Presentation on theme: "Nonstandard Problmes Produced by E. Gretchen Gascon."— Presentation transcript:
Nonstandard Problmes Produced by E. Gretchen Gascon
Concept of Mixed Problems Mixed problems are problems which have both greater than and less than constraints. Mixed problems cannot use the DUEL method for solving (Used in Section 4.3) (Do NOT transpose the matrix to convert to maximization problem) Mixed problems cannot use the Excel SOLVER function. See solving Nonstandard Problem steps page 189. Use slack variables + s, for ≤ and use surplus variable –s for ≥ and use +a (artificial variables) for =. Always eliminate the artificial variable first. (find row in which the artificial variable exists and then pivot on the leftmost positive entry in that row that is not in an artificial variable column. Basic variable – is the nonzero variable in a column of otherwise zeros. Investigating quotients – only use consider quotients >= 0
Problem 4.4 # 15 Step 1 – Convert problem to a maximization problem writing the negative of the w equation. Step 2 – Add slack variables and subtract surplus variables as needed.
Problem 4.4 # 15 con’t Step 3 – Write the initial simplex tableau Step 4 – If any basic variable(s1,s2,, z) has a negative value, locate the nonzero number in that variable’s column, and note what row it is in. (Row 2) Step 5- In the row located in step 4, find the positive entry that is farthest to the left, and note what column it is in. (Col 1)
Problem 4.4 # 15 con’t Step 6 – In the column found in Step 5, choose a pivot by investigating quotients. 150/10 and 100/20, 100/20 is smallest ( Row 2 Col 1 is pivot element) Step 7 – Use row operations to change the other number in the pivot column to 0
Problem 4.4 # 15 con’t (Basic variable are now y1, s1, z) (s2 is no longer a basic variable, because the column now has all non-zero elements and none of the basic variables mentioned are negative, you do not have to repeat steps 4 – 5) Step 8 Once a feasible solution has been found, continue to use the simplex method until the optimal solution is found. ( In this case, Column 3 has a negative in the bottom row. Select that as the pivot column and use investigating quotients to choose row 2 col 3 as the pivot element.
Problem 4.4 # 15 con’t Variable are = 0 which have all non zero values in the column Variable which have only one number can be converted into equations Note: The matrix give the value of z but, w is the opposite of z. (note conversion at the beginning of problem.
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