An Introduction to Management Science
(Ninth edition)
Chapter 9: Network Models
Problem 1:
Find the shortest route from node 1 to each of the other nodes in the transportation network
shown.
5 7 3 6 4 9 2 3 18 3Node Shortest route from
Node 1 Distance 2 1-2 7 3 1-3 9 4 1-2-5-6-4 17 5 1-2-5 12 6 1-2-5-6 14 7 1-2-5-6-7 17 Problem 2:
For the Gorman Construction Company problem (see Figure 9.1), assume that Node 7 is the company’s warehouse and supply center. Several daily trips are commonly made from node 7 to the other nodes or construction sites. Using node 7 as the starting node, find the shortest route from this node to each of the other nodes in the network.
Node Shortest route from
Node 7 Distance 1 7-6-5-3-1 22 2 7-4-2 11 3 7-6-5-3 12 4 7-4 5 5 7-6-5 8 6 7-6 6 Problem 3: In the original Gorman Construction Company problem, we found the shortest distance from the office (node 1) to each of the other nodes or construction sites. Because some of the roads are highways and others are city streets, the shortest-distance routes between the office and the construction sites may not necessarily provide the quickest or shortest-time routes. Shown here is the Gorman road network with travel time values rather than distance values. Find the shortest route from Gorman’s office to each of the construction sites if the objective is to minimize travel time rather than distance.
25
12 11
20 18
6
16 5
15
Problem 4:
Find the shortest route between nodes 1 and 8 in the following network: The shortest route between node 1 and 8:
1-4-5-6-8 3+3+2+2= distance: 10
Problem 5:
Find the shortest route between nodes 1 and 10 in the following network: 1-3-5-8-10 5+6+3+5= distance: 19
Chapter 10: Project scheduling: PERT/CPM
Problem 1:The Mohawk Discount Store is designing a management training program for individuals at its corporate headquarter. The company wants to design the program so that trainees can complete it as quickly as possible. Important precedence relationship must be maintained between assignments or activities in the program. For example, a trainee cannot serve as an assistant to the store manager until the trainee has obtained experience in the credit department and at least one sales department. The following activities are the assignments that must be completed by each trainee in the program. Construct a project network for this problem. Do not perform any further analysis.
Node 7 2 1-2 20 3 1-3 16 4 1-2-4 32 5 1-3-5 31 6 1-3-5-6 36 7 1-2-4-7 43
Activity A B C D E F G H Immediate
Predecessor
- - A A,B A,B C D,F E,G
C
F
A
D
G
Start
Finish
E
H
B
Problem 2:Bridge City Developer is coordinating the construction of an office complex. As part of the planning process, the company generated the following activity list. Draw a project network that can be used to assist in the scheduling of the project activities.
Activity A B C D E F G H I J
Immediate Predecessor
- - - A,B A,B D E C C F,G,H,I
A
D
F
Start
B
E
G
J Finish
H
C I
Problem 3:
Construct a project network for the following project. The project is completed when activities F and G are both completed.
Activity A B C D E F G Immediate Predecessor - - A A C,B C,B D,E
A C
E
D
Start
G
Finish
F
B
Problem 4:Assume that the project in Problem 3 has the following activity times (in months).
Activity A B C D E F G Immediate Predecessor 4 6 2 6 3 3 5 A 0 4 4 0 4
Start
C 4 6 2 5 7 E 6 9 3 7 10 D 4 10 6 4 10 B 0 6 6 1 7 G 5 10 15 10 15 F 6 9 3 12 15Finish
a) Find the critical path.
b) The project must be completed in and year and a half. Do you anticipate difficulty in meeting the deadline? Explain.
a) The critical path A-D-G
b) No, I don’t find a difficulty, because it will be 15 months.
Problem 5:
MDS is a consulting company that specializes in the development of decision support systems. MDS has just obtained a contract to develop a computer system to assist the management of a large company in formulating its capital expenditure plan. The project leader has developed the following list of activities and immediate predecessors. Construct a project network for this problem.
Activity A B C D E F G H I J Immediate Predecessor - - - B A B C,D B,E F,G H
A
E
H
Start
B
D
J Finish
F
I
C G
Chapter 14: Decision Analysis
Problem 1:The following payoff table shows profit for a decision analysis problem with two decisions and three states of natures.
b. If the decision maker knows nothing about the probabilities of the three states of nature, what is the recommended decision, using the optimistic, conservative, and minimax regret approaches?
Decision Alternative State of nature
S1 S2 S3 D1 250 100 25 D2 100 100 75 a. s1 250 D1 s2 100 s3 25 s1 100 s2 100 D2 s3 75 b.
Decision Maximum Profit Minimum profit
S1 S3
D1 250 25
D2 100 75
Optimistic approach (250 is the largest payoff) – Select D1
Conservative approach (75 is the best of the worst possible payoffs) – Select D2 Minimax regret approach D1 – 50; D2- 150; Select D1
Decision Maximum Profit Minimum profit
S1 S3
D1 0 50
Problem 2:
Suppose that a decision maker faced with four decision alternatives and four states of nature develops the following profit payoff table.
Decision Alternative State of nature
S1 S2 S3 S4
D1 14 9 10 5
D2 11 10 8 7
D3 9 10 10 11
D4 8 10 11 13
a. If the decision maker knows nothing about the probabilities of the four states of nature, what is the recommended decision using the optimistic, conservative, and minimax regret approaches? b. Which approach do you prefer? Explain. Is establishing the most appropriate approach before
analyzing the problem important for the decision maker? Explain.
c. Assume that the payoff table provides cost rather than profit payoffs. What is the recommended decision using the optimistic, conservative, and minimax regret approach?
a.
Optimistic approach select D1, 14 Conservative approach select D3, 9 Minimax regret approach select D3, 5
Decision Alternative State of nature
S1 S2 S3 S4
D1 0 1 1 8
D2 3 0 2 6
D3 5 0 1 2
D4 6 0 0 0
b. Yes, it is important to establish the most appropriate approach before analyzing, because in that way you can minimize the loses and maximize the profits. I prefer the minimax regret approach because like this the possibility to lose more is the smallest.
Conservative approach Select D2, D3, 11 Minimax regret approach Select D2
Problem 3:
Southland Corporation’s decision to produce a new line of recreational products has resulted in the need to construct either a small plant or a large plant. The selection of plant size depends on how the
marketplace reacts to the new product line. To conduct an analysis, marketing management has decided to view the possible long-run demand as either low, medium, or high. The following payoff table shows the projected profit in millions of dollars:
Decision Alternative Long-run demand
Low Medium High
Small plant 150 200 200
Large plant 50 200 500
a. Construct a decision tree for this problem
b. Recommend a decision based on the use of the optimistic, conservative, and minimax regret approaches. Low 150 a. Medium 200 Small plant High 200 Low 50 Large plant
Medium 200
High 500
b. Optimistic Approach Choose High long-run demand and Large plant 500 Conservative approach Choose Low long-run demand and Small plant 150 Minimax regret approach Choose Low long-run demand and Large plant 100
Decision Alternative Long-run demand
Low Medium High
Small plant 0 0 300
Large plant 100 0 0
Problem 4:
Investment Advisors, Inc. , has three investment strategies under consideration. Profits from the strategies will depend on what happens to the prime interest rate over the next three months. Payoffs (in thousands of dollars) are shown.
Decision Alternative State of nature
Rate decrease S1 No Change S2 Rate Increase S3
Strategy D1 50 70 40
Strategy D2 55 35 80
Strategy D3 15 60 70
What investment strategy would you recommend based on the use of optimistic, conservative and minimax regret approaches?
Optimistic Strategy D2, 80 Conservative Strategy D1, 40 Minimax regret Strategy D2, 35
Decision Alternative State of nature
Rate decrease S1 No Change S2 Rate Increase S3
Strategy D1 5 0 40
Strategy D2 0 35 0
Problem 5:
The following profit payoff table was presented in Problem 1. Suppose that the decision maker has obtained the probability estimates: P(s1) = 0.65, P(s2) = 0.15, and P(s3) = 0.20. Use the expected value approach to determine the optimal decision.
Decision Alternative State of nature
S1 S2 S3 D1 250 100 25 D2 100 100 75 EV (d1) = 0.65(250) +0.15(100) +0.20(25) = = 162.5 +15 +5 = = 182.5 EV (d2) = 0.65(100) +0.15(100) +0.20(75) = = 65+15+15 = = 95
The optimal decision is D1 with an expected value of 182.5!
Chapter 17: Markov Processes
Problem 1:
In the market share analysis of section 17.1, suppose that we are considering the Markov process associated with the shopping trips of one customer, but we don’t know where the customer shopped during the last week. Thus, we might assume a 0.5 probability that the customer shopped at Murphy’s and 0.5 probability that the customer shopped at Ashley’s at period 0; that is 1(0) = 0.5 and 2(0) = 0.5. Given these initial state probabilities, develop a table
similar to table 17.2 showing the probability of each state in future periods. What do you observe about the long-run probabilities of each state?
Current weekly shopping period
Next weekly shopping period
Murphy’s Food liner 0.9 0.1 Ashley’s Supermarket 0.2 0.8 [ 1(0) 2(0)] = [0.5 0.5] [ 1(1) 2(1)] = [ 1(0) 2(0)] [ ] = [0.5 0.5] [ ] = [0.55 0.45] [ 1(2) 2(2)] = [0.55 0.45] [ ] = [0.585 0.415] [ 1(3) 2(3)] = [0.585 0.415] [ ] = [0.609 0.39] [ 1(4) 2(4)] = [0.609 0.39] [ ] = [0.626 0.373] [ 1(5) 2(5)] = [0.626 0.373] [ ] = [0.638 0.361] [ 1(6) 2(6)] = [0.638 0.361] [ ] = [0.646 0.353] [ 1(7) 2(7)] = [0.646 0.353] [ ] = [0.652 0.347] State probability Period (n) 0 1 2 3 4 5 6 7 1(n) 0.5 0.55 0.585 0.609 0.626 0.638 0.646 0.652 2(n) 0.5 0.45 0.415 0.39 0.373 0.361 0.353 0.347 Problem 2:
Management of the New Fangled Softdrink Company believes that the probability of a customer purchasing Red Pop or the company’s major competition, Super Cola, is based on the customer’s most recent purchase. Suppose that the following transition probabilities are appropriate:
From To
Red Pop Super Cola
Red Pop 0.9 0.1
Super Cola 0.1 0.9
a. Show the two-period tree diagram for a customer who last purchased Red Pop. What is the probability that this customer purchases Red Pop on the second purchase?
b. What is the long-run market share for each of these two products?
c. A RP advertising campaign is being planned to increase the probability of attracting SC customers. Management believes that the new campaign will increase to 0.15 the probability of a customer switching from SC to RP. What’s the projected effect of the advertising campaign on the market shares?
Period 1 Period 2 a. Red Pop (0.9)(0.9) = 0.81 0.9 0.9 0.1 Super Cola (0.9)(0.1) = 0.09 0.1 0.1 Red Pop (0.1)(0.1) = 0.01 0.9 Super Cola (0.1)(0.9) = 0.09 Super Cola = 0.18
Red Pop = 0.81 + 0.01 = 0.82 is the probability the customer purchases Red Pop on the second
Purchase! Customer who last purchased Red Pop Shops at Red Pop Shops at Super Cola
b. *π1(0) π2(0)] = [0.5 0.5] [ 1(1) 2(1)] = [ 1(0) 2(0)] [ ] = [0.5 0.5] [ ] = [0.5 0.5] [ 1(2) 2(2)] = [0.5 0.5]
So, the long-run market share is π1 = 0.5, π2 = 0.5!
c. 1 = 0.9 1 + 0.15 2 2 = 0.1 1 + 0.9 2 1 2 = 1 1 = 0.9 1 + 0.15 1 = 0.9 π1 + 0.15 – 0.15π1 1 = 0.75π1 +0.15 1 = 0.15 1 = 0.6 2 = 1 – π1 2 = 1 - 0.6 2 = 0.4 Problem 3:
a. [ 1(0) 2(0)] = [1 0]
[ 1(1) 2(1)] = [ 1(0) 2(0)] [ ]
= [1 0] [ ] = [0.9 0.10]
The probability of the system being down in the next our of operation: 2(1) = 0.10 b. 1 = 0.9 1 + 0.3 2 2 = 0.10 1 + 0.70 2 1 2 = 1 1 = 0.90 1 + 0.30 1 = 0.90 π1 + 0.30 – 0.30π1 1 = 0.60π1 +0.30 1 = 0.30 1 = 0.75 2 = 1 – π1
2 = 1 - 0.75 2 = 0.25 Problem 4: a. 1 = 0.95 1 + 0.60 2 2 = 0.05 1 + 0.40 2 1 2 = 1 1 = 0.95 1 + 0.60 1 = 0.95 π1 + 0.60 – 0.60π1 1 = 0.35π1 +0.60 1 = 0.60
1 = 0.92 is the steady-state probability of the system being in the running state.
2 = 1 – π1 2 = 1 - 0.92
Problem 5:
a.
From To
No traffic delay Traffic delay
No traffic delay 0.85 0.15 Traffic delay 0.25 0.75 [ 1(0) 2(0)] = [0.5 0.5] [ 1(1) 2(1)] = [ 1(0) 2(0)] [ ] = [0.5 0.5] [ ] = [0.55 0.45]
[ 1(2) 2(2)] = [0.55 0.45] [ ]
= [0.58 0.42]
2(2) = 0.42
The probability for the next 60 minutes the system will be in the delay state is 0.42!
b. 1 = 0.85 1 + 0.25 2 2 = 0.15 1 + 0.75 2 1 2 = 1 1 = 0.85 1 + 0.25 1 = 0.85 π1 + 0.25 – 0.25π1 1 = 0.6π1 +0.25 1 = 0.25
1 = 0.625 – That is the probability that in the long run the traffic will not be in the delay state!
2 = 1 – π1 2 = 1 – 0.625
Chapter 18: Dynamic Programming
Problem 1: Route Value 1-4-5-8-10 22 1-4-5-9-10 25 1-4-6-8-10 27 1-4-6-9-10 23 1-3-5-8-10 19 1-3-5-9-10 22 1-3-6-8-10 26 1-3-6-9-10 22 1-3-7-8-10 22 1-3-7-9-10 21 1-2-5-8-10 22 1-2-5-9-10 25 1-2-6-8-10 24 1-2-6-9-10 20 1-2-7-8-10 251-2-7-9-10 24
The dynamic programming reduces the number of computation by decomposing the large problem into a number of interrelated smaller problems, which are easier to solve.
Problem 2:
a.
Stage 1:
Input Node Arc (decision) Shortest distance to Node 10
7 7-10 8
8 8-10 10
9 9-10 6
Stage 2:
Input Node Arc (decision) Shortest distance to Node 10
5 5-9 11
Stage 3:
Input Node Arc (decision) Shortest distance to Node 10
2 2-5 18 3 3-5 19 4 4-6 21 Stage 4: 1-2 10 1-3 8 1-4 5
So the shortest route from node 1 to node 10 is 1-4-6-9-10 with a distance of (5+4+11+6) = 26. b. In the same way it is: 4-6-9-10 with a distance of (4+11+6) = 21.
c. Route Value 1-2-5-7-10 32 1-2-5-8-10 36 1-2-5-9-10 28 1-3-5-7-10 31 1-3-5-8-10 35 1-3-5-9-10 27 1-3-6-8-10 34 1-3-6-9-10 31 1-4-6-8-10 29 1-4-6-9-10 26
Dynamic programming reduces the number of computation by decomposing the large problem into a number of interrelated smaller problems, which are easier to solve.
