The Decision Sciences department head at a university will be scheduling faculty
ID: 3037214 • Letter: T
Question
The Decision Sciences department head at a university will be scheduling faculty to teach courses during the coming fall semester. Three required courses need to be scheduled. The three courses are at the UG, MBA, and MS levels. Three professors will be assigned to the courses, with each professor receiving one of the courses. Student evaluations of professors are available from previous terms. Based on a rating scale of 5, the average student evaluations for each professor are shown below.
Course
Professor UG MBA MS
A 4.0 3.6 -
B 4.2 3.5 4.3
C 3.8 4.2 4.0
Professor A does not have a Ph. D. and cannot be assigned to teach the MS-level course. If the department head makes teaching assignments based on maximizing the student evaluation ratings over all three courses, what staffing assignments should be made?
PLEASE TYPE ALL ANSWERS-PLEASE NO WRITTEN ANSWERS
Formulate an integer programming model for this assignment problem by determining
(a) The decision variables.
(b) The objective function. What does it represent?
(c) All the constraints. Briefly describe what each constraint represents.
Note: Do NOT solve the problem after formulating.
Explanation / Answer
(a) THE DECISION VARIABLES:
Consider the following table
Each x[i],y[i],z[i] represents 0 or 1 according as the corresponding column professor teaching the corresponding row course.
For example , z[1]=0 (a given constraint) represents the fact that Prof A does not teach MS.
If y[2]=1, that means Prof B teaches the course MBA.
(b) The objective function is to maximize the evaluation of all the concerned professors as per the assignment.
In other words the objection function F = Maximize ( Max(4x[1]+3.6y[1], 4.2x[2]+3.5y[2]+4.3z[2], 3.8x[3]+4.2y[3]+4z[3]))
(c) CONSTRAINTS:
(i) All decision variables take values 0 or 1
(i) Each professor would be assigned exactly one course:
x[1]+y[1]=1, x[2]+y[2]+z[2]=1, x[3]+y[3]+z[3]=1
x[1]+x[2]+x[3]=1 y[1]+y[2]+y[3]=1 (z[1]=0)+z[2]+z[3]=1
A B C UG x[1] x[2] x[3] MBA y[1] y[2] y[3] MS z[1]=0 z[2] z[3]Related Questions
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