Maximize p = 2x + 2y + 2z + 2w subject tox + y + z $ 3y + z + w S 4x + z + w < 5x + y + w< 6x 2 0, y 2 0, z 2 0, w 2 0.p =(x, Y, z, w)=Need Help?Watch ItRead It

Name: Maximize p = 2x + 2y + 2z + 2w subject tox + y + z $ 3y + z + w S 4x
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: Maximize p = 2x + 2y + 2z + 2w subject tox + y + z $ 3y + z + w S 4x + z + w < 5x + y + w< 6x 2 0, y 2 0, z 2 0, w 2 0.p =(x, Y, z, w)=Need Help?Watch ItRead It

Maximize p=2x+2y+2z+2w subject to x+y+z≤3y+z+w≤4x+z+w≤5x+y+w≤6x,y,z,w≥0 Introduce slack variables: Max p=2x+2y+2z+2w+0S1+0S2+0S3+0S4 Subject to: x+y+z+S1=3y+z+w+S2=4x+z+w+S3=5x+y+w+S4=6x,y,z,w, S1,S2,S3,S4≥0 For our convinence , let us change the variables as: Max Z =2x1+2x2+2x3+2x4+0S1+0S2+0S3+0S4 Subject to: x1+x2+x3+S1=3x2+x3+x4+S2=4x1+x3+x4+S3=5x1+x2+x4+S4=6x1,x2,x3,x4, S1,S2,S3,S4≥0 Iteration -1 Cj 2 2 2 2 0 0 0 0 B CB XB x1 x2 x3 x4 S1 S2 S3 S4 Min Ratio XBX1 S1 0 3 (1) 1 1 0 1 0 0 0 3 S2 0 4 0 1 1 1 0 1 0 0 - S3 0 5 1 0 1 1 0 0 1 0 5 S4 0 6 1 1 0 1 0 0 0 1 6 Z = 0 Zj 0 0 0 0 0 0 0 0 Zj-Cj -2 -2 -2 -2 0 0 0 0 Minimum negative in Zj-Cj is -2 and its column index is 1. So, the entering variable is x1. Minimum ratio is 3 and its row index is 1. So, the leaving basis variable is S1. Pivot element is 1. R1(new)=R1(old) R2(new)=R2(old) R3(new)=R3(old)-R1(new) R4(new)=R4(old)-R1(new) Iteration -2 Cj 2 2 2 2 0 0 0 0 B CB XB x1 x2 x3 x4 S1 S2 S3 S4 Min Ratio XBX1 x1 2 3 1 1 1 0 1 0 0 0 - S2 0 4 0 1 1 1 0 1 0 0 4 S3 0 2 0 -1 0 (1) -1 0 1 0 2 S4 0 3 0 0 -1 1 -1 0 0 1 3 Z = 6 Zj 2 2 2 0 2 0 0 0 Zj-Cj 0 0 0 -2 2 0 0 0 Minimum negative in Zj-Cj is -2 and its column index is 4. So, the entering variable is x4. Minimum ratio is 2 and its row index is 3. So, the leaving basis variable is S3. Pivot element is 1. R3(new)=R3(old) R1(new)=R1(old) R2(new)=R2(old)-R3(new) R4(new)=R4(old)-R3(new) Iteration -3 Cj 2 2 2 2 0 0 0 0 B CB XB x1 x2 x3 x4 S1 S2 S3 S4 Min Ratio XBX1 x1 2 3 1 1 1 0 1 0 0 0 3 S2 0 2 0 2 1 0 1 1 -1 0 1 x4 2 2 0 -1 0 1 -1 0 1 0 - S4 0 1 0 (1) -1 0 0 0 -1 1 1 Z = 10 Zj 2 0 2 2 0 0 2 0 Zj-Cj 0 -2 0 0 0 0 0 0 Minimum negative in Zj-Cj is -2 and its column index is 2. So, the entering variable is x2. Minimum ratio is 1 and its row index is 4. So, the leaving basis variable is S4. Pivot element is 1. R4(new)=R4(old) R1(new)=R1(old)-R4(new) R2(new)=R2(old)-2R4(new) R3(new)=R3(old)+R4(new) Iteration -4 Cj 2 2 2 2 0 0 0 0 B CB XB x1 x2 x3 x4 S1 S2 S3 S4 Min Ratio XBX1 x1 2 2 1 0 2 0 1 0 1 -1 1 S2 0 0 0 0 (3) 0 1 1 1 -2 0 x4 2 3 0 0 -1 1 -1 0 0 1 - x2 2 1 0 1 -1 0 0 0 -1 1 - Z = 12 Zj 2 2 0 2 0 0 0 2 Zj-Cj 0 0 -2 0 0 0 0 2 Minimum negative in Zj-Cj is -2 and its column index is 3. So, the entering variable is x3. Minimum ratio is 0 and its row index is 2. So, the leaving basis variable is S2. Pivot element is 3. R2(new)=R2(old)÷3 R1(new)=R1(old)-2R2(new) R3(new)=R3(old)+R2(new) R4(new)=R4(old)+R2(new) Iteration -5 Cj 2 2 2 2 0 0 0 0 B CB XB x1 x2 x3 x4 S1 S2 S3 S4 Min Ratio x1 2 2 1 0 0 0 0.3333 -0.6667 0.3333 0.3333 - x3 2 0 0 0 1 0 0.3333 0.3333 0.3333 -0.6667 - x4 2 3 0 0 0 1 -0.6667 0.3333 0.3333 0.3333 - x2 2 1 0 1 0 0 0.3333 0.3333 -0.6667 0.3333 - Z = 12 Zj 2 2 2 2 0.6667 0.6667 0.6667 0.6667 Zj-Cj 0 0 0 0 0.6667 0.6667 0.6667 0.6667 All Zj-Cj≥0 Optimal solution has been obtained. x1=2, x2=1,x3=0,x4=3. Max p = 2(2)+2(1)+2(0)+2(3) =12 Therefore, p = 12 (x, y, z, w) = (2, 1, 0, 3)