In: Statistics and Probability
Problem 4-09 (Algorithmic)
Epsilon Airlines services predominately the eastern and southeastern United States. A 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 | Employees Needed |
Monday | 60 |
Tuesday | 60 |
Wednesday | 75 |
Thursday | 50 |
Friday | 90 |
Saturday | 45 |
Sunday | 75 |
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 the optimal solution and determine the total number of call-center employees under the optimal solution. Use a software package LINGO or Excel Solver. If your answer is zero, enter "0".
Let Xi = the number of call center employees who start work on day i (i = 1 = Monday, i = 2 = Tuesday…)
Min | X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | ||
s.t. | |||||||||||||||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X2 | + | X2 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1 | + | X2 | + | X3 | + | X4 | + | X5 | + | X6 | + | X7 | |||
X1, X2, X3, X4, X5, X6, X7 ≥ 0 |
Solution:
X1 | = | |
X2 | = | |
X3 | = | |
X4 | = | |
X5 | = | |
X6 | = | |
X7 | = |
Number of excess employees:
Monday | = | |
Tuesday | = | |
Wednesday | = | |
Thursday | = | |
Friday | = | |
Saturday | = | |
Sunday | = |
Total Number of Employees Under the Optimal Solution=
LP model is as follows
Minimize 1*1+1*2+1*3+1*4+1*5+1*6+1*7
show that
1*1+0*2+0*3+1*4+1*5+1*6+1*7 =60
1*1+1*2+0*3+0*4+1*5+1*6+1*7 =60
1*1+1*2+1*3+0*4+0*5+1*6+1*7 =75
1*1+1*2+1*3+1*4+0*5+0*6+1*7 =50
1*1+1*2+1*3+1*4+1*5+0*6+0*7 =90
0*1+1*2+1*3+1*4+1*5+1*6+0*7 =45
0*1+0*2+1*3+1*4+1*5+1*6+1*7 =75
Xi >= 0
Solution by LINDO
Solution
x1=25
x2=0
x3=40
x4=0
x5=25
x6=10
x7=0
Number of Excess employees = Surplus of corresponding day
Monday = (25+0+25+10+0) - 60 =0
Tuesday = (25+0+25+10+0) - 60 =0
Wednesday = (25+0+40+10+0) - 75 =0
Thrusday = (25+0+40+0+0) - 50 =15
Friday = (25+0+40+0+25) - 90 =0
Saturday = (0+40+0+25+10) - 45 =30
Sunday = (40+0+25+10+0) - 75 =0
Total number of employees under the optimal solution = 25+0+40+0+25+10+0 = 100