1. Lecture 5. Basic Feasible Solution
Feng Chen
Department of Industrial Engineering and Logistics
Management
Shanghai Jiao Tong University
Mar 18, 2010
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2. Enumerate Algorithm
min z  cT x
s.t. Ax  b
x0
How to solve LP as above?
If we can find all feasible solutions and the
number of these feasible solutions is finite,
then we can compare these solutions and
select the optimal solution.
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3. Standard LP
•Convert methods
•“<=“ , Slack variables
p •“>=“, excess variables
p’ •Free variable, x=x1-x2
Standard LP with x1 and x2 to be
nonnegative.
Linear
Programming ? For standard LP, can we
get the extreme point by
algebraic calculations.
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4. Standard LP
Any Linear Programming
Converted
Standard Linear Programming
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5. Basic Feasible Solution
There must exists at least one optimal solution at
extreme point.
We need not to search all
points, maybe good to consider
these solutions only related to
extreme points. For this goal,
we introduce another concept
of basic feasible solution.
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6. Constraints Cases
Consider a system, Ax  b A  R m,n
• case 1: m=n • Case 3: m<n
x1  x2  3 x1  x2  3
 x1  2 x2  6  x2  x3  1
• Case 2: m>n We like case 1 & 2, but
no research. If research,
x1  x2  3 we must consider case 3.
 x1  2 x2  6 This case will lead to
x1  4 x2  5 many feasible solutions.
So, we will address this
case.
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7. Which case is hard?
Suppose we have standard constraints: Ax  b A  R m,n
Here m indicates the number of constraints, and n is the number of variables.
• Case 1: m=n Such case is easy to get optimal solution, because there is
x1  x2  3 always unique solution, we need not to do hard job to find the
 x1  2 x2  6 optimal one from many candidates (determined).
• Case 2: m>n Such case seems easy, if we can have a method to determine its
infeasibility. (overdetermined)
x1  x2  3
 x1  2 x2  6 We like case 1 & 2, it seems easy to deal with.
x1  4 x2  5
• Case 3: m<n The case seems hard to get optimal, we do not like, because
x1  x2  3 they give me many solutions, and we need to find the best
 x 2  x3   1 one(undetermined).
This case will lead to many feasible solutions. So, we will address
such difficulty case.
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8. Basic solution
•In case m<n
x1  x2  3
 x2  x3  1
How to find all feasible solutions? Because the number of variable
is larger than the number of equation, we would like to fix some
variables, and solve a normal equation system with left variables.
If we restrict x3 to a value, e.g., x3=0, we can obtain a
solution to the above system as (2,1,0);
 If x3=1, then the corresponding solution is (1,2,1);
 If x3=2, which leads to a solution (0,3,2).
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9. Nonbasic variables
Def. Given Ax=b with n variables and m constraints (n>m), a
solution called basic solution of the system is obtained by
•setting n-m variables equal to 0
•solving for the values of the remaining m variables
:nonbasic varibale :basic variable
Def. If a basic solution is called basic feasible solution(bfs), if
it is also feasible to the prime problem, generally to satisfy the
sign restrictions.
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10. Basic Feasible Solution
(Gia) max z  3x1  2 x2
s.t. 2 x1  x2  x3  100
x1  x2  x4  80
x1  x 5  40
x1 , x2 , x3 , x4 , x5  0
Step 1. Number of Variables =5; number of constraints=3;
Step 1.1 How many of nonbasic variables?
5-3=2
Step 1.2 How many of basic variables?
3
Step 2. Select x3, x4 as the nonbasic variables, and let x3= x4=0, we
get a basic solution of 2 x1  x2  0  100
(20,60,0,0,20) x1  x2  0  80
x1  x 5  40
Mar 18 2010 x1 , x2 ,0,0, x5  0 Page 10
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11. Example
min z  4 x1  3 x2
Basic Nonbasic Basic
s.t. x1  x2  y3  40 variables Variables Feasible
2 x1  x2  y 4  60 Solution
x1 , x2 , y3 , y 4  0 x1, x2 y3,y4 (20,20,0,0)
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12. min z  4 x1  3 x2
Basic Nonbasic Basic
s.t. x1  x2  y3  40 variables Variables Feasible
2 x1  x2  y 4  60 Solution
x1 , x2 , y3 , y 4  0 x1, x2 y3,y4 (20,20,0,0)
x 1 , y3 x2 ,y4 (30,0,10,0)
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13. Basic Nonbasic Basic
min z  4 x1  3 x2 variables Variables Feasible
s.t. x1  x2  y3  40 Solution
2 x1  x2  y 4  60 x1, x2 y3, y4 (20,20,0,0)
x1 , x2 , y3 , y 4  0 x1 , y3 x2 ,y4 (30,0,10,0)
x1 ,y4 x2, y3 (40,0,0,-20)
x2 ,y3 x1, y4 (0,60,-20,0)
x2 ,y4 x1 , y3 (0,40,0,20)
y3 ,y4 x1 , x2 (0,0,40,60)
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14. BFS & Extreme Point
BFS Relationship Extreme Point
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15. BFS & Extreme Point
Theorem For any LP, there is a unique extreme point of the LP’s feasible
region corresponding to each basic feasible solution. Also, there is at least one
bfs corresponding to each extreme point of the feasible region.
Proof.
Given a basic feasible solution
x=(x1,x2,…,xn-m, 0,0,…,0)
The point related to the bfs is an extreme point.
Assume x is not an extreme point, which means the
there exist two different points , y,z
y x
z Contradiction
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16. Homework
pp92. 1,2
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17. The End
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