1 / 38

EMGT 501 HW #1 Solutions Chapter 2 - SELF TEST 18 Chapter 2 - SELF TEST 20

EMGT 501 HW #1 Solutions Chapter 2 - SELF TEST 18 Chapter 2 - SELF TEST 20 Chapter 3 - SELF TEST 28 Chapter 4 - SELF TEST 3 Chapter 5 - SELF TEST 6. Ch. 2 – 18 (a). Ch. 2 – 18 (b). (c) s 1 = 0, s 2 = 0, s 3 = 4/7. Ch. 2 – 20 (a).

danielle-briggs
Download Presentation

EMGT 501 HW #1 Solutions Chapter 2 - SELF TEST 18 Chapter 2 - SELF TEST 20

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. EMGT 501 HW #1 Solutions Chapter 2 - SELF TEST 18 Chapter 2 - SELF TEST 20 Chapter 3 - SELF TEST 28 Chapter 4 - SELF TEST 3 Chapter 5 - SELF TEST 6

  2. Ch. 2 – 18 (a)

  3. Ch. 2 – 18 (b) (c) s1 = 0, s2 = 0, s3 = 4/7

  4. Ch. 2 – 20 (a) Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced

  5. Ch. 2 – 20 (b)

  6. Ch. 2 – 20 (c) The binding constraints are the manufacturing time and the assembly and testing time.

  7. Ch. 3 – 28 (a) Let A = number of shares of stock A B = number of shares of stock B C = number of shares of stock C D = number of shares of stock D To get data on a per share basis multiply price by rate of return or risk measure value.

  8. Solution: A = 333.3, B = 0, C = 833.3, D = 2500 Risk: 14,666.7 Return: 18,000 (9%) from constraint 2

  9. Ch. 3 – 28 (b) Solution: A = 1000, B = 0, C = 0, D = 2500 Risk: 10A + 3.5B + 4C + 3.2D = 18,000 Return: 22,000 (11%)

  10. Ch. 3 – 28 (c) The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333. Obtaining a reasonable return with a lower risk is a preferred strategy in many financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk.

  11. Ch. 4 – 3 x1 = $ automobile loans x2 = $ furniture loans x3 = $ other secured loans x4 = $ signature loans x5 = $ "risk free" securities

  12. Solution Annual Return $188,800 (9.44%)

  13. Ch. 5 – 6 (a)

  14. Ch. 5 – 6 (b)

  15. Ch. 5 – 6 (c) The original basis consists of s1, s2, and s3. It is the origin since the nonbasic variables are x1, x2, and x3 and are all zero. (d) 0 x3 enters because it has the largest negative zj - cj and s2 will leave because row 2 has the only positive coefficient. (e) (f) 30; objective function value is 30 times 25 or 750. (g) Optimal Solution: x1 = 10 s1 = 20 x2 = 0 s2 = 0 x3 = 30 s3 = 0 z = 800.

  16. EMGT 501 HW #2 Chapter 6 - SELF TEST 21 Chapter 6 - SELF TEST 22 Due Day: Sep 27

  17. Ch. 6 – 21 Consider the following linear programming problem: • Write the dual problem. • Solve the dual. • Use the dual solution to identify the optimal solution to the original primal problem. • Verify that the optimal values for the primal and dual problems are equal.

  18. Ch. 6 – 22 A sales representative who sells two products is trying to determine the number of sales calls that should be made during the next month to promote each product. Based on past experience, representatives earn an average $10 commission for every call on product 1 and a $5 commission for every call on product 2. The company requires at least 20 calls per month for each product and not more than 100 calls per month on any one product. In addition, the sales representative spends about 3 hours on each call for product 1 and 1 hour on each call for product 2. If 175 selling hours are available next month, how many calls should be made for each of the two products to maximize the commission?

  19. Formulate a linear program for this problem. • Formulate and solve the dual problem. • Use the final simplex tableau for the dual problem to determine the optimal number of calls for the products. What is the maximum commission? • Interpret the values of the dual variables.

  20. Duality Theory

  21. 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.

  22. 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.

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

  24. Example Dual Problem in Algebraic Form Primal Problem in Algebraic Form Min s.t. Max s.t.

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

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

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

  28. 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.

  29. 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

  30. 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)

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

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

  33. 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.

  34. Min s.t. Min s.t.

  35. Max s.t. Max s.t.

  36. Question 1: Consider the following problem. (b) Work through the simplex method step by step in tabular form. (c) Use a software package based on the simplex method to solve the problem.

  37. Question 2: For each of the following linear programming models, give your recommendation on which is the more efficient way (probably) to obtain an optimal solution: by applying the simplex method directly to this primal problem or by applying the simplex method directly to the dual problem instead. Explain. (a) Maximize subject to (b) Maximize subject to and and

  38. Question 3: Consider the following problem. Maximize subject to and (a) Construct the dual problem. (b) Use duality theory to show that the optimal solution for the primal problem has

More Related