1. L13. Capacitated and uncapacitated network design problems
D. Moltchanov, TUT, Fall 2017
D. Moltchanov, TUT, Spring 2008

2. Network dimensioning problems

3. NDP: network model The general uncapacitated NDP problem
Minimize cost of routing
Given demand volumes, network topology and cost of routing
How much capacity we need over the links?
Example we consider
V = 4 nodes
E = 5 links
D = 3 bidirectional demands
Link rates ce are unknowns!
Demands: hd=15, hd=20, hd=10
Demands d: Pd paths, p = 1,2,…,Pd
sets of path for demand d
flow variables

4. NDP: specifying paths and flows
Demand d = 1, first path
path consists of two links 2 and 4
the set of path for d=1
one flow possible:
Are there other paths?
Sure, there are
We just excluded them!
What is the reason for exclusions
Quite arbitrary, length, external requirements, anything…
In this case: don’t want path longer than 1 hop for this demand
Given topology it is up to us to decide which paths to allow!

5. NDP: specifying paths and flows
Demand d = 2, two paths
set of paths for d = 2:
flow variables:
other paths are disallowed by us!
Demand d = 3: two paths
set of paths for d = 3:

6. NDP: example Let the vectors of flows for demands be
the vector of all allocations is
Flows of a demand must satisfy this demand, thus,
these are demand constraints
In our particular case we have

7. NDP: example Second set of constraints Note the following
links are not exceeded by flows
called capacity constraints
Note the following
LHS: link loads due to flows
We need to know relationship between links and paths
Can be formally specified by link-path incidence coefficients

8. NDP: example Link-path incidence relation table
provides indication which flows appears in LHS
whether path p of demand d uses link e?
Example: d=1 path entries are set to 1
Example: d=3 path entries are set to 1

9. NDP: example Why we introduced ? The link load on link e is given by
Now the capacity constraints are
which is a compact form of
we will also define a vector

10. NDP: example We are interested in minimizing the capacity cost
where is cost of a capacity unit on link e
The whole problem
Minimize
Subject to

11. NDP: example Equivalent to this extended one Minimize Subject to
Demand constraints:
Capacity constraints:
Non-negativity constraints:

12. NDP: example Compare with three nodes example Minimize (routing cost)
subject to flow constraints
and link constraints difference
and positivity constraints

13. NDP: example Compare with three nodes example Three nodes example
Demands were given
Link capacities were given
We minimized the total routing cost
That is called “capacitated design problem”
Current four nodes example
Demands are given
Link capacities are unknown
Link unit costs are given
Minimizing capacity cost required to route demands
This is called “uncapacitated design problem”
Both problems are of linear programming (LP) type. Why?

14. NDP: solution Note the following What are the reasons?
When variables are continuous we have equalities
What are the reasons?
why should we pay for unused capacity?
that is link load by flows should equal the capacity of this link

15. NDP: solution Consider a solution Is this optimal? No…
Just an instance of
First:
Then: via
The link rate vector
total cost
Is this optimal? No…
Path for d=2 carrying
Expensive as
Another path is with cost
Cost path is found in general using

16. NDP: solution What to do? Other observations
Move all the flow from to
That is, set
Savings per unit:
Overall savings:
Other observations
Flow is optimal (only one path!)
Flows are optimal
Why? paths are of the same cost:
Any split of is optimal!
Infinitely many optimal solutions:

17. NDP: short path allocation rule
Remember we had multiple paths?
Demand d=1:
Demand d=2:
Demand d=3:
Uncapacitated
NDP only! 2 1 1 3 1

18. NDP: modification 1 Why not to add for d=1?
Should be better for the cost!
Recall costs
Modifications to constraints and objective function
Demand constraints

19. NDP: modification 1
Capacity constraints

20. NDP: modification 1 Objective function remains the same
Solution to this modified problem
We already have
Paths have costs
Our rule: allocate all to
Can we? Why not? We are dealing with uncapacitated problem!
Savings: per unit overall
Optimal solution
with

21. NDP: non-bifurcated flows
We may request non-bifurcated solution
Splitting of flows are not allowed
One flow for one demand
AKA: unplittable or single path NDP
For our settings
Already only one path allowed for d=1:
For d=2: one out of two allowed
For d=3: one out of two allowed
May not be unique
Bifurcated solution
Non-bifurcated ones:

22. NDP: modular links
We assumed links of any rates! Is this realistic? No!
Modular links needs to be used… e.g. T1/E1/STM-1 etc.
Let’s assume 1 LCU = M DVU
What are the implications?
Allocation may not obey the shortest path allocation rule
One can see it using huge values of M
Optimal link capacity is not the same as optimal link load
So far we used these terms interchangeably
Optimal link capacity optimal link load
Bifurcated solutions are “more optimal” in general
Splitting is good when M is moderate with respect to demands
For huge M non-bifurcated solutions are often obvious
Non-bifurcated with modular links is a complex problem!

23. NDP: modular links Let M = 35, 1 LCU = 35 DVU Non-modular Modular 2 1
Recall
One module at e = 1
One more at e = 5
Non-modular Modular 2 1 35 1 3 1 35

24. NDP: capacitated problem
Uncapacitated design problem
We are given a network, demands, paths, link costs
Link rates are what we need to find
Such that the cost is minimized
Capacitated design problem
We are given a network, demands, paths
Link rates are given (links are installed already)
Flow allocation is what we need to find
Such that routing cost are minimized
Important difference between these two
No link costs in capacitated problem
Cost of using a link could be given (called routing cost)
As a special case: cost of routing for all links may have cost 1

25. NDP: capacitated problem
Given demand constraints
And capacity constraints
Minimize
where is the cost of routing over link e

26. NDP: capacitated problem
Letting in our example
Capacities
routing costs
And allowing one more path for d=1,
There is bifurcated solution
with paths
No non-bifurcated solutions…
This is often the case 10 5 5 10 30

27. NDP: capacitated vs. uncapacitated
minimize subject to
Capacitated
minimize subject to
Solution: LP solvers, e.g. Mathlab, CPLEX, Maple, etc.