Assignment 5 FINAL

# Lower limit current value upper limit 1 145000 180000

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Lower Limit Current Value Upper Limit 1 145.000 180.000 No Upper Limit 2 70.000 100.000 No Upper Limit 3 1500.000 1800.000 2033.333 4 360.000 500.000 560.000 A) What is the optimum production plan for the manufacturer? -By looking underneath the optimal solution heading in output 1, we are able to figure out our optimum production plan. We need: 0 Mini models, 75 Midi models, 0 Mini-Silver models, and 75 Midi-Silver models for an optimum objective function value of \$11,500.000. -The second part of the question which asks if there are alternative optimal solutions will be answered in part h because the same question is asked there. B) What is the marginal value of an additional hour of assembly time? Over what range of assembly time is this marginal value valid? -This question is asking for the dual price of assembly time which is constraint 3 (x 3 ). In output 1, the dual price for constraint 3 is listed as 5. This means that for every additional hour of assembly time, profits will increase by \$5. This dual price is only applicable when the range of assembly time is between 1500 and 2033.333.

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C) Suppose that 80 additional hours of test time could be obtained on the outside for \$4 per hour. Should this be done? What will be the increase in profit? -First we must look at whether this value is in the range for test time hours which is constraint 4 (x 4 ). In output 1, it is seen that the range for test time hours is between 360 and 560. Currently the constraint is 500 and if we obtain 80 additional hours, the dual price will not be applicable. -To answer this question, we must come up with another output from Management Scientist by changing the right hand side of constraint 4 from 500 to 580. - MAX 40x 1 + 60x 2 + 80x 3 + 100x 4 Subject to (s.t.) x 1 + x 2 + x 3 + x 4 ≤ 180 x 3 + x 4 ≤ 100 8x 1 + 10x 2 + 12x 3 + 15x 4 ≤ 1,800 2x 1 + 2x 2 + 4x 3 + 5x 4 ≤ 580 x 1 , x 2 , x 3 , x 4 ≥ 0 -We get the following output: Output 2: OPTIMAL SOLUTION Objective Function Value = 11800.000 Variable Value Reduced Costs X1 0.000 8.000 X2 30.000 0.000 X3 0.000 2.000 X4 100.000 0.000 Constraint Slack/Surplus Dual Prices 1 50.000 0.000 2 0.000 10.000 3 0.000 6.000 4 20.000 0.000 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit X1 No Lower Limit 40.000 48.000 X2 50.000 60.000 66.667 X3 No Lower Limit 80.000 82.000 X4 98.000 100.000 No Upper Limit RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 130.000 180.000 No Upper Limit 2 0.000 100.000 110.000 3 1500.000 1800.000 1900.000 4 560.000 580.000 No Upper Limit
-When this output is examined, it is seen that there is a new optimum production plan which results in an optimum objective function value of \$11,800.000. When we compare the new and old objective function value we see that there is an increase in profit by \$300: Z new – Z old = \$11,800- \$11,500 = \$300. -The cost of these additional hours is: 80 hours x \$4/hour = \$320. -Therefore, we should not obtain 80 additional test time hours because the cost exceeds the profit by \$20.

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