quarter
1
2
3
4
Forecast
1,800
1,100
1,600
900
Production
1,350
1,350
1,350
1,350
 =5,400
Backordered
450
200
450
0
 =1,100
Inventory
0
0
0
0
=0
There is also a cost to hire the workers to move from the 1,300 production level to the 1,350 production level –
50 x $40 = 2,000
Total cost = 5,400 x $30 + 1,100 x $150 + $2,000 = $329,000
Plan C – Mixed – Note that we have to lay off 100 in quarter 1 to get to the 1200 production level
Quarter
1
2
3
4
Forecast
1,800
1,100
1,600
900
Production
1,200
1,100
1,200
900
 = 4,400
Subcontracting
600
Hiring
400
100
 = 1,000
 = 100
Total Cost = 4,400 x $30 + 1,000 x $60 + 100 x $40 + 500 x $80 = $236,000
Layoff
100
100
300
 = 500
Plan C (Mixed) is clearly the least expensive. In this problem, stockouts and storage are relatively expensive
compared to subcontract costs and hiring / firing costs. This leads to the low-cost solution that avoids inventory
– the chase strategy.
8.
Item J
1
2
3
4
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
5
6
7
8
100
200
100
200
100
200
Item K
1
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
2
3
4
5
6
100
10
10
10
40
7
8
7
8
7
8
200
40
0
0
30
60
60
200
200
Item M
1
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
2
3
4
5
6
200
20
20
20
20
400
30
0
0
10
170
400
170
400
Item L
1
Total
Required
On
Hand
Sched.
Receipt
Net
Required
Order
Release
240
2
3
4
5
240
800
240
800
800
6