Lecture on 1-22
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SA305 Spring 2014
Blending Models
Blending various substances together in various ways and mediums can be modeled with linear
programs. We begin with an example.
Example 1.1. Sunco Oil manufactures three types of gasoline (gas 1, gas 2 and gas 3). Each type
is produced by blending three types of crude oil (crude 1, crude 2 and crude 3). The sales price
per barrel of gasoline and the purchase price per barrel of crude oil are given in following table:
Gasoline Sale Price per barrel
Gas 1
$70
$60
Gas 2
Gas 3
$50
Oil
Crude 1
Crude 2
Crude 3
Purchase Price per barrel
$45
$25
$20
Sunco can purchase up to 5000 barrels of each type of crude oil daily.
The three types of gasoline differ in their octane rating and sulfur content. The crude oil blended
to form gas 1 must have an average octane rating of at least 90 and contain at most 1% sulfur.
The crude oil blended to form gas 2 must have an average octane rating of at least 89 and contain
at most 2% sulfur. The crude oil blended to form gas 3 must have an octane rating of at least 87
and contain at most 1% sulfur. The octane rating and the sulfur content of the three types of oil
are given in following table:
Crude
Crude
Crude
Crude
oil
1
2
3
Octane rating
92
86
90
Sulfur content
0.5%
2.0%
3.0%
It costs $4 to transform one barrel of oil into one barrel of gasoline, and Sunco’s refinery can
produce up to 14,000 barrels of gasoline daily.
Sunco’s customers require the following amounts of each gasoline: gas 1: 3000 barrels per day; gas
2:2000 barrels per day; gas 3:1000 barrels per day. The company considers it an obligation to meet
these demands. Formulate an LP that will enable Sunco to maximize daily profits.
Step 1: fix our decision variables.
Set
g(crude type, gas type) = the amount of crude type in barrels we use to produce gas type
so
gij = the amount of crude type i in barrels we use to produce gas type j.
With this let’s make some notes.
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We know that the values of i and j can be 1,2, or 3.
(number of barrels of crude type i used) = gi1 + gi2 + gi3
(number of barrels of gas type j produced) = g1j + g2j + g3j
(total barrels of gas produced) = g11 + g12 + g13 + g21 + g22 + g23 + g31 + g32 + g33
Step 2: Create the objective function.
The big idea here is that
profit=[revenue] - [costs]
which is
profit = [70(g11 + g21 + g31 ) + 60(g12 + g22 + g32 ) + 50(g13 + g23 + g33 )]
− [45(g11 + g12 + g13 ) + 25(g21 + g22 + g23 ) + 20(g31 + g32 + g33 )
+4(g11 + g12 + g13 + g21 + g22 + g23 + g31 + g32 + g33 )] .
This is what we want to maximize.
Step 3: Write down all the constraints.
Obviously we have nonnegativity variable constraints gij ≥ 0 for all i and j.
The purchase constraints on each type of crude is
gi1 + gi2 + gi3 ≤ 5000
for each i = {1, 2, 3}.
The refinery’s production restriction is
g11 + g12 + g13 + g21 + g22 + g23 + g31 + g32 + g33 ≤ 14000
The customer requirements for each type of gas is
g11 + g21 + g31 ≥ 3000
g12 + g22 + g32 ≥ 2000
g13 + g23 + g33 ≥ 1000.
Next let’s work with each gas type octane rating:
Now we must create the blending constraints. These are the difficult ones. First let’s work with
each gas type octane rating:
A key idea is that the ratio of the amount of crude type i used to make gas type j is
number of barrels of crude i used to make gas j
total number of barrels of gas type j made
2
=
gij
g1j + g2j + g3j
And then we can multiply this ratio times the octane level and create a constraint:
92(ratio of crude 1 in gas j)+86(ratio of crude 2 in gas j)+90(ratio of crude 3 in gas j)
≥ (the required octane level for gas j)
So for gas 1 we have
92
g21
g31
g11
+ 86
+ 90
≥ 90
g11 + g21 + g31
g11 + g21 + g31
g11 + g21 + g31
and for gas 2 we have
92
g12
g22
g32
+ 86
+ 90
≥ 89
g12 + g22 + g32
g12 + g22 + g32
g12 + g22 + g32
and for gas 3 we have
92
g23
g33
g13
+ 86
+ 90
≥ 87.
g13 + g23 + g33
g13 + g23 + g33
g13 + g23 + g33
These at a first glance look non-linear! However note that each fraction has the same
denominator. And hence we can multiply both sides the make the linear equations:
92g11 + 86g21 + 90g31 ≥ 90(g11 + g21 + g31 )
92g12 + 86g22 + 90g32 ≥ 89(g12 + g22 + g32 )
92g13 + 86g23 + 90g33 ≥ 87(g13 + g23 + g33 ).
Now the sulfur constraints. Again we use the same idea about ratios to get:
.005g11 + .02g21 + .03g31 ≤ .01(g11 + g21 + g31 )
.005g12 + .02g22 + .03g32 ≤ .02(g12 + g22 + g32 )
.005g13 + .02g23 + .03g33 ≤ .01(g13 + g23 + g33 )
We can subtract these constraints to make writing of these equations shorter. And our model
becomes
3
max
s.t.
31g11 + 41g21 + 46g31 + 11g12 + 31g22 + 36g32 + g13 + 21g23 + 26g33
gij ≥ 0
gi1 + gi2 + gi3 ≤ 5000 for each i = {1, 2, 3}
g11 + g12 + g13 + g21 + g22 + g23 + g31 + g32 + g33 ≤ 14000
g11 + g21 + g31 ≥ 3000
g12 + g22 + g32 ≥ 2000
g13 + g23 + g33 ≥ 1000
2g11 − 4g21 ≥ 0
3g12 − 3g22 + g32 ≥ 0
5g13 − g23 + 3g33 ≥ 0
.005g11 − .01g21 − .02g31 ≥ 0
.015g12 − .01g32 ≥ 0
.005g13 − .01g23 − .02g33 ≥ 0
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