Harkin Electronics is planning its production for the next quarter for its two p

Question

Harkin Electronics is planning its production for the next quarter for its two product lines, relays and capacitors. The profit contribution is $250 for a case of relays and $200 for a case of capacitors. Three resources limit how much of each product the company can produce: labor, stamping capacity, and testing capacity. Next quarter 80,000 labor hours will be available; a case of relays requires 200 labor hours and a case of capacitors requires 150. The stamping machine will be available for 1,200 hours next quarter and a case of relays requires 4 hours on the stamping machine and a case of capacitors requires 2 hours. Relays require 3 hours of testing per case and capacitors require 5. The testing machine will be available for 2,000 hours next quarter. For every case of capacitors, the company can produce no more than two cases of relays. At a minimum the company must produce at least 150 cases of relays.

You must submit your linear programming formulations and show the linear programming solution that you obtained via POM or Solver. If you solved this using another linear programming approach, you may submit that instead of the software solution.   

a) To maximize total profit next quarter, how many cases of relays and capacitors should Harkin Electronics produce during that period? If your answer is in fractional units of cases that is acceptable – do not round to whole number of cases.

b) How much profit will result? As a reminder, provide your linear programming formulation and the software solution to receive any credit for this problem.

c) Using your sensitivity analysis output, answer the following questions. You must indicate the source of your answers from the sensitivity analysis output.

For which resources would it make sense to add additional capacity based on the company’s objective? Why?

If you excluded a resource, why did you exclude it?

For those resources for which you would consider adding additional units, what is the range on the right-hand side values for the shadow prices to remain valid?

How much would you need to change the price of the capacitor so that its value in the optimal solution changes from the current value or in other words when the current optimal solution is no longer valid? Explain.

d) Now suppose that there is a requirement that of the total labor hours used by Harkins, at least 60% must be hours used for producing capacitors. Write the constraint for this requirement. However, you do not need to resolve the problem

Explanation / Answer

Sensitivity Report

a) To maximise profit, 150 cases of relays and 300 cases capacitors need to be produced

b) Maximum Profit = $ 97500

c) Stamping is a binding resource, so it would make sense to add more capacity to stamping operation.

Labor and testing are non-binding resources, so these are excluded.

For stamping resource, the range on the right hand side values are 600 and 1220 for the shadow prices to remain valid.

The price of the capacitor can be reduced by $ 75 , so that the new price of capacitor would be $ 125, at which the current optimal solution is no longer valid.

d) The constraint inequality for this requirement is 150*X2 >= 60%*80000 , or 150*X2 >= 48000

Algebraic formulation of linear pgoramming problem Decision Variables: X1 Number of cases of Relays X2 Number of cases of Capacitors Objective function : Maximise profit contribution: Z = 250*X1 + 200*X2 Subject to constraints : 200*X1 + 150*X2 <= 80000 4*X1 + 2*X2 <= 1200 3*X1 + 5*X2 <= 2000 1*X1 + 0*X2 >= 150 X1, X2 >= 0 Computer output of solution using Solver Maximise profit contribution: 97,500 =CjXj obj fn coefficient (Cj) 250 200 subject to constraints: Items X1 X2 LHS RHS Decision Variable Xj --> 150 300 Labor 200 150 75000 <= 80000 Stamping 4 2 1200 <= 1200 Testing 3 5 1950 <= 2000 Minimum Cases of Relay 1 0 150 >= 150 LHS of constraints is calculated as the sumproduct of the decision variable and the constraint coefficients mentioned in the particular row. Answer: X1 Number of cases of Relays 150 X2 Number of cases of Capacitors 300 Total Profit = $ 97500

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