HW4_Solutions

# HW4_Solutions - ORIE 3300/5300 ASSIGNMENT 4 SOLUTION Fall...

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ORIE 3300/5300 ASSIGNMENT 4 SOLUTION Fall 2008 Problem 2 (b) If we consider the minimum number of employees needed for each day as OUTPUT in the general blending model, the INPUT are diﬀerent types of employees with respect to diﬀerent schedules. There are a total of 10 diﬀerent types as we choose four days out of ﬁve weekdays, and one day out of two weekend day (see data ﬁle schedule.dat for details). The minimum input of each type is 0 and the maximum is any number larger than the total sum of requirement. The minimum output for Monday through Sunday is: 45, 45, 40 ,50, 65, 35, 35, and the maximum is also any number larger than the total sum of requirement. Then the output j produced by input i is 1 if input i work on that output day j, and 0 otherwise. Last, the cost for each input is 1 as we are counting the total number of employees needed. (a) Using the model ﬁle blend.mod and the data ﬁle schedule.dat to solve using AMPL, we get the minimum total employees needed is 70. The num- ber of employees needed for each type is (see data ﬁle for details of each type): fn=0, rn=15, wn=20, tn=0, mn=0, fs=5, rs=5, ws=0, ts=0, ms=25. (Note: it’s easy to verify minimality of this solution since one needs at least 35 + 35 = 70 to satisfy Sat and Sun requirements.) Problem 3 (a) Since ¯ c 3 = 1 > 0, ¯ c 5 = 2 > 0, and ¯ c 1 = - 2 < 0, only x 3 and x 5 can enter. If x 3 enters, ¯ t = min( ¯ b 4 ¯ a 43 , ¯ b 6 ¯ a 63 ) = min( 2 4 , 1 2 ) = min( 1 2 , 1 2 ) = 1 2 so either x 4 or x 6 can leave if x 3 enters. If x 5 enters, ¯ t = min( ¯ b 2 ¯ a 52 , ¯ b 4 ¯ a 54 ) = min( 3 2 , 2 1 ) = 3 2 1

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so x 2 must leave if x 5 enters. Therefore, all possible pairs are (3,4), (3,6), and (5,2). (b) In general, increasing or decreasing of the objective value depends on
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HW4_Solutions - ORIE 3300/5300 ASSIGNMENT 4 SOLUTION Fall...

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