Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms
Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms
4th Edition
ISBN: 9780534423551
Author: Wayne L. Winston
Publisher: Cengage Learning
Expert Solution & Answer
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Chapter 4, Problem 1RP

Explanation of Solution

Optimal solutions:

Consider the following linear programing problem:

  Max Z=5x1+3x2+x3

Subject to the constraints:

  x1+x2+3x36

  5x1+3x2+6x315

  x1,x2,x0

Calculate the optimum solution for the above linear programming problem by using the simplex algorithm as follows:

  • The constraints are ≤ constraints, now it is necessary to convert it to an equality constraint by adding slack variables s1, s2 to the two constraints. The standard form of linear programming problem is as shown below:

  Max Z=5x1+3x2+x3+0s1+0s2

  x1+x2+3x3+s1=6

  5x1+3x2+6x3+s2=15

  x1,x2,x3,s1,s20

The initial simplex table is as follows:

  • Choose base variables by observing that which values form an identity matrix in the table. Here X4, X5 variable values form an identity matrix. Take the corresponding Cj values of X4, X5 as Cb values.

   Zj= CbXb  Zb= (0×6) + (0×15)  Zb= 0Z1= CbX1 Z1= (0×1)+(0×5) Z1= 0

  • Calculate the rest of variables and shown in the initial table:
 Cj 53100
BaseCbXbX1X2X3X4X5
X40611310
X501553601
Zj0000000
Zj-Cj 0-5-3-100
  • From the above simplex table, observe ZjCj values. -5 is the most negative number and therefore the negative entry is -5.
  • Calculate the ratio value by using the following formula.

Ratio=Right hand side value fo the constraintCoefficient of entering variable in the constraint

 Cj 53100 
BaseCbXbX1X2X3X4X5Ratio= Xb/ X1
X40611310(6/1)=6
X501553601(15/5)=3
Zj0000000 
Zj-Cj 0-5-3-100 

R215R2

R1R1R2

The resultant iteration table is as shown below:

 Cj 53100
BaseCbXbX1X2X3X4X5
X40300.41.81-0.2
X15310.61.200.2
Zj 1553601
Zj-Cj 1500501

Since, the last row Zj-Cj contains all positive entries, the solution is optimal. Therefore, the value decision variables and Max Z is,

x1= 5 x2= 0        x3= 0        Max Z = 5x1+3x2+x3 Max Z = (5×0)+(3×5)+(1×0)Max Z = 15

The optimal maximized Z value is 15.

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Chapter 4 Solutions

Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms

Ch. 4.5 - Prob. 1PCh. 4.5 - Prob. 2PCh. 4.5 - Prob. 3PCh. 4.5 - Prob. 4PCh. 4.5 - Prob. 5PCh. 4.5 - Prob. 6PCh. 4.5 - Prob. 7PCh. 4.6 - Prob. 1PCh. 4.6 - Prob. 2PCh. 4.6 - Prob. 3PCh. 4.6 - Prob. 4PCh. 4.7 - Prob. 1PCh. 4.7 - Prob. 2PCh. 4.7 - Prob. 3PCh. 4.7 - Prob. 4PCh. 4.7 - Prob. 5PCh. 4.7 - Prob. 6PCh. 4.7 - Prob. 7PCh. 4.7 - Prob. 8PCh. 4.7 - Prob. 9PCh. 4.8 - Prob. 1PCh. 4.8 - Prob. 2PCh. 4.8 - Prob. 3PCh. 4.8 - Prob. 4PCh. 4.8 - Prob. 5PCh. 4.8 - Prob. 6PCh. 4.10 - Prob. 1PCh. 4.10 - Prob. 2PCh. 4.10 - Prob. 3PCh. 4.10 - Prob. 4PCh. 4.10 - Prob. 5PCh. 4.11 - Prob. 1PCh. 4.11 - Prob. 2PCh. 4.11 - Prob. 3PCh. 4.11 - Prob. 4PCh. 4.11 - Prob. 5PCh. 4.11 - Prob. 6PCh. 4.12 - Prob. 1PCh. 4.12 - Prob. 2PCh. 4.12 - Prob. 3PCh. 4.12 - Prob. 4PCh. 4.12 - Prob. 5PCh. 4.12 - Prob. 6PCh. 4.13 - Prob. 2PCh. 4.14 - Prob. 1PCh. 4.14 - Prob. 2PCh. 4.14 - Prob. 3PCh. 4.14 - Prob. 4PCh. 4.14 - Prob. 5PCh. 4.14 - Prob. 6PCh. 4.14 - Prob. 7PCh. 4.16 - Prob. 1PCh. 4.16 - Prob. 2PCh. 4.16 - Prob. 3PCh. 4.16 - Prob. 5PCh. 4.16 - Prob. 7PCh. 4.16 - Prob. 8PCh. 4.16 - Prob. 9PCh. 4.16 - Prob. 10PCh. 4.16 - Prob. 11PCh. 4.16 - Prob. 12PCh. 4.16 - Prob. 13PCh. 4.16 - Prob. 14PCh. 4.17 - Prob. 1PCh. 4.17 - Prob. 2PCh. 4.17 - Prob. 3PCh. 4.17 - Prob. 4PCh. 4.17 - Prob. 5PCh. 4.17 - Prob. 7PCh. 4.17 - Prob. 8PCh. 4 - Prob. 1RPCh. 4 - Prob. 2RPCh. 4 - Prob. 3RPCh. 4 - Prob. 4RPCh. 4 - Prob. 5RPCh. 4 - Prob. 6RPCh. 4 - Prob. 7RPCh. 4 - Prob. 8RPCh. 4 - Prob. 9RPCh. 4 - Prob. 10RPCh. 4 - Prob. 12RPCh. 4 - Prob. 13RPCh. 4 - Prob. 14RPCh. 4 - Prob. 16RPCh. 4 - Prob. 17RPCh. 4 - Prob. 18RPCh. 4 - Prob. 19RPCh. 4 - Prob. 20RPCh. 4 - Prob. 21RPCh. 4 - Prob. 22RPCh. 4 - Prob. 23RPCh. 4 - Prob. 24RPCh. 4 - Prob. 26RPCh. 4 - Prob. 27RPCh. 4 - Prob. 28RP
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