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IEOR162_SampleMidterm1_sol

# IEOR162_SampleMidterm1_sol - IEOR 162 Spring 2011 Suggested...

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Unformatted text preview: IEOR 162, Spring 2011 Suggested Solution to Sample Midterm 1 1. (a) The graphical solution is shown in the figure below. The feasible region is the shaded zone. The dashed line is the isoprofit line, and the arrow is the improving direction. The problem is unbounded. x x 1 2 ( 1 ) ( 2 ) ( 3 ) i s o p r o f i t i m p r o v i n g d i r e c t i o n (b) The only binding constraint at (2 , 0) is 2 x 1- x 2 ≤ 4. (c) Multiple optimal solutions. Please note that it is “min” rather than “max”. 1 2. We label Sunday as day 1, Monday as day 2, ..., and Saturday as day 7. Then let x ij = number of officers off on days i and j , i = 1 ,..., 7 ,j = i + 1 ,..., 7 . This definition actually defines the following 21 variables: x 12 , x 13 , ..., x 17 , x 23 , x 24 , ..., x 27 , x 34 , ..., x 37 , ..., x 67 . For example, x 12 is the number of officers that are off on Sunday and Monday, x 13 is the number of officers that are off on Sunday and Tuesday, and so on. This set of variables gives us a complete description of the officer schedule. The objective is to minimize the number of officers whose days off are not consecutive, or equivalently, maximizing the number of officers whose days off are consecutive. Therefore, we maximize x 12 + x 23 + ··· + x 67 + x 17 . For Sunday, we need at least 18 officers, which means we may have at most 12 officers off on Sunday. This is achieved by having x 12 + x 13 + ··· + x 17 ≤ 12.12....
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IEOR162_SampleMidterm1_sol - IEOR 162 Spring 2011 Suggested...

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