Question

A garden supplies store prepares various grades of pine bark for mulch: nuggets (x,), mini-nuggets (x and chips (x,). The process requires pine bark, machine time, labor time, and storage space. The following model has been developed. Maximize d 92 Subject to (profit) Sxl + 6x2 + 3 s00 kg Bark Machine 2xd + 42 + Sx3 S 660 minutes Labor 2x1 2+ 3x3 s 480 hours 1xl+ 1x2 + 1 s 1s0 bags Storage a) What is the marginal value of a kilogram of pine bark? Over what range is this price value appropriate? b) What is the maximum price the store would be justified in paying for additional pine bark? c) What is the marginal value of labor? Over what range is this value in effect? d) The manager obtained additional machine time through better scheduling How much additional machine time can be effectively used for this operation? Why? ej if the manager can obtain either additional pine bark or additional storage space, which one should she choose, and how much (assuming additional quantities cost the same as usual? Difa change in the chip operation increased the profit on chips from $6 per bag to $7 per bag, would the optimal quantities change? Would the value of the objective function change? if profits on chips increased to $7 per bag and profits on nuggets decreased by $.60, would the optimal quantities change? Would the value of the objective function change? if so, what would the new value(s) be? h) if the amount of pine bark available decreased by 15 kg, machine time decreased by 27 minutes, and storage capacity increased by S bags, would this fall in the range of feasibility for multiple changes? if so, what would the value of the objective function be?

Answer 1

P.S. The image is slightly unclear. However, we can analyze the sensitivity analysis report to answer most of the question.

Max 9x1 + 9x2 + 6x3

Subject to

5x1 + 6x2 + 3x3 <= 600

1x1 + 4x2 + 5x3 <= 660

2x1 + 4x2 + 3x3 <= 480

1x1 + 1x2 + 1x3 <= 150

The LINDO output is shown below

2 LINDO X File Edit Solve Reports Window Help <untitled> LP OPTIMUM FOUND AT STEP 2 OBJECTIVE FUNCTION VALUE Max 9x1 + 9x2 + 6x3 Subject to 5x1 + 6x2 + 3x3 <= 600 1x1 + 4x2 + 5x3 <= 660 2x1 + 4x2 + 3x3 <= 480 1x1 + 1x2 + 1x3 <= 150 1) 1125.000 VARIABLE x2 VALUE 75.000000 0.000000 75.000000 REDUCED COST 0.000000 1.500000 0.000000 X3 ROW 2) SLACK OR SURPLUS 0.000000 210.000000 105.000000 0.000000 DUAL PRICES 1500000 0.000000 0.000000 1.500000 NO. ITERATIONS= RANGES IN WHICH THE BASIS IS UNCHANGED VARIABLE CURRENT COEF 9.000000 9.000000 6.000000 OBJ COEFFICIENT RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 1.000000 1.000000 1.500000 INFINITY 3.000000 0.600000 X2 X3 пом Е CURRENT RHS 600.000000 660.000000 480.000000 150.000000 RIGHTHAND SIDE RANGES ALLOWABLE ALLOWABLE INCREASE DECREASE 150.000000 105.000000 INFINITY 210.000000 INFINITY 105.000000 19.090910 30.000000

a)

The marginal value of a kilogram of pine bark 1.5 per kg. The range for this is from (600-105) to (600+150). This means from 495 to 750 kg, every 1 kg increase in pine bark will cause the objective function value to rise by 1.5

b)

Taking the above explanation into context, the store should not pay more than 1.5 per kg of pine bark.

c)

Marginal value of labor is 0. The current capacity of labor has not be completely met. There is surplus labor and thus the marginal value (Dual prices) is 0. This is valid in from (480-105) to infinity. This means if the labor availability is anything above 375 hours then there is no influence on the overall solution.

d)

The current machine time available is 660 minutes. However, we have not utilized all 660 minutes. There is a surplus of 210 minutes. That means the utilization has been only 450 minutes. This means there is no point in availing additional machine time when we have not even utilized the available ones.

e)

The marginal values of pine bark and the storage space is 1.5 each. This means there is no difference in terms of unit value. However, there is difference in their ranges. Pine bark can be added up to 750. That means 150 additional. This will give a boost to the objective function by 150*1.5 = 225.

On the other hand storage space can be increased up to 150 + 19.09 = 169.09. This will give a boost of 19.09*1.5 = 28.635.

Thus the manager should choose the pine bark as we can increase that more and earn more profit.

f)

The optimal solution will not change. The range of optimality of chips is from (6-0.6) to (6+3). That means between 5.4 and 9 the optimal solution will remain unchanged. However, the objective function value will change because the profit contribution will be higher by $1 per unit of variable value.