Linear Programming Problem: Production Scheduling Analysis Report

This project presents a comprehensive solution to a linear programming problem focused on optimizing production schedules and minimizing costs for a tire manufacturing company. The problem involves determining the optimal production quantities of nylon and fiberglass tires over a three-month period, considering production capacity constraints for wheeling and regal machines, demand constraints for each month, and the costs associated with production, materials, and inventory. The solution formulates a linear programming model, including an objective function to minimize total costs and a set of constraints representing production capacities, demand requirements, and non-negativity of decision variables. The model aims to determine the optimal production plan, considering factors such as production costs, inventory holding costs, and demand fulfillment. This analysis provides a structured approach to solving complex business problems using mathematical modeling and optimization techniques, offering valuable insights for production planning and resource allocation.

Management Decision Analysis
Linear Programming Problem
Student id & name
[Pick the date]

Linear Programming Problem
Given data and information can be summarized below.
 Company is producing both nylon and fiberglass tyres and the production details for the three
months are as highlighted below:
 The information about the production hours available for the three months for wheeling
machine and regal machine are shown below:
 The production rate for wheeling machine and tyres in terms of hours per tyres are shown
below:
 Variable costs of manufacturing tyres = $5.00 per operating hour
 Inventory carrying charge = $0.10 per tyres per month
 Material costs for nylon = $3.10 per tyre
 Material costs for fiberglass = $3.90 per tyre
 Packaging and shipping costs = $0.23 per tyre
 Price for nylon tyre = $7.00 per nylon
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 Price for fiberglass tyre = $9.00 per nylon
Formulation of linear programming model
Aim
The main aim is to formulate the linear programming model to find the production schedule in
regards to meet the delivery requirement at minimum cost.
Model
Assumption
It is apparent that the linear model needs to be formulated for three months and hence, the list of
variables would be as per the month.
Let
Nylon tyres
xnylon, t= Number of nylon tyres that would be produced through wheeling machine in the month
of t
ynylon ,t= Number of nylon tyres that would be produced through regal machine in the month of t
I nylon, t= Number of nylon tyres that would be kept in inventory at the end of the month of t
Fiberglass tyres
xglass , t= Number of fiberglass tyres that would be produced through wheeling machine in the
month of t
y glass, t= Number of fiberglass tyres that would be produced through regal machine in the month
of t
Iglass ,t = Number of fiberglass tyres that would be kept in inventory at the end of the month of t
It can be seen that there are three months and hence, there would be 6 variables for a month and
thus, total 18 variables. However, it is apparent that at the ending period of august all the
respective tyres would be delivered and thus, the two variables in inventory would be ignored.
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Objective function
(0.75 X nylon, 1 +0.80 Y nylon, 1+0.60 X glass 1+0.70 Y glass 1+0.10 I nylon, 1+0.10 I glass
1)+(0.75 X nylon, 2 +0.80 Y nylon, 2 +0.60 X glass 2 +0.70 Y glass 2 +0.10 I nylon, 2 +0.10 I
glass 2 )+((0.75 X nylon, 3 +0.80 Y nylon, 3 +0.60 X glass 3+0.70 Y glass 3 +0.10 I nylon, 3
+0.10 I glass 3 ))
Or
∑
t =1
3
(¿ 0.75 xnylon,t +0.80 ynylon, t +0.60 I nylon, t +0.70 xglass ,t +0.10 yglass ,t +0.10 I glass ,t ) ¿
It is noteworthy that at the ending period of august all the respective tyres would be delivered
and thus, the two variables in inventory would be ignored that 0.10 I nylon, 3 = 0 , 0.10 I glass 3
= 0
Subject to constraints
 Production capacity constraints for June
For wheeling machine
0.15 xnylon, 1+ 0.12 xglass , 1 ≤ 700
For regal machine
0.16 ynylon ,1 +0.14 y glass ,1 ≤ 1500
 Demand constraints
For nylon for June
xnylon, 1+ ynylon,1 −Inylon ,1=4000
For fiberglass for June
xglass , 1+ y glass ,1−Iglass , 1=1000
For nylon for July
I nylon, 1+xnylon ,2 + ynylon, 2−I nylon,2 =8000
For fiberglass for July
Iglass ,1 + xglass , 2+ yglass ,2−I glass ,2=5000
For nylon for August
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I nylon, 2−xnylon,3 + ynylon ,3=3000
For fiberglass for August
I glass ,2 + xglass , 3+ yglass ,3 =5000
Non- negativity constrains for all the decision variables
xnylon, t ≥0 , ynylon ,t ≥ 0 , Inylon ,t ≥ 0
xglass , t ≥0 , y glass, t ≥ 0 , I glass , t ≥ 0
Final linear programming model
Objective function
MinC=∑
t =1
3
(¿ 0.75 xnylon, t +0.80 ynylon ,t +0.60 Inylon, t +0.70 xglass , t +0.10 yglass , t +0.10 I glass ,t )¿
Subject to constraints
0.15 xnylon, 1+ 0.12 xglass , 1 ≤ 700
0.16 ynylon ,1 +0.14 y glass ,1 ≤ 1500
xnylon, 1+ ynylon,1 −Inylon ,1=4000
xglass , 1+ y glass ,1−Iglass , 1=1000
I nylon, 1+xnylon ,2 + ynylon, 2−I nylon,2 =8000
Iglass ,1 + xglass , 2+ yglass ,2−I glass ,2=5000
I nylon, 2−xnylon,3 + ynylon ,3=3000
I glass ,2 + xglass , 3+ yglass ,3 =5000
xnylon, t ≥0 , ynylon ,t ≥ 0 , Inylon ,t ≥ 0
xglass , t ≥0 , y glass, t ≥ 0 , I glass , t ≥ 0
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