Stelle Office Supplies must fill an order for 2000 modular office dividers. Each

Question

Stelle Office Supplies must fill an order for 2000 modular office dividers. Each divider consists of a frame, a set of legs, and a panel. SOS has limited production and finishing time available and is considering the purchase of some of the components. Letxj,x2. andx3 be the number of frames, leg sets, and panels to make, and x4xz, andx be the mumber of each to buy. The model reflects the costs to be minimized, the amount of production time, the amount of assembly time, and the need for 2000 of each component. Min 20x14x2+15x3 +28x420x25x, st. 30+40x%2 +25x, 180000 15x1 + 10x2 +30xz £ 90000 x1+x4 = 2000 x2+x5 = 2000 s+h"2000 allx, 2 0 The final tableau i:s X6S1 cB20-14 15 28-20 -20 -25 0 0 1 0-033 666.67 0 0 0-17.5 -31.67 01 -833 6666.67 2000 2000 1-5-333 0 0.0331333.33 68667 X6-250 -20 1 0 0 -140 -150 0 X3 z-20 -14-15 -25 -17.33 -25 033 ciZ0 0 0-3-2.67 00-33 a. Calculate the range of optimality for all of the objective function coefficients. b. Calculate the range of feasibility for the first two right-hand sides. c. How much less expensive would it have to be to buy frames before you would consider it? d. How much more expensive would legs have to be to make before you would change your e. What would the total cost be if the cost to make a panel increased by 3.00? f. What would you be willing to pay for more production time? What would happen to the total cost if the amount of assembly time decreased by 2000 hours?

Explanation / Answer

Note the value of Z in the optimal tableau is wrong. It must be  -25 x 666.67 + 0 x 6666.67 - 20 x 2000 - 14 x 2000 - 15 x 1333.33 = -104667

(a)

(b)

(c)

Present cost of x4 = 28

Need to reduce by 3 (note the sensitivity of x4 in part (a)) i.e. when the cost becomes 25 or less, it is beneficial to produce instead of buying.

(d)

Present cost of x2 = 14

Need to increase by 2.67 (note the sensitivity of x2 in part (a)) i.e. when the cost becomes 16.67 or more, it is beneficial to buy instead of producing.

(e)

The Same optimal solution will be found earlier as the change (3 units) is in the range of optimality of x3 and the total cost becomes 20 x 2000 + 14 x 2000 + 18 x 1333.33 + 28 x 0 + 20 x 0 + 25 x 666.67 = 108667.

(f)

Dual price of the production time constraint (b1) is zero. So, nothing should be paid for additional production time.

(g)

Additional cost incurred = Dual price of b2 x 2000 = 0.33 x 2000 = 660

Total Cost = 104667 + 660 = 105327

x1 x2 x3 x4 x5 x6 s1 s2 Basis cB -20 -14 -15 -28 -20 -25 0 0 x6 -25 0 0 0 0.5 0.333 1 0 -0.033 s1 0 0 0 0 -17.5 -31.67 0 1 -0.833 x1 -20 1 0 0 1 0 0 0 0 x2 -14 0 1 0 0 1 0 0 0 x3 -15 0 0 1 -0.5 -0.333 0 0 0.033 Zj -20 -14 -15 -25 -17.33 -25 0 0.33 Cj- Zj 0 0 0 -3 -2.67 0 0 -0.33 (Cj - Zj)/x6 - - - -6 -8.018 0 - 10 (Cj - Zj)/x1 0 - - -3 - - - - (Cj - Zj)/x2 - 0 - - -2.67 - - - (Cj - Zj)/x3 - - 0 6 8.018 - - -10

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