Abstract image of a woman sitting on books and studying on a laptop

Bounded var

Samo Alwatan, profile picture
Samo Alwatan

1) The document describes using the simplex method to solve linear programming problems with bounded variables. It introduces change of variables to reduce the lower bounds to zero, resulting in a problem with non-negative variables. 2) The basis of the transformed problem has m variables satisfying 0 < xj < uj, with the remaining n - m variables either equal to their upper bounds or having their complement equal to the upper bound. 3) It shows that the non-singularity of the basis implies the matrix B is non-singular, thus B forms a basis for the constraint matrix A.

TechnologySports
1 of 37
Downloaded 14 times
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
Simplex algorithm for problems with bounded variables
Simplex method for problems with bounded variables • Consider the linear programming problem with bounded variables min c T g s.t. Ag = h lj ≤ g j ≤ qj j = 1, 2,..., n where g , c, l , q ∈ R n , h ∈ R m , and A is a m × n matrix • Complete the following change of variables to reduce the lower bound to 0 xj = gj – lj (i.e., gj = xj + lj )
Simplex method for problems with bounded variables min c T g s.t. Ag = h lj ≤ g j ≤ qj the problem becomes min c T ( x + l ) s.t. A( x + l ) = h lj ≤ xj + lj ≤ qj j = 1, 2,..., n j = 1, 2,..., n where c, x, l , q ∈ R n , h ∈ R m , and A is a m × n matrix • Complete the following change of variables to reduce the lower bound to 0 xj = gj – lj (i.e., gj = xj + lj )
Simplex method for problems with bounded variables the problem becomes min c T ( x + l ) s.t. A( x + l ) = h lj ≤ xj + lj ≤ qj j = 1, 2,..., n where c, x, l , q ∈ R n , h ∈ R m , and A is a m × n matrix min c T x + c T l s.t. Ax = h − Al lj − lj ≤ xj + lj − lj ≤ qj − lj replacing : uj = qj – lj and j = 1, 2,..., n b = h – Al min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj j = 1, 2,..., n
Simplex method for problems with bounded variables • In this problem c T l + min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj j = 1, 2,..., n 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.
min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj • Consider the explicit formulation of the problem n n min = ∑ c x min z z = ∑jc jjx j j j =1 =1 n s.t. s.t. n x a ∑ ∑xa = b ij j ij j =1 j =1 j i im == i1, 2,...,= 1, 2,..., m i b 0 ≤ x x ≤+ y j = ujj = 1, 2,..., 1, 2,..., n j =n j j uj xj , yj ≥ 0 j = 1, 2,..., n • One way of solving the problem is to introduce slack variables yj, and then use the simplex algorithm. j = 1, 2,..., n
n min z = ∑ c j x j Tableau with m + n rows j =1 n s.t. ∑a x = bi i = 1, 2,..., m xj + yj = uj j = 1, 2,..., n xj , yj ≥ 0 ij j = 1, 2,..., n j j =1 n min z = ∑ c j x j Tableau with m rows j =1 n s.t. ∑a x = bi i = 1, 2,..., m 0 ≤ xj ≤ uj j = 1, 2,..., n ij j j =1 account implicitly
n min z = ∑ c j x j j =1 n s.t. ∑a x ij j = bi j =1 xj + yj = uj xj , yj ≥ 0 i = 1, 2,..., m Non degeneracy: j = 1, 2,..., n all the basic variables are positive at j = 1, 2,..., n each iteration • 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
a) x j basic; y j non basic b) x j non basic; y j basic c) x j basic; y j basic n min z = ∑c x j j j =1 n Sujet à ∑a = bi i = 1,2,..., m xj + yj =uj j = 1,2,..., n xj , yj ≥0 j = 1,2,..., n ij x j j =1 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
a) x j basic; y j non basic b) x j non basic; y j basic c) x j basic; y j basic n min z = ∑c x j j j =1 n Sujet à ∑a = bi i = 1,2,..., m xj + yj =uj j = 1,2,..., n xj , yj ≥0 j = 1,2,..., n ij x j j =1 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
n min z = La base a donc la forme suivante ∑c x j j j =1 n ∑a x s.t. ij = bi j i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n m min z = c T x s.t. Ax + 0 y = b Ix + Iy = u x, y ≥ 0 m  A 0 I I    n To simplify notation, assume the following basic variables: 0 < xi < ui 0 < yi < ui 0 < xj < uj 0 < yj < uj xj=uj yj=uj i = 1,… , m i = 1,…, m xi = ui i = m + 1,… , m + l yi = ui i = m + l + 1,…, m + n
n min z = The basis has the following form ∑c x j j j =1 n ∑a x s.t. ij j = bi i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n m min z = c T x Ax + 0 y = b s.t. Ix + Iy = u min z = c x Sujet à Ax + 0 y = b x, y ≥ 0 T Ix + Iy = u x, y ≥ 0 m 1  A 0 I I    1 n 2 To simplify notation, assume the following basic variables: 3 m 0 < yj < uj m xj=uj yj=uj n−m i = 1,…, m 0 < yi < ui 0 < xj < uj 0 < xi < ui i = 1,…, m xi = ui i = m + 1,…, m + l yi = ui i = m + l + 1,… , m + n
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
The basis has the following form m Basis of A 1 The columns of the basis B of A are those of the variables 0<xj<uj 1 n 2 3 0 < xj < uj 0 < yj < uj xj=uj yj=uj
• Then, we can specify a variant of the simplex method to solve this problem specifically: n min z = ∑ c j x j j =1 n s.t. ∑a x = bi i = 1, 2,..., m 0 ≤ xj ≤ uj j = 1, 2,..., n ij j j =1 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 0 < xj < uj j ∈ IB n – m non basic variables x j = 0 ou u j j ∈ JB
n min z = ∑c x j j j =1 n s.t. ∑a x ij j = bi i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n • At each iteration, we consider a solution (basic) associated with a basis B de A having 0 < xj < uj j ∈ IB m basic variables n – m non basic variables x j = 0 or u j j ∈ JB • 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 x j = 0 ou u j j ∈ JB ∑∑ i = 1, m x ji =xbii = b i − a ij x a ij x j i = 1,2,...,2,..., m j − j j∈JB j∈JB Dependent variables r.h.s. We find similar values as in problems where there are no upper bounds, except for non basic variables x j = 0 or u j j ∈ JB
We find similar values as in problems where there are no upper bounds, except for non basic variables x j = 0 or u j j ∈ JB We have to modify the entering criterion and the leaving criterion accordingly to generate a variant of the simplex algorithm for this problem
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 j ∈ JB if x j = 0 and c j < 0 , it is interesting to increase xj if x j = u j and c j > 0 , it is interesting to decrease xj {c = min { c , −c } Determine c s1 = min j∈JB Let c s s1 s2 j } { : x j = 0 and c s2 = max c j : x j = u j (max { c j∈JB s1 , c s2 } }) If c s ≥ 0, then the solution is optimal, and the algoithm stops. If c s < 0 and c s = c s1 , then the non basic variable xs increases, and go to Step 2.1. If c s < 0 et c s < c s1 , then the non basic variable xs decreases, and go to Step 2.2.
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is  : a is < 0   For all i such that ais > 0, then x ji decreases when xs increases of the value θ . It follows x ji =gi − aisθ ≥ 0 ⇔ aisθ ≤ gi g ⇔ θ≤ i. ais g  Then θ ≤ min  i : ais > 0  1≤ i ≤ m a  is 
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is  : a is < 0  For all i such that ais < 0, then x ji increses Si x increases le problème . It follows when θs = ∞, alors of the value θn’est pas borné inférieurement isθ ≤ u ji − gi x ji =gi − aisθ ≤ u ji ⇔ − aet l’algorithme s’arrête. u j − gi ⇔ θ≤ i . −ais  u j − gi  Then θ ≤ min  i : ais < 0  1≤ i ≤ m  −ais  
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is If θ = ∞, then the problem is not bounded from below, and the algorithm stops.  : a is < 0  
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is If x s = u s , then the set of basic variables is not modified, and the same basis is 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 used at the next iteration. 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr the variable x s remains non basic 1 r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s but its value is modified from 0 to u s . Go to step 1.  : a is < 0  
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is g  = min  i : a is > 0  , a rs 1≤i ≤ m  a is  then the value on the entretring variable x s If θ = gr increases to θ . The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to 0 Pivot on a jr s , and go to step 1.  : a is < 0  
Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is uj − gi  = min  i : a is < 0  , 1≤ i ≤ m − a rs  − a is  then the value on the entretring variable x s If θ = u jr − g r increases to θ . The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to u jr Pivot on a jr s , and go to step 1.  : a is < 0  
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 j ∈ JB if x j = 0 and c j < 0 , it is interesting to increase xj if x j = u j and c j > 0 , it is interesting to decrease xj {c = min { c , −c } Determine c s1 = min j∈JB Let c s s1 s2 j } { : x j = 0 and c s2 = max c j : x j = u j (max { c j∈JB s1 , c s2 } }) If c s ≥ 0, then the solution is optimal, and the algoithm stops. If c s < 0 and c s = c s1 , then the non basic variable xs increases, and go to Step 2.1. If c s < 0 et c s < c s1 , then the non basic variable xs decreases, and go to Step 2.2.
Step 2.2: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in this case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB θ   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  : a is > 0   For all i such that ais < 0, then x ji decreases when xs decreases of the value θ . It follows x ji =gi − ais ( −θ ) ≥ 0 ⇔ − aisθ ≤ gi g ⇔ θ≤ i . − ais  g  Then θ ≤ min  i : ais < 0  1≤ i ≤ m − a  is 
Step 2.2: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB θ   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  : a is > 0  For all i such that ais > 0, then x ji increases when xs decreasess of the value θ . It follows x ji =gi − ais ( −θ ) ≤ u ji ⇔ aisθ ≤ u ji − gi u ji − gi ⇔ θ≤ . ais  u ji − gi  Then θ ≤ min  : ais > 0  1≤ i ≤ m  ais  
Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is If θ = u s , then the set of basic variables is not modified, and the same basis is 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 used at the next iteration. 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr the variable x s remains non basic 1 r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 but its value is modified from u s to 0. Go to step 1.  : a is > 0  
Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  g  = min  i : a is < 0  , − a rs 1≤i ≤ m  − a is  then the value on the entretring variable x s If θ = gr is reduced by θ (i.e., x s ← u s − θ ). The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to 0 Pivot on a jr s , and go to step 1.  : a is > 0  
Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is uj − gi  = min  i : a is > 0  , 1≤ i ≤ m a rs  a is  then the value on the entretring variable x s If θ = u jr − g r is reduced by θ (i.e., x s ← u s − θ ). The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to u jr Pivot on a jr s , and go to step 1.  : a is > 0  
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

More Related Content

PPSX
Bounded variables new
Samo Alwatan
 
PPTX
Differential equations of first order
vishalgohel12195
 
PPTX
Second order homogeneous linear differential equations
Viraj Patel
 
PPT
first order ode with its application
Krishna Peshivadiya
 
PPT
Ch03 3
Rendy Robert
 
PPTX
Automobile 3rd sem aem ppt.2016
kalpeshvaghdodiya
 
PPT
Ch03 1
Rendy Robert
 
PDF
Lecture3 ode (1) lamia
malkinh
 
Bounded variables new
Samo Alwatan
 
Differential equations of first order
vishalgohel12195
 
Second order homogeneous linear differential equations
Viraj Patel
 
first order ode with its application
Krishna Peshivadiya
 
Ch03 3
Rendy Robert
 
Automobile 3rd sem aem ppt.2016
kalpeshvaghdodiya
 
Ch03 1
Rendy Robert
 
Lecture3 ode (1) lamia
malkinh
 

What's hot (20)

PPTX
SERIES SOLUTION OF ORDINARY DIFFERENTIALL EQUATION
Kavin Raval
 
PPT
Differential Equations Lecture: Non-Homogeneous Linear Differential Equations
bullardcr
 
PPTX
Linear differential equation
Pratik Sudra
 
PPT
Ch03 5
Rendy Robert
 
PPT
Differential equations
Charan Kumar
 
PPT
Persamaan Differensial Biasa 2014
Rani Sulvianuri
 
PPT
Ch04 2
Rendy Robert
 
PPT
Linear differential equation with constant coefficient
Sanjay Singh
 
PPTX
First order linear differential equation
Nofal Umair
 
PDF
Partial Differential Equation - Notes
Dr. Nirav Vyas
 
PDF
Simultaneous differential equations
Shubhi Jain
 
PPTX
First order ordinary differential equations and applications
Adani Institute of Infrastructure Engineering College
 
PDF
Higher Differential Equation
Abdul Hannan
 
PPSX
Methods of solving ODE
kishor pokar
 
PPTX
Exact & non differential equation
Siddhi Shrivas
 
PPTX
Differential equations of first order
Uzair Saiyed
 
PDF
Unit1 vrs
Akhilesh Deshpande
 
PPTX
Differential equation and Laplace transform
sujathavvv
 
PPTX
Odepowerpointpresentation1
Pokarn Narkhede
 
PDF
First Order Differential Equations
Itishree Dash
 
SERIES SOLUTION OF ORDINARY DIFFERENTIALL EQUATION
Kavin Raval
 
Differential Equations Lecture: Non-Homogeneous Linear Differential Equations
bullardcr
 
Linear differential equation
Pratik Sudra
 
Ch03 5
Rendy Robert
 
Differential equations
Charan Kumar
 
Persamaan Differensial Biasa 2014
Rani Sulvianuri
 
Ch04 2
Rendy Robert
 
Linear differential equation with constant coefficient
Sanjay Singh
 
First order linear differential equation
Nofal Umair
 
Partial Differential Equation - Notes
Dr. Nirav Vyas
 
Simultaneous differential equations
Shubhi Jain
 
First order ordinary differential equations and applications
Adani Institute of Infrastructure Engineering College
 
Higher Differential Equation
Abdul Hannan
 
Methods of solving ODE
kishor pokar
 
Exact & non differential equation
Siddhi Shrivas
 
Differential equations of first order
Uzair Saiyed
 
Unit1 vrs
Akhilesh Deshpande
 
Differential equation and Laplace transform
sujathavvv
 
Odepowerpointpresentation1
Pokarn Narkhede
 
First Order Differential Equations
Itishree Dash
 
Ad

Viewers also liked (19)

PPTX
Transportation+method
videoaakash15
 
PPTX
T P
dude dude
 
PPT
Foreign Exhange Rate
Nikhil Dhiman
 
PPTX
operation research
ujal70
 
PPT
Foreign exchange rate impact (1)
Mohit Bathineni
 
PPT
Simplex Algorithm
Steve Bishop
 
PPTX
Gross national product
c43300
 
PDF
Supply and Demand
Casey Robertson
 
PPTX
Introducing macroeconomics
boxonomics
 
PPT
Chapter 1 - basic concepts about macroeconomics for BBA
ginish9841502661
 
PPTX
Macroeconomic
shivpal
 
PPT
The Foreign Exchange Market
mayank2012
 
PPT
Foreign Exchange Market
Pradeep Awasare
 
PPTX
Simplex Algorithm
Aizaz Ahmad
 
PPT
Introduction To Macro Economics
Saurabh Goel
 
PPT
Macroeconomics slide
Thao Nguyen
 
PPT
Quantitative Techniques
Deepthy Sai Manikandan
 
PPT
Demand and supply
Kinnar Majithia
 
PPT
Introduction to macroeconomics
Shivam Bindra
 
Transportation+method
videoaakash15
 
T P
dude dude
 
Foreign Exhange Rate
Nikhil Dhiman
 
operation research
ujal70
 
Foreign exchange rate impact (1)
Mohit Bathineni
 
Simplex Algorithm
Steve Bishop
 
Gross national product
c43300
 
Supply and Demand
Casey Robertson
 
Introducing macroeconomics
boxonomics
 
Chapter 1 - basic concepts about macroeconomics for BBA
ginish9841502661
 
Macroeconomic
shivpal
 
The Foreign Exchange Market
mayank2012
 
Foreign Exchange Market
Pradeep Awasare
 
Simplex Algorithm
Aizaz Ahmad
 
Introduction To Macro Economics
Saurabh Goel
 
Macroeconomics slide
Thao Nguyen
 
Quantitative Techniques
Deepthy Sai Manikandan
 
Demand and supply
Kinnar Majithia
 
Introduction to macroeconomics
Shivam Bindra
 
Ad

Similar to Bounded var (20)

PDF
Linear Algebra for AI & ML
Sandip Ladi
 
PPTX
Linear Algebra and Matlab tutorial
Jia-Bin Huang
 
DOCX
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
Rwan Kamal
 
DOCX
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
Abdullaا Hajy
 
DOCX
University of duhok
Rwan Kamal
 
PPTX
MODULE_05-Matrix Decomposition.pptx
AlokSingh205089
 
PDF
UNIT I System of Linear Equations.pdfUNIT I System of Linear Equations.pdf
snavyansh719
 
PPT
Matrix
ST ZULAIHA NURHAJARURAHMAH
 
PDF
Matrices & Determinants
Ishant Jain
 
PPTX
Matrices ppt
aakashray33
 
PDF
Foundations of Machine Learning - Module 1 (LINEAR ALGEBRA )
ashmila21m005
 
PPTX
Bba i-bm-u-2- matrix -
Rai University
 
PPTX
Mathematics I - BSCM103 -Module 4_copy.pptx
SoumiChowdhury7
 
PPT
Linear Algebra and Matrix
itutor
 
PPT
Eigen value , eigen vectors, caley hamilton theorem
gidc engineering college
 
PDF
Lecture_note2.pdf
EssaAlMadhagi
 
PDF
Hi please complete the following with detailed working out Find the .pdf
ezhilvizhiyan
 
PPTX
01_PPT_1_Linear Algebra_Mathematics_Output.pptx
wheneveruseme
 
PPTX
Matrices
Елена Доброштан
 
PPTX
MATRICES-1-35.pptx King of business operations
shivkapoor9000
 
Linear Algebra for AI & ML
Sandip Ladi
 
Linear Algebra and Matlab tutorial
Jia-Bin Huang
 
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
Rwan Kamal
 
Matrices and its Applications to Solve Some Methods of Systems of Linear Equa...
Abdullaا Hajy
 
University of duhok
Rwan Kamal
 
MODULE_05-Matrix Decomposition.pptx
AlokSingh205089
 
UNIT I System of Linear Equations.pdfUNIT I System of Linear Equations.pdf
snavyansh719
 
Matrix
ST ZULAIHA NURHAJARURAHMAH
 
Matrices & Determinants
Ishant Jain
 
Matrices ppt
aakashray33
 
Foundations of Machine Learning - Module 1 (LINEAR ALGEBRA )
ashmila21m005
 
Bba i-bm-u-2- matrix -
Rai University
 
Mathematics I - BSCM103 -Module 4_copy.pptx
SoumiChowdhury7
 
Linear Algebra and Matrix
itutor
 
Eigen value , eigen vectors, caley hamilton theorem
gidc engineering college
 
Lecture_note2.pdf
EssaAlMadhagi
 
Hi please complete the following with detailed working out Find the .pdf
ezhilvizhiyan
 
01_PPT_1_Linear Algebra_Mathematics_Output.pptx
wheneveruseme
 
Matrices
Елена Доброштан
 
MATRICES-1-35.pptx King of business operations
shivkapoor9000
 

Recently uploaded (20)

PPTX
GROUP4NURSINGINFORMATICSREPORT-2 PRESENTATION
ClarisejaneSartillo1
 
PPTX
Benefits of Physical activity for teenagers.pptx
Nokwanda Thabethe
 
PDF
How ambidextrous entrepreneurial leaders react to the artificial intelligence...
IAESIJAI
 
PPTX
Final SEM Unit 1 for mit wpu at pune .pptx
anishtilekar08
 
PPTX
TEXTILE technology diploma scope and career opportunities
kriti38
 
PDF
sbt 2.0: go big (Scala Days 2025 edition)
Eugene Yokota
 
PDF
Convolutional neural network based encoder-decoder for efficient real-time ob...
IAESIJAI
 
DOCX
search engine optimization ppt fir known well about this
StandardCodeLearning
 
PPTX
Custom Battery Pack Design Considerations for Performance and Safety
Epec Engineered Technologies
 
PPTX
MicrosoftCybserSecurityReferenceArchitecture-April-2025.pptx
SilvioHayashi
 
PPTX
AI IN MARKETING- PRESENTED BY ANWAR KABIR 1st June 2025.pptx
HiraJay1
 
PDF
Produktkatalog für HOBO Datenlogger, Wetterstationen, Sensoren, Software und ...
Metrics GmbH
 
PDF
sustainability-14-14877-v2.pddhzftheheeeee
VenkateshGovindaraja9
 
PDF
Hybrid horned lizard optimization algorithm-aquila optimizer for DC motor
IAESIJAI
 
PPTX
Chapter 5: Probability Theory and Statistics
RandyFin
 
PDF
ENT215_Completing-a-large-scale-migration-and-modernization-with-AWS.pdf
ssuser28425f
 
PDF
Zenith AI: Advanced Artificial Intelligence
Biz Preneurs
 
PDF
Improvisation in detection of pomegranate leaf disease using transfer learni...
IAESIJAI
 
PPTX
Build Your First AI Agent with UiPath.pptx
suhanisingh58689
 
PPTX
Configure Apache Mutual Authentication
Antonino Piraino
 
GROUP4NURSINGINFORMATICSREPORT-2 PRESENTATION
ClarisejaneSartillo1
 
Benefits of Physical activity for teenagers.pptx
Nokwanda Thabethe
 
How ambidextrous entrepreneurial leaders react to the artificial intelligence...
IAESIJAI
 
Final SEM Unit 1 for mit wpu at pune .pptx
anishtilekar08
 
TEXTILE technology diploma scope and career opportunities
kriti38
 
sbt 2.0: go big (Scala Days 2025 edition)
Eugene Yokota
 
Convolutional neural network based encoder-decoder for efficient real-time ob...
IAESIJAI
 
search engine optimization ppt fir known well about this
StandardCodeLearning
 
Custom Battery Pack Design Considerations for Performance and Safety
Epec Engineered Technologies
 
MicrosoftCybserSecurityReferenceArchitecture-April-2025.pptx
SilvioHayashi
 
AI IN MARKETING- PRESENTED BY ANWAR KABIR 1st June 2025.pptx
HiraJay1
 
Produktkatalog für HOBO Datenlogger, Wetterstationen, Sensoren, Software und ...
Metrics GmbH
 
sustainability-14-14877-v2.pddhzftheheeeee
VenkateshGovindaraja9
 
Hybrid horned lizard optimization algorithm-aquila optimizer for DC motor
IAESIJAI
 
Chapter 5: Probability Theory and Statistics
RandyFin
 
ENT215_Completing-a-large-scale-migration-and-modernization-with-AWS.pdf
ssuser28425f
 
Zenith AI: Advanced Artificial Intelligence
Biz Preneurs
 
Improvisation in detection of pomegranate leaf disease using transfer learni...
IAESIJAI
 
Build Your First AI Agent with UiPath.pptx
suhanisingh58689
 
Configure Apache Mutual Authentication
Antonino Piraino
 

Bounded var

  • 2. Simplex method for problems with bounded variables • Consider the linear programming problem with bounded variables min c T g s.t. Ag = h lj ≤ g j ≤ qj j = 1, 2,..., n where g , c, l , q ∈ R n , h ∈ R m , and A is a m × n matrix • 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 min c T g s.t. Ag = h lj ≤ g j ≤ qj the problem becomes min c T ( x + l ) s.t. A( x + l ) = h lj ≤ xj + lj ≤ qj j = 1, 2,..., n j = 1, 2,..., n where c, x, l , q ∈ R n , h ∈ R m , and A is a m × n matrix • 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 min c T ( x + l ) s.t. A( x + l ) = h lj ≤ xj + lj ≤ qj j = 1, 2,..., n where c, x, l , q ∈ R n , h ∈ R m , and A is a m × n matrix min c T x + c T l s.t. Ax = h − Al lj − lj ≤ xj + lj − lj ≤ qj − lj replacing : uj = qj – lj and j = 1, 2,..., n b = h – Al min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj j = 1, 2,..., n
  • 5. Simplex method for problems with bounded variables • In this problem c T l + min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj j = 1, 2,..., n 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. min c T x + c T l s.t. Ax = b 0 ≤ xj ≤ uj • Consider the explicit formulation of the problem n n min = ∑ c x min z z = ∑jc jjx j j j =1 =1 n s.t. s.t. n x a ∑ ∑xa = b ij j ij j =1 j =1 j i im == i1, 2,...,= 1, 2,..., m i b 0 ≤ x x ≤+ y j = ujj = 1, 2,..., 1, 2,..., n j =n j j uj xj , yj ≥ 0 j = 1, 2,..., n • One way of solving the problem is to introduce slack variables yj, and then use the simplex algorithm. j = 1, 2,..., n
  • 7. n min z = ∑ c j x j Tableau with m + n rows j =1 n s.t. ∑a x = bi i = 1, 2,..., m xj + yj = uj j = 1, 2,..., n xj , yj ≥ 0 ij j = 1, 2,..., n j j =1 n min z = ∑ c j x j Tableau with m rows j =1 n s.t. ∑a x = bi i = 1, 2,..., m 0 ≤ xj ≤ uj j = 1, 2,..., n ij j j =1 account implicitly
  • 8. n min z = ∑ c j x j j =1 n s.t. ∑a x ij j = bi j =1 xj + yj = uj xj , yj ≥ 0 i = 1, 2,..., m Non degeneracy: j = 1, 2,..., n all the basic variables are positive at j = 1, 2,..., n each iteration • 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
  • 9. a) x j basic; y j non basic b) x j non basic; y j basic c) x j basic; y j basic n min z = ∑c x j j j =1 n Sujet à ∑a = bi i = 1,2,..., m xj + yj =uj j = 1,2,..., n xj , yj ≥0 j = 1,2,..., n ij x j j =1 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. a) x j basic; y j non basic b) x j non basic; y j basic c) x j basic; y j basic n min z = ∑c x j j j =1 n Sujet à ∑a = bi i = 1,2,..., m xj + yj =uj j = 1,2,..., n xj , yj ≥0 j = 1,2,..., n ij x j j =1 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. n min z = La base a donc la forme suivante ∑c x j j j =1 n ∑a x s.t. ij = bi j i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n m min z = c T x s.t. Ax + 0 y = b Ix + Iy = u x, y ≥ 0 m  A 0 I I    n To simplify notation, assume the following basic variables: 0 < xi < ui 0 < yi < ui 0 < xj < uj 0 < yj < uj xj=uj yj=uj i = 1,… , m i = 1,…, m xi = ui i = m + 1,… , m + l yi = ui i = m + l + 1,…, m + n
  • 12. n min z = The basis has the following form ∑c x j j j =1 n ∑a x s.t. ij j = bi i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n m min z = c T x Ax + 0 y = b s.t. Ix + Iy = u min z = c x Sujet à Ax + 0 y = b x, y ≥ 0 T Ix + Iy = u x, y ≥ 0 m 1  A 0 I I    1 n 2 To simplify notation, assume the following basic variables: 3 m 0 < yj < uj m xj=uj yj=uj n−m i = 1,…, m 0 < yi < ui 0 < xj < uj 0 < xi < ui i = 1,…, m xi = ui i = m + 1,…, m + l yi = ui i = m + l + 1,… , m + n
  • 13. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 14. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 15. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 16. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 17. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 18. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 19. B 0 D 0   I1 I 1 0 0  Ξ=  0 0 I2 0    0 0 0 I3    B 0 D 0 I  = 1 0 I    0   a b  = ad − bc det  c d   = d ( a − bd −1c ) where the matrix I is n × n    I1      det ( Ξ ) = det ( I ) det ( B ) − det  [0 D 0] I −1  0     0            = det ( I ) det ( B ) − det 0 { ( )} where the matrix 0 is m × m since [0 D 0] is a m × n matrix and  I1  0  is a n × m matrix 0    Then det ( Ξ ) = det ( I ) det ( B ) . Since Ξ is a basis, then det ( Ξ ) ≠ 0. Consequently det ( B ) = det ( Ξ ) ≠ 0 and then B is non singular. Then B is a basis of A.
  • 20. The basis has the following form m Basis of A 1 The columns of the basis B of A are those of the variables 0<xj<uj 1 n 2 3 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: n min z = ∑ c j x j j =1 n s.t. ∑a x = bi i = 1, 2,..., m 0 ≤ xj ≤ uj j = 1, 2,..., n ij j j =1 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 0 < xj < uj j ∈ IB n – m non basic variables x j = 0 ou u j j ∈ JB
  • 22. n min z = ∑c x j j j =1 n s.t. ∑a x ij j = bi i = 1, 2,..., m j =1 xj + yj = uj xj , yj ≥ 0 j = 1, 2,..., n j = 1, 2,..., n • At each iteration, we consider a solution (basic) associated with a basis B de A having 0 < xj < uj j ∈ IB m basic variables n – m non basic variables x j = 0 or u j j ∈ JB • 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 x j = 0 ou u j j ∈ JB ∑∑ i = 1, m x ji =xbii = b i − a ij x a ij x j i = 1,2,...,2,..., m j − j j∈JB j∈JB Dependent variables r.h.s. We find similar values as in problems where there are no upper bounds, except for non basic variables x j = 0 or u j j ∈ JB
  • 23. We find similar values as in problems where there are no upper bounds, except for non basic variables x j = 0 or u j j ∈ JB We have to modify the entering criterion and the leaving criterion accordingly to generate a variant of the simplex algorithm for this problem
  • 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 j ∈ JB if x j = 0 and c j < 0 , it is interesting to increase xj if x j = u j and c j > 0 , it is interesting to decrease xj {c = min { c , −c } Determine c s1 = min j∈JB Let c s s1 s2 j } { : x j = 0 and c s2 = max c j : x j = u j (max { c j∈JB s1 , c s2 } }) If c s ≥ 0, then the solution is optimal, and the algoithm stops. If c s < 0 and c s = c s1 , then the non basic variable xs increases, and go to Step 2.1. If c s < 0 et c s < c s1 , then the non basic variable xs decreases, and go to Step 2.2.
  • 25. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is  : a is < 0   For all i such that ais > 0, then x ji decreases when xs increases of the value θ . It follows x ji =gi − aisθ ≥ 0 ⇔ aisθ ≤ gi g ⇔ θ≤ i. ais g  Then θ ≤ min  i : ais > 0  1≤ i ≤ m a  is 
  • 26. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is  : a is < 0  For all i such that ais < 0, then x ji increses Si x increases le problème . It follows when θs = ∞, alors of the value θn’est pas borné inférieurement isθ ≤ u ji − gi x ji =gi − aisθ ≤ u ji ⇔ − aet l’algorithme s’arrête. u j − gi ⇔ θ≤ i . −ais  u j − gi  Then θ ≤ min  i : ais < 0  1≤ i ≤ m  −ais  
  • 27. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is If θ = ∞, then the problem is not bounded from below, and the algorithm stops.  : a is < 0  
  • 28. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is If x s = u s , then the set of basic variables is not modified, and the same basis is 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 used at the next iteration. 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr the variable x s remains non basic 1 r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s but its value is modified from 0 to u s . Go to step 1.  : a is < 0  
  • 29. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is g  = min  i : a is > 0  , a rs 1≤i ≤ m  a is  then the value on the entretring variable x s If θ = gr increases to θ . The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to 0 Pivot on a jr s , and go to step 1.  : a is < 0  
  • 30. Step 2.1: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs > 0) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs < 0 ) 0 ≤ x j1 = g − a1s ( 0 + θ ) ≤ u j1 1 0 ≤ x jr = g − ars ( 0 + θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( 0 + θ ) ≤ u jm xs = 0 + θ ≤ u s Let gi = b i − ∑ a ij x j j∈JB θ   gi = min u s , min   1≤ i ≤ m  a is : a is Value of the basic variables x ji   u ji − g i > 0  , min   1≤ i ≤ m  − a is uj − gi  = min  i : a is < 0  , 1≤ i ≤ m − a rs  − a is  then the value on the entretring variable x s If θ = u jr − g r increases to θ . The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to u jr Pivot on a jr s , and go to step 1.  : a is < 0  
  • 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 j ∈ JB if x j = 0 and c j < 0 , it is interesting to increase xj if x j = u j and c j > 0 , it is interesting to decrease xj {c = min { c , −c } Determine c s1 = min j∈JB Let c s s1 s2 j } { : x j = 0 and c s2 = max c j : x j = u j (max { c j∈JB s1 , c s2 } }) If c s ≥ 0, then the solution is optimal, and the algoithm stops. If c s < 0 and c s = c s1 , then the non basic variable xs increases, and go to Step 2.1. If c s < 0 et c s < c s1 , then the non basic variable xs decreases, and go to Step 2.2.
  • 32. Step 2.2: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in this case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB θ   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  : a is > 0   For all i such that ais < 0, then x ji decreases when xs decreases of the value θ . It follows x ji =gi − ais ( −θ ) ≥ 0 ⇔ − aisθ ≤ gi g ⇔ θ≤ i . − ais  g  Then θ ≤ min  i : ais < 0  1≤ i ≤ m − a  is 
  • 33. Step 2.2: Selecting the leaving variable • 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 x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB θ   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  : a is > 0  For all i such that ais > 0, then x ji increases when xs decreasess of the value θ . It follows x ji =gi − ais ( −θ ) ≤ u ji ⇔ aisθ ≤ u ji − gi u ji − gi ⇔ θ≤ . ais  u ji − gi  Then θ ≤ min  : ais > 0  1≤ i ≤ m  ais  
  • 34. Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is If θ = u s , then the set of basic variables is not modified, and the same basis is 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 used at the next iteration. 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr the variable x s remains non basic 1 r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 but its value is modified from u s to 0. Go to step 1.  : a is > 0  
  • 35. Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is  g  = min  i : a is < 0  , − a rs 1≤i ≤ m  − a is  then the value on the entretring variable x s If θ = gr is reduced by θ (i.e., x s ← u s − θ ). The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to 0 Pivot on a jr s , and go to step 1.  : a is > 0  
  • 36. Step 2.2: Selecting the leaving variable • The decrease θ of the entering variable xs is stop by the first of the following three situations happening: θ i) xs reduces to 0 ii) a basic variable x jr decreases to 0 (in this case a rs < 0 ) iii) a basic variable x jr increases to reach its upper bound u jr (in ths case a rs > 0 ) 0 ≤ x j1 = g − a1s ( −θ ) ≤ u j1 1 0 ≤ x jr = g − a rs ( −θ ) ≤ u jr r 0 ≤ x jm = g m − ams ( −θ ) ≤ u jm xs = us − θ ≥ 0 Let gi = b i − ∑ a ij x j j∈JB   g = min  u s , min  i  1≤i ≤ m  − a is : a is Value of the basic variables x ji  u ji − g i  < 0  , min   1≤i ≤ m  a is uj − gi  = min  i : a is > 0  , 1≤ i ≤ m a rs  a is  then the value on the entretring variable x s If θ = u jr − g r is reduced by θ (i.e., x s ← u s − θ ). The entering variable x s becomes a basic variable replacing the leaving variable x jr becoming a non basic variable equal to u jr Pivot on a jr s , and go to step 1.  : a is > 0  
  • 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