Example: Mary’s radiation therapy (Ch. 3.4)
Mary has a tumor in her
bladder that must be
treated with radiation
therapy. From two different
directions her body is
treated with ionizing
radiation. The effects on
healthy and malignant
tissue are known. Which
doses must be
administered to reach a
certain level in the tumor,
while the surrounding
tissue and organs are
spared as much as
possible?
LP model:
Min
s.t.
and
Z = 0.4x1 + 0.5x2
0.3x1 + 0.1x2  2.7,
0.5x1 + 0.5x2 = 6,
0.6x1 + 0.4x2  6,
x1, x2  0.
Solution with simplex method (Ch. 4.6)
Rewrite the model with equalities by introducing slack
(x3), surplus (x5) and auxiliary variables (x4, x6):
Min
s.t.
and
Z = 0.4x1 + 0.5x2 + Mx4 + Mx6
0.3x1 + 0.1x2 + x3
= 2.7,
0.5x1 + 0.5x2
+ x4
= 6,
0.6x1 + 0.4x2
- x5 + x6 = 6,
x1, x2, x3, x4, x5, x6  0.
Note: the coefficients of x4 and x6 in the objective function
are +M, because we are minimizing!
In maximization form:
- Z + 0.4x1 + 0.5x2
+ Mx4
+ Mx6
0.3x1 + 0.1x2 + x3
0.5x1 + 0.5x2
+ x4
0.6x1 + 0.4x2
- x5 + x6
=0
= 2.7
=6
=6
Note: -Z, because it is a minimization problem.
The basic variables are (x3, x4, x6) and must not occur in
the objective function. Gaussian elimination yields:
- Z + (0.4-1.1M)x1 + (0.5-0.9M)x2
+ Mx5
0.3x1
+ 0.1x2 + x3
0.5x1
+ 0.5x2
+ x4
0.6x1
+ 0.4x2
- x5 + x6
= -12M
= 2.7
=6
=6
- Z + (0.4-1.1M)x1 + (0.5-0.9M)x2
+ Mx5
0.3x1
+ 0.1x2 + x3
0.5x1
+ 0.5x2
+ x4
0.6x1
+ 0.4x2
- x5 + x6
= -12M
= 2.7
=6
=6
The simplex tableaus are:
From the last tableau the solution can be read: Basic variables
are x1 = 7.5, x5 = 0.3, x2 = 4.5 (left-most and right-most column),
so the non-basic variables in this corner point are x3 = x4 = 0.
The objective value is (first row): -Z = -5.25, so:
x1 = 7.5, x2 = 4.5 and Z = 5.25.
The feasible region of the original problem (black line
segment), and of the artificial problem (grey), this is a
projection of a 5 dimensional region.
The simplex method finds the points 0, 1, 2, 3. From point
3 the solution is feasible for the original problem. The
solution happens to be optimal (by coincidence).
Two-phase method (Ch. 4.6)
The big-M method will not be implemented, in general,
with an explicit, large value of M, because that value is
problem dependent and because it may lead to numerical
problems (cancellation).
Other approach: The objective function of the big-M
method is:
Min
Z = 0.4x1 + 0.5x2 + Mx4 + Mx6
The simplex method will first try to reduce the
contribution of M for find a feasible point for the original
problem.
Two-phase method:
Phase 1:
Min
Z = x4 + x6
The result of this phase will be that x4 = x6 = 0 (if there is a
feasible solution). Start with this feasible solution and
change the objective function:
Phase 2:
Min
Z = 0.4x1 + 0.5x2
to find the optimal solution.
Example: Mary’s radiation therapy
Phase 1:
-Z
+ x4
+ x6
0.3x1 + 0.1x2 + x3
0.5x1 + 0.5x2
+ x4
0.6x1 + 0.4x2
- x5 + x6
=0
= 2.7
=6
=6
The basic variables are (x3, x4, x6) and must not occur in
the objective function:
- Z -1.1x1 - 0.9x2
+ x5
0.3x1
+ 0.1x2 + x3
0.5x1
+ 0.5x2
+ x4
0.6x1
+ 0.4x2
- x5 + x6
= -12
= 2.7
=6
=6
In the last tableau the objective value 0 has been found so
x4 = x6 = 0. The solution (6, 6, 0.3, 0, 0, 0) is feasible for the
original problem (but may not be optimal).
In phase 2 the objective Z = 0.4x1 + 0.5x2 is minimized. In
the final tableau of phase 1, replace the objective function
and continue with the simplex method:
The optimal solution is (7.5, 4.5, 0, 0, 0.3), Z = 5.25.
Note: The path followed by the two-phase method in this
case differs from the one in the big-M method!
This may occur in the first phase if two equally best
ascent directions where the choice of ascent direction
may be influenced.
What happens if there is no feasible solution?
Min
s.t.
and
Z = 0.4x1 + 0.5x2
0.3x1 + 0.1x2  2.7,
0.5x1 + 0.5x2 = 6,
0.6x1 + 0.4x2  6,
x1, x2  0.
Replace 0.3x1 + 0.1x2  2.7 with 0.3x1 + 0.1x2  1.8. This
yields the following tableaus for the big-M method:
The optimal objective value still contains M. This means
that there is no feasible solution.
The two-phase method will not reach a zero objective
value in its first phase.
Tie breaking in the simplex method (Ch. 4.5)
During the simplex method it may be unclear what to do
at certain moments:
1. Variable entering the basis: Find the largest
coefficient in the objective function. What if two or
more largest coefficients occur?
Example: 3x1 + 3x2.
Answer: Choose one (e.g. the first).
2. Variable leaving the basis: See how large is can
become (minimum ratio test). What if there are
coefficients  0? For example:
2a. 2x1 + x3 = 4, then x1 = 4/2 – x3/2  2 (because x3  0)
Normal situation.
2b. –2x1 + x3 = 4, then x1 = –4/2 + x3/2 –2. This is always
true, so nothing needs to be checked.
2c. 2x1 + x3 = –4, then x1 = –4/2 – x3/2 –2. Not feasible.
Impossible! Calculation error.
2d. –2x1 + x3 = –4, then x1 = 4/2 + x3/2 2. Impossible,
because x1 = 0 was feasible. Calculation error!
Conclusion: Only apply the minimum ratio test when the
pivot element > 0 and the right-hand side  0.
2e. What if no quotient can be computed because all
coefficients in the pivot column are  0? No leaving
variable. The problem is then unbounded. (Usually an
error in the modeling).
3. Two quotients are equal. You run into two constraints
at the same time. This may lead to an infinite loop
(cycling). Solutions for this problem are known.
4. There may be more optimal solutions for the same
objective value. The simplex method will find the first
one and stops. The simplex algorithm can be
restarted in a direction in which the objective function
remains constant (coefficient = 0 for some non-basic
variables). In this way it is possible to find other
optima.
Example of cycling in the simplex method
Max
10x1 – 57x2 – 9x3 – 24x4
s.t.
0.5x1 – 5.5x2 – 2.5x3 + 9x4  0
0.5x1 – 1.5x2 – 0.5x3 + x4  0
x1  1
x1, x2, x3, x4  0
> A:=matrix([[0.5,-5.5,-2.5,9,1,0,0],[0.5,-1.5,-0.5,1,0,1,0],
[1,0,0,0,0,0,1]]);
b:=vector([0,0,1]);
c:=vector([10 ,-57,-9,-24,0,0,0]);
> T1:=initableau(A,b,c);
> T2:=spil(T1,5,1);
> T3:=spil(T2,6,2);
> T4:=spil(T3,1,3);
> T5:=spil(T4,2,4);
> T6:=spil(T5,3,5);
> T7:=spil(T6,4,6);
This tableau is equal to the first tableau. There was no
improvement in Z. All minimum ratios were = 0, so no
steps were actually taken. This cycling will go on forever.
This behavior can be disturbed by choosing another
leaving basic variable:
> T1:=initableau(A,b,c);
> T2:=spil(T1,6,1);
(instead of T2:=spil(T1,5,1))
> T3:=spil(T2,7,3);
This solution is now optimal. Changing the search
direction may help!
Complexity of the simplex method
Max
s.t.
100x1 + 10x2 + x3
x1  1
20x1 + x2  100
200x1 + 20x2 + x3  10000
x1, x2, x3  0
> A:=matrix([[1,0,0,1,0,0],[20,1,0,0,1,0], [200,20,1,0,0,1]]);
b:=vector([1,100,10000]);
c:=vector([100,10,1,0,0,0]);
> T1:=initableau(A,b,c);
> T2:=spil(T1,4,1);
> T3:=spil(T2,5,2);
> T4:=spil(T3,1,4);
> T5:=spil(T4,6,3);
> T6:=spil(T5,4,1);
> T7:=spil(T6,2,5);
> T8:=spil(T7,1,4);
The optimal solution is found after 8 iterations, so the
simplex method travels through all corner points of the
feasible region. This example can be generalized to n
dimensions. This gives an example with
2n constraints and 2n iterations.
This means exponential behaviour (the number of
iterations can grow exponentially in the size of the
problem). Example by Klee and Minty.
The simplex method is not finite (cycling)
but in practice it is
The simplex method is not polynomiaal (Klee, Minty)
but in practice it is:
ca. 1.5m iterations, maximal 3m (m = #constraints)
Computation time  m3.
Interior Point Methods are provably polynomial.
In practice the number of iterations is constant,
ca. 20 – 80.