Consider the Following Problem

Question

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The Answer is (a) $(a) Thus,the optimal solution is (x1,x2)= (4,2)with Z = 16. (b)We introduce the surplus variables x3,x4,slack variable x6, and artificial variables x5 and x7. (Note that these supplementary variables can be numbered in any order.)Using the Big M method,the introduction of these supplementary variables gives the following form of the problem.Minimize Z = 3 x1 + 2 x2 + M x5 + M x7,subject to 2x1 + x2 - x3 + x5 = 10 -3x1 + 2x2 + x6 = 6 x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Converting from minimization to maximization,we next have Maximize Z = - 3 x1 - 2 x2 - M x5 - M x7,subject to (1)2x1 + x2 - x3 + x5 = 10 (2)-3x1 + 2x2 + x6 = 6 (3)x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Let x5,x6 and x7 be the basic variables.The above objective function gives the following Equation (0).(0)Z + 3x1 + 2x2 + Mx5 + Mx7 = 0 Proper form from Gaussian elimination requires that the basic variables x5 and x7 be eliminated from this equation.This is accomplished by subtracting both M times Eq.(1)and M times Eq.(3)from Eg.(0),which yields (0)Z + (-3M + 3)x1 + (-2M + 2)x2 +Mx3 +M x4 = -16 M The corresponding initial simplex tableau is as follows. The initial (artificial)BF solution is (0,0,0,0,10,6,6).The entering basic variable is x1; the leaving basic variable is x5.(c)Iteration 1: The negative coefficients of x1 and x2 in Eq.(0)of the above initial simplex tableau indicates that the initial BF solution is not optimal.After choosing the initial entering basic variable x1 and leaving basic variable x5 as noted above,proper form from Gaussian elimination is restored by dividing Eq.(1)by 2,adding (3M-3)times this new Eq.(1)to Eq.(0),adding 3 times the new Eq.(1)to Eq.(2),and subtracting the new Eq.(1)from Eq.(3).This yields the simplex tableau shown below. The resulting BF solution is (5,0,0,0,0,21,1),which is not optimal because of the negative coefficients of x2 and x3 in Eq.(0).Iteration 2: Comparing the negative coefficients of x2 and x3 in the above Eq.(0),(-0.5M + 0.5)is slightly larger in absolute value than (-0.5M + 1.5),so x2 is chosen as the entering basic variable and x7 as the leaving basic variable,as indicated in the boxes in the above simplex tableau.After restoring proper form from Gaussian elimination,the following simplex tableau is obtained. The current BF solution (4,2,0,0,0,14,0)is optimal,since all the coefficients in Eq.(0)are nonnegative.The resulting optimal solution for the decision variables of the problem is (x1,x2)= (4,2)with Z = 16.$
Thus,the optimal solution is (x₁,x₂)= (4,2)with Z = 16.
(b)We introduce the surplus variables x₃,x₄,slack variable x₆,and artificial variables x₅ and x_7.(Note that these supplementary variables can be numbered in any order.)Using the Big M method,the introduction of these supplementary variables gives the following form of the problem.Minimize Z = 3 x₁ + 2 x₂+ M x₅ + M x₇,subject to 2x₁ + x₂ - x₃ + x₅ = 10 -3x₁ + 2x₂+ x₆ = 6 x₁ + x₂ - x₄ + x₇= 6 and x₁

\ge

0,x₂

\ge

0,x₃

\ge

0,x₄

\ge

0,x₅

\ge

0,x₆

\ge

0,x₇ ≥ 0.Converting from minimization to maximization,we next have Maximize Z = - 3 x₁ - 2 x₂- M x₅ - M x₇,subject to (1)2x₁ + x₂ - x₃ + x₅ = 10 (2)-3x₁ + 2x₂ + x₆ = 6 (3)x₁ + x₂ - x₄ + x₇= 6 and x₁

\ge

0,x₂

\ge

0,x₃

\ge

0,x₄

\ge

0,x₅

\ge

0,x₆

\ge

0,x₇ ≥ 0.Let x₅,x₆ and x₇ be the basic variables.The above objective function gives the following Equation (0).(0)Z + 3x₁+ 2x₂ + Mx₅+ Mx₇= 0 Proper form from Gaussian elimination requires that the basic variables x₅and x₇ be eliminated from this equation.This is accomplished by subtracting both M times Eq.(1)and M times Eq.(3)from Eg.(0),which yields (0)Z + (-3M + 3)x₁+ (-2M + 2)x₂ +Mx₃+M x₄= -16 M The corresponding initial simplex tableau is as follows. $(a) Thus,the optimal solution is (x1,x2)= (4,2)with Z = 16. (b)We introduce the surplus variables x3,x4,slack variable x6, and artificial variables x5 and x7. (Note that these supplementary variables can be numbered in any order.)Using the Big M method,the introduction of these supplementary variables gives the following form of the problem.Minimize Z = 3 x1 + 2 x2 + M x5 + M x7,subject to 2x1 + x2 - x3 + x5 = 10 -3x1 + 2x2 + x6 = 6 x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Converting from minimization to maximization,we next have Maximize Z = - 3 x1 - 2 x2 - M x5 - M x7,subject to (1)2x1 + x2 - x3 + x5 = 10 (2)-3x1 + 2x2 + x6 = 6 (3)x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Let x5,x6 and x7 be the basic variables.The above objective function gives the following Equation (0).(0)Z + 3x1 + 2x2 + Mx5 + Mx7 = 0 Proper form from Gaussian elimination requires that the basic variables x5 and x7 be eliminated from this equation.This is accomplished by subtracting both M times Eq.(1)and M times Eq.(3)from Eg.(0),which yields (0)Z + (-3M + 3)x1 + (-2M + 2)x2 +Mx3 +M x4 = -16 M The corresponding initial simplex tableau is as follows. The initial (artificial)BF solution is (0,0,0,0,10,6,6).The entering basic variable is x1; the leaving basic variable is x5.(c)Iteration 1: The negative coefficients of x1 and x2 in Eq.(0)of the above initial simplex tableau indicates that the initial BF solution is not optimal.After choosing the initial entering basic variable x1 and leaving basic variable x5 as noted above,proper form from Gaussian elimination is restored by dividing Eq.(1)by 2,adding (3M-3)times this new Eq.(1)to Eq.(0),adding 3 times the new Eq.(1)to Eq.(2),and subtracting the new Eq.(1)from Eq.(3).This yields the simplex tableau shown below. The resulting BF solution is (5,0,0,0,0,21,1),which is not optimal because of the negative coefficients of x2 and x3 in Eq.(0).Iteration 2: Comparing the negative coefficients of x2 and x3 in the above Eq.(0),(-0.5M + 0.5)is slightly larger in absolute value than (-0.5M + 1.5),so x2 is chosen as the entering basic variable and x7 as the leaving basic variable,as indicated in the boxes in the above simplex tableau.After restoring proper form from Gaussian elimination,the following simplex tableau is obtained. The current BF solution (4,2,0,0,0,14,0)is optimal,since all the coefficients in Eq.(0)are nonnegative.The resulting optimal solution for the decision variables of the problem is (x1,x2)= (4,2)with Z = 16.$
The initial (artificial)BF solution is (0,0,0,0,10,6,6).The entering basic variable is x₁; the leaving basic variable is x₅.(c)Iteration 1: The negative coefficients of x₁and x₂ in Eq.(0)of the above initial simplex tableau indicates that the initial BF solution is not optimal.After choosing the initial entering basic variable x₁and leaving basic variable x₅ as noted above,proper form from Gaussian elimination is restored by dividing Eq.(1)by 2,adding (3M-3)times this new Eq.(1)to Eq.(0),adding 3 times the new Eq.(1)to Eq.(2),and subtracting the new Eq.(1)from Eq.(3).This yields the simplex tableau shown below. $(a) Thus,the optimal solution is (x1,x2)= (4,2)with Z = 16. (b)We introduce the surplus variables x3,x4,slack variable x6, and artificial variables x5 and x7. (Note that these supplementary variables can be numbered in any order.)Using the Big M method,the introduction of these supplementary variables gives the following form of the problem.Minimize Z = 3 x1 + 2 x2 + M x5 + M x7,subject to 2x1 + x2 - x3 + x5 = 10 -3x1 + 2x2 + x6 = 6 x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Converting from minimization to maximization,we next have Maximize Z = - 3 x1 - 2 x2 - M x5 - M x7,subject to (1)2x1 + x2 - x3 + x5 = 10 (2)-3x1 + 2x2 + x6 = 6 (3)x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Let x5,x6 and x7 be the basic variables.The above objective function gives the following Equation (0).(0)Z + 3x1 + 2x2 + Mx5 + Mx7 = 0 Proper form from Gaussian elimination requires that the basic variables x5 and x7 be eliminated from this equation.This is accomplished by subtracting both M times Eq.(1)and M times Eq.(3)from Eg.(0),which yields (0)Z + (-3M + 3)x1 + (-2M + 2)x2 +Mx3 +M x4 = -16 M The corresponding initial simplex tableau is as follows. The initial (artificial)BF solution is (0,0,0,0,10,6,6).The entering basic variable is x1; the leaving basic variable is x5.(c)Iteration 1: The negative coefficients of x1 and x2 in Eq.(0)of the above initial simplex tableau indicates that the initial BF solution is not optimal.After choosing the initial entering basic variable x1 and leaving basic variable x5 as noted above,proper form from Gaussian elimination is restored by dividing Eq.(1)by 2,adding (3M-3)times this new Eq.(1)to Eq.(0),adding 3 times the new Eq.(1)to Eq.(2),and subtracting the new Eq.(1)from Eq.(3).This yields the simplex tableau shown below. The resulting BF solution is (5,0,0,0,0,21,1),which is not optimal because of the negative coefficients of x2 and x3 in Eq.(0).Iteration 2: Comparing the negative coefficients of x2 and x3 in the above Eq.(0),(-0.5M + 0.5)is slightly larger in absolute value than (-0.5M + 1.5),so x2 is chosen as the entering basic variable and x7 as the leaving basic variable,as indicated in the boxes in the above simplex tableau.After restoring proper form from Gaussian elimination,the following simplex tableau is obtained. The current BF solution (4,2,0,0,0,14,0)is optimal,since all the coefficients in Eq.(0)are nonnegative.The resulting optimal solution for the decision variables of the problem is (x1,x2)= (4,2)with Z = 16.$
The resulting BF solution is (5,0,0,0,0,21,1),which is not optimal because of the negative coefficients of x₂and x₃ in Eq.(0).Iteration 2: Comparing the negative coefficients of x₂and x₃ in the above Eq.(0),(-0.5M + 0.5)is slightly larger in absolute value than (-0.5M + 1.5),so x₂ is chosen as the entering basic variable and x₇as the leaving basic variable,as indicated in the boxes in the above simplex tableau.After restoring proper form from Gaussian elimination,the following simplex tableau is obtained. $(a) Thus,the optimal solution is (x1,x2)= (4,2)with Z = 16. (b)We introduce the surplus variables x3,x4,slack variable x6, and artificial variables x5 and x7. (Note that these supplementary variables can be numbered in any order.)Using the Big M method,the introduction of these supplementary variables gives the following form of the problem.Minimize Z = 3 x1 + 2 x2 + M x5 + M x7,subject to 2x1 + x2 - x3 + x5 = 10 -3x1 + 2x2 + x6 = 6 x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Converting from minimization to maximization,we next have Maximize Z = - 3 x1 - 2 x2 - M x5 - M x7,subject to (1)2x1 + x2 - x3 + x5 = 10 (2)-3x1 + 2x2 + x6 = 6 (3)x1 + x2 - x4 + x7 = 6 and x1 \ge 0,x2 \ge 0,x3 \ge 0,x4 \ge 0,x5 \ge 0,x6 \ge 0,x7 ≥ 0.Let x5,x6 and x7 be the basic variables.The above objective function gives the following Equation (0).(0)Z + 3x1 + 2x2 + Mx5 + Mx7 = 0 Proper form from Gaussian elimination requires that the basic variables x5 and x7 be eliminated from this equation.This is accomplished by subtracting both M times Eq.(1)and M times Eq.(3)from Eg.(0),which yields (0)Z + (-3M + 3)x1 + (-2M + 2)x2 +Mx3 +M x4 = -16 M The corresponding initial simplex tableau is as follows. The initial (artificial)BF solution is (0,0,0,0,10,6,6).The entering basic variable is x1; the leaving basic variable is x5.(c)Iteration 1: The negative coefficients of x1 and x2 in Eq.(0)of the above initial simplex tableau indicates that the initial BF solution is not optimal.After choosing the initial entering basic variable x1 and leaving basic variable x5 as noted above,proper form from Gaussian elimination is restored by dividing Eq.(1)by 2,adding (3M-3)times this new Eq.(1)to Eq.(0),adding 3 times the new Eq.(1)to Eq.(2),and subtracting the new Eq.(1)from Eq.(3).This yields the simplex tableau shown below. The resulting BF solution is (5,0,0,0,0,21,1),which is not optimal because of the negative coefficients of x2 and x3 in Eq.(0).Iteration 2: Comparing the negative coefficients of x2 and x3 in the above Eq.(0),(-0.5M + 0.5)is slightly larger in absolute value than (-0.5M + 1.5),so x2 is chosen as the entering basic variable and x7 as the leaving basic variable,as indicated in the boxes in the above simplex tableau.After restoring proper form from Gaussian elimination,the following simplex tableau is obtained. The current BF solution (4,2,0,0,0,14,0)is optimal,since all the coefficients in Eq.(0)are nonnegative.The resulting optimal solution for the decision variables of the problem is (x1,x2)= (4,2)with Z = 16.$
The current BF solution (4,2,0,0,0,14,0)is optimal,since all the coefficients in Eq.(0)are nonnegative.The resulting optimal solution for the decision variables of the problem is (x₁,x₂)= (4,2)with Z = 16.

Consider the Following Problem

Definitions: