BUS 202 Quantitative Applications

Solutions to Assignment 11

Problem 5-18
Problem 5-19
Problem 5-25
Problem 5-31 (not assigned)
Problem 6-2
Problem 6-3 thru 6 (not assigned)
Problem 6-18

5-18.

 MIN     5 X11 + 7 X12 + 6 X13 + 8 X21 + 9 X22 + 10 X23 + 4 X31 + 3 X32 + 11X33
  SUBJECT TO
           X11 + X12 + X13  <=   50
           X21 + X22 + X23  <=   275
           X31 + X32 + X33  <=   175
           X11 + X21 + X31 >=    100
           X12 + X22 + X32 >=   250
           X13 + X23 + X33 >=   150

Solution
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5-19.

The relevant cost that must be minimized is the transportation and packaging costs. Thus the cell costs are the cost of packaging one unit at a location and shipping it to a warehouse.
Since we have more capacity than demand, minimizing just the transportation cost, may force the use of costlier capacity more fully and leave some inexpensive capacity unused, leading to sub-optimization. If we had just enough capacity to meet all demand, simply minimizing the transportation costs would have been appropriate since we would need all the capacity-- expensive and cheap.
Solution
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5-25.

This is a straightforward assignment problem with a square cost matrix;
Solution
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5-31.

Since there are more developers than plots one of the developers will not get an assignment. We can handle the situation in two ways: 1) Create a dummy plot (whoever gets the dummy plot does not get a real plot) or 2) Enter the constraints for each developer as "<=" to allow not assigning a plot to one of the developers as follows.

MAX     19 A1 + 19 A2 + 29 A3 + 23 A4 + 24 A5 + 23 B1 + 21 B2 + 27 B3
       + 19 B4 + 25 B5 + 19 C1 + 19 C2 + 22 C3 + 20 C5 + 23 D1 + 19 D3 + 21 D4
       + 18 D5
  SUBJECT TO
            A1 + A2 + A3 + A4 + A5 =    1
            B1 + B2 + B3 + B4 + B5 =    1
            C1 + C2 + C3 + C4 + C5 =    1
            D1 + D2 + D3 + D4 + D5 =    1
            A1 + B1 + C1 + D1 <=   1
            A2 + B2 + C2 + D2 <=   1
            A3 + B3 + C3 + D3 <=   1
            A4 + B4 + C4 + D4 <=   1
            A5 + B5 + C5 + D5 <=   1

Solution with a dummy plot
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6-2.

a) x1=3.5; x2=4.5; OV= 12.5
b) There are 29 of them. Blue dots in the green area.
c) There are two optimal solutions x1=0; x2=6 and x1=2; x2=5, both have OV=12
d) We get x1=3; x2=4; OV=11. clearly not optimal.

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6-3.

For a Min problem the LP relaxation would provide a Lower Bound . Since the LP relaxation is obtained by ignoring some constraints(integrality) the solution value would be better (lower) than if we had enforced those constraints.

6-4.

For a Min problem a feasible solution (obtained by rounding the LP solution or any other means) gives an Upper Bound . Since we are not interested in any solution whose value exceeds the value of this known solution.

6-5.

For a Max problem the LP provides an Upper Bound .

6-6.

For a Max problem rounding (to get a feasible solution would give a Lower Bound .

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6-18.

Let Aj, Bj and Cj the shipments from warehouse A, B and C to customer j and YA, YB, and YC zero/one switch variables denoting whether warehouse A, B, and C will be used. Since warehouses can be assumed to have unlimited capacity, for modeling purposes we might consider the capcities (if operated) to be equal to the total demand, the most that might ever be shipped from a single warehouse.

  MIN 15A1 + 32A2 + 21A3 + 9B1 + 7B2 + 6B3 + 11C1 + 18C2
    + 5C3 + 500YA + 750YB + 600YC
  SUBJECT TO
            A1 + B1 + C1 >= 200
            A2 + B2 + C2 >= 150
            A3 + B3 + C3 >= 175
            A1 + A2 + A3 <= 525YA 
            B1 + B2 + B3 <= 525YB 
            C1 + C2 + C3 <= 525YC 
            YA + YB + YC >= 2
            Aj, Bj, Cj>=0, YA, YB, YC =0,1

Excel solution
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