In: Operations Management
A company manufactures mountain bikes, hybrid bikes and road bikes. Each bicycle passes through three departments: fabrication, painting, and final assembly. The total labor hours available for fabrication are 100, for painting are 120 and for final assembly are 140. The labor requirements and profit are given in the following table.
Type |
Fabrication (hr) |
Painting (hr) |
Assembly (hr) |
Profit ($) |
Mountain |
3 |
5 |
3 |
90 |
Hybrid |
4 |
2 |
7 |
110 |
Road |
3 |
6 |
3 |
70 |
Formulate the linear programming model (be sure to state what the variables stand for.
How many bicycles of each type should the company make to maximize its profit (record your answer to the nearest whole number)? What is the maximum profit?
Find the sensitivity ranges for the three constraint amounts.
Following information given in the question :-
(1) available Labour hours for Fabrication:-100
(2) available Labor hours for Painting:- 120
(3) available hours for Assembly:- 140
Data table provided
Type | Fabrication hr | Painting Hr | Assembly Hr | Profit ($) | |
Mountain | 3 | 5 | 3 |
90 |
|
Hybrid | 4 | 2 | 7 |
110 |
|
Road | 3 | 6 | 3 |
70 |
Decision Variables :- Maximum profit depended upon total no of cycles made in available hrs
(1) X1 = Total no cycle made of Montain
(2) X2 = Total no of cycles made of Hybrid
(3) X3 = Total no cycles made of Road
Objective Function :-
let Z be the maximum profit; Z = 90X1 +110 X2 + 70 X3
For maximizing the profit we have to utilize maximum labor hrs.
Constraints :-
From the given data table it is clear that
hrs has taken to manufacture one cycle * total no of cycles = Total available hrs.
Therefore ;
3X1+4X2+3X3=100 -----(1)
5X1+2X2+6X3 =120------(2)
3X1+7X2+3X3 =70------ (3)
Now Solving the above three equations using determinants or linear method we can find the values of X1 , X2 , X3 as
X1 = 10 , X2 =15 X3 = 23
Threfore maximum profit
Z= 90 X1 + 110 X2 + 70 X3 =90 * 10 + 110* 15 + 70* 23 = 4160
Process of Solving above three equations:-
From (1) X1 = (100-4x2-3x3)/3
Putting the value of X1 in equations 2 we get an equation in X2 AND X3
14 X2- 3 X3=140 ---------------(4)
NOW putting value of x1 from equation 3 in equation 1 we directly get the value of X3 = 23
Similarly putting the value of X3 in equation 4 we get X2 = 15
And putting the values of X3 and X2 in equation 1 we get X1 =10