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The linear programming problem using simplex method. The optimal solution is x 1 = 50 , x 2 = 100 , s 1 = 0 , z = 150 _ . Procedure used: “ Selecting Basic and Non-basic Variables for the Simplex Method Given a simplex tableau, Step 1: Numbers of variables- Determine the number of basic variables and the number of non-basic variables. These numbers do not change during the simplex process. Step 2: Selecting basic variables - A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element (usually 1) and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. This procedure always selects P as a basic variable, since the P column never changes during the simplex process. Step 3: Selecting non-basic variables - After the basic variables are selected in step 2, the remaining variables are selected as the non-basic variables. The tableau columns under the non-basic variables usually contain more than one nonzero element. Selecting the Pivot Element Step 1: Locate the most negative indicator in the bottom row of the tableau to the left of the P column (the negative number with the largest absolute value). The column containing this element is the pivot column. If there is a tie for the most negative indicator, choose either column. Step 2: Divide each positive element in the pivot column above the dashed line into the corresponding element in the last column. The pivot row is the row corresponding to the smallest quotient obtained. If there is a tie for the smallest quotient, choose either row. If the pivot column above the dashed line has no positive elements, there is no solution, and stop. Step 3: The pivot (or pivot element) is the element at the intersection of the pivot column and pivot row. Performing a Pivot Operation A pivot operation, or pivoting, consists of performing row operations as follows: Step 1: Multiply the pivot row by the reciprocal of the pivot element to transform the pivot element into a 1. (If the pivot element is already a 1, omit this step.) Step 2: Add multiples of the pivot row to other rows in the tableau to transform all other nonzero elements in the pivot column into 0's”. Calculation: The given linear programming problem is; Maximize z = x 1 + x 2 subject to 3 x 1 + 4 x 2 ≤ 550 5 x 1 + 4 x 2 ≤ 650 x 1 , x 2 ≥ 0 To convert the problem constraints into a system of equations slack variables is used. Since there are m = 2 problem constraints, so it needs three slack variables s 1 , s 2 to convert the inequality into equality. 3 x 1 + 4 x 2 + s 1 = 550 4 x 1 + 4 x 2 +s 2 = 650 − x 1 − x 2 + P = 0 Form the simplex tableau: z x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 3 4 1 0 550 0 5 4 0 1 650 1 − 1 − 1 0 0 0 ] A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element (usually 1) and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. So, the basic variables are s 1 , s 2 , s 3 , P . And non-basic variables are x 1 , x 2 . To select the pivot column, select the most negative number in the bottom row of the tableau and to the left of the P column. The column containing this element is the pivot column . If there is a tie for the most negative indicator, choose either column. Here is a tie for the choice of the first pivot column. So, let choose the column 1 as the first pivot column. To select the pivot row, check for the smallest quotient for each row above the dashed line. If there is a tie for the smallest quotient, choose either row. 550 3 ≈ 183.33 650 5 = 130 ( Smallest ) So, the second row is selected as the pivot row. The element at the intersection of the pivot column and the pivot row is called the pivot element. z x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 3 4 1 0 550 0 5 4 0 1 650 1 − 1 − 1 0 0 0 ] So, the element enclosed in rectangle 5 is the pivot element. The variable on the top of the pivot column is called as the entering variable. the variable to the left of the pivot row is the exiting variable. So, x 1 is the entering variable and s 2 is the exiting variable. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 3 4 1 0 550 5 4 0 1 650 − 1 − 1 0 0 0 ] x 1 x 2 s 1 s 2 P s 1 s 2 P [ 3 4 1 0 550 5 4 0 1 650 0 − 0.2 0 0.2 130 ] R 3 + 1 5 R 1 ~ [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] R 1 − 3 5 R 2 Note that there are negative indicators in the bottom row. Here is a tie for the choice of the second pivot column. So, let choose the column 2 as the second pivot column. To select the pivot row, check for the smallest quotient for each row above the dashed line. If there is a tie for the smallest quotient, choose either row. 160 1.6 = 100 ( Smallest ) 650 4 = 162.5 So, the first row is selected as the pivot row. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] So, the element enclosed in rectangle 1.6 is the pivot element. The variable on the top of the pivot column is called as the entering variable. the variable to the left of the pivot row is the exiting variable. So, x 2 is the entering variable and s 1 is the exiting variable. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] ~ [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 0 0.125 0.125 150 ] R 3 + 1 8 R 1 ~ [ 0 1.6 1 − 0.6 160 5 0 − 2.5 2.5 250 0 0 0.125 0.125 150 ] R 2 − 2.5 R 1 Since there are no more negative indicators in the bottom row, the optimal solution has been found. Therefore, the corresponding entries in Row 3 is zero, thus x 1 , x 2 are basic while s 1 and s 2 are non-basic. Hence, the optimal solution is x 1 = 50 , x 2 = 100 , s 1 = 0 , z = 150 _ .

The linear programming problem using simplex method. The optimal solution is x 1 = 50 , x 2 = 100 , s 1 = 0 , z = 150 _ . Procedure used: “ Selecting Basic and Non-basic Variables for the Simplex Method Given a simplex tableau, Step 1: Numbers of variables- Determine the number of basic variables and the number of non-basic variables. These numbers do not change during the simplex process. Step 2: Selecting basic variables - A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element (usually 1) and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. This procedure always selects P as a basic variable, since the P column never changes during the simplex process. Step 3: Selecting non-basic variables - After the basic variables are selected in step 2, the remaining variables are selected as the non-basic variables. The tableau columns under the non-basic variables usually contain more than one nonzero element. Selecting the Pivot Element Step 1: Locate the most negative indicator in the bottom row of the tableau to the left of the P column (the negative number with the largest absolute value). The column containing this element is the pivot column. If there is a tie for the most negative indicator, choose either column. Step 2: Divide each positive element in the pivot column above the dashed line into the corresponding element in the last column. The pivot row is the row corresponding to the smallest quotient obtained. If there is a tie for the smallest quotient, choose either row. If the pivot column above the dashed line has no positive elements, there is no solution, and stop. Step 3: The pivot (or pivot element) is the element at the intersection of the pivot column and pivot row. Performing a Pivot Operation A pivot operation, or pivoting, consists of performing row operations as follows: Step 1: Multiply the pivot row by the reciprocal of the pivot element to transform the pivot element into a 1. (If the pivot element is already a 1, omit this step.) Step 2: Add multiples of the pivot row to other rows in the tableau to transform all other nonzero elements in the pivot column into 0's”. Calculation: The given linear programming problem is; Maximize z = x 1 + x 2 subject to 3 x 1 + 4 x 2 ≤ 550 5 x 1 + 4 x 2 ≤ 650 x 1 , x 2 ≥ 0 To convert the problem constraints into a system of equations slack variables is used. Since there are m = 2 problem constraints, so it needs three slack variables s 1 , s 2 to convert the inequality into equality. 3 x 1 + 4 x 2 + s 1 = 550 4 x 1 + 4 x 2 +s 2 = 650 − x 1 − x 2 + P = 0 Form the simplex tableau: z x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 3 4 1 0 550 0 5 4 0 1 650 1 − 1 − 1 0 0 0 ] A variable can be selected as a basic variable only if it corresponds to a column in the tableau that has exactly one nonzero element (usually 1) and the nonzero element in the column is not in the same row as the nonzero element in the column of another basic variable. So, the basic variables are s 1 , s 2 , s 3 , P . And non-basic variables are x 1 , x 2 . To select the pivot column, select the most negative number in the bottom row of the tableau and to the left of the P column. The column containing this element is the pivot column . If there is a tie for the most negative indicator, choose either column. Here is a tie for the choice of the first pivot column. So, let choose the column 1 as the first pivot column. To select the pivot row, check for the smallest quotient for each row above the dashed line. If there is a tie for the smallest quotient, choose either row. 550 3 ≈ 183.33 650 5 = 130 ( Smallest ) So, the second row is selected as the pivot row. The element at the intersection of the pivot column and the pivot row is called the pivot element. z x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 3 4 1 0 550 0 5 4 0 1 650 1 − 1 − 1 0 0 0 ] So, the element enclosed in rectangle 5 is the pivot element. The variable on the top of the pivot column is called as the entering variable. the variable to the left of the pivot row is the exiting variable. So, x 1 is the entering variable and s 2 is the exiting variable. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 3 4 1 0 550 5 4 0 1 650 − 1 − 1 0 0 0 ] x 1 x 2 s 1 s 2 P s 1 s 2 P [ 3 4 1 0 550 5 4 0 1 650 0 − 0.2 0 0.2 130 ] R 3 + 1 5 R 1 ~ [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] R 1 − 3 5 R 2 Note that there are negative indicators in the bottom row. Here is a tie for the choice of the second pivot column. So, let choose the column 2 as the second pivot column. To select the pivot row, check for the smallest quotient for each row above the dashed line. If there is a tie for the smallest quotient, choose either row. 160 1.6 = 100 ( Smallest ) 650 4 = 162.5 So, the first row is selected as the pivot row. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] So, the element enclosed in rectangle 1.6 is the pivot element. The variable on the top of the pivot column is called as the entering variable. the variable to the left of the pivot row is the exiting variable. So, x 2 is the entering variable and s 1 is the exiting variable. x 1 x 2 s 1 s 2 P s 1 s 2 P [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 − 0.2 0 0.2 130 ] ~ [ 0 1.6 1 − 0.6 160 5 4 0 1 650 0 0 0.125 0.125 150 ] R 3 + 1 8 R 1 ~ [ 0 1.6 1 − 0.6 160 5 0 − 2.5 2.5 250 0 0 0.125 0.125 150 ] R 2 − 2.5 R 1 Since there are no more negative indicators in the bottom row, the optimal solution has been found. Therefore, the corresponding entries in Row 3 is zero, thus x 1 , x 2 are basic while s 1 and s 2 are non-basic. Hence, the optimal solution is x 1 = 50 , x 2 = 100 , s 1 = 0 , z = 150 _ .

ADVANCED ENGINEERING MATH.>CUSTOM<

ADVANCED ENGINEERING MATH.>CUSTOM<

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Chapter 22 Solutions

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Ch. 22.1 - Prob. 1P Ch. 22.1 - Prob. 2P Ch. 22.1 - Prob. 3P Ch. 22.1 - Prob. 4P Ch. 22.1 - Prob. 5P Ch. 22.1 - Prob. 6P Ch. 22.1 - Prob. 7P Ch. 22.1 - Prob. 8P Ch. 22.1 - Prob. 9P Ch. 22.1 - CAS EXPERIMENT. Steepest Descent. (a) Write a...

Ch. 22.2 - Prob. 1P Ch. 22.2 - Prob. 2P Ch. 22.2 - Prob. 3P Ch. 22.2 - Prob. 4P Ch. 22.2 - Prob. 5P Ch. 22.2 - Prob. 6P Ch. 22.2 - Prob. 8P Ch. 22.2 - Prob. 9P Ch. 22.2 - Prob. 10P Ch. 22.2 - Prob. 11P Ch. 22.2 - Prob. 12P Ch. 22.2 - Prob. 13P Ch. 22.2 - Prob. 14P Ch. 22.2 - Prob. 15P Ch. 22.2 - Prob. 16P Ch. 22.2 - Prob. 17P Ch. 22.2 - Prob. 18P Ch. 22.2 - Prob. 19P Ch. 22.2 - Prob. 20P Ch. 22.2 - Prob. 21P Ch. 22.2 - Prob. 22P Ch. 22.3 - Prob. 1P Ch. 22.3 - Prob. 2P Ch. 22.3 - Prob. 3P Ch. 22.3 - Prob. 4P Ch. 22.3 - Prob. 5P Ch. 22.3 - Prob. 6P Ch. 22.3 - Prob. 7P Ch. 22.3 - Prob. 8P Ch. 22.3 - Prob. 9P Ch. 22.3 - Prob. 10P Ch. 22.3 - Prob. 11P Ch. 22.3 - Prob. 12P Ch. 22.3 - Prob. 13P Ch. 22.3 - Prob. 14P Ch. 22.4 - Prob. 1P Ch. 22.4 - Prob. 2P Ch. 22.4 - Prob. 3P Ch. 22.4 - Prob. 4P Ch. 22.4 - Prob. 6P Ch. 22.4 - Prob. 7P Ch. 22.4 - Prob. 8P Ch. 22.4 - Prob. 9P Ch. 22 - Prob. 1RQ Ch. 22 - Prob. 2RQ Ch. 22 - Prob. 3RQ Ch. 22 - Prob. 4RQ Ch. 22 - Prob. 5RQ Ch. 22 - Prob. 7RQ Ch. 22 - Prob. 8RQ Ch. 22 - Prob. 9RQ Ch. 22 - Prob. 10RQ Ch. 22 - Prob. 11RQ Ch. 22 - Prob. 13RQ Ch. 22 - Prob. 14RQ Ch. 22 - Prob. 15RQ Ch. 22 - Prob. 16RQ Ch. 22 - Prob. 17RQ Ch. 22 - Prob. 18RQ Ch. 22 - Prob. 19RQ Ch. 22 - Prob. 20RQ

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