Question

Problem 7-41

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.

1. Formulate a linear programming model that can be used to determine the number of gallons of regular gasoline and the number of gallons of premium gasoline that should be produced in order to maximize total profit contribution. If required, round your answers to two decimal places.
   

   Let R	=	number of gallons of regular gasoline produced
   P	=	number of gallons of premium gasoline produced

   Max	R	+	P
   s.t.
   	R	+	P	≤		Grade A crude oil available
   	R	+	P	≤		Production capacity
   			P	≤		Demand for premium
   			R, P	≥

2. What is the optimal solution?
   

   Gallons of regular gasoline
   Gallons of premium gasoline
   Total profit contribution	$

3. What are the values and interpretations of the slack variables?
   

   Constraint	Value of Slack Variable	Interpretation
   1
   2
   3

4. What are the binding constraints?
   

   Grade A crude oil available
   Production capacity
   Demand for premium

Answer 1

Answer:

Given data,

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

Z_{A(0, 20,000)}= 0.3(0)+0.5(20000) =10000

Z_{A(20000, 20,000)} = 0.3(20000)+0.5(20000)=16000

Z_{A(40000, 10,000)} = 0.3(40000) + 0.5(10000) =17000

Z_D(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.