Question

[P1](15pts) IE Department at WSU admits Undergraduate (UG), Masters (MS), and Doctoral (PhD) students to its programs every year. In order to have a smooth admission period, the department head wants to know how many employees to hire. Each UG, MS, PhD student requires 3, 4, and 5 hours of employee time to complete the admission process, respectively. In order to sustain the three programs, the department head decided to admit at least 15 UG, 50 MS, and 10 PhD students. Moreover, there are 55 UG, 180 MS, and 40 PhD applicants this year. Let x1, x2, and x3 represent the number of UG, MS, and PhD students admitted to the programs and let y represent the employee hiring decision: y is an integer valued variable indicating the number of employees. Every employee increases labor capacity by 160 hours. Salary for the additional employees is $60,000. Department makes a revenue of $4,000 for each UG, $5,000 for every MS student, and $400 for each PhD student. If the department wants to maximize its profit (revenue from students – salary of additional employees), how many students of each category would be admitted, and how many employees would be hired?

Answer 1

Decision Variables:

Number of UG students = x1

Number of MS students = x2

Number of PhD students = x3

Number of employees = y

Objective Function:

We have to maximize the revenue generated.

The revenue generated is equal to the revenue from students – salary given to employees.

Revenue from UG students = $4000 per student

Revenue from MS students = $5000 per student

Revenue from PhD students = $400 per student

Salary to employee = $60000

MAXIMIZE 4000*x1 + 5000*x2 + 400*x3 – 60000*y

Constraints:

Department decides to admit at least 10 UG, 50 MS and 10 PhD students.

x1 >= 15

x2 >= 50

x3 >= 10

Time required should not exceed the capacity of hours available with department.

Each UG student admission requires = 3 hours

Each MS student admission requires = 4 hours

Each PhD student admission requires = 5 hours.

Each employee can give = 160 hours

3*x1 + 4*x2 + 5*x3 <= 160*y

The number of students admitted should not exceed the number of applicants.

UG applicants = 55

MS applicants = 180

PhD applicants = 40

x1 <= 55

x2 <= 180

x3 <= 40

Putting everything in excel:

Solver Parameters:

Optimal Solution:

Here, number of employees cant be in fraction. Hence we will round it off to the nearest whole number (6)

Hence we have total time of 960 hours.

As the admission in UG and MS is equal to the applicants, the number of students wont increase. Only number of PhD students can be increased in order to compensate the extra time.

Extra time = 960 – 935 = 25 hours

Each PhD student takes 5 hours.

Hence increase in PhD students = 25/5 = 5

So the optimal solution is:

UG	MS	PhD	Employees
x1	x2	x3	y
55	180	15	6

The Optimal Revenue is = $766000

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