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Problem 4-09 (Algorithmic) Epsilon Airlines services predominately the eastern and southeastern United States. The vast majority of Epsilon’s customers make reservations through Epsilon’s website, but a small percentage of customers make reservations via phone. Epsilon employs call-center personnel to handle these reservations along with any problems with the website reservation system and for the rebooking of flights for customers if their plans change or their travel is disrupted. Staffing the call center appropriately is a challenge for Epsilon’s management team. Having too many employees on hand is a waste of money, but having too few results in very poor customer service and the potential loss of customers Epsilon analysts have estimated the minimum number of call-center employees needed by day of week for the upcoming vacation season (June, July, and the first two weeks of August). These estimates are as follows: Minimum Number of Day Monday Tuesday Wednesday Thursday Friday Saturday Sunday Employees Needed 80 50 75 45 60 50 The call-center employees work five consecutive days and then have two consecutive days off. An employee may start work any day of the week. Each call-center employee receives the same salary. Assume that the schedule cycles and ignore start-up and stopping of the schedule. Develop a model that will minimize the total number of call-center employees needed to meet the minimum requirements. Find
Expert Answer
| Let | x1 is number of employees that start on Monday | ||||
| x2 is number of employees that start on Tuesday | |||||
| x3 is number of employees that start on Wednesday | |||||
| x4 is number of employees that start on Thursday | |||||
| x5 is number of employees that start on Friday | |||||
| x6 is number of employees that start on Saturday | |||||
| x7 is number of employees that start on Sunday | |||||
So objective function will be
Minimise Z =x1+x2+x3+x4+x5+x6+x7
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Subject to constraints
Monday workforce-Those starting on Tuesday and Wednesday will be on leave
x1+x4+x5+x6+x7>=55
Tuesday workforce-Those starting on Wednesday and Thursday will be on leave
x1+x2+x5+x6+x7>=80
Wednesday workforce-Those starting on Thursday and Friday will be on leave
x1+x2+x3+x6+x7>=50
Thursday workforce-Those starting on Friday and Saturday will be on leave
x1+x2+x3+x4+x7>=75
Similarly for Friday
x1+x2+x3+x4+x5>=45
Saturday
x2+x3+x4+x5+x6>=60
Sunday
x3+x4+x5+x6+x7>=50
Solution obtained from solver is
Z=90 , minimum employee to be kept
The number of employees starting each day is obtained as
| Monday | 5 |
| Tuesday | 35 |
| Wednesday | 0 |
| Thursday | 10 |
| Friday | 0 |
| Saturday | 15 |
| Sunday | 25 |
