Business Mathematics MTH-367 Lecture 16

Chapter 11 The Simplex and Computer Solutions Methods continued

Last Lecture’s Summary The Simplex Method for Maximization problem with all ≤ constraints Maximization problem with mixed constraints

Today’s topics Minimization problems Special phenomena 1.Alternative Optimal Solutions 2.No Feasible Solution 3.Unbounded Solutions Examples

The simplex procedure is slightly different when minimization problems are solved. Aside from assigning artificial variables objective function coefficients of +M, the only difference relates to the interpretation of row (0) coefficients. The following two rules are modifications of rule 1 and rule 2. These apply for minimization problems. Minimization Problems

Rule 1A: Optimality Check in Minimization Problem. In a minimization problem, the optimal solution has been found if all row (0) coefficients for the variables are less than or equal to 0. If any row (0) coefficients are positive for non- basic variables, a better solution can be found by assigning a positive quantity to these variables.

Rule 2A: New basic variable in minimization problem. In a minimization problem, the non-basic variable which will replace a current basic variable is the one having the largest positive row (0) coefficient. Ties may be broken arbitrarily.

Example Solve the following linear programming problem using the simplex method. Minimize z = 5x 1 + 6x 2 subject to x 1 + x 2 ≥ 10 2x 1 + 4x 2 ≥ 24 x 1, x 2 ≥ 0

Solution: This problem is first rewritten with the constraints expressed as equations, as follows: Minimize z = 5x 1 + 6x 2 + 0E 1 + 0E 2 + MA 1 + MA 2 subject to x 1 + x 2 – E 1 + A 1 = 10 2x 1 + 4x 2 – E 2 + A 2 = 24 x 1, x 2, E 1, E 2, A 1, A 2 ≥ 0

Basic Variable s zx1x1 x2x2 E1E1 E2E2 A1A1 A2A2 Need to Be Transformed to Zero b i Row Number 1–5–600–M 0(0) A1A1 011–101010(1) A2A2 0240–10124(2) The initial tableau for this problem appears in above table. Note that the artificial variables are the basic variables in this initial solution. These –M coefficient in row (0) must be changed to 0 using row operations if the value of z is to be read from row (0).

z Key Column Transformed to zero Row b i Number ratio x1x1 x2x2 E1E1 E2E2 A1A1 A2A2 1–5–3M–6+5M–M 0034M(0) A1A1 011–101010(1) 10/1 = 10 A2A2 0240–10124(2) 24/4 = 6*

In this initial solution the non-basic variables are x 1, x 2, E 1, and E 2. The basic variables are the two artificial variables with A 1 = 10, A 2 = 24, and z = 34M. Applying rule 1A, we conclude that this solution is not optimal. The x 2 column becomes the new key column.

Basic Variable s z Key Column b i Ratio x1x1 x2x2 E1E1 E2E2 A1A1 A2A2 1–8+2M0–4M-6+M06-5M144+16M(0) A1A1 020–41416 x2x2 01/210–1/401/46 Applying rule 1A, we see that this solution is not optimal. Continuing with the same procedure we arrive at the following solution.

Applying rule 1A, we conclude that this solution is optimal. Therefore, z is minimized at a value of 52 when x 1 = 8 and x 2 = 2. Basic Variable s z Row b i Number x1x1 x2x2 E1E1 E2E2 A1A1 A2A /24-M½-M52(0) x1x1 010–2½2–½8(1) x2x2 0011–½–1½2(2)

In Chapter 10, certain phenomena which can arise when solving LP problems were discussed. Specifically, the phenomena of alternative optimal solutions, no feasible solution, and unbounded solutions were presented. In this section we discuss the manner in which these phenomena occur when solving by the simplex method. Special Phenomena

Alternative optimal solutions, results when the objective function is parallel to a constraint which binds in the direction of optimization. In two-variable problems we are made aware of alternative optimal solutions with the corner- point method when a “tie” occurs for the optimal corner point. When using the simplex methods, alternative optimal solutions are indicated when 1- an optimal solution has been identified, 2- the row (0) coefficient for a non-basic variable equals zero. Alternative Optimal Solutions

The first condition confirms that there is no better solution than the present solution. The presence of a 0 in row (0) for a non-basic variable indicates that the non-basic variable can become a basic variable (can become positive) and the current value of the objective function (known to be optimal) will not change. Consider the LP problem: Maximizez = 6x 1 + 4x 2 subject to x 1 + x 2 ≤ 5 3x 1 + 2x 2 ≤ 12 x 1, x 2 ≥ 0 Alternative Optimal Solutions cont’d

Basic Variable s z Key Column Row b i Number Ratio x1x1 x2x2 S1S1 S2S2 1–6–4000(0) S1S (1) S2S (2) Above table presents the initial simplex with S 1 = 5 and S 2 = 12 being the basic variable; x 1 and x 2 are non-basic variables. By rule 1 we see that a better solution exists. Using rule 2, x 1 is selected as the new basic variable and the x 1 column becomes the key column. The b i /a ik ratios are computed with S 2 identified as the departing variable.

The elements in the x 1 column are transformed from using row operations, with the resulting simplex tableau shown in the following table. In this new solution, x 1 has replaced S 2 in the basis; the basic variables are S 1 = 1 and x 1 = 4 with a resulting objective functions value of z = 24. Basic Variable s z Row b i Number x1x1 x2x2 S1S1 S2S (0) S1S1 001/31–1/31(1) x1x1 012/301/34(2)

If in the above table we treat the x 2 column as a key column associated with the entry of the new basic variables x 2, b i /a ik ratios are computed as usual and S 1 is identified as the departing variable. The following indicates the new solution. The optimality check (rule 1) indicates that the solution is optimal with x 1 = 2, x 2 = 3, and z = 24. Basic Variable s z Row b i Number x1x1 x2x2 S1S1 S2S (0) x1x1 0013–12(1) x2x2 010–213(2)

A problem has no feasible solution if there are no values for the variables which satisfy all the constraints. The condition of no feasible solution is signaled in the simplex method when an artificial variable appears in an optimal basis at a positive value. Let’s solve the following LP problem, which, by inspection, has no feasible solution. Maximizez = 10x x 2 subject to x + x 2 ≤ 5 x 1 + x 2 ≥ 20 x 1, x 2 ≥ 0 No Feasible Solutions

Basic Variable s z Key Column Needs to be set equal to zero Row b i Number x1x1 x2x2 S1S1 E2E2 A2A2 1–10–2000M0(0) R 0 S1S (1) R 1 A2A2 0110–1120(2) R 2 Basic Variable s z Key Column Transformed to zero Row b i Number b i /a ik x1x1 x2x2 S1S1 E2E2 A2A2 1–10–M–20–M0M0–20M(0) S1S (1) 5/1 = 5* A2A2 0110─1120(2) 20/1 =

Basic Variable s zx1x1 x2x2 S1S1 E2E2 A2A2 Row b i Number 110 0M+20M 0–15M+100(0) x2x (1) A2A2 00 0–1 115(2) The above 3 tables present the simplex iterations in solving this problem. In the above table, all row (0) coefficient are greater than or equal to 0, indicating an optimal solution. However, one of the basic variables is A 2 and it has a value of 15; that is, the optimal solution is x 2 = 5, A 2 = 15, and z = -15M Since an artificial variable have no meaning in a LP problem, and the assignment of A 2 = 15 signals no feasible solution to the problem.

Unbounded solutions exist when 1. there is an unbounded solution space 2. Improvement in the objective function occurs with movement in the direction of the unbounded portion of the solution space. 3.If at any iteration of the simplex method the a ik values are all 0 or negative for the variable selected to become the new basic variable, there is an unbounded solution for the LP problem. Unbounded Solutions

Consider the following problem: Maximize z = –2x 1 + 3x 2 subject to x 1 ≤ 10 2x 1 – x 2 ≤ 30 x 1, x 2 ≥ 0 Basic Variables z Key Column bibi x1x1 X2X2 S1S1 S2S2 12–3000 S1S S2S2 02–10130

The above table presents the initial simplex tableau. Row (0) indicates that the non-basic variable x 2 will result in an improved value of z. However, the values are 0 and –1, these suggest that for each unit introduced of x 2, S 1 will not change and S 2 will increase by 1 unit. Neither of these basic variables will be driven to zero. This signals an unbounded solution Basic Variables z Key Column bibi x1x1 X2X2 S1S1 S2S2 12–3000 S1S S2S2 02–10130

Review Minimization problems Special phenomena 1.Alternative Optimal Solutions 2.No Feasible Solution 3.Unbounded Solutions Examples

Next Lecture The Dual Problem Formulation of the Dual Problem Primal-Dual Solutions Properties Examples