SIMPLEX METHOD
Geometric idea: Consider corner points only.
- Start in a corner
- Find an edge with maximal increase of the objective
function
- Find a better CP at the end of this edge
- Stop if improvement is possible
Practical implementation cannot be geometric, but must
be algebraic
Inqualities are not suitable for algebra!
Start from the standard form:
Max
Z = c1x1 + c2x2 + … + cnxn
s.t.
a11x1 + a12x2 + … + a1nxn ≤ b1
a21x1 + a22x2 + … + a2nxn ≤ b2
…
…
…
…
…
…
…
…
am1x1 + am2x2 + … + amnxn ≤ bm
and
x1, x2 …, xn ≥ 0
Here all bi ≥ 0, so x1 = x2 = … = xn = 0 is a feasible starting
solution (CPF).
Rewrite all inequalities as equalities:
x1 + 4x2 ≤ 17
⇔
0 ≤ 17 - x1 - 4x2
Introduce a new extra variable: x3 := 17 - x1 - 4x2
x1 + 4x2 ≤ 17
⇔
x1 + 4x2 + x3 = 17
∧
x3 ≥ 0
x3 is called slack variable.
x3 indicates how “well” the inequality x1 + 4x2 ≤ 17 holds
(x3 > 0) or does not hold (x3 < 0), e.g. how much the
production is under or over its upper bound.
For each ≤ inequality we get an extra non-negative (slackvariable.
n variables with m constraints (≤) ⇒
n + m variables with m constraints (=)
A feasible CP solution is called basic solution.
In a corner point n (slack) variables are = 0, so we have a
maximum of m positive variables. They are called the
basic variables of that CP.
Simplex method algebraically:
1 Starting solution: feasible basic solution.
(= zero solution of the original problem)
2 Create a new basic solution that differs one basic
variable: One non-basic variable becomes > 0 while one
other basic variable becomes = 0. (This corresponds to
walking to a new vertex along an edge of the feasibility
region).
The objective value must improve.
The new solution is the feasible basic solution.
3 Continue with 2 until no improvement is possible. Stop.
In an LP problem a local optimum is also global, so it is
correct that the algorithm will stop.
Standard example from the book: Wyndor Glass Co.
In three factories doors with aluminum and wooden
framings are constructed. Aluminum frames need 1 hour
in factory 1 and 3 hours in plant 3 per batch, the profit is
$3000. Wooden frames use 2 hours in plant 2 and 3 hours
in factory 3 per batch, the profit is $5000. The plants have
available for production: 4, 12, and 18 hours, respectively.
LP model:
Max
s.t.
Z = 3x1 + 5x2
x1
≤
2x2 ≤
3x1 + 2x2 ≤
x1, x2 ≥ 0
4
12
18
(capacity plant 1)
(capacity plant 2)
(capacity plant 3)
(quantities)
Replace the inequalities by equalities (Augmented form):
Max
s.t.
Z = 3x1 + 5x2
x1
+ x3
=
2x2
+ x4
=
3x1 + 2x2
+x5 =
x1, x2, x3, x4, x5 ≥ 0
4
12
18
Simplex algorithm:
Start with the origin of the original problem:
(0, 0) → (0, 0, 4, 12, 18)
(Initial Basic Feasible solution (BF))
Is this solution optimal?
The “free” (non-basic) variables are x1 and x2. The
objective function is Z = 3x1 + 5x2 and it increases if x1 or
x2 increases. The increase is maximal for x2 (largest
coefficient), so x2 must be increased
How large can x2 become?
Constraints:
x3 = 4 – x1
x4 = 12 – 2x2
x5 = 18 – 3x1 – 2x2
x1 = 0, x2 must increase:
x3 = 4
x4 = 12 – 2x2
x5 = 18 – 2x2
≥0
≥0
≥0
⇒ x2 ≤ 6
⇒ x2 ≤ 9
Conclusion: x2 can be 6, maximal.
In this case x4 = 0.
Starting solution:
after one iteration:
(0, 0, 4, 12, 18) basis: x3, x4, x5
(0, 6, 4, 0, 6)
basis: x3, x2, x5
What next?
Z
-3x1 - 5x2
x1
+ x3
+ x4
2x2
+x5
3x1 + 2x2
x1, x2, x3, x4, x5 ≥ 0
=
=
=
=
0
4
12
18
Can the non-basic variables x1, x4 be increased?
What is the effect on the objective function?
Rewrite the system. Take care that the coefficient of x2 in
the x4 equation becomes 1 and remove all other
coefficients from the x2 column:
Z
-3x1 - 5x2
x1
2x2
3x1 + 2x2
Z
-3x1 - 5x2
x1
x2
3x1 + 2x2
Z
+ x3
+ ½ x4
+ 5/2x4
-3x1
x1
+ x3
x2
3x1
=
+ x3
=
+ x4
=
+x5 =
+ ½ x4
- x4
0
4
12
18
=
=
=
+x5 =
0
4
6
18
=
=
=
+x5 =
30
4
6
6
Now it is clear that Z increases as a function of x1,
decreases with x4.
How far can x1 increase (if x4 = 0)?
x3 =
x2 =
x5 =
4 - x1 ⇒ x1 ≤ 4
6
6 - 3x1 ⇒ x1 ≤ 2
Conclusion: x1 can be 2, maximal. Then x5 = 0.
New BF solution is (2, 6, 2, 0, 0), basis: x3, x2, x1
Next iteration: rewrite the system:
Z
+ 5/2x4
-3x1
x1
+ x3
+ ½ x4
- x4
x2
3x1
=
=
=
+x5 =
30
4
6
6
becomes:
Z
+ 5/2x4
x1
+ ½ x4
- 1/3x4 +1/3x5
=
=
=
=
30
4
6
2
x1
+ 3/2x4 + x5
x3 + 1/3x4 - 1/3x5
+ ½ x4
- 1/3x4 +1/3x5
=
=
=
=
36
2
6
2
-3x1
x1
+ x3
x2
Z
x2
The objective function now decreases if x4 or x5 are
increased.
Local optimum = global optimum.
BF solution is (2, 6, 2, 0, 0), so x1 = 2, x2 = 6, Z = 36
Simplex method in tabular form