Computational Methods for
Management and Economics
Carla Gomes
Module 6c
Initialization in Simplex
(Textbook – Hillier and Lieberman)
Initialization
Adapting to other forms
•
•
•
•
Equality constraints
Negative Right-hand sides
Constraints ≥
Minimization
Artificial Variable Approach
• What if our problem is not in standard form – i.e., it may
include functional constraints of type = and ≥ and also
negative RHS?
• How can we identify an initial BFS?
• Approach – artificial variable technique:
– Construct an artificial problem:
• introduce a dummy variable – an artificial variable, s.t., the usual
non-negativity constraints, just for the purpose of being initial basic
variable -- into each of the constraints that are non standard;
• Objective function – modify it to impose an exorbitant penalty on the
artificial variables
– Simplex method – forces the artificial variables to become 0, one
at a time, until they are all gone, providing an initial basic solution
for the original problem.
Handling Equality Constraints
• ai1 x1 + ai2 x2 + … + ain xn = bi
• Can be transformed into:
ai1 x1 + ai2 x2 + … + ain xn ≥ bi
ai1 x1 + ai2 x2 + … + ain xn ≤ bi
But we still would have to deal with one constraint
in non-standard form.
Wyndor Glass
• Requirement: plant 3must be uses at full
capacity  3 x1 + 2 x2 = 18;
• Algebraic form:
Z - 3 x1 - 2 x2
= 0;
x1
+ x3
= 4;
2 x2
+ x3 = 12;
3 x1 + 2 x2
= 18;
What’s the problem with this initial system from the simplex’s
point of view?
Coefficients of Variables
Bas Var
Z
x3
x4
?
equation
Z
0
1
2
3
x1
1
0
0
0
x2
-3
1
0
3
x3
-5
0
2
2
What is the basic variable in the 3rd equation?
No obvious BFS
x4
0
1
0
0
RHS
0
0
1
0
0
4
12
18
Procedure
Construct an artificial problem that has the same optimal
solution as the real problem by making two modifications to
the real problem:
1.
Apply the artificial variable technique by introducing a non-negative artificial
variable into the equality constraint as if it were a slack variable.
3 x1 + 2 x2 + x5 = 18;
2.
x = 18 - 3 x1 + 2 x2
5
Assign an overwhelming penalty to having
x 0
by changing the objective function to:
5
Z = 3 x 1 + 2 x2 - M
x
5
M is a huge positive number (Big M Method).
x 18
5
Note: because x5 plays the role of the slack variable in the
3rd constraint, this constraint becomes equivalent to
3 x1 + 2 x2 <=18; therefore the feasible region is identical to
the feasible region of the original Wyndor Glass problem.
3. Converting Z equation into proper form
Z - 3 x1 - 2 x2
+M x
5
x1
+ x3
2 x2
+ x4
3 x 1 + 2 x2
+ x5
=0
= 4;
= 12;
= 18;
Is this form the right form for the simplex? Why?
convert Z equation into the proper form by subtracting
from Eq(0), Eq(3) times M
Converting Z equation into proper form
Coefficients of Variables
Bas Var
Z
x3
x4
x5
Z
x3
x4
x5
equation
Z
x1
x2
0
1
2
3
1
0
0
0
-5
0
2
2
0
1
0
0
x5
0M
0
1
0
0
1
2
3
1 -(3M+3) -(2M+5)
0
1
0
0
0
2
0
3
2
0
1
0
0
0
0
1
0
-3
1
0
3
x3
x4
RHS
0
0
1
0
4
12
18 -M
0 -18M
0
0
1
4
12
18
Now we are ready to use the Simplex method.
Note: Think of M as a huge number. M only appears in Eq 0 in
Expressions of the form aM + b; because M is huge it is enough to
Compare the multiplicative factor a.
What variable will enter the basis? What variable will leave the basis?
Z
x3
x4
x5
0
1
2
3
1 -(3M+3) -(2M+5)
0
1
0
0
0
2
0
3
2
0
1
0
0
0
0
1
0
0 -18M
0
0
1
4
12
18
4
6
What operations do we need to perform to this tableau?
Answer: Multiply pivot row by (-3) and add it to eq (3);
Multiply pivot row by (3M+3) and add it to eq (0);
Z
x1
x4
x5
0
1
2
3
1
0
0
0
0 -(2M+5) 3M+3
1
0
0
2
0
2
1
0
-3
0
0
1
0
0 -6M+12
0
4
0
12
1
6
Is this a feasible solution for the original problem? Optimal?Why?
What variable will enter the basis? What variable will leave the basis?
Z
x1
x4
x5
0
1
2
3
1
0
0
0
0 -(2M+5) 3M+3
1
0
0
2
0
2
1
0
-3
0
0
1
0
0 -6M+12
0
4
0
12
1
6
6
3
What operations do we need to perform to this tableau?
Answer: Divide pivot row by 2
Multiply new pivot row by (-2) and add it to eq (2);
Multiply new pivot row by (2M+5) and add it to eq (0);
Z
x1
x4
x2
0
1
2
3
1
0
0
0
0
1
0
0
0
0
0
1
-4 1/2
1
3
-1 1/2
0 M+ 5/2
0
0
1
-1
0
1/2
27
4
6
3
Is this a feasible solution for the original problem? Optimal?Why?
What variable will enter the basis? What variable will leave the basis?
Z
x1
x4
x2
0
1
2
3
1
0
0
0
0
1
0
0
0
0
0
1
-4 1/2
1
3
-1 1/2
0 M+ 5/2
0
0
1
-1
0
1/2
27
4
6
3
Z
x1
x4
x2
0
1
2
3
1
0
0
0
0
1
0
0
0
0
0
1
0
0
1
0
1 1/2 M+ 1
- 1/3
1/3
1/3
- 1/3
1/2
0
36
2
2
6
Is this a feasible solution for the original problem? Optimal?Why?
4
2
Graphical Visualization of Wyndor Glass
Problem
x1=2,x2=5,x3=2,x4=0,x5=0
x1=4,x2=3,x3=0,x4=6,x5=0
x1=4,x2=0,x3=0,x4=12,x5=18
x1=0,x2=0,x3=4,x4=12,x5=18
Handling Negative Right Hand Sides
x1
-x1
+ x3
- x3 ≥ 1;
<=-1;
Handling ≥ constraints
• x1 + x2 >= 6 start by introducing a surplus
variable
• x1 + x2 - x3 = 6 now we know how to deal
with equality constraints by adding an
artificial variable, x3 ≥ 0;
• x1 + x2 - x3 + x4= 6 x4 ≥ 0 is an artificial
variable
Note: x1 + x2 - x3 + x4= 6  x1 + x2 = 6 + x3 - x4
Since x3 >=0 and x4 >=0 that means that x1 + x2 can now take
any value (below or above the constraint) so this procedure in fact
eliminates the constraint from the artificial problem.
Minimization
• Minimize Z = c1 x1 + c2 x2 + … + cn xn

• Maximize - Z = - c1 x1 - c2 x2 - … - cn xn
The Big M Method
Description of the Big M Method
1. Modify the constraints so that the RHS of each constraint is
nonnegative. Identify each constraint that is now an = or ≥ constraint.
2. Convert each inequality constraint to augmented form (add a slack
variable for ≤ constraints, add an surplus (excess) variable for ≥
constraints).
3. For each ≥ or = constraint, add artificial variables. Add sign restriction
ai ≥ 0.
4. Let M denote a very large positive number. Add (for each artificial
variable) Mai to min problem objective functions or -Mai to max
problem objective functions.
5. Since each artificial variable will be in the starting basis, all artificial
variables must be eliminated from row 0 before beginning the simplex.
Remembering M represents a very large number, solve the
transformed problem by the simplex.
Math Programming and
Radiation Therapy
Radiation Therapy Overview
• High doses of radiation (energy/unit mass) can kill
cells and/or prevent them from growing and
dividing
– True for cancer cells and normal cells
• Radiation is attractive because the repair
mechanisms for cancer cells is less efficient than
for normal cells
(slides adapted from James Orlin’s)
Radiation Therapy Overview
• Recent advances in radiation therapy now make it
possible to
– map the cancerous region in greater detail
– aim a larger number of different beamlets with greater
specificity
• This has spawned the new field of tomotherapy
• “Optimizing the Delivery of Radiation Therapy to
Cancer patients,” by Shepard, Ferris, Olivera, and
Mackie, SIAM Review, Vol 41, pp 721-744, 1999.
• Also see http://www.tomotherapy.com/
Conventional Radiotherapy
Relative Intensity of Dose Delivered
Conventional Radiotherapy
Relative Intensity of Dose Delivered
Conventional Radiotherapy
• In conventional radiotherapy
– 3 to 7 beams of radiation
– radiation oncologist and physicist work
together to determine a set of beam angles and
beam intensities
– determined by manual “trial-and-error” process
Goal: maximize the dose to the
tumor while minimizing dose to the
critical area
Critical Area
Tumor area
With a small number of beams, it is difficult to
achieve these goals.
Recent Advances
• More accurate map of tumor area
– CT -- Computed Tomography
– MRI -- Magnetic Resonance Imaging
• More accurate delivery of radiation
– IMRT: Intensity Modulated Radiation Therapy
– Tomotherapy
Tomotherapy: a diagram
Radiation Therapy: Problem
Statement
• For a given tumor and given critical areas
• For a given set of possible beamlet origins and
angles
• Determine the weight of each beamlet such that:
– dosage over the tumor area will be at least a target level
gL .
– dosage over the critical area will be at most a target
level gU.
Display of radiation levels
Linear Programming Model
• First, discretize the space
– Divide up region into a 2D (or 3D) grid of pixels
More on the LP
• Create the beamlet data for each of
p = 1, ..., n possible beamlets.
• Dp is the matrix of unit doses delivered by beam p.
p
=
ij
D
unit dose
delivered to pixel (i,j)
by beamlet p
Linear Program
• Decision variables w = (w1, ..., wp)
• wp = intensity weight assigned to beamlet p
for p = 1 to n;
• Dij = dosage delivered to pixel (i,j)
Dij   p1 D w p
n
p
ij
An LP modeltook 4 minutes to
minimize

D
ij
(i, j)
solve.
Dij   p1 D w p
n
p
ij
Dij  g L
for ( i , j )  T
Dij  g U
for ( i , j )  C
wp  0
for all p
In an example reported in the paper, there were
more than 63,000 variables, and more than 94,000
constraints (excluding upper/lower bounds)
What to do if there is no feasible
solution
• Use penalties: e.g., Dij  gL – yij
and then penalize y in the objective.
• Consider non-linear penalties (e.g., quadratic)
• Consider costs that depend on damage rather than
on radiation
• Develop target doses and penalize deviation from
the target
Optimal Solution for the LP
An Optimal Solution to an NLP
Further considerations
• Minimize damage to critical tissue
• Maximize damage to tumor cells
• Minimize time to carry out the dosage
• LP depends on the technology
Mary’s radiation therapy: simplified version
• Goal – design and select the combination of beamlets to
be used and the intensity of each one, to generate the best
possible dose distribution (units: kilorads)
• Decision variables:
– x1 - dose at the entry point for beamlet 1
– x2 - dose at the entry point for beamlet 1
• Objective function:
– Z – total dosage reaching healthy anatomy
min z = 0.4 x1 + 0.5 x2
st
0.3x1 + 0.1x2 ≤ 2.7 critical tissues
0.6x1 + 0.4x2 = 6 tumor region
0.5x1 + 0.5x2 ≥ 6
x1, x2, > 0
center of tumor
min z = 0.4 x1 + 0.5 x2
st
0.3x1 + 0.1x2 ≤ 2.7
0.6x1 + 0.4x2 = 6
0.5x1 + 0.5x2 ≥ 5
x1, x2, > 0
Getting the problem into augmented
form (canonical form) for simplex
st
1
equation
z
0
1
2
3
-1
0
0
0
Simplex Tableaux
x1
x2
x3
-1.1M+0.4 -0.95M+0.5
0.3
0.1
0.5
0.5
0.6
0.4
x4
0
1
0
0
x5
0M
0
1
0
Optimal?
Entering Variable?
x1 enters the base and x3 leaves the base
x6
0
0
-1
rhs
ratio
0 -12M
0
2.7
0
6
1
6
ero
equation
z
0
1
2
3
equation
z
0
1
2
3
equation
x1
x2
x3
-1 -1.1M+0.4 -0.9M+0.5
0
0.3
0.1
0
0.5
0.5
0
0.6
0.4
0
1
0
0
x1
x2
x3
-1 -1.1M+0.4 -0.9M+0.5
0
0.3
0.1
0
0.5
0.5
0
0.6
0.4
0
1
0
0
z
0
1
2
3
x1
-1
0
0
0
z
x1
-1
0
0
0
z
z
x4
x5
x2
x3
x4
0 -16/30M+11/30 11/3M-4/3
1
1/3
10/3
0
1/3
-5/3
0
0.2
-2
0
0
1
0
x1
rhs
ratio
0 -12M
0
2.7
0
6
1
6
x6
rhs
0 -2.1M-3.6
0
9
0
1.5
1
0.6
x6
rhs
0 -2.1M-3.6
0
9
0
1.5
1
0.6
M
0
0
-1
x5
M
0
0
-1
x3
x4
0 -5/3M+7/3
1
20/3
0
5/3
0
-10
x5
x6
rhs
0 -5/3M+11/6
-8/3M-11/6 -0.5M-4.7
0
5/3
-5/3
8
1
5/3
-5/3
0.5
0
-5
5
3
x2
x3
x4
0 -5/3M+7/3
1
20/3
0
5/3
0
-10
x5
x6
rhs
0 -5/3M+11/6
8/3M-11/6 -0.5M-4.7
0
5/3
-5/3
8
1
5/3
-5/3
0.5
0
-5
5
3
x2
0
1
0
0
x6
x2
0
1
0
0
x1
-1
0
0
0
rhs
ratio
0 -12M
0
2.7
0
6
1
6
M
0
0
-1
x5
tie on -5/3
z
x6
M
0
0
-1
0
0
1
0
0
0
1
0
0
1
0
0
-1
0
0
0
x5
0
0
1
0
x2
x3
x4
0 -16/30M+11/30 11/3M-4/3
1
1/3
10/3
0
1/3
-5/3
0
0.2
-2
x1
-1
0
0
0
x4
x3
0
1
0
0
27
4.5
3
>4
0.3
tie on -5/3
x4
0.5
5
1
-5
9
12
10
x5
M-1.1
-1
0.6
3
x6
0
0
1
0
rhs
M
0
-1
0
-5.25
7.5
0.3
4.5
When do we get a feasible solution for the original problem?
What the optimal solution? x1 = ? x2 = ? Z = ? x4 = ?
x5 = ? x6 = ?
From Big M Method to the TwoPhase Method
• The Big-M Method can be thought of as having
two phases:
– Phase 1 – all the artificial variables are driven to 0
(because of the penalty M) in order to reach an initial
BF solution to the real problem;
– Phase 2 – all the artificial variables are kept to 0
(because of the penalty M) while the simplex method
generates a sequence of BF solutions for the real
problem that leads to an optimal solution.
Two phase method – streamlined procedure for
performing the two-phases directly, without introducing M
explicitly;
Two Phase Method: Mary’s radiation therapy
• Real problem’s objective function: min z = 0.4 x1 + 0.5 x2
• Big M method’s objective function: min z = 0.4 x1 + 0.5
x2 + M x4 + M x6
• Since the two first coefficients are negligible
compared to M, the two phase method drops M by
using the following objective functions:
– Phase 1: minimize z = x4 +
x6 (until x4 = , x6 =0)
– Phase 2: min z = 0.4 x1 + 0.5 x2 (with x4 = , x6 =0)
Phase 1: Mary’s radiation therapy
min z = x4 + x6
st
0.3x1 + 0.1x2 + x3
0.6x1 + 0.4x2
0.5x1 + 0.5x2
= 2.7
+ x4
=6
- x5 + x6 = 6
x1, x2, x3, x4 , x5, x6 >= 0
The solution of phase 1 is the initial solution for phase 2, assuming
That the optimal solution of phase 1 is z=0 and x4 , x6 = 0;
Phase 2: Mary’s radiation therapy
min z = 0.4x1 + 0.5x2
st
0.3x1 + 0.1x2 + x3
= 2.7
0.6x1 + 0.4x2
=6
0.5x1 + 0.5x2
- x5 = 6
x1, x2, x3, x5 >= 0
Note: the solution from phase 1 is the initial BFS for phase 2.
Tableau for Phase 1: Mary’s radiation therapy
min z = x4 + x6
st
0.3x1 + 0.1x2 + x3
0.6x1 + 0.4x2
0.5x1 + 0.5x2
= 2.7
+ x4
=6
- x5 + x6 = 6
x1, x2, x3, x4 , x5, x6 >= 0
What is the basic variable in equation 1? Equation 2? Equation 3?
What operations do we need to perform in order to get the canonical
form for the simplex?
(multiply row 2 by –1 and row 3 by –1; add new row 2 and new row 3
to row 0)
Tableau for Phase 1: Mary’s radiation therapy
equation
BV
0
1 x3
2 x4
3 x6
z
x1
-1
0
0
0
x2
-1.1
0.3
0.5
0.6
x3
-0.9
0.1
0.5
0.4
x4
0
1
0
0
x5
0
0
1
0
x6
1
0
0
-1
rhs
0
0
0
1
ratio
-12
2.7
6
6
note that we are using -z
equation
BV
0
1 x3
2 x4
3 x6
equation
z
x1
-1
0
0
0
z
0
1 x1
2 x4
3 x6
equation
x1
-1
0
0
0
z
0
1 x1
2 x4
3 x2
equation
x2
x1
x3
x2
x4
x3
x5
x4
x6
x5
x6
9
12
12
-2.1
9
1.5
0.6
27
4.5
4
rhs
8/3
-5/3
-5/3
5
x6
0
-5
1
5
ratio
-12
2.7
6
6
rhs
0
0
0
1
-5/3
5/3
5/3
-5
x5
1
-4
3/5
6
rhs
0
0
0
1
1
0
0
-1
0
0
1
0
x4
0
0
1
0
x6
1
0
0
-1
0
0
1
0
-5/3
20/3
5/3
-10
x3
0
0
0
1
x5
0
0
1
0
11/3
10/3
-5/3
-2
0
0
0
1
x2
0
1
0
0
x4
0
1
0
0
-16/30
1/3
1/3
0.2
0
1
0
0
x1
-1
0
0
0
x3
-0.9
0.1
0.5
0.4
0
1
0
0
-1
0
0
0
z
0
1 x1
2 x3
3 x2
x2
-1.1
0.3
0.5
0.6
-0.5
8
0.5
3
>4
0.3
rhs
1
5
-1
-5
0
6
0.3
6
What’s the initial solution for phase 2? x1=6; x3 = 0.3; x2=6, x4,x5,x6=0
Phase 2: Mary’s radiation therapy
min z = 0.4x1 + 0.5x2
st
0.3x1 + 0.1x2 + x3
= 2.7
0.6x1 + 0.4x2
=6
0.5x1 + 0.5x2
- x5 = 6
x1, x2, x3, x5 >= 0
Using phase 1 solution, how do we get the 1st tableau for phase 2?
Preparing for Phase 2
equation
z
x1
0
1 x1
2 x3
3 x2
x2
-1
0
0
0
x3
0
1
0
0
x4
0
0
0
1
x5
0
0
1
0
1
-4
3/5
6
0
-5
1
5
Drop the artificial variables
equation
z
0
1 x1
2 x3
3 x2
x1
-1
0
0
0
x2
0
1
0
0
x3
0
0
0
1
x5
0
0
1
0
rhs
0
-5
1
5
0
6
0.3
6
Substitute phase 2 objective function
equation
z
0
1 x1
2 x3
3 x2
x1
-1
0
0
0
x2
0.4
1
0
0
x3
0.5
0
0
1
x5
0
0
1
0
rhs
0
-5
1
5
0
6
0.3
6
Restore canonical form for simplex for phase 2
equation
z
0
1 x1
2 x3
3 x2
x1
-1
0
0
0
x2
0
1
0
0
x3
0
0
0
1
x5
0
0
1
0
rhs
-0.5
-5
1
5
x6
-5.4
6
0.3
6
rhs
1
5
-1
-5
0
6
0.3
6
Phase 2: Mary’s radiation therapy
equation
z
0
1 x1
2 x3
3 x2
x1
-1
0
0
0
x2
0
1
0
0
x3
0
0
0
1
x5
0
0
1
0
rhs
-0.5
-5
1
5
-5.4
6
0.3
6
What variable leaves the basis and what variable enters the basis?
equation
z
x1
0
1 x1
2 x3
3 x2
equation
-1
0
0
0
z
0
1 x1
2 x3
3 x2
x2
0
1
0
0
x1
-1
0
0
0
x3
0
0
0
1
x2
0
1
0
0
x5
0
0
1
0
x3
0
0
0
1
rhs
-0.5
-5
1
5
x5
0.5
5
1
-5
-5.4
6
0.3
6 >1
rhs
0
0
1
0
Is it optimal? What’s the solution? And Z?
-3.25
7.5
0.3
4.5
0.3
Graphical Visualization of Phase I and
Phase II
Big M Method vs. Two-phase Method
The two-phase method streamlines the Big M method by
using only the multiplicative factors in phase 1 and by
dropping the artificial variables in phase 2.

Two-phase method is commonly used in computational
implementations.
No Feasible Solutions
• If the original problem has no feasible
solutions, then either the Big M method or
the phase 1 of the two-phase method yields
a final solution that has at least one
artificial variable greater than zero.
No Feasible Solutions
min z = 0.4 x1 + 0.5 x2
st
0.3x1 + 0.1x2 ≤ 1.8
0.6x1 + 0.4x2 = 6
0.5x1 + 0.5x2 ≥ 5
x1, x2, > 0
No Feasible Solutions
1 x3
2 x4
3 x5
equation
equation
0
0
0
BV
0
1 x3
2 x4
3 x5
z
BV
z
0
1 x1
2 x4
3 x6
equation
equation
0.3
0.5
0.6
x1
x2
x3
-1 -1.1M+0.4 -0.9M+0.5
0
0.3
0.1
0
0.5
0.5
0
0.6
0.4
x1
-1
0
0
0
BV
0
1 x1
2 x4
3 x6
z
BV
0
1 x1
2 x2
3 x6
z
0.1
0.5
0.4
x1
x4
0
1
0
0
x5
x2
x3
x4
0 -16/30M+11/30 11/3M-4/3
1
1/3
10/3
0
1/3
-5/3
0
0.2
-2
0
0
1
0
x2
0
1
0
0
0
0
-1
0
0
1
0
0
0
1
0
x1
-1
0
0
0
0
1
0
x2
x3
x4
-16/30M+11/30 11/3M-4/3
1/3
10/3
1/3
-5/3
0.2
-2
0
1
0
0
-1
0
0
0
1
0
0
0
0
1
x6
M
0
0
-1
x5
x5
x3
x4
x5
0 M+0.5
1/6M-1.1
0
5
-1
1
-5
3
0
-1
-0.5
rhs
ratio
0 -12M
0
1.8
0
6
1
6
x6
M
0
0
-1
0
0
0
1
x6
M
0
0
-1
x6
M
0
0
-1
1.8
6
6
6
12
10
rhs
-5.4M-2.4
6
3
2.4
rhs
0 -2.1M-3.6
0
6
0
3
1
2.4
rhs
0 -0.6M-5.7
0
3
0
9
1
0.6
Why is it an infeasible solution to the original problem?
24
9
12
Handling variables allowed to be negative
• In several situation we need to consider variables
that can have negative values – e.g., rates of
increase.
We consider two cases:
– Variables with a bound on the negative values
allowed
– Variables with no bound on the negative values
allowed
Variables with a bound on the negative values
allowed
• Let’s consider the lower bound L, where L is a negative
constant;
• Let’s assume the constraint:
• x ≥L
• x ≥ L  x’ = x - L , x’ ≥ 0  x = x’ + L
So x and the constraint
x’ ≥ 0.
x ≥ L can be replace with x’ + L and
Wyndor Glass
• x1 – rate of increase in production (does not go below –10)
Let
x1 = rate of increase of the number of doors to produce
x2 = the number of windows to produce
Maximize Z = 3 x1+ 5 x2
subject to
x1 ≤ 4
2x2≤ 12
3x1 + 2x2 ≤ 18
and
x1 ≥ -10, x2 ≥0.
Maximize Z = 3 (x’-10) + 5 x2
subject to
(x’-10) ≤ 4
2 x2 ≤ 12
3 (x’-10) + 2x2 ≤ 18
and
(x’-10) ≥ -10, x2 ≥ 0.
Maximize Z = -30 + 3x’ + 5 x2
subject to
x’≤ 14
2x’ ≤ 12
3x’+ 2x2 ≤ 48
and
x’ ≥ 0, x2 ≥ 0.
Note: A similar technique can be used when L is positive and a variable
is subject to the constraint: x ≥ L
Variables with no bound on the negative values
allowed
• Let’s consider a variable x that can have arbitrary values
• What transformation do we perform to use the LP model?
• x = x+ - x- where x+ ≥ 0; x- ≥ 0 (note that x can have arbitrary values
since x+ and x- can only have positive values)
• What’s the interpretation of this transformation?
• Any BF solution has the property that either x+ = 0 or x- = 0 (or both)
• Therefore at the optimal solution we have:
x if x > 0
x+ =
x+ is the positive part
0
otherwise
|x| if x < 0
x+ =
x- is the negative part
0 otherwise
Example: if x=10 then x+=10 and x-= 0
Wyndor Glass
• x1 – rate of increase in production (can have arbitrary values))
Let
x1 = rate of increase of the number of doors to produce
x2 = the number of windows to produce
Maximize Z = 3 x1+ 5 x2 Maximize Z = 3 (x+ - x-) + 5 x2
subject to
subject to
x1 ≤ 4
(x+ - x- ) ≤ 4
2x2≤ 12
2 x2 ≤ 12
3x1 + 2x2 ≤ 18
3 (x+ - x- ) + 2x2 ≤ 18
and
x2 ≥0.
and
x+ ≥0, x- ≥0 , x2 ≥ 0.
Maximize Z = 3x+
subject to
x+ - x-
- 3 x- + 5 x2
≤4
2 x2 ≤ 12
3x+
- 3 x-
+ 2x2 ≤ 18
and
x+ ≥0, x- ≥0 , x2 ≥ 0.
Note: A similar technique can be used when L is positive and a variable
is subject to the constraint: x ≥ L
• From a computational view point this approach
has the disadvantage of introducing new variables.
If all the variables can have arbitrary values the
transformed model will have twice as many
variables. Can we do better?
• Yes – by introducing only one additional variable.
All variables xj are replaced by:
– with xj = xj’-x’’ where xj’ ≥0, x’’ ≥ 0
(x’’ is the same for all
variables; -x’’ corresponds to the largest (in absolute terms) negative
original variable; xj’is the amount by which xj exceeds x’’.)