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IE220_F08_hw1sol

# IE220_F08_hw1sol - IE220 Introduction to Operations...

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IE220: Introduction to Operations Research Fall 2008 Prof. Imre P´olik Homework 1 Solutions 3.1-8 Let x 1 denote the number of units of special risk insurance and x 2 the number of units for mortgages. The optimization problem is max 5 x 1 + 2 x 2 3 x 1 + 2 x 2 2400 x 2 800 2 x 1 1200 x 1 , x 2 0 The feasible set is shown in the following figure: 200 400 600 800 200 400 600 800 1000 1200 x 1 x 2 3 x 1 + 2 x 2 2400 2 x 1 1200 x 2 800 max 5 x 1 + 2 x 2 (600 , 300) After graphing the constraints the optimal solution turns out to be at the intersection of the constraints 2 x 1 = 1200 and 3 x 1 + 2 x 2 = 2400, which implies x 1 = 600, x 2 = 300. The optimal objective function value is Z = 3600. 1

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3.1-12 The point (2 , 3) satisfies the constraint kx 1 + x 2 2 k +3 with equality for all values of k , thus this constraint boundary always passes through the point. The other end of this constraint x 1 x 2 1 2 3 4 5 6 7 8 1 2 3 4 5 x 2 3 - x 1 + x 2 2 kx 1 + x 2 2 k + 3 max x 1 + 2 x 2 ( 2 + 3 k , 0 ) boundary (see the figure) is on the x 1 axis, at (2+ 3 k , 0), (and there is no intersection if k = 0,
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