|
  
|
|
|
|
|
|
|
|
|
Operations Research - Linear Programming
Egwald's popular web pages are provided without cost to users.
Graphical Solution |
Simplex Algorithm |
Primal Simplex Tableaux Generator |
Dual Simplex Algorithm |
Dual Simplex Tableaux Generator
firm's least cost decision | sensitivity analysis | dual simplex method | dual simplex tableaux The Egwald's Coffee Import Decision. Egwald's Coffee, a purveyor of speciality coffee, imports Yemen Mocha and Ethopian Harrar coffee beans from the Middle East, and Java and Sumatra coffee beans from Indonesia. Egwald has a standing order with coffee merchants for at least one hundred bags (60 kilos per bag) of "green coffee beans" every three months for each single-origin coffee, plus additional coffee beans as required, all at the prevailing prices. After a secret process of aging, roasting, and blending, Egwald markets this coffee, generically known as Mocha-Java, under the trade name of MoJo Coffee Egwald. Egwald blends the roasted beans according to four formulae. Formula 1 combines 1 part Mocha with 2 parts Java; Formula 2 combines 1 part Mocha with 2 parts Sumatra; Formula 3 combines 1 part Harrar with 2 parts Java; Formula 4 combines 1 part Harrar with 2 parts Sumatra. At the beginning of each quarter of the year, Egwald estimates the demand for MoJo Coffee Egwald, and orders quantities of green coffee beans to replenish inventories based on the prevailing coffee bean prices, taking into consideration the differentials in direct costs associated with processing beans according to each formula. Egwald's decision involves minimizing the cost of producing MoJo - a least cost formulation problem - subject to expected quarterly demand from coffee lovers for MoJo, the price of green coffee beans, and processing costs for each blend of MoJo coffee. Define the variables of the problem according to the following table:
The first line in the table reads that the formula for blend 1 of MoJo combines Mocha and Java roasted coffee beans; the second line reads that the formula for blend 2 combines Mocha and Sumatra roasted coffee beans; etc. The second table illustrates Egwald Coffee's quarterly standing default order from each single-origin coffee supplier: Dual Simplex Solution. The primal objective of the Egwald's Coffee is to minimize costs, by ordering appropriate amounts of coffee beans. The firm's primal production decision is captured in the firm's Primal Linear Programming Problem, where I multiplied the coefficients of the objective function, in $ per sack, by -1 to turn the minimization problem into a maximization problem.
subject to the constraints
The standard form for the initial dual simplex tableau is:
In the dual simplex algorithm, a sequence of tableaux are calculated. A given Tableauk has the form:
after consolidating the center 2 by 2 block of matrices.
The Tableaux of the Dual Simplex Method. Phase 0: Drive the artificial variables from the basis.
Basis for Tableau0: [s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, s11, s12, s13, ]. Value of Objective Function = 0. Proceed to the next tableau as follows: Phase 0: Drive the artificial variables from the basis. A. In Tableau0: 1. Select an artificial variable in the basis: s*13. 2. Select a nonzero element in row F4 as pivot: x13,4 = -2. The pivot column = 4. B. To create Tableau1: 3. Compute row F41 = F40 / (-2). 4. Subtract multiples of row F41 from all other rows of Tableau0 so that column x14 = e13 in Tableau1.
Basis for Tableau1: [s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, s11, s12, x4, ]. Value of Objective Function = 0. Proceed to the next tableau as follows: Phase 0: Drive the artificial variables from the basis. A. In Tableau1: 1. Select an artificial variable in the basis: s*12. 2. Select a nonzero element in row F3 as pivot: x12,3 = -2. The pivot column = 3. B. To create Tableau2: 3. Compute row F32 = F31 / (-2). 4. Subtract multiples of row F32 from all other rows of Tableau1 so that column x23 = e12 in Tableau2.
Basis for Tableau2: [s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, s11, x3, x4, ]. Value of Objective Function = 0. Proceed to the next tableau as follows: Phase 0: Drive the artificial variables from the basis. A. In Tableau2: 1. Select an artificial variable in the basis: s*11. 2. Select a nonzero element in row F2 as pivot: x11,2 = -2. The pivot column = 2. B. To create Tableau3: 3. Compute row F23 = F22 / (-2). 4. Subtract multiples of row F23 from all other rows of Tableau2 so that column x32 = e11 in Tableau3.
Basis for Tableau3: [s1, s2, s3, s4, s5, s6, s7, s8, s9, s10, x2, x3, x4, ]. Value of Objective Function = 0. Proceed to the next tableau as follows: Phase 0: Drive the artificial variables from the basis. A. In Tableau3: 1. Select an artificial variable in the basis: s*10. 2. Select a nonzero element in row F1 as pivot: x10,1 = -2. The pivot column = 1. B. To create Tableau4: 3. Compute row F14 = F13 / (-2). 4. Subtract multiples of row F14 from all other rows of Tableau3 so that column x41 = e10 in Tableau4. Phase II: Goal: get ß >= 0.
Basis for Tableau4: [s1, s2, s3, s4, s5, s6, s7, s8, s9, x1, x2, x3, x4, ]. Value of Objective Function = 0. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau4: 1. Select a pivot row, row, with b4row < 0: row = 9 associated with b49 = -1000. 2. Compute the ratios -Ø / MJ as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 12 associated with 9.3. Thus x9,12 = -1 is the pivot; variable s9 will leave the basis; variable x12 will enter the basis.B. To create Tableau5: 3. Compute row MJ5 = MJ4 / (-1). 3. Subtract multiples of row MJ5 from all other rows of Tableau4 so that column x12 = e9 in Tableau5.
Basis for Tableau5: [s1, s2, s3, s4, s5, s6, s7, s8, x12, x1, x2, x3, x4, ]. Value of Objective Function = -9282. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau5: 1. Select a pivot row, row, with b5row < 0: row = 8 associated with b58 = -1000. 2. Compute the ratios -Ø / B4 as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 10 associated with 1.3. Thus x8,10 = -1 is the pivot; variable s8 will leave the basis; variable x10 will enter the basis.B. To create Tableau6: 3. Compute row B46 = B45 / (-1). 3. Subtract multiples of row B46 from all other rows of Tableau5 so that column x10 = e8 in Tableau6.
Basis for Tableau6: [s1, s2, s3, s4, s5, s6, s7, x10, x12, x1, x2, x3, x4, ]. Value of Objective Function = -10608. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau6: 1. Select a pivot row, row, with b6row < 0: row = 6 associated with b66 = -1000. 2. Compute the ratios -Ø / B2 as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 11 associated with 1.3. Thus x6,11 = -1 is the pivot; variable s6 will leave the basis; variable x11 will enter the basis.B. To create Tableau7: 3. Compute row B27 = B26 / (-1). 3. Subtract multiples of row B27 from all other rows of Tableau6 so that column x11 = e6 in Tableau7.
Basis for Tableau7: [s1, s2, s3, s4, s5, x11, s7, x10, x12, x1, x2, x3, x4, ]. Value of Objective Function = -11934. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau7: 1. Select a pivot row, row, with b7row < 0: row = 7 associated with b77 = -1000. 2. Compute the ratios -Ø / B3 as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 9 associated with 2.7. Thus x7,9 = -1 is the pivot; variable s7 will leave the basis; variable x9 will enter the basis.B. To create Tableau8: 3. Compute row B38 = B37 / (-1). 3. Subtract multiples of row B38 from all other rows of Tableau7 so that column x9 = e7 in Tableau8.
Basis for Tableau8: [s1, s2, s3, s4, s5, x11, x9, x10, x12, x1, x2, x3, x4, ]. Value of Objective Function = -14586. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau8: 1. Select a pivot row, row, with b8row < 0: row = 5 associated with b85 = -1000. 2. Compute the ratios -Ø / B1 as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 6 associated with 110.1. Thus x5,6 = -1.5 is the pivot; variable s5 will leave the basis; variable x6 will enter the basis.B. To create Tableau9: 3. Compute row B19 = B18 / (-1.5). 3. Subtract multiples of row B19 from all other rows of Tableau8 so that column x6 = e5 in Tableau9.
Basis for Tableau9: [s1, s2, s3, s4, x6, x11, x9, x10, x12, x1, x2, x3, x4, ]. Value of Objective Function = -124644. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau9: 1. Select a pivot row, row, with b9row < 0: row = 1 associated with b91 = -100. 2. Compute the ratios -Ø / M as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 5 associated with 21.2. Thus x1,5 = -0.5 is the pivot; variable s1 will leave the basis; variable x5 will enter the basis.B. To create Tableau10: 3. Compute row M10 = M9 / (-0.5). 3. Subtract multiples of row M10 from all other rows of Tableau9 so that column x5 = e1 in Tableau10.
Basis for Tableau10: [x5, s2, s3, s4, x6, x11, x9, x10, x12, x1, x2, x3, x4, ]. Value of Objective Function = -126765.6. Proceed to the next tableau as follows: Phase 0: Complete. Phase I: Complete. Phase II: Goal: get ß >= 0. A. In Tableau10: 1. Select a pivot row, row, with b10row < 0: row = 4 associated with b104 = -100. 2. Compute the ratios -Ø / S as per the last row. Discard ratios which are not positive and ratios associated with artificial variables. Select the column with the least positive ratio as the pivot column: col = 7 associated with 0.7. Thus x4,7 = -1 is the pivot; variable s4 will leave the basis; variable x7 will enter the basis.B. To create Tableau11: 3. Compute row S11 = S10 / (-1). 3. Subtract multiples of row S11 from all other rows of Tableau10 so that column x7 = e4 in Tableau11.
Basis for Tableau11: [x5, s2, s3, x7, x6, x11, x9, x10, x12, x1, x2, x3, x4]. Value of Objective Function = -126831.9. Phase 0: Complete. Phase I: Complete. Phase II: Complete. Primal Solution: [x5, s2, s3, x7, x6, x11, x9, x10, x12, x1, x2, x3, x4] = [100, 133.3, 466.7, 100, 466.7, 700, 150, 150, 0, 50, 50, 233.3, 0]; P = 126831.9. (Primal x variables not in the basis have a value of 0). Dual Solution: [yM, yH, yJ, yS, yB1, yB2, yB3, yB4, yMJ, yF1, yF2, yF3, yF4] = [21.2, 0, 0, 0.7, 110.1, 114, 112.7, 115.4, 124.6, 17.2, 15.2, 19.9, 18.6]; D = 126831.9. Egwald Coffee's Least Cost Import Solution. Sacks of coffee imported and blends produced.
Purchasing and processing costs
Post Optimality Analysis. Over what ranges of the values of the coefficients of the model, such as costs ( c1, c2, c3, c4, c5, c6, c7, c8, c9, c10, c11, c12), or the constraints on inputs and outputs ( bM, bH, bJ, bS, bB1, bB2, bB3, bB4, bMJ, bF1, bF2, bF3, bF4), will the above basic feasible solution remain optimal? By the phrase "remain optimal," I mean that the variables in the basis of the current optimal basic solution remain in the basis of the new optimal basic solution, after the values of the coefficients are altered. As long as the list of variables in each basis is the same, even though the values of the variables may change with the changes in the coefficients of the model, the current basic solution is said to remain optimal. The primal variables, x1, x2, x3, x4, x5, x6, x7, x8, x9, x10, x11, x12, take on the values 50, 50, 233.33, 0, 100, 466.67, 100, 0, 150, 150, 700, 0, with per bag costs of 165.75, 165.75, 152.49, 152.49, 92.82, 92.82, 99.45, 99.45, 14.59, 10.61, 11.93, 9.28, and per pound costs of 1.25, 1.25, 1.15, 1.15, 0.7, 0.7, 0.75, 0.75, 0.11, 0.08, 0.09, 0.07. The basis variables associated with rows (1 to 13) of Tableau11 are the variables ( x5, s2, s3, x7, x6, x11, x9, x10, x12, x1, x2, x3, x4 ) whose associated objective function coefficients are ( 92.82, 0, 0, 99.45, 92.82, 11.93, 14.59, 10.61, 9.28, 165.75, 165.75, 152.49, 152.49). Objective Function Coefficients - NonBasic Variables By how much can one decrease the cost (increase the profit) of x8 before this variable will enter the basis of the new basic optimal solution. To determine the entry cost for x8, one computes the opportunity cost of producing one unit of x8 = the increase in costs from producing one unit of x8 = z8, using standard L.P. notation. The value of z8 is calculated by weighting each entry of the column associated with x8 in Tableau11, xi, 8, by the objective function coefficient associated with the row of that entry:
z8 =
-92.82 * -1 + 0 * 0 + 0 * 0 + -99.45 * 1 + -92.82 * 1 + -11.93 * 1.5 + -14.59 * -1.5 + -10.61 * 1.5 + Per bag costs for x8 can decrease from 99.45 to 97.46, and the cost per pound of x8 can decrease from 0.75 to 0.735, before x8 will enter the basis in the new optimal primal solution. Objective Function Coefficients - Basic Variables By how much can one increase the cost of variable x5 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. To determine the basis upset price for x5, one needs to know the increase in costs that result from increasing the value of each nonbasic variable by one unit, i.e. the opportunity cost, zj of producing one unit of each nonbasic variable variable, xj. One also needs to know by how much the value of each nonbasis variable will increase from a one unit decrease in the basic variable x5, i.e. the trade-off rate x1, j, from row 1 of Tableau11 associated with basis variable x5, between the variable x5 and each nonbasis variable, xj. Call zcj the difference between zj and the objective function coefficient cj for variable xj, divided by x1, j from Tableau11. (For slack variables, the objective function coefficient is 0.) If zcj > 0, then the nonbasic variable xj is a candidate to replace x5, when the cost of x5 (or equivalently, the objective function coefficient, c5) is increased by zcj. Taking the least positive zcj over all candidate nonbasic variables yields the possible increase in the cost of x5 before this variable leaves the basis of the existing optimal solution, i.e. a different primal or slack variable will be associated with row 1. For the basic variable x5 associated with row 1 of Tableau11, the candidate nonbasic variables — primal variables xj, or slack variables sj — and their zcj values are given by:
xj: zcj = (zj - cj) / x1, j, or The zcj values for x5 associated with row 1 of Tablueau11 are:
s4: zc16 = (0.66 - 0) /
1 = 0.66 The least positive zcj is zc16 = 0.66. Consequently, costs per bag for x5 can increase from 92.82 to 93.48, and the costs per pound of x5 can increase from 0.7 to 0.705, before the existing basis is upset. (x5 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x7 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x7 associated with row 4 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
1 = 1.99 The least positive zcj is zc8 = 1.99. Consequently, costs per bag for x7 can increase from 99.45 to 101.44, and the costs per pound of x7 can increase from 0.75 to 0.765, before the existing basis is upset. (x7 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x6 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x6 associated with row 5 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
1 = 1.99 The least positive zcj is zc8 = 1.99. Consequently, costs per bag for x6 can increase from 92.82 to 94.81, and the costs per pound of x6 can increase from 0.7 to 0.715, before the existing basis is upset. (x6 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x11 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x11 associated with row 6 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
1.5 = 1.33 The least positive zcj is zc8 = 1.33. Consequently, costs per bag for x11 can increase from 11.93 to 13.26, and the costs per pound of x11 can increase from 0.09 to 0.1, before the existing basis is upset. (x11 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x9 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x9 associated with row 7 of Tablueau11 are:
s4: zc16 = (0.66 - 0) /
1.5 = 0.44 The least positive zcj is zc16 = 0.44. Consequently, costs per bag for x9 can increase from 14.59 to 15.03, and the costs per pound of x9 can increase from 0.11 to 0.113, before the existing basis is upset. (x9 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x10 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x10 associated with row 8 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
1.5 = 1.33 The least positive zcj is zc8 = 1.33. Consequently, costs per bag for x10 can increase from 10.61 to 11.93, and the costs per pound of x10 can increase from 0.08 to 0.09, before the existing basis is upset. (x10 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). The basic variable x12 is degenerate. No analysis is available here. Similarily, by how much can one increase the cost of variable x1 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x1 associated with row 10 of Tablueau11 are:
s4: zc16 = (0.66 - 0) /
0.5 = 1.33 The least positive zcj is zc16 = 1.33. Consequently, costs per bag for x1 can increase from 165.75 to 167.08, and the costs per pound of x1 can increase from 1.25 to 1.26, before the existing basis is upset. (x1 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x2 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x2 associated with row 11 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
0.5 = 3.98 The least positive zcj is zc8 = 3.98. Consequently, costs per bag for x2 can increase from 165.75 to 169.73, and the costs per pound of x2 can increase from 1.25 to 1.28, before the existing basis is upset. (x2 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). Similarily, by how much can one increase the cost of variable x3 before a new optimal basic solution is obtained, i.e. the current list of basis variables is replaced by a different list. The zcj values for x3 associated with row 12 of Tablueau11 are:
x8: zc8 = (-97.46 - -99.45) /
0.5 = 3.98 The least positive zcj is zc8 = 3.98. Consequently, costs per bag for x3 can increase from 152.49 to 156.47, and the costs per pound of x3 can increase from 1.15 to 1.18, before the existing basis is upset. (x3 might possibly be in the basis (possibly as a degenerate variable) of the new optimal primal basic solution). The basic variable x4 is degenerate. No analysis is available here. If you change the data on the form in the sensitivity analysis section above and click "Submit L.P.", this web page will display the tableaux for the new optimal solution. Continue this process until you convince yourself of the validity of the postoptimality analysis. Economic Interpretation of the Dual Solution. The firm's dual problem is to determine the costs, denoted by yM, yH, yJ, yS, and yMJ per minimum number of sacks of Mocha, Harrar, Java, Sumatra, and MoJo required, maximizing the total value of coffee beans priced at these shadow costs, yet remain within the costs per sack of for each variable, x1 to x12, valued at each shadow price. The firm's dual production decision is captured in the firm's Dual Linear Programming Problem.
Dual Solution: [yM, yH, yJ, yS, yB1, yB2, yB3, yB4, yMJ, yF1, yF2, yF3, yF4] = [21.2, 0, 0, 0.7, 110.1, 114, 112.7, 115.4, 124.6, 17.2, 15.2, 19.9, 18.6]; D = 126831.9. End of the Linear Programming Dual Simplex Method Linear Programming References.
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
