LP practice-1
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Northeastern University *
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6230
Subject
Industrial Engineering
Date
Dec 6, 2023
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Uploaded by GrandExploration12563
LP practice
knitr::opts_chunk$set(echo = TRUE, comment = NA)
LP Model: example 1
Comfortable Hands is a company that features a product line of winter gloves for the entire family, men, women
and children. They are trying to decide what mix of these three types of gloves to produce. The company’s labor
force is unionized. Each full-time employee works a 40-hour week. By union contract, the number of full-time
employees can never drop below 20. Non-union part-time workers can also be hired with the following restrictions:
i. Each part-time worker works 20 hours per week.
ii. There must be at least 2 full-time employees for each part-time employee.
All three types of gloves are made out of the same genuine cowhide leather. Comfortable Hands has a long-term
contract with a leather supplier and receives a 5000 ft2 shipment of the material each week. The material and
labor requirements along with the gross profit per glove sold are given in the following table.
Table showing the allocation of resources for producing gloves
data<-matrix(c(2,30,"$8",1.5,45,"$10",1,40,"$6"),ncol=3,byrow=TRUE)
# Specifying the column names and row names
colnames(data)<-c("Material","Labor","Profit")
rownames(data)<-c("Men", "Women","Children")
table=as.table(data)
table
Material Labor Profit
Men
2
30
$8
Women
1.5
45
$10
Children 1
40
$6
Each full-time employee earns $13 per hour, while each part-time employee earns $10 per hour. Management
wishes to know what mix of each of the three types of gloves to produce per week, as well as how many full-time
and part-time workers to employ. They would like to maximize their net profit - their gross profit from sales minus
their labor costs.
a. Clearly define decision variables
b. What is the objective function?
c. What are the constraints?
d. Write down the full mathematical formulation for this LP problem.
Answer to the question
Suppose,
The number of men’s glove
The number of men’s glove
The number of men’s glove
The number of full-time labor
The number of part-time labor
a. Thus, the decision variables of the company are:
b. The objective function is to maximize the net profit which is the difference between the gross profit and the
labor cost. So,
c. Constraints: Material Constraint:
Labor constraint:
For each part-time labor, there is at least 2 full-time labor, so
d. The complete LP model of the given question is then
Subject to the constraints:
Material Constraint:
Labor constraint:
And non-negativity of the decision variables:
LP Model: example 2
Slim-Down Manufacturing makes a line of nutritionally complete, weight-reduction beverages. One of their
products is a strawberry shake which is designed to be a complete meal. The strawberry shake consists of several
ingredients. Some information about each of these ingredients is given below.
data<-matrix(c(1,50,20,3,"$10",
75,100,0,8,"$8",
0,0,50,1,"$25",
0,120,0,2,"$15",
30,80,2,25,"$6"),ncol=5,byrow=TRUE)
# Specifying the column names and row names
colnames(data)<-c("Cal_fat","Total Cal","Vitamin Cont","Thick","Costs")
rownames(data)<-c("Strawberry flav", "Cream","Vitamin suppl",
"Artificial sweetener","Thickening agent")
table=as.table(data)
table
Cal_fat Total Cal Vitamin Cont Thick Costs
Strawberry flav
1
50
20
3
$10
Cream
75
100
0
8
$8
Vitamin suppl
0
0
50
1
$25
Artificial sweetener 0
120
0
2
$15
Thickening agent
30
80
2
25
$6
The nutritional requirements are as follows. The beverage must total between 380 and 420 calories (inclusive). No
more than 20% of the total calories should come from fat. There must be at least 50 milligrams (mg) of vitamin
content. For taste reasons, there must be at least two tablespoons (tbsp) of strawberry flavoring for each tbsp of
artificial sweetener. Finally, to maintain proper thickness, there must be exactly 15 mg of thickeners in the
beverage. Management would like to select the quantity of each ingredient for the beverage which would minimize
cost while meeting the above requirements.
a. Clearly define decision variables
b. What is the objective function
c. What are the constraints
d. Write down the full mathematical formulation for this LP problem
Answer
a. The decision variables can be denoted and defined as follows:
S = Tablespoons of strawberry flavoring,
C = Tablespoons of cream,
V = Tablespoons of vitamin supplement,
A = Tablespoons of artificial sweetener,
T = Tablespoons of thickening agent.
b. Objective function is
c. Subject to
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and non-negativity of the decision variables
Solving LP practice example 1 in R
library
(lpSolve)
Set matrix corresponding to coefficients of constraints by rows. Do not consider the non-negative decision
variables; it is automatically assumed
Objective function
f.obj <- c(8,10,6,-520,-200)
Set constraints
f.con <- matrix(c(2,1.5,1,0,0,
30,45,40,-2400,-1200,
0,0,0,-1,2,
0,0,0,-1,0),nrow=4, byrow = TRUE)
Set inequality signs
f.dir <- c("<=",
"<=",
"<=",
"<=")
Set right hand side coefficients
f.rhs <- c(5000,0,0,20)
Get the final value (z)
lp("max", f.obj, f.con, f.dir, f.rhs)
Success: the objective function is 4500
Success: the objective function is 4500
Variables final values of the decisionn variables
lp("max", f.obj, f.con, f.dir, f.rhs)$solution
[1] 2500.0
0.0
0.0
25.0
12.5
Example 2
library
(lpSolve)
Set matrix corresponding to coefficients of constraints by rows.
Do not consider the non-negative decision variables; it is automatically assumed
obj <- c(10,8,15, 25,6)
constraint <- matrix(c(50,100,120,0,80,
50,100,120,0,80,
-9,55,-24,0,14,
20,0,0,50,2,
1,0,-2,0,0,
3,8,2,1,25),nrow=6, byrow = TRUE)
Set inequality signs
direction <- c(">=",
"<=",
"<=",
">=",
">=",
"=")
Set right hand side coefficients
rhs_coeff <- c(380,420,0,50,0,15)
Get the final value (C)
minimized_cost=lp("min", obj, constraint, direction, rhs_coeff)
minimized_cost$objval
[1] 58.3125
Get the values of the decision variables
minimized_cost$solution[1]
[1] 3.208333
minimized_cost$solution[2]
[1] 0.2708333
minimized_cost$solution[3]
[1] 1.604167
minimized_cost$solution[4]
[1] 0
minimized_cost$solution[5]
[1] 0
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