Formulation of Problem #1 and Soln to Prob #2

# Formulation of Problem #1 and Soln to Prob #2 - Setting x 1...

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CSA 273 Solutions to Homework Set #7 Problem #1 Let x i = Fraction of the portfolio invested in Bond i; i=1,2,3,4 Maximize Total Expected Rate of Return = .13x 1 + .08x 2 + .12x 3 + .14x 4 Such that x 1 + x 2 + x 3 + x 4 = 1 (Fractions must total 1) .06x 1 + .08x 2 + .10x 3 + .09x 4 .08 (Worst case return constraint) 3x 1 + 4x 2 + 7x 3 +9x 4 6 ("Duration" constraint) x 1 .40 x 2 .40 x 3 .40 (Restrictions on a single bond investment) x 4 .40 x i 0 for i=1,2,3,4 Solution is on accompanying spreadsheet. The optimal strategy is to invest 40% of the total portfolio or \$400,000 in Bond 1 and 30% or \$300,000 into each of Bond 3 and Bond 4. This investment strategy will yield a maximum expected rate of return of 13% (or an expected annual rate of return of \$130,000). Problem #2 a. To convert to standard form, subtract a surplus variable,e 1 , on LHS of first constraint, and add a slack variable, s 2 ,to LSH of second constraint. Min z= 3x 1 + x 2 s.t. x 1 - e 1 =2 (1) x 1 + x 2 + s 2 = 4 (2) 2x 1 - x 2 =3 (3) x 1 , x 2, e 1, s 2 ≥0 b. Solution 1: Select x 1 to be the nonbasic variable.

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Unformatted text preview: Setting x 1 =0 we get e 1 =-2 from (1), x 2 =-3 from (3) and s 2 = 7 from (2). Basic solution 1: x 1 =0, x 2 = -3, e 1 = -2, s 2 = 7 Solution 2: Select x 2 to be the nonbasic variable. Setting x 2 =0, we get : x 1 =3/2 from (3), e 1 =-1/2 from (1), and s 2 = 5/2 from (2) Basic solution 2: x 1 =3/2, x 2 =0, e 1 = -1/2, s 2 = 5/2 Solution 3: Select e 1 to be the nonbasic variable. Setting e 1 =0, we get : x 1 =2 from (1) , x 2 =1 from (3), and s 2 = 1 from (2) Basic solution 3: x 1 =2, x 2 =1, e 1 =0, s 2 = 1 Solution 4: Select s 2 to be the nonbasic variable. Setting s 2 =0, we get : x 1 =7/3 and x 2 =5/3 from (2) and (3), and e 1 =1/3 from (1). Basic solution 4: x 1 =7/3, x 2 =5/3, e 1 =1/3, s 2 =0 c. Solutions 3 and 4 are feasible. d. For solution 3, z=3(2)+ 1 =7 For solution 4, z=3(7/3) + 5/3 = 8 2/3 Thus solution 3 is optimal since the objective is to minimize z. e. See graph on Excel chart....
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## This note was uploaded on 04/17/2008 for the course CSA 273 taught by Professor Patton during the Spring '08 term at Miami University.

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Formulation of Problem #1 and Soln to Prob #2 - Setting x 1...

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