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Math Advanced Math Custom Kreyszig: Advanced Engineering Mathematics

The linear programming problem using simplex method. The optimal solution is x 1 = 0.8 , x 2 = 2.1 , x 3 = 1.9 , z = 9.8 _ . 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 Z as a basic variable, since the Z 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 first negative indicator in the bottom row of the tableau to the left of the Z column. The column containing this element is the pivot column. If there is a tie for the 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: 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 optimal solution is given as: Maximize f = 2 x 1 + 3 x 2 + x 3 subject to x 1 + x 2 + x 3 ≤ 4.8 10 x 1 + x 3 ≤ 9.9 x 2 − x 3 ≤ 0.2 x 1 , x 2 ≥ 0 It is known that in order to convert the problem constraints into a system of equations slack variables is used. Since there are m = 3 problem constraints, the problem needs three slack variables s 1 , s 2 , s 3 to convert the inequality into equality. x 1 + x 2 + x 3 + s 1 = 4.8 10 x 1 + x 3 + s 2 = 9.9 x 2 − 3 x 3 + s 3 = 0.2 − 2 x 1 − 3 x 2 − x 3 + z = 0 Form the simplex tableau: z x 1 x 2 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] To select the pivot column, select the first negative number in the first row of the tableau. The column containing this element is the pivot column. In this problem, select second column as 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. 4.8 1 = 4.8 9.9 10 = 0.99 ( Smallest ) So, the third 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 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 10 is the pivot element. z x 1 x 2 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ~ [ 1 0 − 3 − 4 5 0 1 5 0 1.98 0 0 1 9 10 1 − 1 10 0 3.81 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ( R 1 → R 1 + 2 10 R 3 R 2 → R 2 − 1 10 R 3 ) Note that there is another negative indicator in third 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. 3.81 1 = 3.81 0.2 1 = 0.2 ( Smallest ) So, the fourth 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. [ 1 0 − 3 − 4 5 0 1 5 0 1.98 0 0 1 9 10 1 − 1 10 0 3.81 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 1 is the pivot element. [ 1 0 0 − 19 5 0 1 5 3 2.58 0 0 0 19 10 1 − 1 10 − 1 3.61 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ( R 1 → R 1 + 3 R 4 R 2 → R 2 − R 4 ) Note that there is another negative indicator in fourth 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. 3.61 × 10 19 = 1.9 ( Smallest ) 9.9 1 = 9.9 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. [ 1 0 0 − 19 5 0 1 5 3 2.58 0 0 0 1.9 1 − 1 10 − 1 3.61 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 1.9 is the pivot element. [ 1 0 0 0 2 0 1 9.8 0 0 0 1.9 1 − 1 10 − 1 3.61 0 10 0 0 − 10 19 20 19 10 19 8 0 0 1 0 10 19 − 1 19 9 19 2.1 ] ( R 1 → R 1 + 2 R 2 R 3 → R 3 − 10 19 R 2 R 4 → R 4 + 10 19 R 2 ) Note that there are no more negative indicators in the first row. Therefore, the optimal solution has been found. From the above matrix x 1 , x 2 , x 3 are basic variables and s 2 , s 1 , s 3 are non-basic variables. The value of x 1 is calculated as follows. 10 x 1 = 8 x 1 = 0.8 Also, the value of x 2 , x 3 is 2.1 , 1.9 respectively. Hence, the optimal solution is x 1 = 0.8 , x 2 = 2.1 , x 3 = 1.9 , z = 9.8 _ .

The linear programming problem using simplex method. The optimal solution is x 1 = 0.8 , x 2 = 2.1 , x 3 = 1.9 , z = 9.8 _ . 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 Z as a basic variable, since the Z 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 first negative indicator in the bottom row of the tableau to the left of the Z column. The column containing this element is the pivot column. If there is a tie for the 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: 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 optimal solution is given as: Maximize f = 2 x 1 + 3 x 2 + x 3 subject to x 1 + x 2 + x 3 ≤ 4.8 10 x 1 + x 3 ≤ 9.9 x 2 − x 3 ≤ 0.2 x 1 , x 2 ≥ 0 It is known that in order to convert the problem constraints into a system of equations slack variables is used. Since there are m = 3 problem constraints, the problem needs three slack variables s 1 , s 2 , s 3 to convert the inequality into equality. x 1 + x 2 + x 3 + s 1 = 4.8 10 x 1 + x 3 + s 2 = 9.9 x 2 − 3 x 3 + s 3 = 0.2 − 2 x 1 − 3 x 2 − x 3 + z = 0 Form the simplex tableau: z x 1 x 2 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] To select the pivot column, select the first negative number in the first row of the tableau. The column containing this element is the pivot column. In this problem, select second column as 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. 4.8 1 = 4.8 9.9 10 = 0.99 ( Smallest ) So, the third 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 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 10 is the pivot element. z x 1 x 2 x 3 s 1 s 2 s 3 [ 1 − 2 − 3 − 1 0 0 0 0 0 1 1 1 1 0 0 4.8 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ~ [ 1 0 − 3 − 4 5 0 1 5 0 1.98 0 0 1 9 10 1 − 1 10 0 3.81 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ( R 1 → R 1 + 2 10 R 3 R 2 → R 2 − 1 10 R 3 ) Note that there is another negative indicator in third 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. 3.81 1 = 3.81 0.2 1 = 0.2 ( Smallest ) So, the fourth 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. [ 1 0 − 3 − 4 5 0 1 5 0 1.98 0 0 1 9 10 1 − 1 10 0 3.81 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 1 is the pivot element. [ 1 0 0 − 19 5 0 1 5 3 2.58 0 0 0 19 10 1 − 1 10 − 1 3.61 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] ( R 1 → R 1 + 3 R 4 R 2 → R 2 − R 4 ) Note that there is another negative indicator in fourth 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. 3.61 × 10 19 = 1.9 ( Smallest ) 9.9 1 = 9.9 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. [ 1 0 0 − 19 5 0 1 5 3 2.58 0 0 0 1.9 1 − 1 10 − 1 3.61 0 10 0 1 0 1 0 9.9 0 0 1 − 1 0 0 1 0.2 ] So, the element enclosed in rectangle 1.9 is the pivot element. [ 1 0 0 0 2 0 1 9.8 0 0 0 1.9 1 − 1 10 − 1 3.61 0 10 0 0 − 10 19 20 19 10 19 8 0 0 1 0 10 19 − 1 19 9 19 2.1 ] ( R 1 → R 1 + 2 R 2 R 3 → R 3 − 10 19 R 2 R 4 → R 4 + 10 19 R 2 ) Note that there are no more negative indicators in the first row. Therefore, the optimal solution has been found. From the above matrix x 1 , x 2 , x 3 are basic variables and s 2 , s 1 , s 3 are non-basic variables. The value of x 1 is calculated as follows. 10 x 1 = 8 x 1 = 0.8 Also, the value of x 2 , x 3 is 2.1 , 1.9 respectively. Hence, the optimal solution is x 1 = 0.8 , x 2 = 2.1 , x 3 = 1.9 , z = 9.8 _ .

Custom Kreyszig: Advanced Engineering Mathematics

Custom Kreyszig: Advanced Engineering Mathematics

10th Edition

ISBN: 9781119166856

Author: Kreyszig

Publisher: JOHN WILEY+SONS INC.CUSTOM

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Custom Kreyszig: Advanced Engineering Mathematics

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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