Assignment 4 Part 2: There are 2 questions each of it worth 25 points. Show your
ID: 2250748 • Letter: A
Question
Assignment 4 Part 2: There are 2 questions each of it worth 25 points. Show your calculations step by step. Form Lagrange function and write first order conditions and solve. Please include snapshots of written pages of solution in Microsoft word and email it to me. Name: Instructor's name 1. Formulate Linear Programming Problem (represent mathematically) for Diet Problem given In slides. First define decision variables then write objective function and constraints. ve given LP pro using simp maximum value of objective function. Write steps. max z= 4x1 + 6x2 s.t 2 5r290 r20 i-1,2Explanation / Answer
There is no information for Problem 1 since slides are not present, i am solving the problem 2 with simplex method
Q2)
Maximize p = 4x1 + 6x2 subject to
-x1 + x2 <= 11
x1 + x2 <= 27
2x1 + 5x2 <= 90
x1 >= 0
x2 >= 0
Now start setting up the simplex table with the above constraints
Tableau #1
x1 x2 s1 s2 s3 s4 s5 p
-1 1 1 0 0 0 0 0 11
1 1 0 1 0 0 0 0 27
2 5 0 0 1 0 0 0 90
1 0 0 0 0 -1 0 0 0
0 1 0 0 0 0 -1 0 0
-4 -6 0 0 0 0 0 1 0
Tableau #2
x1 x2 s1 s2 s3 s4 s5 p
-1 1 1 0 0 0 0 0 11
1 1 0 1 0 0 0 0 27
2 5 0 0 1 0 0 0 90
-1 0 0 0 0 1 0 0 0
0 1 0 0 0 0 -1 0 0
-4 -6 0 0 0 0 0 1 0
Tableau #3
x1 x2 s1 s2 s3 s4 s5 p
-1 1 1 0 0 0 0 0 11
1 1 0 1 0 0 0 0 27
2 5 0 0 1 0 0 0 90
-1 0 0 0 0 1 0 0 0
0 -1 0 0 0 0 1 0 0
-4 -6 0 0 0 0 0 1 0
Tableau #4
x1 x2 s1 s2 s3 s4 s5 p
-1 1 1 0 0 0 0 0 11
2 0 -1 1 0 0 0 0 16
7 0 -5 0 1 0 0 0 35
-1 0 0 0 0 1 0 0 0
-1 0 1 0 0 0 1 0 11
-10 0 6 0 0 0 0 1 66
Tableau #5
x1 x2 s1 s2 s3 s4 s5 p
0 1 0.285714 0 0.142857 0 0 0 16
0 0 0.428571 1 -0.285714 0 0 0 6
1 0 -0.714286 0 0.142857 0 0 0 5
0 0 -0.714286 0 0.142857 1 0 0 5
0 0 0.285714 0 0.142857 0 1 0 16
0 0 -1.14286 0 1.42857 0 0 1 116
Tableau #6
x1 x2 s1 s2 s3 s4 s5 p
0 1 0 -0.666667 0.333333 0 0 0 12
0 0 1 2.33333 -0.666667 0 0 0 14
1 0 0 1.66667 -0.333333 0 0 0 15
0 0 0 1.66667 -0.333333 1 0 0 15
0 0 0 -0.666667 0.333333 0 1 0 12
0 0 0 2.66667 0.666667 0 0 1
As per the optimization, the maximum value occurs when x1=15 and x2=12
Therefore max value Z = 4x1 + 6x2 = 4(15) + 6(12) = 132