In: Statistics and Probability
Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for regular gasoline and $0.50 per gallon for premium gasoline. Each gallon of regular gasoline contains 0.3 gallons of grade A crude oil and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil. For the next production period, Southern has 18,000 gallons of grade A crude oil available. The refinery used to produce the gasolines has a production capacity of 50,000 gallons for the next production period. Southern Oil's distributors have indicated that demand for the premium gasoline for the next production period will be at most 20,000 gallons.
Let R |
= |
number of gallons of regular gasoline produced |
P |
= |
number of gallons of premium gasoline produced |
|
R |
+ |
P |
|||
s.t. |
||||||
R |
+ |
P |
|
Grade A crude oil available |
||
R |
+ |
P |
|
Production capacity |
||
P |
|
Demand for premium |
||||
R, P |
|
Gallons of regular gasoline |
|
Gallons of premium gasoline |
|
Total profit contribution |
$ |
|
Value of Slack Variable |
|
1 |
||
2 |
||
3 |
Grade A crude oil available |
|
Production capacity |
|
Demand for premium |
|
OR FOR BETTER UNDERSTANDING
Given that,
Southern Oil Company produces two grades of gasoline: regular and premium. The profit contributions are $0.30 per gallon for regular gasoline and $0.50 per gallon for premium gasoline. Each gallon of regular gasoline contains 0.3 gallons of grade.
A crude oil and each gallon of premium gasoline contains 0.6 gallons of grade A crude oil. For the next production period, Southern has 18,000 gallons of grade A crude oil available. The refinery used to produce the gas lines has a production capacity of 50,000 gallons for the next production period.
Southern Oil's distributors have indicated that demand for the premium gasoline for the next production period will be at most 20,000 gallons.
(a).
Write a linear program.A linear program is a mathematical model with a linear objective function and a set of linear constraints where variables are non negative:
Let R represent the number of gallons of regular gasoline prod.ed and P represent the number of gallons of premium gasoline produced.
The Profit per each regular gasoline is given as 0 3.
Therefore, for R. gallons of regular gasoline, the profit will be 0.3R.
Similarly, the profit per each premium gasoline is given as 0.5
Therefore, for P gallons of regular gasoline, the profit will be 0.5R
Now the total profit will be 0.3R + 0.5P
This implies, Z = 0.3R + 0.5P
The profit function should be maximized.
Hence, the objective [unction is Max Z =0.3R+ 0.5P
Now from constrains:
Each gallon of regular gasoline contains
Hence, the formulated linear programming problem is shown below:
Max Z=0.3R+0.5R=P
Subjected to constraints:
0.3R+0.6P 18,000
R+P 50,000
P 20,000
R,P 0
(b).
To find the optimal solution, use graphical method to solve the LPP obtained:
Step 1:
Consider the constraints as equations.
Therefore,
0.3R+0.6P=18,000
R+P=50,000
P=20,000
Step 2:
Find the co-ordinates for the first equation:
When R=0, the points of the equation 0.3R+0.6P=18,000 are shown below.
0.3R+0.6P=18,000
0.3(0)+0.6P=18,000
p=30,000
When p=0,
0.3R+0.6(0)=18,000
R=18000/0.3
R=60,000
Therefore, the points for the equation 0.3R+0.6P=18,000 are (0, 30,000) and (60,000, 0).
Join the points to get the straight line.
Step 3:
Find the co-ordinates for the second equation:
For the second equation R+P=50,000, the points are shown below:
When R=0
R+P= 50,000
0+P=50000
P=50000
When P=0
R+P= 50,000
R+0=50000
R=50000
Therefore, the points for the equation R+P= 50,000 are (0,50000) and (50000, 0).
Join the two points to get the straight line.
Step 4:
The third equation co-ordinates are (0, 20,000) which is parallel to are Axis "X".
The graph is shown below:
The values of "p" are taken on Axis "Y" and the values of "R" are taken on Axis "X".
From the above graph it is observed that points O, A, B, C forms a feasible region.
Now consider the points of feasible region and substitute in the profit function.
Max Z = 0.3R+ 0.5P
Therefore,
Z(0,0)= 0.3(0)+0.5(0)= 0
ZA(0, 20,000)= 0.3(0)+0.5(20000) =10000
ZA(20000, 20,000) = 0.3(20000)+0.5(20000)=16000
ZA(40000, 10,000) = 0.3(40000) + 0.5(10000) =17000
ZD(5)=0.3(50000)+0.5(0)=15000
The maximum value of Z is at the point "C" .
Therefore, the optimal solution at the point is "C".
(c)
To obtain the values of the variables and maximum profit, substitute the point "C" which (40,000 ,10,000) in the objective function.
Hence, the maximize profit is Z =17,000 at R =40,000 and P= 10,000
(d).
A binding constraint is a constraint whose value satisfies the
optimal solution and that any changes in its value changes the
optimal solution. Therefore, from the graph it Is observed that the
intersection of the two equations 0.3R+0.6P=18,000 and R+P=50,000
named as point "C" satisfies the optimal solution.
Hence, the binding constraints are 0.3R+ 0.6P
18,000 and R+P
50,000.
NOTE:: I HOPE YOUR HAPY WITH MY ANSWER......***PLEASE SUPPORT ME WITH YOUR RATING.....
***PLEASE GIVE ME "LIKE".....THANK YOU