LP Sensitivity 1

7/30/2019 LP Sensitivity 1

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Linear Programming:

Sensitivity Analysis


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Using LINGO for sensitivity Analysis


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Resource-

I

Resource-II


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Right-hand Side Changes

What is the economic value of having additionalresources?

What would we do with additional resources, ifwe have them?

First let us solve the Dual of the Original Problem.


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What is the value of one additional unit ofResource-I.

Interpret it in terms of primal-LP

What is the optimal profit with

(a) 7 units of resource-I

(b) 8 units of resource-I


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Marginal Economic value

0 1 2 3 4 5 6 7 8 9 10 11

Rs. 16

Rs. 4

Number of units of Resource-I

MarginalEconomic Value


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Changes to the Objective functioncoefficients

Identify the ranges of objective function coefficients forthe current solution to remain optimal.

How much the profit from first product has to increasefrom Rs. 20 to make us produce one more unit of that

product (from the current optimal of 2)

Ans: Rs. 12 (20 - 2*16)

How much the profit from first product has to decreasefrom Rs. 20 to make us produce one less unit of thatproduct (from the current optimal of 2)

Ans: Rs. 4 (1*16 20)


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Range ofCurrentoptimal


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Reduced Costs

When a decision variable has an optimal valueofzerothe allowable increase for the objectivefunction coefficient is also called the reducedcost.


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What happens when the reduced cost for thesecond variable is included in the objectivefunction?


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Example: LAWN CARE Inc.Source: Liberatore and Nydick John Wiley & Sons Inc.

Lawn Care produces three lawn fertilizers that arecomposed of Nitrate, Phosphate, and Potash indifferent proportions. Fertlizer-1 (F1) is 5% Nitrate,5% Phosphate and 10% Potash; Fertlizer-2 (F2) is10% Nitrate, 10% Phosphate and 5% Potash;Fertlizer-3 (F3) is 10% Nitrate, 5% Phosphate and5% Potash. Lawn care currently 1050 tons ofNitrate, 1000 tons of Phosphate and 1580 tons ofPotash available. The fertilizer profits per ton are

$17, $20 and $ 12 respectively. We assume forsimplicitys sake that all the fertilizers of each typethat can be produced can be sold.


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1. How much tons of each fertilizer should LawnCare produce and what profit will begenerated?

Ans:14,400 (F1); 2800 (F2), 0 (F3) $ 300,800

2. How much each resource is unused if the

optimal solution is implemented?Ans: 50 (Nitrate)

3. Current production level of F3 is zero. At what

objective function coefficient value would it bebeneficial to start producing F3?

Ans: Increase from 12 to 12.33


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1. If the objective function coefficient for X1 increases by$25 per ton will the optimal value of X1, X2 and X3increase, decrease, or remain unchanged?

Ans:X1 would increase (25 > 23), X2 will decrease, X3 canincrease/decrease/remain unchanged

2. Suppose that Lawn care is considering introducing a

new fertilizer that uses zero Nitrate, zero Phosphateand 20% potash. The new fertilizer will generate a profitof $25 per ton. If additional potash could not bepurchased, how much if any of the new fertilizer should

Lawn Care produce?Ans: Atleast 5*1080 = 5400 tons of F4 need to be

produced.


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1. Suppose that additional Phosphate and Potash can besecured if needed. Each additional ton of Phosphatewill cost $ 135.50. For each additional ton of Potash, up

to 200, a cost of $92.75 is incurred. Additional tons ofPotash beyond 200 will cost $ 89.55 per ton. If only oneincrease can be made, which additional resource(phosphate or potash), if any, should be secured? How

many additional tons should be obtained?

Ans: Phosphate the profit increase will be 50*(153.33-135.50) = $891.50

Potash 200*(93.33-92.7) + 220*(93.33 89.55) = $947.60Therefore Potash should be increased by atleast 420 tons.

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LP Sensitivity 1