麻省理工 网络优化 课程课件20lagrangian_relaxation_1

A
0 or 1 for all ( i , j )
The Lagrangian relaxation was obtained by penalizing the constraint and then eliminating (relaxing) the constraint. Theorem. L( ) z( ) z*
j
( i , j ) A ij
j
xij
x ji
1 if i = s 1 if i = t 0 otherwise A
xij
0 or 1 for all ( i , j )
7
What is the shortest path from node 1 to node 6?
2
(1,10)
(1,1) (2,3)
c xij
j
j
xij
t xij
x ji
T
1 if i = s 1 if i = t 0 otherwise
Complicating constraint
( i , j ) A ij
xij
0 or 1 for all (i, j)
A
4
Example
Find the shortest path from node 1 to node 6 with a transit time at most 10 2
1
(10,3)
(10,1)
6
To reduce the transit time of the shortest path, we put a penalty proportional to transit time. Suppose that we charge a toll of $1 per unit transit time.
(1,10) (1,1) (2,3)
i (cij,tij) j
4
(1,7)
1
(1,2) (5,7)
(10,1) (2,2)
6
(10,3)
3
(12,3)
5
5
Shortest Paths with Transit Time Restrictions
Shortest path problems are easy. Shortest path problems with transit time restrictions are NP-hard. We say that constrained optimization problem Y is a relaxation of problem X if Y is obtained from X by eliminating one or more constraints. We will “relax” the complicating constraint, and then use a “heuristic” of penalizing too much transit time. We will then connect it to the theory of Lagrangian relaxations.
4
(1,7)
1
(10,3)
(1,2) (5,7)
(10,1) (2,2)
6
3
(12,3)
5
8
The shortest path is 1-2-4-6
The transit time of 1-2-4-6 is 18. 2
(1,10) (1,1) (2,3) (1,2) (5,7) (2,2)
4
(1,7)
( Ax b )
P( )
X
Lemma 16.1.
For all vectors , L( )
z*.
20
The Lagrangian Multiplier Problem (with equality constraints)
L( ) min cx
s.t. x Lemma 16.1.
( Ax b )
9
3
(12,3)
5
What is the new shortest path from 1 to 6?
2
11
2 5
4
11
8
The modified cost of 1-2-5-6 is 20. The transit time is 15. The cost is 5. 6 1-2-5-6 is still not feasible. We increase the toll to $2
13
Weighted Sum 120 100 80 60 40 20 0 0 2 4 Toll 6 8 10
Parametric analysis of Tolls 30 25 20 15 10 5 0 0 2 4 Toll 6 8 10 Transit Time Weighted Sum 10 * Toll Cost
P( )
X
z*.
For all vectors , L( )
A bound for a minimization problem is better if it is higher. The problem of finding the best bound is called the Lagrangian multiplier problem.
6
Shortest Paths with Transit Time Restrictions
Step 1. (A relaxation approach). Solve the problem without the complicating constraint. If the solution satisfies the complicating constraint, then it is optimal for the original problem. c xij
In solving network flow problems, we not only solve the problem, but we provide a guarantee that we solved the problem. Guarantees are one of the major contributions of an optimization approach. But what can we do if a minimization problem is too hard to solve to optimality? Sometimes, the best we can do is to offer a lower bound on the best objective value. If the bound is close to the best solution found, it is almost as good as optimizing.
3
An Example: Constrained Shortest Paths
Given: a network G = (N,A) cij cost for arc (i,j) tij traversal time for arc (i,j)
Z* = Min s. t.
( i , j ) A ij
15.082 and 6.855J
Lagrangian Relaxation I never missed the opportunity to remove obstacles in the way of unity.
—Mohandas Gandhi
On bounding in optimization
2
Overview
Decomposition based approach. Start with Easy constraints Complicating Constraints. Put the complicating constraints into the objective. We will obtain a lower bound on the optimal solution for minimization problems. In many situations, this bound is close to the optimal solution value.
The modified costs of 1-2-5-6 is 35. 2
21 3 8 5 19 6 12
4
15
1-3-2-5-6 is also optimal 6 And it is optimal for the original problem as well! The transit time of 1-3-2-5-6 is 10. The cost is 15.
cP( ) =
tij]
Bounding Principle. Suppose that cP( ). Then cP( ) -
0 and that P is
the minimum cost path with respect to modified costs T is a lower bound on the length
16
of the constrained shortest path.
Towards the Lagrangian Relaxation
z* = min s.t.
( i , j ) A ij
c xij
j
j
xij
x ji T
1 if i = s 1 if i = t 0 otherwise A
( i , j ) A ij
t xij
xij For each
0 or 1 for all (i, j)
0, replace the objective by:
( i , j ) A ij
z( ) = min
c xij
(
t ij ) xij
( i , j ) A ij
t xij T )
(i, j) A
(cij
T t xij T
c ( P*) c( P*)
Then P* is optimal. Proof.
T
c ( P*)
TL( )z* c源自 P *) c ( P*)T.
19
The Lagrangian Relaxation Technique
z* min cx
s.t. Ax b x X
P
L( ) min cx
s.t. x
10
1
13
3 12
4
3
15
5
What is the new shortest path from 1 to 6?
The path 1-2-5-6 is still an optimal path. 2
21 3 8 5 19 6 12
4
15
1
16
6
3
18
5
11
And there is an alternative shortest path.
17
Then z( )
z* for
0 because
( i , j ) A ij
The Lagrangian Relaxation
L( ) = min
(i, j) A
(cij
t ij ) xij x ji
T
s.t. xij
j
xij
j
1 if i = s 1 if i = t 0 otherwise
15
On Bounding
We could charge any non-negative toll on transit times.
For a path P, let
cP = tP =
(i,j) P cij (i,j) P tij (i,j) P [cij +
For fixed
and P, let
12
1
16
3
18
5
A parametric analysis
Toll 0 1 2 3 4 5 6 7 8 9 10 Weighted Sum 3 20 35 45 55 64 72 80 88 96 104 Cost 3 5 5 15 15 24 24 24 24 24 24 Transit Time 18 15 15 10 10 8 8 8 8 8 8 Weighted Sum - 10 * Toll 3 10 15 15 15 14 12 10 8 6 4
18
Application to constrained shortest path
L( ) = min
(i, j) A
(cij
t ij ) xij
T
Let c(P) be the cost of path P Let c ( P ) be the modified cost of path P Corollary. Suppose P* is a shortest path for the problem with modified costs, and suppose