Network optimization problems D Moltchanov TUT Spring 2008 D Moltchanov TUT Spring 2014 Network dimensioning problems NDP network model The general uncapacitated NDP problem Minimize cost of ID: 224898
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Slide1
L12. Network optimization problems
D. Moltchanov, TUT, Spring 2008
D.
Moltchanov
, TUT, Spring 2014Slide2
Network dimensioning problemsSlide3
NDP: network model
The general uncapacitated NDP problemMinimize cost of routingGiven demand volumes, network topology and cost of routingHow much capacity we need over the links?
Example we consider
V
= 4 nodes
E
= 5 links
D = 3 bidirectional demandsLink rates ce are unknowns!Demands: hd=15, hd=20, hd=10Demands d: Pd paths, p = 1,2,…,Pdsets of path for demand dflow variablesSlide4
NDP: specifying paths and flows
Demand d = 1, first pathpath consists of two links 2 and 4the set of path for d=1one 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 demandGiven topology it is up to us to decide which paths to allow!Slide5
NDP: specifying paths and flows
Demand d = 2, two pathsset of paths for d = 2: flow variables:other paths are disallowed by us!Demand d = 3: two paths
set of paths for
d
= 3:
flow
variables:
other paths are disallowed by us!Slide6
NDP: example
Let the vectors of flows for demands bethe vector of all allocations isFlows of a demand must satisfy this demand, thus,these are demand constraintsIn our particular case we haveSlide7
NDP: example
Second set of constraintslinks are not exceeded by flowscalled capacity constraintsNote the followingLHS: link loads due to flows
We need to know relationship between links and paths
Can be formally specified by link-path
incidence coefficientsSlide8
NDP: example
Link-path incidence relation tableprovides indication which flows appears in LHSwhether 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 1Slide9
NDP: example
Why we introduced ?The link load on link e is given byNow the capacity constraints arewhich is a compact form of
we will also define a vector Slide10
NDP: example
We are interested in minimizing the capacity costwhere is cost of a capacity unit on link eThe whole problemMinimize
Subject to Slide11
NDP: example
Equivalent to this extended oneMinimizeSubject toDemand constraints: Capacity constraints:
Non-negativity constraints: Slide12
NDP: example
Compare with three nodes exampleMinimize (routing cost)subject to flow constraintsand link constraints difference
and positivity
constraintsSlide13
NDP: example
Compare with three nodes exampleThree nodes exampleDemands were givenLink capacities were givenWe minimized the total routing costThat is called “capacitated design
problem”
Current four nodes example
Demands are given
Link capacities are
unknown
Link unit costs are givenMinimizing capacity cost required to route demandsThis is called “uncapacitated design problem”Both problems are of linear programming (LP) type. Why?Slide14
NDP: solution
Note the followingWhen variables are continuous we have equalitiesWhat are the reasons?why should we pay for unused capacity?that is link load by flows should equal the capacity of this linkSlide15
NDP: solution
Consider a solutionJust an instance ofFirst: Then: viaThe link rate vectortotal costIs this optimal? No…
Path for
d
=2 carrying
Expensive as
Another path is with cost
Cost path is found in general usingSlide16
NDP: solution
What to do?Move all the flow from to That is, set Savings per unit: Overall savings:Other observationsFlow is optimal (only one path!)Flows are optimal
Why? paths are of the same cost:
Any split of is optimal!
Infinitely many optimal solutions: Slide17
NDP: short path allocation rule
Remember we had multiple paths?Demand d=1:Demand d=2:
Demand
d
=3:
1
1
1
2
3
Uncapacitated
NDP
only!Slide18
NDP: modification 1
Why not to add for d=1?Should be better for the cost!Recall costsModifications to constraints and objective functionDemand constraintsSlide19
NDP: modification 1
Capacity constraintsSlide20
NDP: modification 1
Objective function remains the sameSolution to this modified problemWe already havePaths have costsOur rule: allocate all to Can we? Why not? We are dealing with uncapacitated problem!
Savings: per unit overall
Optimal solution
w
ith Slide21
NDP: non-bifurcated flows
We may request non-bifurcated solutionSplitting of flows are not allowedOne flow for one demandAKA: unplittable or single path NDPFor our settingsAlready 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 uniqueBifurcated solutionNon-bifurcated ones: Slide22
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 DVUWhat are the implications?Allocation may not obey the shortest path allocation rule One can see it using huge values of MOptimal link capacity is not the same as optimal link loadSo far we used these terms interchangeablyOptimal 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 obviousNon-bifurcated with modular links is a complex problem!Slide23
NDP: modular links
Let M = 35, 1 LCU = 35 DVU Recall One module at e = 1One more at e = 5Non-modular Modular
1
1
1
2
3
35
35Slide24
NDP: capacitated problem
Uncapacitated design problemWe are given a network, demands, paths, link costsLink rates are what we need to findSuch that the cost is minimizedCapacitated design problemWe are given a network, demands, pathsLink rates are given (links are installed already)Flow allocation is what we need to findSuch 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 1Slide25
NDP: capacitated problem
Given demand constraintsAnd capacity constraints
Minimize
w
here is the cost of routing over link
eSlide26
NDP: capacitated problem
Letting in our exampleCapacities routing costsAnd allowing one more path for d=1, There is bifurcated solutionwith paths
No non-bifurcated solutions…
This is often the case
30
5
10
5
10Slide27
NDP: capacitated vs. uncapacitated
Uncapacitatedminimize subject toCapacitatedminimize subject to
Solution: LP solvers, e.g.
Mathlab
, CPLEX, Maple, etc.