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Practice Exam
DS 523
Spring 2008
1. Let A, B, and C be the amounts invested in companies A, B, and C. If no more than 50% of the total investment can be in company B, then
|a. |B ≤ 5 |
|b. |A − .5B + C ≤ 0 |
|c. |.5A − B − .5C ≤ 0 |
|d. |−.5A + .5B − .5C ≤ 0 |
ANS: D
2. The problem which deals with the distribution of goods from several sources to several destinations is the
|a. |maximal flow problem |
|b. |transportation problem |
|c. |assignment problem |
|d. |shortest-route problem |
ANS: B
3. The objective of the transportation problem is to
|a. |identify one origin that can satisfy total demand at the destinations and at the same time minimize total shipping cost.|
|b. |minimize the number of origins used to satisfy total demand at the destinations. |
|c. |minimize the number of shipments necessary to satisfy total demand at the destinations. |
|d. |minimize the cost of shipping products from several origins to several destinations. |
ANS: D
4. The assignment problem constraint x31 + x32 + x33 + x34 ≤ 2 means
|a. |agent 3 can be assigned to 2 tasks. |
|b. |agent 2 can be assigned to 3 tasks. |
|c. |a mixture of agents 1, 2, 3, and 4 will be assigned to tasks. |
|d. |there is no feasible solution. |
| |ANS: A |
5. Constraints in a transshipment problem
|a. |correspond to arcs. |
|b. |include a variable for every arc. |
|c. |require the sum of the shipments out of an origin node to equal supply. |
|d. |All of the alternatives are correct. |
ANS: B
6. A constraint with a positive slack value
|a. |will have a positive dual price. |
|.b. |will have a negative dual price. |
|c. |will have a dual price of zero. |
|d. |has no restrictions for its dual price. |
ANS: C
7. The amount by which an objective function coefficient can change before a different set of values for the decision variables becomes optimal is the
|a. |optimal solution. |
|b. |dual solution. |
|c. |range of optimality. |
|d. |range of feasibility. |
ANS: C
8. The 100% Rule compares
|a. |proposed changes to allowed changes. |
|b. |new values to original values. |
|c. |objective function changes to right-hand side changes. |
|d. |dual prices to reduced costs. |
ANS: A
9. A section of output from The Management Scientist is shown here.
|Variable |Lower Limit |Current Value |Upper Limit |
|1 |60 |100 |120 |
What will happen to the solution if the objective function coefficient for variable 1 decreases by 20?
|a. |Nothing. The values of the decision variables, the dual prices, and the objective function will all remain the same. |
|b. |The value of the objective function will change, but the values of the decision variables and the dual prices will |
| |remain the same. |
|c. |The same decision variables will be positive, but their values, the objective function value, and the dual prices will |
| |change. |
|d. |The problem will need to be resolved to find the new optimal solution and dual price. |
ANS: B
10. The amount that the objective function coefficient of a decision variable would have to improve before that variable would have a positive value in the solution is the
|a. |dual price. |
|b. |surplus variable. |
|c. |reduced cost. |
|d. |upper limit. |
ANS: C
1. Tots Toys makes a plastic tricycle that is composed of three major components: a handlebar-front wheel-pedal assembly, a seat and frame unit, and rear wheels. The company has orders for 12,000 of these tricycles. Current schedules yield the following information.
| |Requirements | | |Cost to |Cost to |
|Component |Plastic |Time |Space |Manufacture |Purchase |
|Front |3 |10 |2 |8 |12 |
|Seat/Frame |4 |6 |2 |6 |9 |
|Rear wheel (each) |.5 |2 |.1 |1 |3 |
| Available |50000 |160000 |30000 | | |
The company obviously does not have the resources available to manufacture everything needed for the completion of 12000 tricycles so has gathered purchase information for each component. Develop a linear programming model to tell the company how many of each component should be manufactured and how many should be purchased in order to provide 12000 fully completed tricycles at the minimum cost.
ANS:
|Let |FM = number of fronts made |
| |SM = number of seats made |
| |WM = number of wheels made |
| |FP = number of fronts purchased |
| |SP = number of seats purchased |
| |WP = number of wheels purchased |
| | |
|Min |8FM + 6SM + 1WM + 12FP + 9SP + 3WP |
| | |
|s.t. |3FM + 4SM + .5WM ≤ 50000 |
| |10FM + 6SM + 2WM ≤ 160000 |
| |2FM + 2SM + .1WM ≤ 30000 |
| |FM + FP ≥ 12000 |
| |SM + SP ≥ 12000 |
| |WM + WP ≥ 24000 |
| |FM, SM, WM, FP, SP, WP ≥ 0 |
2. Canning Transport is to move goods from three factories to three distribution centers. Information about the move is given below. Give the network model and the linear programming model for this problem.
|Source |Supply |Destination |Demand |
|A |200 |X |50 |
|B |100 |Y |125 |
|C |150 |Z |125 |
Shipping costs are:
| |Destination |
|Source |X |Y |Z |
|A |3 |2 |5 |
|B |9 |10 |-- |
|C |5 |6 |4 |
| |(Source B cannot ship to destination Z) |
ANS:
[pic]
|Min |3XAX + 2XAY + 5XAZ + 9XBX + 10XBY + 5XCX + 6XCY + 4XCZ |
| | |
|s.t. |XAX + XAY + XAZ ≤ 200 |
| |XBX + XBY ≤ 100 |
| |XCX + XCY + XCZ ≤ 150 |
| |XDX + XDY + XDZ ≤ 50 |
| |XAX + XBX + XCX + XDX = 250 |
| |X AY + XBY + XCY + XDY = 125 |
| |XAZ + XBZ + XCZ + XDZ = 125 |
| | Xij ≥ 0 |
PTS: 1 TOP: Transportation problem
3. The binding constraints for this problem are the first and the horizontal axis.
|Min |x1 + 2x2 |
| | |
|s.t. |x1 + x2 ≥ 300 |
| |2x1 + x2 ≥ 400 |
| |2x1 + 5x2 ≤ 750 |
| |x1 , x2 ≥ 0 |
|a. |Keeping c2 fixed at 2, over what range can c1 vary before there is a change in the optimal solution point? |
|b. |Keeping c1 fixed at 1, over what range can c2 vary before there is a change in the optimal solution point? |
|c. |If the objective function becomes Min 1.5x1 + 2x2, what will be the optimal values of x1, x2, and the objective function?|
|d. |If the objective function becomes Min 7x1 + 6x2, what constraints will be binding? |
|e. |Find the dual price for each constraint in the original problem. |
ANS:
|a. |.8 ≤ c1 ≤ 2 |
|b. |1 ≤ c2 ≤ 2.5 |
|c. |x1 = 300, x2 = 0, z = 450 |
|d. |Constraints 1 and 3 will be binding. |
|e. |Dual prices are 1, 0, 0. |
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