L13. Shortest path routing D. Moltchanov, TUT, Spring 2008 D. Moltchanov, TUT, Spring 2014

Outline Network dimensioning problems Delay optimization Shortest-path routing problems Fair networks Topological design Restoration design

NDP: delay optimization

Packet delay on a single link Example: let Average packet size: Kbyte Link rate: Mbps The service rate of a link (how many packets per second) Now consider single node performance Mean delay in M/M/1 queue as we know is For arrival rate packets per second (pps) delay is

Packet delay on a single link Important observations Delay bound 15ms: utilization should be less than 64%! Delay is highly non-linear Utilization determine delay We may work with just link utilization not delays!

Packet delay on a single link 11 ms is really small! Should we care? First reason: a lot of hops! 16*11 = 176ms. + propagation delay (D/C) + protocols Second reason: this is just queuing delay… Other delay components are unavoidable, e.g. propagation delay

Packet delay on a single link All these hypotheses Self-similar traffic Long-range depends traffic Non-stationary traffic Pareto distribution of transfer sizes Heavy-tailed distributions of transfer sizes … Are having one thing in-common: high variability High variability Traffic no longer smoothes our like Poisson traffic Statistical multiplexing gains are not that huge Poisson is still a good approximation

Packet delay on a single link The following are really important M/M/1 is just an approximation Often M/M/1 is too optimistic (mean=variance for Poisson traffic) Keep utilization less than 50% (not 64%) to get within 15ms. delay

NDP: delay minimization Uncapacitated minimize subject to Capacitated minimize subject to Solution: convex solvers, e.g. Mathlab, CPLEX, Maple, etc.

NDP: getting rid of convexity Piecewise linear approximation

Shortest path routing

SPR: shortest path routing Shortest path routing (SPR) is what IGP do Intermediate system to intermediate system (IS-IS) Open shortest path first (OSPF) Do not confuse with shortest path first allocation of NDP Shortest path routing For each demand its volume is realized over SP SP is with respect to some link weight system Not with respect to the minimum number of hops! Values are sometimes called “link costs” Why the difference? Routing cost is not the same as routing over SPs You’ll see it in what follows Major one: in SPR rules are enforced by SPR protocols (e.g. OSPF)

SPR: four nodes Let link weights in our example to be then the following path are will be used for entire demands while the rest of unique as SPs are unique happens to be non-bifurcated This is what OSPF will do! Solution is denoted as To highlight dependence on w 1 =1 w 2 =3 w 4 =2 w 5 =4 w 3 =1

SPR: four nodes single path SPR Are there any problems with solution, Recall the link rates we used: Link load vector It does not fit the network at all What if you’ll run OSPF? Trivial solution: make not always possible… This is single path SPR problem w 1 =1 w 2 =3 w 4 =2 w 5 =4 w 3 =1 c 1 =5c 2 =10 c 3 =10 c 5 =30 c 4 =5

SPR: four nodes single path SPR Single path SPA allocation problem Demand constraints: Capacity constraints: Non-negativity: Very complex problem Non-bifurcated single path solution may not exist Hard to get even when exists Also: the weight system inducing the solution may not exist!!!

SPR: special problem Consider the following problem Demand d = 1 between nodes 1 and 7 Demand d = 2 between nodes 2 and 6 Demands volumes Link capacities are all 1, Analyzing the problem Two path for each demand Solution is evident: allocate Is there a link weight system inducing it? No! Not possible for single shortest path! What is about splitting flows?

SPR: four nodes ECMP rule Let’s get back to four nodes example Assume the link weight system is Shortest path with respect to the number of hops now! Very natural for networks Consider demand d = 1 Two shortest paths! Which path should OSPF use? ECMP (Equal Cost Multipath Rule) Split between shortest path Half to w 1 =1 w 2 =1 w 4 =1 w 5 =1 w 3 =1 c 1 =5c 2 =10 c 3 =10 c 5 =30 c 4 =5

SPR: ECMP rule Consider the problem on the right Demand d = 1 from node 6 to 7, Link weight system: Three shortest path from 6 to 7 What is the allocation according to OSPF? What is about d = 2, from 7 to 6, ? Same three paths Allocation Think why they are different…

SPR: special problem Analyzing the problem Two path for each demand Solution is evident: allocate No weight system inducing single path! ECMP rule allocation? Weight 1 to links 3-4, 4-5 Weight 2 to other links Costs become 6 for all paths ECMP of OSPF can be enforced

SPR: general problems Forward problem: Given a set of link weights Find the shortest path(s) from one node to another Classic graph theory problem you already studied Dijkstra and Bellman-Ford algorithms Algorithms for all shortest path (for ECMP) Backward problem For given link capacities and demand volumes Find a link weight system Such that flow allocation is feasible Setting up link weights The latter is what we are going to study