A Small Oil Company Has a Refining Budget of $200,000

question 18

Essay

A small oil company has a refining budget of $200,000 and would like to determine the optimal production plan for profitability.The following table lists the costs associated with its three products.
$$\begin{array} { | c | c | c | c | } \hline \text { Product } & \begin{array} { c } \text { Refining } \\\text { Budget } \\\text { (\$ per unit of } \\\text { output) }\end{array} & \begin{array} { c } \text { Marketing } \\\text { Budget } \\\text { (\$er unit of } \\\text { output) }\end{array} & \begin{array} { c } \text { Material } \\\text { Availability } \\\text { (gallons of } \\\text { crude per unit } \\\text { of output) }\end{array} \\\hline \text { Gasoline } & 0.40 & 0.10 & 10 \\\text { Heating Oil } & 0.10 & 0.05 & 5 \\\text { PlasticResin } & 0.60 & 0.07 & 20 \\\hline\end{array}$$
Marketing has a budget of $50,000,and the company has 750,000 gallons of crude oil available.Each gallon of gasoline contributes 14 cents of profits,heating oil provides 10 cents,and plastic resin 30 cents per unit.The refining process results in a ratio of two units of heating oil for each unit of gasoline produced.This problem has been modeled as a linear programming problem and solved on the computer.The output follows:
Solution
$\begin{array}{rrrr}\text { Variable } & \begin{array}{r}\text { Variable } \\\text { Vabel }\end{array} & \begin{array}{r}\text { Original } \\\text { Value }\end{array} & \begin{array}{r}\text { Coefficient } \\\text { Sensitivity }\end{array} \\\text { Var1 } & 0.000 \mathrm{0} & 0.140 \mathrm{0}&0 \\\operatorname{Var}^{2} & 150000.000 \mathrm{0} & 0.100 \mathrm{0}&0 \\\operatorname{Var}^{3} & 0.000 \mathrm{0} & 0.300 \mathrm{0}&0\end{array}$
$\begin{array}{rrrr}\begin{array}{r}\text { Constraint } \\\text { Label }\end{array} & \begin{array}{r}\text { Original } \\\text { RHV }\end{array} & \begin{array}{r}\text { Slack or } \\\text { Surplus }\end{array} & \begin{array}{r}\text { Shadow } \\\text { Price }\end{array} \\\text { Const1 } & 20000 0 & 18500 0 & 0 \\\text { Const2 } & 5000 \mathrm{0} & 4250 \mathrm{0} & 0 \\\text { Const3 } & 75000 \mathrm{0} & 0 & 0.0200\end{array}$

Objective Function
Value:                                                                                        15000
Sensitivity Analysis and Ranges
Objective Function Coefficients
$$\begin{array} { r r r r } \text { Variable } & \text { Lower } & \text { Original } & \text { Upper } \\\text { Label } & \text { Limit } & \text { Coefficient } & \text { Limit } \\\\\text { Var1 } & \text { No Limit } & 0.14 & 0.2 \\\text { Var2 } & 0.075 & 0.1 & \text { No Limit } \\\text { Var3 } & \text { No Limit } & 0.3 & 0.4\end{array}$$
Right-Hand-Side Values
$$\begin{array} { r r r r } \text { Constraint } & \text { Lower } & \text { Original } & \text { Upper } \\\text { Label } & \text { Limit } & \text { Value } & \text { Limit } \\\\\text { Const1 } & 1500 0 & 20000 0 & \text { No Limit } \\\text { Const2 } & 750 0 & 5000 0 & \text { No Limit } \\\text { Const3 } & 0 & 75000 C & 5000000\end{array}$$
a.Give a linear programming formulation for this problem.Make the variable definitions and constraints line up with the computer output.
b.What product mix maximizes the profit for the company using its limited resources?
c.How much gasoline is produced if profits are maximized?
d.Give a full explanation of the meaning of the three numbers listed following.
First Number: Slack or surplus of 42500 for constraint 2.
Second Number: Shadow price of 0 for constraint 1.
Third Number: An upper limit of "no limit" for the right-hand-side value constraint 1.

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Definitions:

Ordinary Simple Interest Rate

The standard interest rate applied to a loan or investment, calculated on the principal amount without compounding over a specific period.

360-Day Year

A financial calculation assumption where the year is considered to have 360 days for simplifying interest calculations.

Exact Simple Interest

Interest calculated precisely on the principal amount over a specific period, normally based on a 365-day year.

365-Day Year

A method of calculating interest that uses a fixed calendar year of 365 days for computations, commonly used in financial contexts.