Description

#install.packages(“lpSolve”) library(“lpSolve”) #Activating the lpSolve Package
Further going down in order to solve the problem we need to define the objective, constraints, direction and constants
Objective Function
The Objective function is to Max Z = 420(L1 + L2 + L3) + 360(M1 + M2 + M3) + 300(S1 + S2 + S3) which can be written as Z = 420L1 + 360M1 + 300S1 + 420L2 + 360M2 + 300S2 + 420L3 + 360M3 + 300S3
Subject to the following constraints
L1 + M1 + S1 ≤ 750
L2 + M2 + S2 ≤ 900
L3 + M3 + S3 ≤ 450 20L1 + 15M1 + 12S1 ≤ 13000
20L2 + 15M2 + 12S2 ≤ 12000 20L3 + 15M3 + 12S3 ≤ 5000
L1 + L2 + L3 ≤ 900
M1 + M2 + M3 ≤ 1200
S1 + S2 + S3 ≤ 750
Non Negativity Constraints
L1,L2,L3,M1,M2,M3,S1,S2,S3 ≥ 0
The above constraints can be written as below
L1 + M1 + S1 + 0L2 + 0M2 + 0S2 + 0L3 + 0M3 + 0S3 ≤ 750
0L1 + 0M1 + 0S1 + L2 + M2 + S2 + 0L3 + 0M3 + 0S3 ≤ 900
0L1 + 0M1 + 0S1 + 0L2 + 0M2 + 0S2 + L3 + M3 + S3 ≤ 450
20L1 + 15M1 + 12S1 + 0L2 + 0M2 + 0S2 + 0L3 + 0M3 + 0S3 ≤ 13000
0L1 + 0M1 + 0S1 + 20L2 + 15M2 + 12S2 + 0L3 + 0M3 + 0S3 ≤ 12000 0L1 + 0M1 + 0S1 + 0L2 + 0M2 + 0S2 + 20L3 + 15M3 + 12S3 ≤ 5000
L1 + 0M1 + 0S1 + L2 + 0M2 + 0S2 + L3 + 0M3 + 0S3 ≤ 900
0L1 + M1 + 0S1 + 0L2 + M2 + 0S2 + 0L3 + M3 + 0S3 ≤ 1200
0L1 + 0M1 + S1 + 0L2 + 0M2 + S2 + 0L3 + 0M3 + S3 ≤ 750
Defining the Objective Function – f.obj
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f.obj <- c(420,360,300,420,360,300,420,360,300)
Defining the Constraints – f.con
f.con <- matrix(c(1,1,1,0,0,0,0,0,0, 0,0,0,1,1,1,0,0,0, 0,0,0,0,0,0,1,1,1,
20,15,12,0,0,0,0,0,0,
0,0,0,20,15,12,0,0,0,
0,0,0,0,0,0,20,15,12, 1,0,0,1,0,0,1,0,0, 0,1,0,0,1,0,0,1,0,
0,0,1,0,0,1,0,0,1), nrow = 9, byrow=T)
Defining the Direction of the constraints – f.dir
f.dir <- c(‘<=’, ‘<=’, ‘<=’,
‘<=’, ‘<=’,
‘<=’, ‘<=’, ‘<=’, ‘<=’)
Defining the constants i.e. the right hand side values – f.rhs
f.rhs <- c(750,900,450,13000,12000,5000,900,1200,750)
Calling the lp function to solve the problem basing the objective function i.e. to maximize the profits
lp(‘max’,f.obj,f.con,f.dir,f.rhs)
## Success: the objective function is 708000
Calling the lp function again to get the values for the variables defined above
lp(‘max’,f.obj,f.con,f.dir,f.rhs)$solution
## [1] 350.0000 400.0000 0.0000 0.0000 400.0000 500.0000 0.0000 133.3333 ## [9] 250.0000
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