{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW4_Solutions

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 the reduced cost ¯ c k of the entering variable. The objective value goes up (down) if ¯ c k > 0( < 0) AS LONG AS the new value of x k is positive.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern