# Individual assignment 5.docx - Individual assignment#5 Unit...

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Individual assignment #5 Unit 6 Course: Busi 2013 individual assignment Chapter 8 Question 28 Answer: a. Let G = amount invested in growth stock fund S = amount invested in income stock fund M = amount invested in money market fund Max 0.20 G + 0.10 S + 0.06 M s.t. 0.10 G + 0.05 S + 0.01 M £ (0.05) (300,000) Hartmann's max risk G ³ (0.10) (300,000) Growth fund min. S ³ (0.10) (300,000) Income fund min. M ³ (0.20) (300,000) Money market min, G + S + M £ 300,000 Funds available G , S , M ³ 0 b. The solution to Hartmann's portfolio mix problem is given. Optimal Objective Value 36000.00000 Variable Value Reduced Cost G 120000.00000 0.00000 S 30000.00000 0.00000 M 150000.00000 0.00000 Constraint Slack Shadow Price 1 0.00000 1.55556 2 90000.00000 0.00000 3 0.00000 -0.02222
4 90000.00000 0.00000 5 0.00000 0.12000 Objective Allowable Allowable Coefficient Increase Decrease 0.20000 Infinite 0.05000 0.10000 0.02222 1.20000 0.06000 0.14000 0.04000 RHS Allowable Allowable Value Increase Decrease 0.00000 8100.00000 8100.00000 0.00000 90000.00000 Infinite 0.00000 162000.00000 30000.00000 0.00000 90000.00000 Infinite 300000.00000 Infinite 300000.00000 c. These are given by the objective coefficient ranges. The portfolio above will be optimal as long as the yields remain in the following intervals: Growth stock 0.15 £ c 1 £ 0.60 Income stock No Lower Limit < c 2 £ 0.122 Money Market 0.02 £ c 3 £ 0.20 d. The shadow price for the first constraint provides this information. A change in the risk index from 0.05 to 0.06 would increase the constraint RHS by 3000 (from 15,000 to 18,000). This is within the right-hand-side range, so the shadow price of 1.556 is applicable. The value of the optimal solution would increase by (3000) (1.556) = 4668. Hartmann's yield with a risk index of 0.05 is 36,000 / 300,000 = 0.12 His yield with a risk index of 0.06 would be 40,668 / 300,000 = 0.1356 e. This change is outside the objective coefficient range so we must re-solve the problem. The solution is shown below.
LINEAR PROGRAMMING PROBLEM MAX .1G + .1S + .06M S.T. 1) .1G + .05S + .01M < 15000 2) G > 30000 3) S > 30000 4) M > 60000 5) G + S + M < 300000 OPTIMAL SOLUTION Optimal Objective Value 27600.00000 Variable Value Reduced Cost G 30000.00000 0.00000 S 210000.00000 0.00000 M 60000.00000 0.00000 Constraint Slack Shadow Price 1 900.00000 0.00000 2 0.00000 0.00000 3 180000.00000 0.00000 4 0.00000 -0.04000 5 0.00000 0.09200 Objective Allowable Allowable Coefficient
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