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Berkeley ELENG 228A - Optimization

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Optimization - Part 2ContentsHighly Optimized ToleranceHighly Optimized ToleranceGibbs SystemsRoutingComplexity Complexity Halting TheoremCook’s TheoremComplexity Routing with Multiple ConstraintsRouting with Multiple ConstraintsRouting with Multiple ConstraintsRouting with Multiple ConstraintsRouting with Multiple ConstraintsRouting with Multiple ConstraintsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsRouting and Wavelength AssignmentsOptimization: SummaryOptimization -Part 2Jean WalrandUniversity of California, Berkeley2ContentsnMotivationnOptimization of networksnMathematicsnTechnologynConvex ProgrammingnConvexitynNetwork ProvisioningnDualitynCongestion ControlnHighly Optimized TolerancenGibbs systemsnRoutingçnComplexitynMultiple ObjectivesnWavelength AssignmentnSummary3Congestion Control …01020304050601285582109136163190217244271298325352379406433460487Rates equalize àfair shareABxDEyTCP Congestion Control: AIMD4Congestion Control …CxyABxC D EyLimit rates:x = yTCP Congestion Control: AIMDChiu and Jain, 19885Congestion Control …ABxC D EyAdjust rates based on estimated backlog. Roughly,X ~ 1/q where q is backlog in router.Then, one can show that x ~ y. The intuition is that the flows will have similar backlogs.Two types of proof: •Lyapunov•Show that algorithm is a gradient projection algorithm for a convex problem [àconverges if …]TCP Congestion Control: Vegas6Congestion Control …1: x12: x2q: q1, q2d1 = round-trip delayp1 = propagation delayd2 = round-trip delayp2 = propagation delay7Congestion Control …1: x12: x2q: q1, q2d1 = round-trip delayp1 = propagation delayd2 = round-trip delayp2 = propagation delay8Congestion Control …9Highly Optimized TolerancenIdea: Systems optimized for typical situationsnAs a consequence, they are vulnerable to extreme situationsnWe explore this effect for the WWW and other models.John Doyle, Caltech…10Highly Optimized TolerancenHeavy TailsDC = length of codewordsafter data compression[exponential]FF = size of forest fires[heavy]WWW = file lengths[heavy]11Highly Optimized TolerancePower laws, Highly Optimized Tolerance and generalized source codingJohn Doyle, J.M. Carlson, 200012Gibbs SystemsnMotivationnStudy of systems with many interacting componentsnConnection between networks, statistical mechanics, etc…13Gibbs Systems14Gibbs Systems15Notes on reversibility16RoutingnEasy Problem:nShortest Path (Dijkstra, Bellman-Ford)nDifficult Problems:nFinding path with bound on cost and delaynAssigning wavelengths to paths to maximize the number of accepted pathsnOnline and OfflinenFirst two words about complexity17Complexity nExamples of ProblemsnSSP: Subset-Sum Problem: Given a set of n positive integers, is there a subset of them that add up to N? [Note: easy to verify a solution; to see if there is one, one must essentially try the 2nsubsets.]nKP: Knapsack Problem: Given a set of items each with a weight, find the number of each item to include to achieve the maximum weight less than a given value.nHP: Halting Problem: Given an algorithm a and its input i, will it stop or will it run forever? nSAT: Boolean SatisfiabilityProblem: Given a Boolean expression, is there an assignment (true or false) of the variables that makes the expression true?18Complexity nComplexity measures the time (number of steps or operations) or the space (memory) required to solve a problem by the best possible algorithmnP (Polynomial): Problems whose solution can be found on a sequential deterministic machine in a time that is polynomial in the size of the inputnNP (Nondeterministic Polynomial): Problems whose solution can be verified in polynomial time on a sequential deterministic machine; equivalently, whose solution can be found on a sequential nondeterministic machine in polynomial time. [Guess and check.] nOpen problem: P = NP?nNP-Hard: A decision problem H such that any decision problem in NP can be reduced to H in polynomial timenNP-Complete: A problem in NP to which every other problem in NP can be reduced in polynomial time = both NP-Hard and NPnSAT, KP, SSP are NP-Complete, HP is NP-Hard but is not NP19Halting TheoremTheorem (Turing): There is no algorithm that can decide, for any algorithm a and any input iwhether the algorithm a(i)halts.Proof:nAssume there is an algorithm with h(a, i) = true if halts and h(a, i) = falseotherwise. nConsider the algorithm tdefined so that t(i)is as follows: if h(i, i) = false, then return true; if h(i, i) = true, loop forever.nDoes the algorithm t(t)halt? If it does, then h(t, t)= false, which says that the algorithm twith input tdoes not halt. On the other hand, if t(t)does not halt, then h(t, t)must be true, so that t(t)should halt.nHence, there cannot be an algorithm h.20Cook’s TheoremTheorem (Stephen Cook):SAT is NP-CompleteProof:nSAT is NP because a nondeterministic Turing machine (NTM) can guess an assignment of variables and check. nConsider a problem that can be solved in p(n) by a NTM.nThe NTM corresponds to a tape with n symbols, transition rules, and an acceptance rule. Every instance of the problem can be encoded into a Boolean expression B that is satisfied iffthe NTM accepts.nThe variables describe the contents of the tape and the position of the head at the different steps; the Boolean expression states that the rules of the NTM are followed and that the machine finishes in an accepting state. Counting shows that B is of size p(n).21Complexity nEXPTIME-Complete: Problems that require an exponential timenCurrently, all know algorithms for solving an NP-complete problem require an exponential-time.nSome problems require e^(e^n) time; other problems (e.g., halting) cannot be solved in finite timenNote that a problem that requires a time 106n1000is P, yet intractablenThe average time of an NP problem might be P22Routing with Multiple ConstraintsnProblem: Given a graph where each link has an integer cost and an integer delay, find a path with cost £C and delay £D.nExample:nFind a path from S to D with cost £4 and delay £3 èNonenFind a path from S to D with cost £5 and delay £3 èSADnFind a path from S to D with cost £3 and delay £4 èSBDnFind a path from S to D with cost £5 and delay £4 èSAD or SBDSABD(3, 1)(1, 2)(2, 1)(1, 2)(Cost, Delay)23Routing


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Berkeley ELENG 228A - Optimization

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