chapter 6 Sensitivity Analysis

T T c B N − ( cN + ∆cN ) = cB B −1 N − cN − ∆cT N T B −1 T
= ( 3 1 2 ) − ( ∆c2 = ( 3 − ∆c2 1 2 )
0 0)
If the optimal basis remain unchanged, we have 3 − ∆c2 ≥ 0 So if the coefficient of x2 changed in ( −∞，]，the original 4 optimal basis will remain optimal。 that is：∆c2 ≤ 3，c2 + ∆c2 ≤ 4
It is known that x2 and x3 are basic varibles in opitmal tableau, then find the optimal tableau
Consider the following LP and its optimal simplex tableau has been found as follows, then find the values of all unknown coefficients and their changing ranges in which the optimal basis can remain unchanged
In the same way to analyze the coefficient of x2 c2 ∆c2 if c2 → c2 + ∆c2 , then cN → c5 + 0 = cN + ∆cN c 0 6 and the test indicators are changed to be：
How the change in the right-hand sides affect the optimal solution
T b + ∆ → B −1b, cB B −1b
Example: Find out the changing range of b1 in which the optimal basis remain unchanged max s.t. S = x1 + x2 + 3 x3 x1 + x2 + 2 x3 + x4 =40 x1 + 2 x2 + x3 + x5 = 20 x2 + x3 + x6 = 15 x≥0
Sensitivity Analysis
1.What is the main content of Sensitivity Analysis 2.Changes in the objective function coefficients 3.Changes in the right-hand sides 4.Adding a new variable or a new constraint 5.Exercises
T T T cB + ∆ → cB B −1 N − cN , cB B −1b T T cN + ∆ → cB B −1 N − cN
Example: Determine the changing ranges of the objective coefficients of x1 and x2 , in which the optimal basis remain unchanged max s.t. S = x1 + x2 + 3 x3 x1 + x2 + 2 x3 + x4 =40 x1 + 2 x2 + x3 + x5 = 20 x1 + x3 + x6 = 15 x≥0
What is the main content of Sensitivity Analysis
Explore how changes in an LP’s objective function coefficients change the optimal solution Explore how changes in an LP’s right hand sides change the optimal solution Explore how changes the optimal solution when adding a new variable or a new constraint
−∆c1 )
−∆c1 )
To make the optimal basis remain unchanged, we have 3 + ∆c1 ≥ 0 1 + ∆c1 ≥ 0 i.e. − 1 ≤ ∆c1 ≤ 2, 0 ≤ c1 + ∆c1 ≤ 3 2 − ∆c ≥ 0 1 In other words, if the coefficient of x1 change in [ 0,3] , then the optimal basis will remain optimal
T cB B −1 N − cT = ( 3 1 2 ) N
c1 ∆c1 if c1 → c1 + ∆c1 , then cB → c3 + 0 = cB + ∆cB c 0 4 and the test indicators are altered to be：
max S = k1 x1 + k2 x2 + k3 x3 s.t. x + x + 2 x ≤ b 1 1 2 3 x1 + 2 x2 + x3 ≤ b2 x2 + x3 ≤ b3 x≥0
1 2 1 0 1 −1 5 −1 −1 B = 1 1 0 , B = 0 0 1 , B b = 15 0 1 0 1 −1 −1 5
b1 b1 ∆b1 Assume b1 → b1 + ∆b1 , then b = b2 → b2 + 0 = b + ∆b b b 0 3 3 and the basic solution is changed to be B −1 ( b + ∆b ) = B −1b + B −1∆b 5 0 1 −1 ∆b1 5 + ∆b1 = 15 + 0 0 1 0 = 15 5 1 −1 −1 0 5 To make the optimal basis unchanged,we have 5 + ∆b1 ≥ 0 i.e. ∆b1 ≥ −5，b1 + ∆b1 ≥ 35 That is,if b1 changes in will remain optimal。 then [35, + ∞ )， the optimal basis
∆c1
−∆c1 )
T T cB + ∆cB ) B −1 N − cT = cB B −1 N − cT + ∆cB B −1 N ( N N T
= c B N − c + ( ∆c1
T B −1 T N
∆c1 2 − ∆c1 )
= ( 3 1 2 ) + ( ∆c1
∆c1
= ( 3 + ∆c1 1 + ∆c1
⇒ x* = (15 0 5 15 0 10 0 ) , S * = 30
T
or x* = (15 0 5 ) , S * = 30
T
Exercises
Consider the following LP
max S = − x1 + x2 s.t. 2x + x + x = 4 1 2 3 x1 + x2 ≤ 2 x≥0
0 1 −1 1 c B ∆N − c7 = (1 3 0 ) 0 0 1 1 − c7 = 3 − c7 ≥ 0 1 −1 −1 1
T B −1
Add a constraint to LP
max S = x1 + x2 + 3x3 max S = x1 + x2 + 3 x3 s.t. x + x + 2 x ≤ 40 1 2 3 s.t. x + x + 2 x ≤ 40 1 2 3 x1 + 2 x2 + x3 ≤ 20 x1 + 2 x2 + x3 ≤ 20 ⇒ x2 + x3 ≤ 15 x2 + x3 ≤ 15 3x2 + 2 x3 ≤ 10 x≥0 x≥0
Use the lingo system to verify the results
Adding a New Variable
Example： Add a variable x7 to LP and 1.Find out the allowable changing range of c7 , in which the optimal basis remain unchanged？ 2. How to increase the opimal objective value by alter the value of c7 max S = x1 + x2 + 3x3 max S = x1 + x2 + 3x3 + c7 x7 s.t. x + x + 2 x + x =40 s.t. x + x + 2 x + x +x =40 1 2 3 4 1 2 3 4 7 x1 + 2 x2 + x3 + x5 = 20 ⇒ x1 + 2 x2 + x3 + x5 + x7 = 20 x2 + x3 + x6 = 15 x2 + x3 + x6 + x7 = 15 x≥0 x≥0