Q 2-31 Min 3A + 4B s.t. 1A + 3B ≧ 6 B = - 1/3A + 2 1A + 1B ≧ 4

Objective Function Value = 3(3) + 4(1) = 13 B = - A + 4 Feasible Region Objective Function = 3A + 4B Optimal Solution A = 3, B = 1 Objective Function Value = 3(3) + 4(1) = 13 B = - 1/3A + 2 A

Q 2-38 a. Let S = yards of the standard grade material / frame P = yards of the professional grade material / frame Min 7.50S + 9.00P s.t. 0.10S + 0.30P ≧ 6 carbon fiber (at least 20% of 30 yards) 0.06S + 0.12P ≦ 3 kevlar (no more than 10% of 30 yards) S + P = 30 total S, P ≧ 0

P Total Extreme Point S = 10, P = 20 Feasible region is the line segment Kevlar Carbon fiber Extreme Point S = 15, P = 15 S

Q 2-38 c. Extreme Point Cost (15, 15) 7.50(15) + 9.00(15) = 247.50 (10, 20) 7.50(10) + 9.00(20) = 255.00 255.00 > 247.50 Therefore, the optimal solution is S = 15, P = 15

Q 2-38 d. Min 7.50S + 8.00P s.t. 0.10S + 0.30P ≧ 6 carbon fiber Changing is only this value (From 9.00 to 8.00) Min 7.50S + 8.00P s.t. 0.10S + 0.30P ≧ 6 carbon fiber (at least 20% of 30 yards) 0.06S + 0.12P ≦ 3 kevlar (no more than 10% of 30 yards) S + P = 30 total S, P ≧ 0 Therefore, optimal solution does not change: S = 15, and P =15 New optimal solution value = 7.50(15) + 8(15) = $232.50

Q 2-38 e. At $7.40 per yard, the optimal solution is S = 10, P = 20. So, the optimal solution is changed. The value of the optimal solution is reduced to 7.50 (10) + 7.40 (20) = 223.00. A lower price for the professional grade will change the optimal solution.

Q 3-7 a. From figure 3.14, Optimal solution: U = 800, H = 1200 Estimated Annual Return 3(800) + 5(1200) = $8400

Q 3-7 b. Constraints 1 and 2 because of 0 at Slack/Surplus column. All funds available are being fully utilized and the maximum risk is being incurred.

Q 3-7 c. Constraint Dual Prices Funds Avail 0.09 Risk Max 1.33 U.S. Oil Max A unit increase in the RHS of Funds Avail makes 0.09 improvement in the value of the objective function. Risk Max also makes 1.33 improvement in the value of the objective function. A unit increase in the RHS of U.S. Oil Max makes no improvement in the objective function value.

Q 3-7 d. NO A unit increase in the RHS of U.S. Oil Max makes no improvement in the objective function value.

Q 3-10 a. From figure 3.16, Optimal solution: S = 4,000, M = 10,000 Estimated total risk = 8(4,000) + 3(10,000) = 62,000

Q 3-10 b. Variable Range of Optimality S 3.75 to No Upper Limit M No Lower Limit to 6.4

Q 3-10 c, d, e, f (c) 5(4,000) + 4(10,000) = $60,000 (d) 60,000/1,200,000 = 0.05 (e) 0.057 (f) 0.057(100) = 5.7

Theory of Simplex Method

A two-variable linear programming problem Max s.t. and For any LP problem with n decision variables, each CPF (Corner Point Feasible) solution lies at the intersection of n constraint boundaries; i.e., the simultaneous solution of a system of n constraint boundary equations.

Max s.t.

Original Form Augmented Form Max s.t. Max s.t. (1) Matrix Form (2)

Matrix Form (2) is Max s.t. (3) or Max s.t. (4) where

Then, we have Max s.t. (5) (6) where

Eq. (6) becomes (7) Putting Eq. (7) into (5), we have (8) So, (9) (10)

Eq. (10) can be expressed by (11) From Eq. (2), (12)

Thus, initial and later simplex tableau are

The Overall Procedure 1. Initialization: 2. Iteration: Same as for the original simplex method. 2. Iteration: Step 1 Determine the entering basic variable: Same as for the Simplex method.

Step 2 Determine the leaving basic variable: Same as for the original simplex method, except calculate only the numbers required to do this [the coefficients of the entering basic variable in every equation but Eq. (0), and then, for each strictly positive coefficient, the right-hand side of that equation].

3. Optimality test: Step 3 Determine the new BF solution: Derive and set 3. Optimality test: Same as for the original simplex method, except calculate only the numbers required to do this test, i.e., the coefficients of the nonbasic variables in Eq. (0).

Fundamental Insight Z RHS Z 1 Row0 Row1~N

Coefficient of: Right Side Iteration BV -3 -5 0 0 0 0 1 0 1 0 0 4 0 2 0 1 0 12 3 2 0 0 1 18

1 Coefficient of: Right Side Iteration BV 0 0 0 1 0 1 0 0 4 0 1 0 0 6 0 0 0 1 0 1 0 0 4 1 0 1 0 0 6 3 0 0 -1 1 6

1 Coefficient of: Right Side Iteration BV -3 0 0 0 1 0 1 0 0 4 -3 0 0 0 1 0 1 0 0 4 1 0 1 0 0 6 3 0 0 -1 1 6

1 Coefficient of: Right Side Iteration BV -3 0 0 0 1 0 1 0 0 4 -3 0 0 0 1 0 1 0 0 4 1 0 1 0 0 6 3 0 0 -1 1 6

1 so Coefficient of: Right Side Iteration BV -3 0 0 0 30 1 0 1 0 0 4 -3 0 0 0 30 1 0 1 0 0 4 1 0 1 0 0 6 3 0 0 -1 1 6

The most negative coefficient 4 6 minimum The most negative coefficient Coefficient of: Right Side Iteration BV -3 0 0 0 30 1 0 1 0 0 4 1 0 1 0 0 6 3 0 0 -1 1 6

2 Coefficient of: Right Side Iteration BV 0 0 0 0 0 1 2 0 1 0 0 6 0 0 0 0 0 1 2 2 0 1 0 0 6 1 0 0 2

2 Coefficient of: Right Side Iteration BV 0 0 0 1 0 0 1 2 0 1 0 0 6 0 0 0 1 0 0 1 2 2 0 1 0 0 6 1 0 0 2

2 so Coefficient of: Right Side Iteration BV 0 0 0 1 36 0 0 1 2 0 0 0 1 36 0 0 1 2 2 0 1 0 0 6 1 0 0 2

Duality Theory

One of the most important discoveries in the early development of linear programming was the concept of duality. Every linear programming problem is associated with another linear programming problem called the dual. The relationships between the dual problem and the original problem (called the primal) prove to be extremely useful in a variety of ways.

Primal and Dual Problems Primal Problem Dual Problem Max s.t. Min s.t. for for for for The dual problem uses exactly the same parameters as the primal problem, but in different location.

In matrix notation Primal Problem Dual Problem Maximize subject to Minimize subject to Where and are row vectors but and are column vectors.

Example Primal Problem Dual Problem Max s.t. Min s.t.

Primal Problem in Matrix Form Dual Problem in Matrix Form Max s.t. Min s.t.

Primal-dual table for linear programming Primal Problem Coefficient of: Right Side Coefficient of: Objective Function Coefficients for (Minimize) Dual Problem VI VI VI Right Side Coefficients for Objective Function (Maximize)

Relationships between Primal and Dual Problems One Problem Other Problem Constraint Variable Objective function Right sides Minimization Maximization Variables Constraints Unrestricted Constraints Variables Unrestricted

The feasible solutions for a dual problem are those that satisfy the condition of optimality for its primal problem. A maximum value of Z in a primal problem equals the minimum value of W in the dual problem.

Rationale: Primal to Dual Reformulation Lagrangian Function Max cx s.t. Ax b x 0 L(X,Y) = cx - y(Ax - b) = yb + (c - yA) x = c-yA Min yb s.t. yA c y 0

The following relation is always maintained yAx yb (from Primal: Ax b) yAx cx (from Dual : yA c) From (1) and (2), we have cx yAx yb At optimality cx* = y*Ax* = y*b is always maintained. (1) (2) (3) (4)

“Complementary slackness Conditions” are obtained from (4) ( c - y*A ) x* = 0 y*( b - Ax* ) = 0 xj* > 0 y*aj = cj , y*aj > cj xj* = 0 yi* > 0 aix* = bi , ai x* < bi yi* = 0 (5) (6)

Any pair of primal and dual problems can be converted to each other. The dual of a dual problem always is the primal problem.

Converted to Standard Form Dual Problem Min W = yb, s.t. yA c y 0. Max (-W) = -yb, s.t. -yA -c y 0. Converted to Standard Form Its Dual Problem Max Z = cx, s.t. Ax b x 0. Min (-Z) = -cx, s.t. -Ax -b x 0.

Min s.t. Min s.t.

Max s.t. Max s.t.

Home Work Ch 4: Problem 1 Ch 4: Problem 10 Additional Problems: A-1 and A-2 (See the proceeding PPS) Due Date: September 16

A – 1 Theory of Simplex Method: Consider the following problem. Let x5 and x6 denote slack variables for the two constraints. After you apply the simplex method, a portion of the final simplex tableau is as follows:

A – 1 cont’d Solve the problem. Basic Variable Coefficient of: Right Side Eq. Z x1 x2 x3 x4 x5 x6 (0) 1 (1) -1 (2) 2 Solve the problem. What is B-1 ? How about B-1b and CBB-1b ? If the right hand side is changed from (5, 4) to (5, 5), how is and optimal solution changed? How about an optimal objective value?

A – 2 Linear Programming: Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of their products is a strawberry shake which is designed to be a complete meal. The strawberry shake which is designed to be a complete meal. The strawberry shake consists of several ingredients. Some information about each of these ingredients is given below.

A – 2 cont’d Ingredient Calories from Fat (per tbsp) Total Calories (per tbsp) Vitamin Content (mg/tbsp) Thickeners (mg/tbsp) Cost (¢/tbsp) Strawberry flavoring 1 50 20 3 10 Cream 75 100 8 Vitamin supplement 25 Artificial sweetener 120 2 15 Thickening agent 30 80 6

A – 2 cont’d The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No more than 20 percent of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin content. For taste reasons, there must be at least 2 tablespoons (tbsp) of strawberry flavoring for each tablespoon of artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the beverage. Management would like to select the quantity of each ingredient for the beverage which would minimize cost while meeting the above requirements.

A – 2 cont’d Formulate a linear programming model for this problem. Show its dual formulation. Solve this model by your computer and confirm complementary slackness condition.