Answer to Question #251137 in Operations Research for VINCENT

Question

 Solve the following linear programming problem using the simplex method.

𝑀𝑖𝑛𝑖𝑚𝑖𝑧𝑒 𝑃 = 2100𝑦1 + 2400𝑦2 + 10𝑦3 − 70𝑦4

𝑠𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜

25𝑦1 + 15𝑦2 + 𝑦3 ≥ 250

20𝑦1 + 30𝑦2 − 𝑦3 − 𝑦4 ≥ 300

𝑦1 ≥ 0, 𝑦2 ≥ 0, 𝑦3 ≥ 0 , 𝑦4 ≥ 0 

Answer 1

Solution:

Given problem is

Min z=2100y1+2400y2+10y3-70y4

subject to

25y1+15y2+y3≥250

20y1+30y2-y3-y4≥300

y1≥0, y2≥0, y3≥0, y4≥0

and y1,y2,y3,y4≥0;

After introducing surplus,artificial variables

Min z=2100y1+2400y2+10y3-70y4+0S1+0S2+0S3+0S4+0S5+0S6+MA1+MA2+MA3+MA4+MA5+MA6

subject to

25y1+15y2+y3-S1+A1=250

20y1+30y2-y3-y4-S2+A2=300

y1-S3+A3=0

y2-S4+A4=0

y3-S5+A5=0

y4-S6+A6=0

and y1,y2,y3,y4,S1,S2,S3,S4,S5,S6,A1,A2,A3,A4,A5,A6≥0

Since all Zj-Cj≤0

Hence, optimal solution is arrived with value of variables as :

y1=0,y2=16.6667,y3=0,y4=200

Min z=26000