Module 2
1Topic 2:
Relationship between extreme points and corresponding BFS:
We now relate the extreme point of the feasible set with BFS of the system of constraint describing the feasible set.
Theorem 2.1: Every extreme point of a set of feasible solution of the LPP (P) max z = cTx
subject to Ax = b, x≥0 is a BFS and vice-versa.
Proof: Let x be a BFS of LPP, i.e, x = [xB,0] ,where 0 is a null vector of n−m component and xB is the vector of m basic variable in BFS then xB = B−1b. Now to the contrary let x be not an extreme point of the feasible set S of (P). Then there exist x1, x2 ∈S, x1 6=x2 and λ∈(0,1) such that
x = (1−λ)x1+λx2
"
xB 0
#
= (1−λ)
"
u1 v1
# + λ
"
u2 v2
# ,
where u1 and u2 respectively are the vector values in x1 and x2 corresponding to the basic variablex and v1, v2 are vectors of those values variables which are zero in x. We have,
λB = (1−λ)u1+λu2 0 = (1−λ)v1+λv2
The second equality yieldv1 = 0, v2= 0 , sincev1, v2 ≥0 and λ∈(0,1). Thus x1 =
"
u1
0
#
, x2 =
"
u2
0
#
Nowx1, x2 ∈S, so,
Bu1=b=Bu2
implying
B−1(Bu1) = B−1(Bu2) i.e u1 = u2.
But this contradicts the fact that x1 6=x2. Sox is an extreme point of S. We next prove that ifx∈S is an extreme point of the feasible set S of (P) thenx is a BFS.
Suppose firstkvariable in x are non-zero,k≤m. That is, x = (x1, x2, . . . xk,0, . . . ,0)T ∈S , and
k
X
i=1
aixi= b.
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We will prove that the set {a1, . . . , ak} in <m is a linearly independent set. Let us suppose that this set is linearly dependent. Then there exist scalars λ1, . . . , λk such that
k
X
i=1
λiai= 0 (1)
with at least one λi 6= 0.
Define µ = min n
xi
|λi|, λi6= 0o
.Then µ >0 . Next chooseε, 0< ε < µ. Then xi+ελi>0 ∀i= 1,2,3, . . . , k
xi−ελi>0 ∀i= 1,2,3, . . . , k
To explain it , if λi > 0 then obviously xi+ελi > 0 and as ε < µ ≤ xλi
i. So, xi −ελi > 0 . Similarly if λi <0, thenxi−ελi >0 and as ε < µ≤ −λxi
i , soxi+ελi >0.
Define a vector λ whose first k-component are λ1, . . . , λk , and rest n−m components are zero.Take
ˆ
x = x+ελ∈ <n
˜
x = x−ελ∈ <n.
Then ˆx ≥ 0. Also as Aλ = 0 in view of (1), so, Aˆx = A˜x = Ax = b, giving ˆx,x˜ ∈ S.
Moreover,ε >0 and at least oneλi6= 0 so ˆx6= ˜x , withx= 12(ˆx+ ˜x). But this contradicts that x is an extreme point of S. Hence the set {a1, a2, ....ak} is linearly independent set. Ifk =m, then by taking a bases matrix as
B = [a1, a2, . . . ak]
We have x is a BFS of LPP. If k < m, then since rank(A) = m so the set {a1, a2, . . . ak} can be extended by adding k−m columns from A to form basis of<m. Hence x is a BFS of LPP.
Only thing to realize is that in this case xis a degenerate BFS.
Example 2.5: Consider the following LPP.
max x1+x2
subject to x1−x2 ≥0 x1+x2 ≤4 x1, x2≥0
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C( 22, )
B(4,0) >
<
A(0,0)
x1
x2
Fig 1.1
For A (0,0) the BFS is (0,0,0,4) while for B (4,0) and C (2,2), the BFS are (4,0,4,0) and (2,2,0,0) respectively. Note that while talking of BFS, the constraints of LPP are converted to equations by introducing additional variables as
x1−x2−x3 = 0
x1+x2+x4 = 4 (2)
x1, x2, x3, x4 ≥ 0
This example also shows that for an extreme point A(0,0) there are three bases namely B1 = {x3, x4}, B2 = {x1, x4} and B3 = {x2, x4}. This is because the BFS (0,0,0,4) is a degenerate BFS. Also, if we take variablesx1= 0, x4 = 0 then the system becomes
−x2−x3 = 0 x2 = 4
which does not yield a BFS asx3 =−4<0.So, among the five possible BFS of system (2), the other two yield extreme points B and C.
Remark 2.1: If the LPP has a degenerate BFS then more than one basis correspond to the same extreme point of the feasible set LPP. This fact is important and should be kept in mind as later on we would be using it.
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Example 2.6: Consider the following LPP
min z=x1+x2 subject tox1+ 2x2 ≤0 x1+x2 ≥0 x2 ≤0 x1, x2 ≥0 The feasible set is shown in the figure.
D: 2, )( 4
B:(1,0)
x
1x
2E:(0,4)
0 C:(10,0)
A:(0,1)
Fig 1.2
It has five extreme points A, B, C, D, E. Now to see their correspondence with BFS, we convert the system of inequalities representing the constraints to equation by introducing additional variables in them. Consider
x1+ 2x2+x3 = 10 x1+x2−x4 = 1
x2+x5 = 4 x1, x2, x3, x4, x5 ≥ 0 For A(0,1), the BFS is (0,1,8,0,3)
For B(1,0), the BFS is (1,0,9,0,4) For C(10,0), the BFS is (10,0,0,9,4) For D(2,4), the BFS is (2,4,0,5,0) For E(0,4), the BFS is (0,4,2,3,0).
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