Operations Research - Linear Programming

Economics Application - Profit Maximization

by

Elmer G. Wiens

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Graphical Solution | Simplex Algorithm | Primal Simplex Tableaux Generator | Dual Simplex Algorithm | Dual Simplex Tableaux Generator
Economics Application - Profit Maximization | Economics Application - Least Cost Formula

firm's production decision | sensitivity analysis | dual simplex method | dual simplex tableaux
profit maximizing solution | postoptimality analysis | dual solution interpretation | references

The Firm's Production Decision.

A firm can produce a product using three production processes. Let the variables x₁, x₂, and x₃ stand for the product and the quantities of the product produced by activities 1, 2, and 3, respectively. Each process uses labour (L), capital (K), and materials and supplies (M) to produce the product. The variable capital stands for various types of machinery and equipment. Assume that the use of labour and capital is measured in hours, and that M is an index number for the quantities of materials and supplies used. The availability of inputs of labour, capital, and materials and supplies, and the amounts of each input used to produce one unit of product by each production activity, are categorized by the coefficients of the following table:

The first column of the table means that 25 hours of labour (L), 35 hours of capital (K), and 30 units of materials and supplies (M) are used by process 1 to produce one unit of x₁. The first row of the table means that 25, 32, and 43 units of labour are used to produce one unit of x₁, x₂, and x₃, respectively. Furthermore, the total hours of labour used can not exceed 850 hours. The other columns and rows read similarly.

The unit costs of L, K, and M are:

w_L = 14, w_K = 34, and w_M = 12,

and the costs to produce one unit of x₁, x₂, and x₃ are:

cost₁ = 1900 = 14 * 25 + 34 * 35 + 12 * 30,

cost₂ = 1730 = 14 * 32+ 34 * 25 + 12 * 36,

cost₃ = 1762 = 14 * 43+ 34 * 20 + 12 * 40.

While the items produced by each process are similar, they are of differing quality. Consumers will pay a premium for higher quality items. I assumed that each item's quality depends on the "amount of capital equipment" used in its production. You can change this assumption by changing the prices in the user-friendly form below, i.e. perhaps it is the labour intensive item that commands a price premium for the product of interest. The new web page will recompute the relevant variables.

Consumers' tastes are reflected in the prices of one unit of x₁, x₂, and x₃:

price₁ = 2610, price₂ = 2400, and price₃ = 2250

The profits to the firm of producing one unit of x₁, x₂, and x₃ given by:

profit = price - cost:

are:

profit₁ = 710, profit₂ = 670, and profit₃ = 488

The primal objective of the firm is to maximize profits, by producing appropriate amounts of products x₁, x₂, and x₃. The firm's primal production decision is captured in the firm's Primal Linear Programming Problem.

Solution Sensitivity Analysis.

The form below permits you to change the data that specify the firm's production decision. You can perform a sensitivity analysis of the firm's linear programming problem by changing the data, and clicking "Submit L.P.". A new web page will display with all the firm's data updated with your new data, and the firm's new optimal production program. Compare the new results with the results of the current web page.

After you fill in your data and click "Submit L.P.", the program will automatically calculate a sequence of tableaux that solves the primal and dual l.p. problems. Examine the tableaux that follow to see how the dual simplex method proceeds to find the solution.

Dual Simplex Method.

The standard form for the initial dual simplex tableau is:

In the dual simplex algorithm, a sequence of tableaux are calculated. A given Tableau^k has the form:

after consolidating the center 2 by 2 block of matrices.

The Tableaux of the Dual Simplex Method.

Phase I: Goal: get Ø >= 0.

Basis for Tableau⁰: [s₁, s₂, s₃, ]. Value of Objective Function = 0.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Goal: get Ø >= 0.

A. In Tableau⁰:

1. Select a target column, tcol, with Ø_tcol < 0: Ø⁰₁ = -710, tcol = 1.

2. Select any row, r, with a positive entry in tcol = 1 as the pivot row: row = 2 associated with x_2,1 = 35 and constraint K.

B. To create Tableau¹:

4. Compute row K¹ = K⁰ / (35).

5. Subtract multiples of row K¹ from all other rows of Tableau⁰ so that column x₁ = e₂ in Tableau¹.

Basis for Tableau¹: [s₁, x₁, s₃, ]. Value of Objective Function = 16228.57.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Goal: get Ø >= 0.

A. In Tableau¹:

1. Select a target column, tcol, with Ø_tcol < 0: ع₂ = -162.857, tcol = 2.

2. Select any row, r, with a positive entry in tcol = 2 as the pivot row: row = 3 associated with x_3,2 = 14.571 and constraint M.

B. To create Tableau²:

4. Compute row M² = M¹ / (22.857).

5. Subtract multiples of row M² from all other rows of Tableau¹ so that column x₃ = e₃ in Tableau².

Basis for Tableau²: [s₁, x₁, x₃, ]. Value of Objective Function = 17288.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Goal: get Ø >= 0.

A. In Tableau²:

1. Select a target column, tcol, with Ø_tcol < 0: ز₂ = -110.4, tcol = 2.

2. Select any row, r, with a positive entry in tcol = 2 as the pivot row: row = 3 associated with x_3,2 = 0.638 and constraint M.

B. To create Tableau³:

4. Compute row M³ = M² / (0.638).

5. Subtract multiples of row M³ from all other rows of Tableau² so that column x₂ = e₃ in Tableau³.

Phase II: Goal: get ß >= 0.

Basis for Tableau³: [s₁, x₁, x₂, ]. Value of Objective Function = 19517.65.

Proceed to the next tableau as follows:

Phase 0: Complete.

Phase I: Complete.

Phase II: Goal: get ß >= 0.

A. In Tableau³:

1. Select a pivot row, row, with b³_row < 0: row = 1 associated with b³₁ = -7.059.

B. To create Tableau⁴:

3. Compute row L⁴ = L³ / (-0.971).

3. Subtract multiples of row L⁴ from all other rows of Tableau³ so that column s₃ = e₁ in Tableau⁴.

Basis for Tableau⁴: [s₃, x₁, x₂]. Value of Objective Function = 19436.36.

Phase 0: Complete.

Phase I: Complete.

Phase II: Complete.

Primal Solution: [s₃, x₁, x₂] = [7.273, 8.788, 19.697];   P = 19436.364.

(Primal x variables not in the basis have a value of 0).

Dual Solution: [y_L, y_K, y_M] = [11.515, 12.061, 0];   D = 19436.364.

Firm's Profit Maximizing Solution.

Quantities of the product produced by each production process, and the accompanying revenues, costs, and profits.

Post Optimality Analysis.

Over what ranges of the values of the coefficients of the model, such as profits (c₁, c₂, c₃), or the availability of inputs (b_L, b_K, b_M), 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, and x₃, take on the values 8.79, 19.7, and 0 with per unit profits of 710, 670, and 488 at per unit prices of 2610, 2400, and 2250. The basis variables associated with rows ( 1, 2, 3) of Tableau⁴ are the variables ( s₃, x₁, x₂ ) whose associated objective function coefficients are ( 0, 710, 670).

Objective Function Coefficients - NonBasic Variables

By how much can one increase the price of x₃ before this variable will enter the basis of a new basic optimal solution. To determine the entry price for x₃, one computes the opportunity cost of producing one unit of x₃ = the profits given up to produce one unit of x₃ = z₃, using standard L.P. notation. The value of z₃ is calculated by weighting each entry of the column associated with x₃ in Tableau⁴ by the objective function coefficient associated with the row of that entry:

z₃ = 0 * -6.73 + 710 * -0.88 + 670 * 2.03 = 736.36.

Per unit profits for x₃ can increase from 488 to 736.36, and the price per unit of x₃ can increase from 2250 to 2498.36 before x₃ will enter the basis in the new optimal primal solution.

Objective Function Coefficients - Basic Variables

By how much can one decrease the price 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. To determine the basis upset price for x₁, one needs to know the decrease in profits that result from increasing the value of each nonbasic variable by one unit, i.e. the opportunity cost, z_j of producing one unit of each nonbasic variable variable, x_j.

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 x₁, i.e. the trade-off rate x_{2, j}, from row 2 of Tableau⁴ associated with basis variable x₁, between the variable x₁ and each nonbasis variable, x_j.

Call zc_j the difference between z_j and the objective function coefficient c_j for variable x_j, divided by x_{2, j} from Tableau⁴. (For slack variables, the objective function coefficient is 0.) If zc_j > 0, then the nonbasic variable x_j is a candidate to replace x₁, when the price of x₁ (or equivalently, the objective function coefficient, c₁) is reduced by zc_j. Taking the least positive zc_j over all candidate nonbasic variables yields the possible decrease in the price of x₁ before this variable leaves the basis of the existing optimal solution, i.e. a different primal or slack variable will be associated with row 2.

For the basic variable x₁ associated with row 2 of Tableau⁴, the candidate nonbasic variables — primal variables x_j, or slack variables s_j — and their zc_j values are given by:

x_j: zc_j = (z_j - c_j) / x_{2, j}, or

s_j: zc_j = (z_j - 0) / x_{2, j}.

The zc_j values for x₁ associated with row 2 of Tablueau⁴ are:

s₂: zc₅ = (12.06 - 0) / 0.06 = 186.56

The least positive zc_j is zc₅ = 186.56. Consequently, profits per unit for x₁ can decrease from 710 to 523.44, and the price per unit of x₁ can decrease from 2610 to 2423.44 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 decrease the price 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 zc_j values for x₂ associated with row 3 of Tablueau⁴ are:

x₃: zc₃ = (736.36 - 488) / 2.03 = 122.33

s₁: zc₄ = (11.52 - 0) / 0.07 = 162.86

The least positive zc_j is zc₃ = 122.33. Consequently, profits per unit for x₂ can decrease from 670 to 547.67, and the price per unit of x₂ can decrease from 2400 to 2277.67 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).

Postoptimality Analysis.

RHS Coefficients - b-vector Constraints

Over what ranges of values for the availability of labour (L), capital (K), and materials and supplies (M), will the existing set of basic variables remain in the basis of the optimal solution, after these variables values are adjusted?

In the postoptimality analysis for the objective function coefficients, I used the the columns in Tableau⁴associated with each nonbasis variable x_j or s_j, or / and the row of Tableau⁴ associated with the basis variable.

The postoptimaility analysis for the resource constraints associated with the L, K, and M values, I use the matrix formed from the columns of with the slack variables in Tableau⁴.

On the simplex algorithm page, I showed how this matrix, called B, is the inverse of the matrix A_B from Tableau⁰ associated with the basis variables in the optimal primal solution. Moreover, the ß column of the final tableau, Tableau⁴, is equal to the product of the B matrix and the original vector of resource constraints, b, i.e. ß = B * b, which also yields the values of the basis variables, x_basis in the optimal primal solution.

The range of values of the coefficients of the b-vector that will maintain the current basis depends on the following observation. If I change the ith coefficient b_i to b_i, then the new vector, b, must obey:

x_basis = ß = B * b >= 0

since the values of the new optimal x_basis coefficients must be nonnegative for the exisiting basis to continue as the basis in the the new optimal basic primal solution.

The B matrix for the current optimal basic solution is:

With b = (850, 800, 980), the perturbed RHS coefficients must obey:

Since these inequalities must be satisfied simultaneously for each b_i considered separately, the upper and lower bounds are:

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 values y_L, y_K, and y_M per unit of input L, K and M that will minimize the total value of resources used, yet maintain each product's level of profit. The firm's dual production decision is captured in the firm's Dual Linear Programming Problem.

Dual Solution: [y_L, y_K, y_M] = [11.515, 12.061, 0];   D = 19436.364.

The variables, y_L, y_K, and y_M, measure the increase in profits, P in the primal program, from a one unit increase in the availability of labour (L), capital (K), and materials and supplies (M), respectively. (See what happens to profits when you increase the resource constraint of an input by one unit on the form in the sensitivity analysis section above and click "Submit L.P.".)

A value of zero for a variable y_i, associated with a positive value for slack variable s_i in the optimal solution of the primal program, means that excess amounts of that input are available.

The variables, y_L, y_K, and y_M, are referred to as opportunity prices, or shadow prices for the inputs L, K, and M.

End of the Linear Programming Dual Simplex Method

Linear Programming References.

• Sposito, Vincent. Linear and Nonlinear Programming. Ames, Iowa: Iowa UP, 1975.
• Restrepo, Rodrigo A. Theory of Games and Programming: Mathematics Lecture Notes. Vancouver: University of B.C., 1967.
• Restrepo, Rodrigo A. Linear Programming and Complementarity. Vancouver: University of B.C., 1994.
• Taha, Hamdy A. Operations Research: An Introduction. 2nd ed. New York: MacMillan, 1976.
• Wu, Nesa, and Richard Coppins. Linear Programming and Extensions. New York: McGraw-Hill, 1981.