大学最优化课件

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
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 ⇒ x1 + x3 ≤ 15 x1 + x3 ≤ 15 3x2 + 2 x3 ≤ 10 x≥0 x≥0
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 x1 + x3 + x6 = 15 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
How the change in the objective function Coefficients affect the optimal solution
Suppose we have solved an LP and have found the optimal simplex tableau, according to the following table, analyze how changes the optimal simplex tableau as some coefficient in objective function is altered
T T cB + ∆cB ) B −1 N − cT = cB B −1 N + ∆cB B −1 N − cT ( N N T T T = cB B −1 N − cT + ∆cB B −1 N N
T ∆cB B −1 N = ( ∆c1
1 1 −1 0 0 ) 1 0 1 = ( ∆c1 −2 −1 −1

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 x1 + x3 + x6 = 15 x1 + x3 + x6 + x7 = 15 x≥0 x≥0
∆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
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 2 3 1 x1 + 2 x2 + x3 ≤ b2 x1 + x3 ≤ b3 x≥0
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
⇒ 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
It is known that x2 and x3 are basic varibles in opitmal tableau, then find the optimal tableau