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Please show all work so it can be easily understood. This must be done by hand method. NO EXCEL PLEASE. ] The Borrough Garment manufactures mens shirts and womens blouses for WolMak. In an agreement, Wolmak will accept all production supplied by Borrough Garment. The production process includes cutting, sewing, and packing. Borrough employs 25 workers in the cutting department, 31 in the sewing department, and 5 in the packing department. The factory works on 8-hour shift, 5 days a week. The following table gives the time requirements and profits per unit for the two garments Hours Per Unit Garments Cuttin Sewin 60 40 Packin 12 4 Profit(S/unit) 80 100 Shirts 20 Blouses 40 a. [10 points] What is the best weekly production schedule for Borrough? Formulate the LP and solve it by Simplex Method (no point will be given if solved by graphical method) [5 points] Is there any worker who will always be idling if the optimal schedule is used in practice? What is your suggestion to the human resource manager? b. c. [4 points] If you are the human resource manager, which department would you like to hire a new worker given the fact the company have extra money to hire only one more people. (Show your reason.) [3 points] If you are allowed to change the number of workers in the sewing department, what is the range of changes that you can make while keeping the optimal tableau obtained in (a) except its right hand side d. e. [3 points] If the profit of the mens shirt increases by $25 per unit, what is the optimal total profit?

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Answer #1

Working hours available in Cutting dept = 25*8*5 = 1000 hours

Working hours available in Sewing dept = 31*8*5 = 1240 hours

Working hours available in Packing dept = 5*8*5 = 200 hours

a) LP model is following

Let S and B be the weekly production of Shirts and Blouses

Max 80S + 100B

s.t.

20S+40B <= 1000

60S+40B <= 1240

12S+4B <= 200

S, B >= 0

Solution by simplex method is following:

SIMPLEX METHOD Initial Simplex Tableau Maximization 80 100 0 0 Basic Ratio s3 Solution (non-negative) СВі 0 Variable S 20 60 12 40 40 1000 1240 200 25 31 50 Optimality Test: Cj - Zj 20, so solution is not optimal yet Entering Variable: max value of (Cj-Zj), entering variable is B Leaving Variable: min-ratio, leaving variable is s1 0 0 lteration 1 100 0 25 240 100 2500 50 0.5 1 0.025 0 40 10 Optimality Test: Cj - Zj 2 0, so solution is not optimal yet Entering Variable: max value of (Cj-Zj), entering variable is S Leaving Variable: min-ratio, leaving variable is s2 0 0 0.1 0 Cj -Zj 30 0 2.5 0 10 teration 2 100 80 0 0 1 0.038 -0.01 0 0 -0.03 0.025 0 0 0.15 -0.25 1 0 Optimality Test: Cj - Zj S 0, so optimal solution is reached 40 2680 1.75 -0.75 0 We see that values of Cj - Zj satisfy the optimality condition Cj -Zj<0 Therefore, we have reached the optimal solution Optimal Solution Objective Value Z 2680

b) In the optimal tableau, s3 = 40, which means there will be 40 idle hours in packing dept. So 1 worker will always be idling if the optimal schedule is put in practice. So, we suggest human resource dept to use that worker in cutting or sewing instead of packing.

c) In the optimal tableau, we see that the dual price of s1 and s2 are 1.75 and 0.75 respectively. (see the last row and column s1 and s2 of optimal tableau. Dual prices are negative of these values)

Dual price of s1 is higher (1.75) , therefore hiring a worker in Cutting dept will give more incremental profit. Hence, new worker should be hired in cutting dept.

d) We see that the optimal tableau will change if the minimum ratio corresponding to s2 in iteration 1 becomes more than 10, that will happen, if the RHS becomes 400 from 240. So allowable increase is 160. The minimum ratio will also change if the intermediate solution value for s2 in interation 1 becomes 0. So allowable decrease is 240

Therefore, range of feasibility is from  1000 (=1240-240)  to 1400 (=1240+160)

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