Question

16.   A dairy farmer’s cows need three nutrients (A, B, and C) to subsist and produce milk each day. Each cow must receive the equivalent of 100 units of nutrient A, 200 units of nutrient B, and 50 units of nutrient C in order to maximize milk output. The farmer can use any combination of three feeds ( f1, f2, and f3) in meeting these minimum requirements. The local feed dealer sells all three feeds, which have the following cost per pound and nutrient equivalents (for A, B, and C) per pound.

Nutrient Content Per Pound

Feed A B C ($/pound) f1 5 22 3 0.25 f2 10 25 2 0.50 f3 7 12 5 0.27

Cost

Assume that the farmer’s objective is to minimize the cost per cow of buying any com- bination of these three feeds that satisfy the daily nutrient requirements of the cow.

a.        Write the standard form of this LP problem.

b.        Solve this problem using the simplex method.

Answer 1

B = 2271+ 2542 +1243 @ let the farmer use foifa and fa in the satio & The total cost of the diet consisting of a write ad alta ordy 72 units of fa and X3 units off I given by - C= 0.25%,+ 0.50 X2 +0-2743 The total amount of nutrient A= 5x14 1082 478 B= 22x+ 25%. - " C = 371 + 2x2 -15% The mom srequumont of nowe At 100, nedzie Bis 200 and nutrient e les 30. Therefore, we have the following constraints Gdy + 10X2+3x3 >100 223)+257 +1273 > 200 3X1 + 2% +503 >ão The standard form of this LP is giron by - Man @= 0.25X1 +0.5042 +0.2783 Bubject to 5X1+ 10x 27723 7,100 2241+28 72 +1273 > 200 3X1 + 2X2 + 8X3 > 50 230, 7>0 & X>0 ♡ The initial mateix is writion as - ra no 711007 22 25 12 200 3 2 50 (0-23 0.50 0:27 The Hangose of this mattix is - f522 3 0.237 10 28 20-60 7 12 ع 100 200 301

The dual Babiem Le civon de May: P = 100x + 200x2 4 50x3 subject to 57, + 2272+ 33 % 0.25 10x + 2512 +243 5050 7114 12 H2 + 5x3 40.27 !., 22, X37,0 QOOO noor m dro willing the equations with stack variables - ax, + 2272 +373 +3,= 0.25 1001+ 25%2+273 +82=0.50 7X1+12X2 + 5x3 + 82 -0.27 - 100%,- 20042 - 50x3 +P=0 The initial mattis fe given as 1. a. x2 xz si se 83 P I b 22 3 0 0.2 10 23 2 0 0.50 7 12 a 0 0 0 1 0.27 -100 -200 -50 0 0 1 0 Most -ve =-200, Indo cable + 0.25, 0:50, 0:27 0:27 bing the smallest non negative indicatos, an 12 of the pivot value. The motor becomes n. x2 X3 & Sz 83 PI b -220-100 (+385-50) (2xD*) 12 112-83 7/12 5/12 o o 1/2 o 1 0.27/12 16067 0 33-33 0 0 16.67 x = (ve value in old tableau pirct column) x (value in new . tableau piret row)+ old tableau valie z (-22) +(-100) = Similarly use compute. The last row of above matrix 12=0 sito