Epsilon Airlines services predominantly 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 phones. Epsilon employs call center personnel to handle these reservations and to deal with website reservation system problems and for the rebooking of flights for customers whose plans have changed or whose 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 the week for the upcoming vacation season (June, July, and the first two weeks of August). These estimates are as follows:
Day Minimum Number of Employees Needed
Monday 75
Tuesday 90
Wednesday 50
Thursday 60
Friday 45
Saturday 75
Sunday 50
The call center employees work for five consecutive days and then have two consecutive days off. An employee may start work on 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.
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 + X4 + X5 + X6 + X7
X1 + X2 + X5 + X6 + X7
X1 + X2 + X3 + X6 + X7
X1 + X2 + X3 + X4 + X7
X1 + X2 + X3 + X4 + X5
X2 + X3 + X4 + X5 + X6
X3 + X4 + X5 + X6 + X7
X1, X2, X3, X4, X5, X6, X7
Find the optimal solution.
X1 =
X2 =
X3 =
X4 =
X5 =
X6 =
X7 =
Total Number of Employees =
Give the number of call center employees that exceed the minimum required.
Excess employees:
Monday =
Tuesday =
Wednesday =
Thursday =
Friday =
Saturday =
Sunday =

Respuesta :

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

Explanation:

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

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

The optimal solution would be as follows:

[tex]X1 =[/tex] 5

[tex]X2 =[/tex] 35

[tex]X3 =[/tex] 0

[tex]X4 =[/tex] 10

[tex]X5 =[/tex] 0

[tex]X6 =[/tex] 15

[tex]X7 =[/tex] 25

Total employees [tex]= 90[/tex]

Employees

As per the question,

Assuming

[tex]X1[/tex] as the no. of employees beginning on Monday

[tex]X2[/tex] as the no. of employees beginning on Tuesday

[tex]X3[/tex] as the no. of employees beginning on Wednesday

[tex]X4[/tex] as the no. of employees beginning on Thursday

[tex]X5[/tex] as the no. of employees beginning on Friday

[tex]X6[/tex] as the no. of employees beginning on Saturday

[tex]X7[/tex] as the no. of employees beginning on Sunday

So,

[tex]Z = X1+X2+X3+X4+x=X5+X6+X7[/tex]

Constraints

On Monday = One who begins the work on Tuesday and Wednesday would be on leave.

[tex]X1+X4+X5+X6+X7>=55[/tex]

On Tuesday = One who begins the work on Wednesday and Thursday would be on leave.

[tex]X1+X2+X5+X6+X7>=80[/tex]

On Wednesday One who begins the work on Thursday and Friday would be on leave.

[tex]X1+X2+X3+X6+X7>=50[/tex]

On Thursday, One who begins the work on Friday and Saturday would be on leave.  

[tex]X1+X2+X3+X4+X7>=75[/tex]

On Friday also,

[tex]X1+X2+X3+X4+X5>=45[/tex]

Saturday

[tex]X2+X3+X4+X5+X6>=60[/tex]

Sunday

[tex]X3+X4+X5+X6+X7>=50[/tex]

So,

[tex]Z = 90[/tex]

The no. of employees would be:

[tex]X1 = Monday = 5\\X2 = Tuesday = 35\\X3 = Wednesday = 0\\X4 = Thursday = 10\\X5 = Friday = 0\\X6 = Saturday = 15\\X7 = Sunday = 25[/tex]

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