Linear Algebra, Theory And Applications, 2012a

6.3. THE SIMPLEX ALGORITHM 145
by 2/3 and then using this to zero out the entries below it,
⎛
⎞
3
2
0 − 1 2
0 1 0 7
⎜ − 3 1
2
1
2
0 0 0 1
⎟
⎝ 1
1
2
0
2
1 0 0 3 ⎠ .
3
1
2
0
2
0 0 1 3
Now all the numbers on the bottom left row are nonnegative so the process stops. Now
recall the variables and columns were ordered as x 2 ,x 4 ,x 5 ,x 1 ,x 3 . The solution in terms of
x 1 and x 2 is x 2 = 0 and x 1 = 3 and z =3. Note that in the above, I did not worry about
permuting the columns to keep those which go with the basic variables on the left.
Here is a bucolic example.
Example 6.3.3 Consider the following table.
F 1 F 2 F 3 F 4
iron 1 2 1 3
protein 5 3 2 1
folic acid 1 2 2 1
copper 2 1 1 1
calcium 1 1 1 1
This information is available to a pig farmer and F i denotes a particular feed. The numbers
in the table contain the number of units of a particular nutrient contained in one pound of
the given feed. Thus F 2 has 2 units of iron in one pound. Now suppose the cost of each feed
in cents per pound is given in the following table.
F 1 F 2 F 3 F 4
2 3 2 3
A typical pig needs 5 units of iron, 8 of protein, 6 of folic acid, 7 of copper and 4 of calcium.
(The units may change from nutrient to nutrient.) How many pounds of each feed per pig
should the pig farmer use in order to minimize his cost?
His problem is to minimize C ≡ 2x 1 +3x 2 +2x 3 +3x 4 subject to the constraints
x 1 +2x 2 + x 3 +3x 4 ≥ 5,
5x 1 +3x 2 +2x 3 + x 4 ≥ 8,
x 1 +2x 2 +2x 3 + x 4 ≥ 6,
2x 1 + x 2 + x 3 + x 4 ≥ 7,
x 1 + x 2 + x 3 + x 4 ≥ 4.
where each x i ≥ 0. Add in the slack variables,
x 1 +2x 2 + x 3 +3x 4 − x 5 = 5
5x 1 +3x 2 +2x 3 + x 4 − x 6 = 8
x 1 +2x 2 +2x 3 + x 4 − x 7 = 6
2x 1 + x 2 + x 3 + x 4 − x 8 = 7
x 1 + x 2 + x 3 + x 4 − x 9 = 4