Shift planning with Linear Programming

问题

The Modeling and Solving Linear Programming with R book has a nice example on planning shifts in Sec 3.7. I am unable to solve it with R. Also, I am not clear with the solution provided in the book.

Problem

A company has a emergency center which is working 24 hours a day. In the table below, is detailed the minimal needs of employees for each of the six shifts of four hours in which the day is divided.

     Shift    Employees
00:00 - 04:00    5
04:00 - 08:00    7
08:00 - 12:00   18
12:00 - 16:00   12
16:00 - 20:00   15
20:00 - 00:00   10

R solution

I used the following to solve the above.

library(lpSolve)
obj.fun <- c(1,1,1,1,1,1)
constr <- c(1,1,0,0,0,0,
            0,1,1,0,0,0,
            0,0,1,1,0,0,
            0,0,0,1,1,0,
            0,0,0,0,1,1,
            1,0,0,0,0,1)
constr.dir <- rep(">=",6)
constr.val <-c (12,25,30,27,25,15)
day.shift <- lp("min",obj.fun,constr,constr.dir,constr.val,compute.sens = TRUE)

And, I get the following result.

> day.shift$objval
[1] 1.666667
> day.shift$solution
[1] 0.000000 1.666667 0.000000 0.000000 0.000000 0.000000

This is nowhere close to the numerical solution mentioned in the book.

Numerical solution

The total number of solutions required as per the numerical solution is 38. However, since the problem stated that, there is a defined minimum number of employees in every period, how can this solution be valid?

s1 5 s2 6 s3 12 s4 0 s5 15 s6 0

回答1:

Your mistake is at the point where you initialize the variable constr, because you don't define it as a matrix. Second fault is your matrix itself. Just look at my example.

I was wondering why you didn't stick to the example in the book because I wanted to check my solution. Mine is based on that.

library(lpSolve)
obj.fun <- c(1,1,1,1,1,1)
constr <- matrix(c(1,0,0,0,0,1,
        1,1,0,0,0,0,
        0,1,1,0,0,0,
        0,0,1,1,0,0,
        0,0,0,1,1,0,
        0,0,0,0,1,1), ncol = 6, byrow = TRUE)
constr.dir <- rep(">=",6)
constr.val <-c (5,7,18,12,15,10)
day.shift <- lp("min",obj.fun,constr,constr.dir,constr.val,compute.sens = TRUE)

day.shift$objval
# [1] 38
day.shift$solution
# [1]  5 11  7  5 10  0

EDIT based on your question in the comments:

This is the distribution of the shifts on the periods:

shift | 0-4 | 4-8 | 8-12 | 12-16 | 16-20 | 20-24
---------------------------------------------------
20-4  | 5   | 5   |      |       |       |
0-8   |     | 11  | 11   |       |       |
4-12  |     |     | 7    | 7     |       |
8-16  |     |     |      | 5     | 5     |
12-20 |     |     |      |       | 10    | 10
18-24 |     |     |      |       |       |
----------------------------------------------------
sum   | 5   | 16  | 18   | 12    | 15    | 10
----------------------------------------------------
need  | 5   | 7   | 18   | 12    | 15    | 10
---------------------------------------------------

来源：https://stackoverflow.com/questions/43153099/shift-planning-with-linear-programming