determine a feeding plan that utilizes all the available grain 2. A farmer must

Question

determine a feeding plan that utilizes all the available grain 2. A farmer must determine a plan to feed his stock during the coming winter. He has two types of stock, each with distinct nutritional requirements. To feed the stock the farmer has available 1000 lb of grain harvested over the summer. How- ever, this supply of grain is not adequate to meet the nutritional demands of the entire stock over the winter, and so the farmer must supplement this supply with feeds purchased from the local coop. Determine a feeding plan that utilizes all the available grain, satisfies the nutritional demands of the stock, and minimizes the amount spent on the supplementary feeds. The data follow. Nutritional demands (minimal number of units required/winter). Element B 360 700 Elemeni C 650 450 Element A Stock 1 Stock 2 150 90 Nutritional contents (units/lb) and costs (cents/lb) of the grain and the two avail- able feeds:

Explanation / Answer

Solution

Let the quantity (lb) to be supplemented be x for Feed 1and y for Feed 2.

Then, the Linear Programming Problem is:

Minimize Z = 15x + 23y [total cost of Feed 1 and Feed 2 supplement],

subject to:

200 + x + 3y 240 or x + 3y 40 …… (A)[constraint on element A]

900 + 5x + 7y 1060 or 5x + 7y 160 …… (B)[constraint on element B]

800 + 10x + 13y 1100 or 10x + 13y 300 …… (C)[constraint on element C]

x, y 0 [non-negativity constraint]

The above LPP can be solved by graphical method as explained below:

Treating the constraints as equations, each representing a straight line,

A cuts x-axis at 40 and y-axis at 40/3 (13.3);

B cuts x-axis at 32 and y-axis at 160/7 (22.86);

C cuts x-axis at 30 and y-axis at 300/13 (23.07)

Since all constraints are ‘’ type, the feasible region is above all the three lines.

Solving A, B, C equations pair-wise, the points of intersections are:

S - A&B: (25, 5),

R - A&C: (380/17, 100/17),

Q - C&B: (4, 20).

Thus, the feasible region is a right-open polygon YPQRSTX where P = (0, 200/13) and T = (40, 0).

Substituting the coordinates of points P, Q, R, S and T as x and y of the objective function Z, we have:

ZP =530.76,     

ZQ =520,

ZR =470.58,

ZS =490,

ZT =600.

Clearly, ZR is the least.

=> optimum solution is:

(x = 380/17, y = 100/17) and the corresponding total cost is 470.58. ANSWER

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