1. problems with bounded variables
Simplex algorithm
for
problems with bounded variables

2. Simplex method for problems with bounded variables
Consider the linear programming problem with bounded variables
Complete the following change of variables to reduce the lower bound to 0
xj = gj – lj (i.e., gj = xj + lj )

3. Simplex method for problems with bounded variables
the problem becomes
Complete the following change of variables to reduce the lower bound to 0
xj = gj – lj (i.e., gj = xj + lj )

4. Simplex method for problems with bounded variables
the problem becomes
replacing : uj = qj – lj and b = h – Al

5. Simplex method for problems with bounded variables
In this problem
since cTl is a constant, we can eliminate it from the minimisation without
modifying the optimal solution.
Then in the rest of the presentation we consider the problem without this constant.

6. Consider the explicit formulation of the problem
One way of solving the problem is to introduce slack variables yj,
and then use the simplex algorithm.

8. Consider a basic feasible solution of this problem
Because of the constraints xj + yj = uj, at least one of the variables xj or yj is basic, j = 1,2,…,n.
Then for all j = 1,2,…,n, one of the three situations holds:
a) xj = uj is basic and yj = 0 is non basic
b) xj = 0 is non basic and yj = uj is basic
c) 0 < xj < uj is basic and 0 < yj < uj is basic
Non degeneracy:
all the basic variables
are positive at
each iteration

9. m + n basic variables required
There are n variables yj
There are at least m variables xj
that are basic
Exactly m variables xj satisfying
0 < xj < uj.
For contradiction, if m0 m variables xj
satisfy the relation, then the
m0 corresponding variables yj would be
basic.
Furthermore, for the n – m0 other indices j,
either xj = uj (case a) or yj = uj (case b)
would be verified.
Then the number of basic variables
would be equal to
2m0 + (n – m0) = m0 + n m + n

10. m + n basic variables required
There are n variables yj
There are at least m variables xj
that are basic
Exactly m variables xj satisfying
0 < xj < uj.
For contradiction, if m0 m variables xj
satisfy the relation, then the
m0 corresponding variables yj would be
basic.
Furthermore, for the n – m0 other indices j,
either xj = uj (case a) or yj = uj (case b)
would be verified.
Then the number of basic variables
would be equal to
2m0 + (n – m0) = m0 + n m + n

11. La base a donc la forme suivante
0 < xj < uj
0 < yj < uj
xj=uj
yj=uj

12. The basis has the following form
0 < xj < uj
0 < yj < uj
xj=uj
yj=uj

20. The basis has the following form
Basis of A
The columns of
the basis B of A
are those of
the variables
0<xj<uj n
0 < xj < uj
0 < yj < uj
xj=uj
yj=uj

21. Then, we can specify a variant of the simplex method to solve this problem
specifically:
by dealing implictly with the upper bound uj. At each iteration, we consider a solution (basic) associated with a basis B de A having
m basic variables
n – m non basic variables

22. n – m non basic variables
At each iteration, we consider a solution (basic) associated with a basis B de A having
m basic variables
n – m non basic variables
Denote the indices of the basic variables IB = {j1, j2, …, jm} where ji is the index of the basic variable in the ith row, then
We find similar
values as in problems
where there are no
upper bounds, except for
non basic variables

23. upper bounds, except for
We have to modify the entering criterion
and the leaving criterion accordingly to
generate a variant of the simplex algorithm
for this problem
We find similar
values as in problems
where there are no
upper bounds, except for
non basic variables

24. Step 1: Selecting the entering variable
The criterion to select the entering variable must be modified to account for the non basic variables xj being equal to their upper bounds uj since these variables can be reduced.
Hence, for an index
if , it is interesting to increase xj
if , it is interesting to decrease xj

25. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

26. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let
Si θ = ∞, alors le problème n’est pas
borné inférieurement et l’algorithme
s’arrête.

27. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let
If θ = ∞, then the problem is not
bounded from below, and the
algorithm stops.

28. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

29. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

30. Step 2.1: Selecting the leaving variable
Value of the
basic variables
The increase θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reach its upper bound us
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

31. Step 1: Selecting the entering variable
The criterion to select the entering variable must be modified to account for the non basic variables xj being equal to their upper bounds uj since these variables can be reduced.
Hence, for an index
if , it is interesting to increase xj
if , it is interesting to decrease xj

32. Step 2.2: Selecting the leaving variable
Value of the
basic variables
The decrease θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reduces to 0
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in this case )
Let

33. Step 2.2: Selecting the leaving variable
Value of the
basic variables
The decrease θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reduces to 0
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

34. Step 2.2: Selecting the leaving variable
Value of the
basic variables
The decrease θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reduces to 0
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

35. Step 2.2: Selecting the leaving variable
Value of the
basic variables
The decrease θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reduces to 0
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

36. Step 2.2: Selecting the leaving variable
Value of the
basic variables
The decrease θ of the entering variable xs is stop by the first of the following three situations happening:
i) xs reduces to 0
ii) a basic variable decreases
to 0 (in this case )
iii) a basic variable
increases to reach its upper
bound (in ths case )
Let

37. References
M.S. Bazaraa, J.J. Jarvis, H.D. Sherali, “ Linear Programming and Network Flows”, 3rd edition, Wiley-Interscience (2005), p. 217
F.S. Hillier, G.J. Lieberman, “Introduction to Operations Research”, Mc Graw Hill (2005), Section 7.3
D. G. Luenberger, “ Linear and Nonlinear Programming ”, 2nd edition, Addison-Wesley (1984), Section 3.6