EMGT 501 HW #1 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 Due Day: Sep 13 Ch. 2 – 18 For the linear program Max 4 x1 1x2 s.t. 10 x1 2 x2 30 3x1 2 x2 12 2 x1 2 x2 10 x1 , x2 0 a. Write this linear program in standard form. b. Find the optimal solution using the graphical solution procedure. c. What are the values of the three slack variables at the optimal solution? Ch. 2 – 20 Embassy Motorcycle (EM) manufactures two lightweight motorcycles designed for easy handling and safety. The EZRider model has a new engine and a low profile that make it easy to balance. The Lady-Sport model is slightly larger, uses a more traditional engine, and is specifically designed to appeal to women riders. Embassy produces the engines for both models at its Des Moines, Iowa, plant. Each EZRider engine requires 6 hours of manufacturing time and each Lady-Sport engine requires 3 hours of manufacturing time. The Des Moines plant has 2100 hours of engine manufacturing time available for the next production period. Embassy’s motorcycle frame supplier can supply as many EZ-Rider frames as needed. However, the Lady-Sport frame is more complex and the supplier can provide only up to 280 Lady-Sport frames for the next production period. Final assembly and testing requires 2 hours for each EZ-Rider model and 2.5 hours for each Lady-Sport model. A maximum of 1000 hours of assembly and testing time are available for the next production period. The company’s accounting department projects a profit contribution of $2400 for each EZ-Rider produced and $1800 for each Lady-Sport produced. a. Formulate a linear programming model that can be used to determine the number of units of each model that should be produced in order to maximize the total contribution to profit. b. Find the optimal solution using the graphical solution procedure. c. Which constraints are binding. Ch. 3 – 28 National Insurance Associates carries an investment portfolio of stocks, bonds, and other investment alternatives. Currently $200,000 of funds are available and must be considered for new investment opportunities. The four stock options National is considering and the relevant financial data are as follows: Stock A B C D Price per share $100 $50 $80 $40 Annual rate of return 0.12 0.08 0.06 0.10 Risk measure per dollar invested 0.10 0.07 0.05 0.08 The risk measure indicates the relative uncertainty associated with the stock in terms of its realizing the projected annual return; higher values indicate greater risk. The risk measures are provided by the firm’s top financial advisor. National’s top management has stipulated the following investment guidelines: the annual rate of return for the portfolio must be at least 9% and no one stock can account for more than 50% of the total dollar investment. a. Use linear programming to develop an investment portfolio that minimizes risk. b. If the firm ignores risk and uses a maximum return-on-investment strategy, what is the investment portfolio? c. What is the dollar difference between the portfolios in parts (a) and (b)? Why might the company prefer the solution developed in part (a)? Ch. 4 – 3 The employee credit union at State University is planning the allocation of funds for the coming year. The credit union makes four types of loans to its members. In addition, the credit union invests in risk-free securities to stabilize income. The various revenue-producing investments together with annual rates of return are as follows: Type of Loan/Investment Automobile loans Furniture loans Other secured loans Signature loans Risk-free securities Annual Rate of Return (%) 8 10 11 12 9 The credit union will have $2,000,000 available for investment during the coming year. State laws and credit union policies impose the following restrictions on the composition of the loans and investments. • Risk-free securities may not exceed 30% of the total funds available for investment. • Signature loans may not exceed 10% of the funds invested in all loans (automobile, furniture, other secured, and signature loans). • Furniture loans plus other secured loans may not exceed the automobile loans • Other secured loans plus signature loans may not exceed the funds invested in risk-free securities. How should the $2,000,000 be allocated to each of the loan/investment alternatives to maximize total annual return? What is the projected total annual return? Ch. 5 – 6 Basis cB x1 5 2 0 3 x2 x3 20 25 1 0 2 1 0 -1/2 s1 0 1 0 0 s2 0 0 1 0 s3 0 0 0 1 40 30 15 zj cj zj a. Complete the initial tableau. b. Write the problem in tableau form. c. What is the initial basis? Does this basis correspond to the origin? Explain. d. What is the value of the objective function at this initial solution? e. For the next iteration, which variable should enter the basis, and which variable should leave the basis? f. How many units of the entering variable will be in the next solution? Before making this first iteration, what do you think will be the value of the objective function after the first iteration? g. Find the optimal solution using the simplex method. 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) Max 4x1 + 1x2 + 0s1 + 2x2 + 1s1 + 0s2 + 0s3 s.t. 10x1 3x1 + 2x2 2x1 + 2x2 x1 , = 30 + 1s2 = 12 + 1s3 x2, s1, s2, s3 0 = 10 x2 Ch. 2 – 18 (b) 14 12 10 8 6 Optimal Solution x1 = 18/7, x2 = 15/7, Value = 87/7 4 2 (c) s1 = 0, s2 = 0, s3 = 4/7 x1 0 2 4 6 8 10 Ch. 2 – 20 (a) Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced Max 2400E + 1800L 6E + 3L s.t. L 2E + 2.5L 2100 Engine time 280 Lady-Sport maximum 1000 Assembly and testing E, L 0 L Ch. 2 – 20 (b) 700 Number of EZ-Rider Produced 600 Engine Manufacturing Time 500 400 Frames for Lady-Sport 300 Optimal Solution E = 250, L = 200 Profit = $960,000 200 100 Assembly and Testing 0 E 100 200 300 400 Number of Lady-Sport Produced 500 Ch. 2 – 20 (c) The binding constraints are the manufacturing time and the assembly and testing time. 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. Min 10A + 3.5B + 4C + 3.2D 100A + 50B + 80C + 40D = 200,000 12A + 4B + 4.8C + 18,000 100,000 100,000 100,000 100,000 s.t. 4D 100A 50B 80C 40D A, B, C, D (9% of 200,00) 0 Solution: A = 333.3, B = 0, C = 833.3, D = 2500 Risk: 14,666.7 Return: 18,000 (9%) from constraint 2 Ch. 3 – 28 Max (b) 12A + 4B + 4.8C + 4D s.t. 100A + 50B + 80C + = 200,000 100,000 100,000 100,000 100,000 40D 100A 50B 80C 40D A, Solution: Risk: Return: B, C, D0 A = 1000, B = 0, C = 0, D = 2500 10A + 3.5B + 4C + 3.2D = 18,000 22,000 (11%) 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. Ch. 4 – 3 0.08x1 Max x1 = $ automobile loans x2 = $ furniture loans x3 = $ other secured loans x4 = $ signature loans x5 = $ "risk free" securities + 0.10x2 + 0.11x3 + 0.12x4 + 0.09x5 s.t. x5 600,000 x4 or or -0.10x1 - x1 - + - 0.10x3 x2 + x3 x1 x2 + x3 0 or x1 + 0.10(x1 + x2 + x3 + x4) 0.10x2 x2 + [1] 0.90x4 0 [2] [3] x3 + x4 + x3 + x4 - x5 0 [4] + x3 + x4 + x5 = 2,000,000 [5] x1 , x2, x3, x5 x4, x5 0 Solution Automobile Loans (x1) = $630,000 Furniture Loans (x2) = $170,000 Other Secured Loans (x3) = $460,000 Signature Loans (x4) = $140,000 Risk Free Loans (x5) = $600,000 Annual Return $188,800 (9.44%) Ch. 5 – 6 (a) x1 x2 x3 s1 s2 s3 Basis cB 5 20 25 0 0 0 s1 0 2 1 0 1 0 0 40 s2 0 0 2 1 0 1 0 30 s3 0 3 0 -1/2 0 0 1 15 Z -cj + zj 0 -5 -20 -25 0 0 0 Ch. 5 – 6 (b) Max 5x1 + 20x2 2x1 + + 25x3 + 0s1 + 0s2 + 0s3 s.t. 1x2 2x2 3x1 + 1s1 + 1x3 = 40 + 1s2 - = 30 + 1s3 1/2x3 x1, x 2, x3, s1, s2, s3, 0. = 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 (e) and s2 will leave because row 2 has the only positive coefficient. (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.