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    Sensitivity analysis problems

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    4.13 Consider the Win Big Gambling Club media selection example discussed in section 3.3 (page 88) of Chapter 3. Use the Sensitivity Report for this LP model (shown in Screenshot 4-6) to answer the following questions. Each question is independent of the others.

    The original Win Big Gambling Club media selection example is as follows:
    Objective Function (in num. audience reached)
    Max 5000T + 8500N + 2400P + 2800A
    Subject to the constraints:
    Budget is $8000
    800T + 925N + 290P + 380A < 8000
    At Least 5 Radio Spots per Week
    P + A > 5

    No More Than $1800 per Week for Radio
    290P + 380A < 1800
    Max Number of Ads per Week
    T < 12 P < 25
    N < 5 A < 20

    Finally non negativity T, N, P, A > 0

    The optimal solution found to be:
    T = 1.97 television spots.
    N = 5.00 newspaper ads.
    P = 6.21 30 second prime time radio spots.
    A = 0.00 1-minute afternoon radio spots.

    This produces an audience exposure of 67,240 contacts.
    (a) What would be the impact if management approved spending $200 more on radio advertising each week?
    It can be seen from the sensitivity report that the shadow price for Maximum radio $ is 2.03 and the maximum increase allowed is $1575. This means that each additional dollar spent in radio advertising (up to $1575) will increase the audience by 2.03. Thus, if the management approves spending $200 per week extra on radio advertising, the audience coverage would increase by 2.03*200 = 406 to 67,646.

    (b) Would it help Win Big if it could get out of the contractual agreement to place at least five radio spots each week?
    No, it does not help as in optimal solution, 6.21 radio spots are already placed. Thus, this contractual agreement is not at all binding constraint.
    (c) The radio station manager agrees to run the after- noon radio spots during some of the more popular programs. He thinks this will increase the audience reached per ad to 3,100. Will this change the optimal solution? Why or why not?
    The after-noon radio spots are currently not in the optimal solution. This means that currently, the afternoon radio spots are not used. The reduced cost for the corresponding variable is -344.83. This means that the objective function co-efficient (Number of audience reached) for this variable has to increase at least by 344.83 to 3144.83 (=2800+344.83).
    This proposal from the radio station manager promises the increase to be to 3100. This is less than the minimum requirement according to the sensitivity report. Thus, this proposal would not change the current optimal solution.
    (d) There is some uncertainty in the audience reached per TV spot. For what range of values for this OFC will the current solution remain optimal?
    TV spots currently reach 5000 contacts. Looking at the sensitivity report, it can be seen that the range of TV spots over which the solution remains optimal is 0(=5000-5000) to 6620.69(5000+1620.69). Thus the range of values for this OFC over which this optimal solution remain optimal is 0 to 6620.69.

    4.19 Kathy Roniger¡¦s diet problem first presented in Chapter 3 as Problem 3-35 on page 133. Use Solver to create the Sensitivity Report for this LP problem. Then answer the following questions using this report.
    Each question is independent of the others.
    (a) Interpret the shadow prices for the carbohydrates and iron constraints.
    (b) What will happen to total cost if Kathy insists on using milk in her diet?
    (c) What will be the maximum amount Kathy could pay for beans to make it a cost effective item for inclusion in her diet?
    (d) Is the solution to this problem a unique optimal solution? Justify your answer.
    The following is the formulation
    Objective: Minimize total cost = $0.60M + 2.35G + 1.15C + 2.25F + 0.58B + 1.17S + 0.33P

    Subject to:
    295M + 1,216G + 394C + 358F + 128B + 118S + 279P ?T 1,500 (Max calories)
    295M + 1,216G + 394C + 358F + 128B + 118S + 279P ?d 900 (Min calories)
    0.2M + 0.2G +.4.3C + 3.2F + 3.2B + 14.1S + 2.2P ?d 4 (Iron)
    16M + 96G +9C + 0.5F + 0.8B + 1.4S + 0.5P ?T 50 (Fat)
    16M +81G + 74C + 83F + 7B + 14S + 8P ?d 26 (Protein)
    22M + 28B + 19S + 63P ?T 50 (Carbohydrates)
    All variables ?d 0 (Non-negativity)
    Optimal solution:
    Milk(M) Meat(G) Chicken(C ) Fish(F) Beans(B) Spinach(S) Potatoes(P)
    0.000 0.499 0.173 0.000 0.000 0.105 0.762

    Total cost: $1.75
    The sensitivity report for the problem is given below:
    Final Reduced Objective Allowable Allowable
    Cell Name Value Cost Coefficient Increase Decrease
    $B$4 Number of pounds Milk 0.000 0.148 0.600 1E+30 0.148
    $C$4 Number of pounds Meat 0.499 0.000 2.350 0.227 11.064
    $D$4 Number of pounds Chicken 0.173 0.000 1.150 0.973 0.072
    $E$4 Number of pounds Fish 0.000 1.232 2.250 1E+30 1.232
    $F$4 Number of pounds Beans 0.000 0.261 0.580 1E+30 0.261
    $G$4 Number of pounds Spinach 0.105 0.000 1.170 0.227 0.980
    $H$4 Number of pounds Potatoes 0.762 0.000 0.330 0.410 0.753

    Final Shadow Constraint Allowable Allowable
    Cell Name Value Price R.H. Side Increase Decrease
    $I$7 Min calories 900.000 0.002 900.000 90.390 47.378
    $I$8 Min iron 4.000 0.074 4.000 33.245 1.382
    $I$9 Min protein 60.780 0.000 26.000 34.780 1E+30
    $I$10 Max calories 900.000 0.000 1500.000 1E+30 600.000
    $I$11 Max fat 50.000 -0.003 50.000 3.741 7.214
    $I$12 Max carbohydrates 50.000 -0.007 50.000 11.066 44.195

    (a) Interpret the shadow prices for the carbohydrates and iron constraints.
    Shadow price for iron is $0.074. This means that each additional gram of iron allowed in the meal will increase the meal cost by $0.074.
    Shadow price for carbohydrate is -$0.007. This means that each additional gram of carbohydrate allowed in the meal will reduce the meal cost by $0.007.
    (b) What ...

    Solution Summary

    This posting contains solutions to following sensitivity analysis problems.