EXPLANATION FILE OF PROGRAM SIMPLEX
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LINEAR PROGRAMMING
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The Simplex Method
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1. Kinds of problems
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The goal of linear programming is to optimize an economic function, linear
with respect to independant variables and subjected to constraints.
So we can solve all the problems that can be formulated by a function to be
optimized, with constraints.
For example:
- determine the best distribution of product components.
- dispatch machine working times
- dispatch investments in the best way.
- dispatch production quantities.
- Maximize benefits by optimizing saling prices.
- Minimize production costs, etc.
2. The Simplex Model
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We must maximize (or minimize) an economic function, EF:
EF = a x + a x + ... + a x
1 1 2 2 n n
where variables, x1, x2 ... xn are independant and subjected to constraints,
also linear with respect to these independant variables.
Here is an example of linear constraint:
a x + a x + ... + a x <= b
i1 1 i2 2 in n
where a are real constants and b are real right-side constants.
ij j
When we want to maximize (or minimize) an economic function for which cons-
traints are defined by upper limits, and when the independant variables are
non-negative, we choose a particular method of linear programming, called the
"Simplex" method.
This latter allows obtaining the best solution after some iterations, and
gives the optimal value of EF with the corresponding values of the independant
variables, fulfilling all the constraints.
3. Particular Cases
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The theory of the Simplex method is beyond the scope of this explanation
file, we will only give a few particular applications, noting peculiarities
that may crash the Simplex program:
3.1 More than one solution
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It may happen that two variables interchange between them selves, causing an
endless loop and no optimal solution is found.
3.2 The n-uplet is not a solution
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If the solution (x1, x2 ... xn) = (0, 0 ... 0) is not an obvious solution of
the problem, you just have to add a supplementary x to fulfill the null
condition. n+1
3.3 Modify the constraints
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Sometimes we want a constraint to have a strict value without being lower
than a given value. As before, for this kind of problem, a supplementary va-
riable will be necessary.
3.4 Contracditory variables
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Again, a supplementary variable can often solve the problem.
3.5 Minimization
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This is taken into account in program Simplex. You just have to take -EF
instead of EF and then solve normally.
3.6 Dual problems
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Let us consider a maximization problem with three independant variables:
| MAX EF = a1 x1 + a2 x2 + a3 x3
| a11 x1 + a12 x2 + a13 x3 <= b1
| a21 x1 + a22 x2 + a23 x3 <= b2
| a31 x1 + a32 x2 + a33 x3 <= b3
This the "primal" form; We can define the "dual" form as follows:
| MIN EF = b1 x1 + b2 x2 + b3 x3
| a11 x1 + a12 x2 + a13 x3 >= a1
| a21 x1 + a22 x2 + a23 x3 >= a2
| a31 x1 + a32 x2 + a33 x3 >= a3
The two problems have the same solutions. This a way to transform a maximi-
zation problem into a minimization one and inversely.
Concerning the economic meaning of duality, this is still under discussion.
4. Example
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4.1 Wording
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A builder of houses can make 3 kinds of them with various profits: 15000$,
17000$ and 20000$.
Each kind must respect following conditions:
1) for supply reasons, the number of houses of kind 2 built each month must
not exceed the number of kind 3 by more than two units.
2) for staff reasons, the buider can make each month up to 5, 5 and 3, res-
pectively of kinds 1, 2 and 3.
3) for organization reasons, the builder can make up to 8 houses monthly of
kinds 1,2 and 3, respectively in the proportions of 3, 2 and 1.
The builder, having these data, wants to maximize his monthly profit by
knowing the number oh houses of each kind to build monthly.
4.2 The Simplex model
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Calling X1, X2, X3 the number of houses of each kind built each month, we
want to maximize
EF = 15000 X1 + 17000 X2 + 20000 X3
let it be
MAX [EF] = 15 X1 + 17 X2 + 20 X3
Concerning the constraints, we will take building amounts for one month:
So we have:
- supply conditions ==> 0 X1 + 1 X2 - 1 X3 <= 2
- staff conditions ==> 0.2 X1 + 0.2 X2 + 0.3333 X3 <= 1
- organization conditions ==> 3 X1 + 2 X2 + 1 X3 <= 8
Finally, we have the Simplex problem:
| MAX [EF] = 15 X1 + 17 X2 + 20 X3
| X2 - X3 <= 2
| 3 X1 + 3 X2 + 5 X3 <= 15
| 3 X1 + 2 X2 + X3 <= 8
Program Simplex gives the results:
0.3333 first kind
3 second kind (for one month)
1 third kind
EF = 76
The maximum monthly profit is 76 000 $.
From [Modèles pratiques de décision Tome 2, By Jean-Pierre Blanger,
PSI Editions, France, 1982].
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End of file Simplex.txt