线性规划的班次计划 [英] Shift planning with Linear Programming

问题描述

使用R建模和求解线性规划本书在第3.7节中有一个很好的示例，说明了计划转移.我无法用R解决.而且，我对本书中提供的解决方案也不清楚.

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.

一家公司设有急救中心，该中心一天24小时不间断地工作.在 下表详细列出了每个员工的最低需求 每天四班制的六班制.

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解决方案

我使用以下方法解决了上述问题.

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)

然后，我得到以下结果.

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

这与本书中提到的数值解决方案相去甚远.

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

根据数值解所需的解总数为38.但是，由于问题表明，每个时期都有规定的最小雇员人数，因此该解决方案如何有效?

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

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

推荐答案

您的错误在于初始化变量constr的时候，因为您没有将其定义为矩阵.第二个错误是矩阵本身.看看我的例子.

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.

我想知道为什么您不坚持书中的示例，因为我想检查我的解决方案.我的基于此.

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

根据您在评论中的问题进行

EDIT based on your question in the comments:

这是周期内班次的分布:

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
---------------------------------------------------

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