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Homework 4

# Homework 4 - solutions to the LP If an LP has a feasible...

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x 2 100 x 1 100 #182 ISE 2404 Homework 4 1. Page 139, Question 3 max z = 2x 1 + 2.5x 2 s.t. x 1 + 2x 2 350 2x 1 + x 2 400 x 1 , x 2 0 The corner points of the LP define the feasible region. All basic feasible solutions lie within the feasible region. 2. Page 149, Question 1 max z = 3x 1 + 2x 2 s.t. 2x 1 + x 2 100 x 1 + x 2 80 x 1 40 x 1 , x 2 0 - - 3 200002110010011010801000140 - 0 20 - - 031200110 2200101 1401000140 002 - - - 0 11600110 22000 111201000140 001 - - - 1018001 1206000 11120101 1020 x 1 = 60 x 2 = 20 The maximum profit that Giapetto can expect is \$180

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3. Page 149, Question 3 max z = 2x 1 – x 2 + x 3 s.t. 3x 1 + x 2 + x 3 60 x 1 – x 2 + 2x 3 10 x 1 + x 2 – x 3 20 x 1 , x 2 0 - - 21 10 - - 000311100601 120101011 100120 - 0 130 - - - - - 202004 51 30301 120101002 30 1110 / 003 20 / / - - / / / - / - / / 3 21 2250011 1 210101 201 21 21501 3 20 1 21 25 x 1 = 15 x 2 = 5 x 3 = 0 z = 25 4. Page 154, Question 7 The set of optimal solutions to an LP is a convex set because all optimal solutions to an LP must also be feasible
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Unformatted text preview: solutions to the LP. If an LP has a feasible region, it must be convex. 5. Page 158, Question 1 max z = 2x 2 s.t. x 1 – x 2 4 ≤-x 1 + x 2 1 ≤ x 1 , x 2 ≥-0 200--01 1104 11011-2002-200115 11011-2002-200115 11011 x 1 should enter the basis, but the ratio test fails to identify a row to leave the basis. Therefore, we conclude that the LP is unbounded. x 1 = 10,000 x 2 = 10,000 z = 20,000 10,000 ≥ 6. Page 158, Question 3 Upon the next iteration of the simplex algorithm, the ratio test will fail to identify a row to leave the basis. 7. Page 178, Question 2 Upon the first iteration of the simplex algorithm, x 2 will identified as the variable to enter the basis. However, the ratio will fail to identify a row to leave the basis....
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Homework 4 - solutions to the LP If an LP has a feasible...

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