Please solve the following Operation Research (OR) question.?

Question

A restaurant opens seven days a week. The number of workers needed to operate the restaurant is as follows. Mon: 14; Tue: 13; Wed: 15; Thur: 16; Fri: 19; Sat: 18 and Sun: 11. Every worker works five consecutive days and take two days off. This pattern will repeat indefinitley. How can you minimize the number of workers that staff the restaurant?

Answer 1

let
a = # starting work on Monday
b = # starting work on Tuesday
c = # starting work on Wednesday
d = # starting work on Thursday
e = # starting work on Friday
f = # starting work on Saturday
g = # starting work on Sunday
Then
a + 0 + 0 + d + e + f + g = 14
a + b + 0 + 0 + e + f + g = 13
a + b + c + 0 + 0 + f + g = 15
a + b + c + d + 0 + 0 + g = 16
a + b + c + d + e + 0 + 0 = 19
0 + b + c + d + e + f + 0 = 18
0 + 0 + c + d + e + f + g = 11
b - d = - 1, c - e = 2, d - f = 1, e - g = 3, b + c + d + e + f + 0 = 18, 0 + c + d + e + f + g = 11

c - e = 2, d - f = 1, e - g = 3, c + 2d + e + f + 0 = 17
c + d + e + f + g = 11

d - f = 1, e - g = 3, 2d + 2e + f + 0 = 15
d + 2e + f + g = 9

e - g = 3, 2e + 3f + 0 = 13, 2e + f + g = 8

2e + 3f = 13, 3e + f = 11

2e + 33 - 9e = 13, 7e = 20, e = 20/7
60/7 + f = 77/7, f = 17/7
40/7 + 17/7 + g = 56/7, g = -1/7
d - 17/7 = 7/7, d = 24/7
c =14/7 + 20/7, c = 34/7
b = 24/7 - 7/7, b = 17/7
a = 17/7 + 7/7, a = 24/7
Rounding up, you have
a = 4, b = 3, c = 5, d = 4, e = 3, f = 3, g = 0
Checking,
4 + 0 + 0 + 4 + 3 + 3 + 0 = 14 Monday
4 + 3 + 0 + 0 + 3 + 3 + 0 = 13 Tuesday
4 + 3 + 5 + 0 + 0 + 3 + 0 = 15 Wednesday
4 + 3 + 5 + 4 + 0 + 0 + 0 = 16 Thursday
4 + 3 + 5 + 4 + 3 + 0 + 0 = 19 Friday
0 + 3 + 5 + 4 + 3 + 3 + 0 = 18 Saturday
0 + 0 + 5 + 4 + 3 + 3 + 0 = 16 Sunday
This has you overstaffed by 5 on Sunday, which you can accept, or work out an arrangement for "extra" days off. The "neatest" solution is to have the 5 people starting on Wednesday work a 4-day week.

Answer 2

You need at least 19. You are not going to hire extra workers who are a drain on the business, so 19 is also the maximium.

How many non-working 'slots' are there ? Well, Friday everyone works. Sat 1 can be off, Sun 8, Mon 5, Tue 6, Wed 4 and Thur 3. This gives you 27 slots.

This means 11 workers will 1 day off and 8 workers get 2 days off.

You won't be paying people to take days off, so you have 2 'types' of worker - 6 day a week workers and 5 day a week workers.

Hire 19 workers. Offer 11 work for 6 days a week, and 8 work 5 days. You decide which days they work when you hire them.

Anyone who wants to swap their day off has to get some-one else to agree to the swap.

Fire anyone who complains.

Answer 3

This is a linear programming problem.

Use these 7 variables:
x1 = the number of workers who start on Monday
x2 = the number of workers who start on Tuesday
etc.
x7 = the number of workers who start on Sunday.

Build one constraint corresponding to each day of the week, for example the Monday constraint is:
x1 + x4 + x5 + x6 + x7 >= 14.

For Tuesday: x1 + x2 + x5 + x6 + x7 >= 13.

Do this for every day of the week and you will have 7 constraints. Minimize x1+x2+x3+x4+x5+x6+x7 subject to these constraints.

P.S.: And sic the labor code police on Steve B!!!

Answer 4

Close on Mon.
::

Answer 5

close on Saturday