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# CMU ISM 95760 - 2017 Midterm Exam, MORNING

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90-722: Management Science I, MORNING SECTION Midterm, Spring 2017Do not turn the page until you are told to do so. Write your answers below each question.Note how much each problem is worth and budget your time. Total points possible: 100Name: ______________________________Q1: _________ out of 8 possible pointsQ2: _________ out of 12 possible pointsQ3: _________ out of 15 possible pointsQ4: _________ out of 12 possible pointsQ5: _________ out of 20 possible pointsQ6: _________ out of 12 possible pointsQ7: _________ out of 12 possible pointsQ8: _________ out of 9 possible pointsTotal Possible Points: 1001Question #1: (8 Points): On the first day of class, we created a 2D data table showing how the size of a loan payment depended on its term and interest rate. Show how the (“world’s worst”) dialog box was filled in to get Excel to complete the table.2Question #2: (12 Points): a) How many decision variables and constraints (besides non-negativity) are there in a standard transshipment problem with 3 sources, 4 transshipment nodes, and 5 sinks, where every transshipment node is connected to every source and sink, but no source is connected directly to any sink? (Recall that source is a synonym for “supply node” and sink is a synonym for “destination” or “demand node”.) b) In a standard make vs. buy problem, what would you expect to be the relationship between the objective function coefficients of the variables M3 and B3?c) Suppose you solved a shortest path problem as a transshipment problem using the trick we discussed in class, shipping one unit at least cost from the origin to the destination. If the optimal solution had 7 decision variables whose value at optimality was 1, what would that tell you? 3Question #3: (15 Points): I’ve pasted below the graphical solution to a standard two-variable product mix problem whose objective function is Max Z = X1 + X2. a) What is the optimal solution value?b) The second constraint is X1 ≤ 3. What is its allowable decrease?c) What is the shadow price on constraint #2?d) Add to the diagram a constraint that would create degeneracy. e) What would the solution be if the objective function coefficient on X2 increased to 4?4Question #4: (12 Points): Recall the Backpacks to Beyond (B2B) problems you worked for HW.Here is the spreadsheet after the optimization in HW #2, Problem #10.a) What type of problem is this?b) What is the solution and solution value?c) What constraints are binding?d) What does the 6 in Cell D10 mean?5Question #5: (20 Points): Write the algebraic formulation of the LP in the previous problem.6Question #6: (12 Points): I’ve printed the sensitivity report for this problem below. Use it to answer these questions.a) What would the objective function coefficient for artisanal backpacks have to be in order for itto be optimal for B2B to make and sell some of them?b) For what range of right-hand side values on the materials constraint would the nature of the solution remain the same?c) If B2B could acquire 100 hours of stitching time, how much should they be willing to pay for that additional resource?d) If B2B had to make one basic or one artisanal backpack, which would it rather make and why? 7Question #7: (12 Points): I ran a solver table to explore the implications of how the solution and solution value of this LP depends on the amount of material available (i.e., right hand side of the first constraint).a) What does the 50 under the Artisanal column mean?b) Why does it make sense that the first half dozen rows of the table are dominated by Flashy and Primo?8c) What is the shadow price for material when there is between 525 and 550 units of it available?d) Why are the last six rows of the table all the same? 9Question #8: (9 Points): Draw the network diagram corresponding to this transshipment problem. Label arcs with their costs. Put the node’s ID number inside its circle and put the associated demand or supply (if any) in a box next to the node. Minimize Z = X12 + 3 X13 + 5 X24 + 3 X25 + 5 X34 + 9 X35 + X45 Subject toX12 + X13 = 10X12  X24  X25 = 0X13  X34  X35 = 0X24 + X34  X45 = 0X25 + X35 + X45 = 10Xij ≥ 0 for all

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