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Unformatted text preview: 1. First convert the constraint on average maturity to linear form: 4x1 + 0x2 – 2x3 <= 0. Then add a fictitious constraint defining risk, which will be treated (by us) as a second objective function. Below x5 is the slack variable on the average maturity constraint and x6 is the slack variable on the risk row. Output from The Simplex Assistant: Iteration #1:Do Gaussian elimination with pivot= row 2, column 1. The resulting tableau is: Iteration #2: Do Gaussian elimination with pivot=row 1, column 3. The resulting tableau is: This is optimal for maximizing the return, i.e. for θ =0. If we write this in the tableau format that we are used to, with the objective (return)- θ (risk), we get the following tableau: -z x1 x2 x3 x4 x5 RHS 1 0 -0.0033333 0 -0.043333+2 θ-0.0091667+0.5 θ-0.43333+20 θ 0 0 0.66667 1 0.66667 -0.16667 6.6667 0 1 0.33333 0 0.33333 0.16667 3.3333 In order for this solution to be optimal, all the coefficients in the objective row of the tableau have to be nonpositive, i.e. we have: 0.5 0.0091667- 20 0.43333- ≤ + ≤ + θ θ This gives the following bounds on θ : 0.021667 ≤ θ and 0.018333 ≤ θ .Therefore this solution will be optimal for 0.018333 ≤ ≤ θ . The return can be easily read off from the objective row: 0.43333, as well as the risk: 20. The optimal(return)- θ (risk)= 0.43333-20 θ for 0.018333 ≤ ≤ θ...
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This note was uploaded on 03/08/2012 for the course ORIE 3300 at Cornell.