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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