MGMT 650 – Management Science and Decision Analysis
Final Preparation Guide
1. Key concepts of chapter 3
a. Sensitivity Analysis; Interpreting Management Scientist output reports
i. Range of optimality
ii. 100% rule
iii. Range of feasibility
iv. Dual price – what it means? How it can be used?
2. Key concepts of chapter 7
a. Transshipment problem and its Linear Programming formulation
3. Key concepts of chapter 8
a. Formulation of integer programs
b. Multiple choice constraints
c. K out of N alternatives constraints
d. Conditional and co-requisite constraints
4. Key concepts of chapter 9
a. Understanding of the shortest route problem
i. How can it be used for optimizing replacement policies (in-class example) and scheduling purposes (HW #5 problem)
5. Key concepts of chapter 10
a. Understanding of activity slacks, critical activities, critical path, how much activities can be delayed?
b. How is slack calculated?
c. Ability to interpret to Management Scientist outputs for project networks
d. Ability to set up project network diagrams with precedence relationships between activities
6. Key concepts of chapter 11
a. Understanding of the EOQ model
i. What dies the EOQ signify?
ii. What are holding, ordering cost? And their formulas.
iii. Total cost formula
iv. Cycle time, number of orders per year, etc.
v. Ability to interpret to Management Scientist outputs
7. Key concepts of chapter 16
a. Moving averages, Exponential smoothing – how do they work?
b. Trend analysis and use of the linear trend equation
c. Interpretation of seasonal indices
d. Ability to interpret to Management Scientist outputs
e. Ability to recognize different types of time series data
Please review all HW questions and solutions carefully.
Please make sure that you understand various Management Scientist outputs.
Sample Questions
1. (Chapter 3 – Sensitivity Analysis) Use the following Management Scientist output to answer the questions.
LINEAR PROGRAMMING PROBLEM
MAX 31X1+35X2+32X3
S.T.
1) 3X1+5X2+2X3>90
2) 6X1+7X2+8X3<150
3) 5X1+3X2+3X3<120
OPTIMAL SOLUTION
Objective Function Value = 763.333
Variable Value Reduced Costs
------
X1 13.333 0.000
X2 10.000 0.000
X3 0.000 10.889
Constraint Slack/Surplus Dual Prices
------
1 0.000 -0.778
2 0.000 5.556
3 23.333 0.000
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
------
X1 30.000 31.000 No Upper Limit
X2 No Lower Limit 35.000 36.167
X3 No Lower Limit 32.000 42.889
RIGHT HAND SIDE RANGES
Constraint Lower Limit Current Value Upper Limit
------
1 77.647 90.000 107.143
2 126.000 150.000 163.125
3 96.667 120.000 No Upper Limit
a. Give the solution to the problem.
b. Which constraints are binding?
c. What would happen if the coefficient of x1 increased by 3?
d. What would happen if the right-hand side of constraint 1 increased by 10?
Answers:
a. x1 = 13.33, x2 = 10, x3 = 0, s1 = 0, s2 = 0, s3 = 23.33, z = 763.33
b. Constraints 1 and 2 are binding.
c. The value of the objective function would increase by 40.
d. The value of the objective function would decrease by 7.78.
2. (Chapter 3 – Sensitivity Analysis) Use the following Management Scientist output to answer the questions.
MIN 4X1+5X2+6X3
S.T.
1) X1+X2+X3<85
2) 3X1+4X2+2X3>280
3) 2X1+4X2+4X3>320
Objective Function Value = 400.000
Variable Value Reduced Costs
------
X1 0.000 1.500
X2 80.000 0.000
X3 0.000 1.000
Constraint Slack/Surplus Dual Prices
------
1 5.000 0.000
2 40.000 0.000
3 0.000 -1.250
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
------
X1 2.500 4.000 No Upper Limit
X2 0.000 5.000 6.000
X3 5.000 6.000 No Upper Limit
RIGHT HAND SIDE RANGES
Constraint Lower Limit Current Value Upper Limit
------
1 80.000 85.000 No Upper Limit
2 No Lower Limit 280.000 320.000
3 280.000 320.000 340.000
a. What is the optimal solution, and what is the value of the profit contribution?
b. Which constraints are binding?
c. What are the dual prices for each resource? Interpret.
d. Compute and interpret the ranges of optimality.
e. Compute and interpret the ranges of feasibility.
Answers
a. x1 = 0, x2 = 80, x3 = 0, s1 = 5, s2 = 40, s3 = 0, Z = 400
b. Constraint 3 is binding.
c. Dual prices are 0, 0, and -1.25.
They measure the improvement in Z per unit increase in each right-hand side.
d. 2.5 c1 < ¥
0 c2 6
5 c3 < ¥
As long as the objective function coefficient stays within its range, the current optimal solution point will not change, although Z could.
e. 80 b1 < ¥
-¥ < b2 320
280 b3 340
As long as the right-hand side value stays within its range, the currently binding constraints will remain so, although the values of the decision variables could change. The dual variable values will remain the same.
3. (Chapter 10 – Project Scheduling) A set of activities in a project, their immediate predecessors and expected times are listed below. A part of the Management Scientist output has also been provided.
(a) Some activity slacks are given. Determine the slacks of the remaining activities.
(b) Which activities are critical and why?
(c) What is the project completion time?
(d) Can activity ‘I” be delayed without delaying the project’s completion? Justify your answer.
Answers
(a) Determine the slack of each activity: slack of an activity can be calculated as LS-ES or LF-EF
A – 3, C – 8, D – 3, F – 0, G – 0, I – 8
(b) Which activities are critical and why?
B, E, F, G – because their slacks are 0.
(c) What is the project completion time? 19 days – completion time of activity G.
(d) Can activity I be delayed without delaying the project’s completion? Justify your answer.
Yes, it has a positive slack of 8 days and hence can be delayed by a maximum of 8 days without delaying the project.
4. (Chapter 8 – Linear Integer Programming) Grush Consulting has five projects to consider. Each will require time in the next two quarters according to the table below.
Project / Time in first quarter / Time in second quarter / RevenueA / 5 / 8 / 12000
B / 3 / 12 / 10000
C / 7 / 5 / 15000
D / 2 / 3 / 5000
E / 15 / 1 / 20000
Revenue from each project is also shown. Develop a model whose solution would maximize revenue, meet the time budget of 25 in the first quarter and 20 in the second quarter, and not do both projects C and D.
Answers
Let A = 1 if project A is selected, 0 otherwise; same for B, C, D, and E
Max 12000A + 10000B + 15000C + 5000D + 20000E
s.t. 5A + 3B + 7C + 2D + 15E £ 25
8A + 12B + 5C + 3D + 1E £ 20
C + D £ 1
A,B,C,D,E={0,1}
5. A car rental agency uses 96 boxes of staples a year. It costs $10 to order staples, and carrying costs are $0.80 per box on an annual basis.
Based on the Economic Order Quantity (EOQ) as determined by Management Scientist and other information provided below, answer the following questions?
INVENTORY POLICY
****************
OPTIMAL ORDER QUANTITY 48.99
ANNUAL ORDERING COST $19.60
TOTAL ANNUAL COST $39.19
MAXIMUM INVENTORY LEVEL 48.99
AVERAGE INVENTORY LEVEL 24.49
NUMBER OF ORDERS PER YEAR 1.96
CYCLE TIME (DAYS) 186.26
(a) What is the EOQ?
(b) What is the significance of EOQ?
(c) What is the annual inventory holding cost?
(d) How frequently does the company place order for staples?
(e) What is the annual expenditure for ordering 60 boxes of staples? How much is saved every year by ordering the EOQ?
Answers
(a) What is the EOQ?
About 49 boxes (48.99 rounded up)
(b) What is the significance of EOQ?
It is that quantity which minimizes the sum of annual inventory holding/carrying cost and annual ordering cost; in this case, 49 boxes.
(c) What is the annual inventory holding cost at EOQ?
At EOQ, the annual inventory holding cost equals the annual ordering cost (given above). So is the annual inventory holding cost at EOQ is $19.60.
(d) How frequently does the company place order for staples?
Every 186.26 days (given by the cycle time).
(e) What is the annual expenditure for ordering 60 boxes of staples? Is ordering 60 boxes a good idea for this company?
If Q =60, annual ordering cost = (D/Q).C0 = (96/60).10 = $16 which is less than the annual ordering cost at EOQ ($19.60).
But the annual holding cost = ½ (Q.Ch) = ½(60x0.80) = $24, which is greater than the annual holding cost at EOQ ($19.60).
Thus total annual cost at Q=60 boxes is $16+$24 = $40, but the total cost at EOQ = $39.19. Hence its perhaps a good idea to order 60 boxes (though savings are very small).
The main idea that I ant to illustrate here is that a sub-optimal order quantity will result in higher costs.
6. Chapter 8, problem 7
Answers
a. x1 + x3 + x5 + x6 = 2
b. x3 - x5 = 0
c. x1 + x4 = 1
d. x4 £ x1
x4 £ x3
e. x4 £ x1
x4 £ x3
x4 ³ x1 + x3 - 1
7. (Open-ended question) In each of the following scenarios, what in your opinion (among Trend, Seasonality, Cycles and Irregular Variations) would be an accurate representation of the following demand behavior? And why?
a) Demand for Mother’s day greeting cards
b) Demand for vacations on the moon
c) Demand for toothpaste at your local Stater Bros.
d) Demand for printer cartridges at Office Depot over the entire country
e) Demand for motel rooms inland in the wake of a hurricane alert along the Florida Keys.
8. A manager is using the equation below to forecast quarterly demand for a product:
Yt = 6,000 + 80t where t = 0 at Q1 of this year.
What forecasts are appropriate for the last quarter of this year and the first quarter of next year?
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