Multi commodity flows. Idan Maor Tel Aviv University. interdiction. G=(N,A) is a directed graph capacity for every i,j V: If (i,j) A then = 0 . k pairs of distinguished vertices, (s1, t1),…(sk, tk). the cost of sending 1 unit of commodity k over ( i ,j).
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(, ,)
(5,1,4)



(7,2,4)

(10,6,3)
(10,0,0)
(4,0,0)



(7,4,2)
(5,4,1)

=(5*1)+(7*2)+(7*6)+(7*0)+
(5*1)+(3*2)+(3*3)+(3*0)+(3*0)=85
(∞,1,1)
(∞,1,1)
the flow can be fractional, Even if the cost and capacities are integers
1 unit flow
From s1 to t1.
Result 4.
the flow can be fractional, Even if the cost and capacities are integers
1 unit flow
From s2 to t2.
Result 4.
the flow can be fractional, Even if the cost and capacities are integers
1 unit flow
From s3 to t3.
Result 4.
the flow can be fractional, Even if the cost and capacities are integers
1/2 unit flow
From s1 to t1
From s2 to t2
From s3 to t3
Result 6.
Using the dual variables the Reduce cost for the problem
The dual linear program is :
In order to simplify the problem, in this part will add a new assumptions:
(p)=(p) =
=.
=
[]=
.
The exponential number of variables are not a pitfall for this approach, since the linear pogromming structure promise us that there exits an optimal solution with at most m+ K paths .
there is no need to represents all the columns (paths), all the time we can use lazy approach and generate only when nodded.
=+
Just state that if we don’t use the total capacity of some arc, then the arc price can be zero.
From , we get that :
If (x,r) is feasible in the resource directive problem , then r is feasible in the resource allocation problem and .
Proof:
it’s easy to see that if (x,r) is feasible in the resource directive problem then r is also feasible in the resource allocation problem, since the problems share the same resources contrarians
For the second part since x is feasible for every commodity we get :
z(r) = ≤.
According to the definition of the recourse allocation problem.
If r is a feasible in the resource allocation problem then for some vector x, (x,r) is a feasible solution for the resource directive problem and z(r)=cx.
Proof:
If r is a feasible solution for the resource allocation problem then by the definition of =cx for some vector x.
So x there a vector x that satisfies z(r)=c(x).
a sub gradient ϒ of z(r) at point r= is any vector that satisfies )+ϒ(r).
For all with
 the set of all resource allocations for commodity k, such that the sub problem is feasible .
Claim :
l be a sub gradient of at the point , then =(,,….,) is a sub gradient of z(r) at =(, , ….).
Proof:
Lets look on the network flow problem:
, q is a vector of upper bounds on the arc flows.
the optimal solution when =q.
={
Claim:
For any non negative vector q’,for which the problem is feasible, (q’).
i.e. is a sub gradient of the objective function.
Proof:
According to the definition of reduce cost cx=
==.
The last equality follows from the optimality condition :
=0 if And the definition of
={
Proof:
Let x’ be the solution of the problem when q=q’.
Then =q’.
Since and .
We got :
q’
=
++q’
+(q’).
After receiving the new resource allocation vector r, we need to verify the feasibility.
If it’s not feasible we need to move to another point r’ in order to preserve the feasibility.
one approach that ensures that the algorithm converges is to choose the closet point to r, in the sense to minimize
We can solve the problem in the following way :