Apps

Sunco Gasoline Blending Optimization

Gasoline Blending


Problem

Sunco oil has three different processes that can be used to manufacture various types of gasoline. Each process involves blending oils in the company's catalytic cracker.

Process 1

Running process 1 for an hour costs $5 and requires 2 barrels of crude oil 1 and 3 barrels of crude oil 2. The output from running process 1 for an hour is 2 barrels of gas 1 and 1 barrel of gas 2.

Process 2

Running process 2 for an hour costs $4 and requires 1 barrel of crude 1 and 3 barrels of crude 2. The output from process 2 for an hour is 3 barrels of gas 2.

Process 3

Running process 3 for an hour costs $1 and requires 2 barrels of crude 2 and 3 barrels of gas 2. The output from running process 3 for an hour is 2 barrels of gas 3.

Each week, 200 barrels of crude 1, at $2/ barrel, and 300 barrels of crude 2 at $3/barrel, may be purchased. All gas produced can be sold at the following per-barrel prices: gas 1, $9; gas 2, $10; gas 3, $24. Formulate an LP whose solution will maximize revenues less costs. Assume that only 100 hours of time on the catalytic cracker are available each week.

  • Let x[i] = no. of hours process i is run per week (where i =1,2,3)
  • Let o[i] = no. of barrels of oil i that is purchased per week (i =1,2)
  • Let g[i] = no. of barrels of gas i sold per week (i=1,2,3)

Model

Model sunco
  Variables
    x[1:3] = 30,  >=0
    o[1]   = 100, >=0, <=200
    o[2]   = 100, >=0, <=300
    g[1:3] = 100, >=0
    obj
    profit
  End Variables

  Equations
    ! minimize (-profit) = maximize (profit)
    obj = -profit

    ! profit per week = revenue - costs
    profit = 9*g[1]+10*g[2]+24*g[3]-5*x[1]-4*x[2]-x[3]-2*o[1]-3*o[2]

    ! consumption of crude 1
    2*x[1] + x[2] = o[1]

    ! consumption of crude 2
    3*x[1] + 3*x[2] + 2*x[3] = o[2]

    ! generation of gas 1
    2*x[1] = g[1]

    ! generation (and consumption) of gas 2
    x[1] + 3*x[2] - 3*x[3] = g[2]

    ! generation of gas 3
    2*x[3] = g[3]

    ! cat cracker available 100 hours per week
    x[1] + x[2] + x[3] <= 100
  End Equations
End Model

Solution

Run process 2 for 100 hours/week = $1500/week

If gas 1 price rises above $11.5/barrel, the optimal solution is to run process 1.

If gas 3 price rises above $26/barrel, the optimal solution is to run processes 2 and 3 for equal periods of time (50 hours).