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Resource Allocation: Deterministic Analysis

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Traffic Model Stochastic Different sample paths with different properties Expected case analysis Deterministic Properties applying to every sample path worst case analysis

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System Description Single server service rate C Single buffer Finite Infinite Arrival stream Departure Stream Output Queued System

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Reichs Equation A(t) Total traffic arriving in interval (0, t) D(t) Total traffic departing in interval (0, t) X(t) Queue length at time t X(t) = sup 0 s t (A(t) – A(s) – C(t-s)) D(t) = A(t) – X(t) = A(t) - sup 0 s t (A(t) – A(s) – C(t-s)) = inf 0 s t (A(s) + C(t-s))

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Convolution D(t) = inf 0 s t (A(s) + C(t-s)) = inf 0 s t (A(s) + B(t-s)) A*B(t) = inf 0 s t (A(s) + B(t-s)) (t) = 0 for t 0 = otherwise A* (t) = A A* d (t) = A(t-d) d (t)= (t-d) For causal B, B , thus A* B A* = A

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Service curve for network elements A(t) D(t) S(t) S(t) S(t) and S(t) are nonnegative, non-decreasing causal functions such that D A* S and D A* S The first is minimum service curve and the second the maximum service curve If D = A*F for some non-negative, non-decreasing causal F, then F is the service curve

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Examples coder packetizer Coder emits bursts of bytes at rate r Packetizer packetizes L bits in a packet, if less than L bits are available, then packetizes whatever available Maximum delay is L/r D A(t – L/r) Minimum service curve is S = L/r

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Constant rate server S(t) = C max(t, 0) Coder + Packetizer Constant rate server S DpDp D A D = D p * S (A * L/r ) * S = A * ( L/r * S) Minimum service curve is L/r * S

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Delay in a network element Let u be the minimum time such that all the data arriving in [0, t] depart by time u Then u = inf{s: D(s) A(t)} FIFO service Maximum delay max = sup t 0 {u – t: inf{s: D(s) A(t)}} This is the maximum horizontal distance between A and D D (A * max ) If A curve is shifted to the right by max it is below D

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Envelope A function E(t) is an envelope for the arrival function A(t) if A(t) – A(u) E(t-u) for all u and t, 0 u t That is A A*E For causal E, E , thus A* E A* = A A*E Thus A = A*E An envelope is sub-additive if for all u t, E(t) E(u) + E(t-u) Thus E E*E for a sub-additive envelope

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Regulator A network element is a regulator with envelope E, if for any input arrival process A, the departure process D satisfies D D*E Output of a source is typically statistically characterizable Source output can be upper-bounded by envelopes by passing it through a regulator

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Buffered Leaky Bucket Regulator There exists a fictitious token bucket which obtains token at the rate . The bucket can hold up to tokens A packet generated from a source can be released only if the token bucket has a token, and the token is removed after releasing the packet

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D(t) – D(u) + (t-u) In [u, t] the maximum amount of data that can be transported is the total number of tokens generated in [u, t] + the total number of tokens accumulated at u. The first quantity is (t-u) The second is Thus D D*E where E(t) = + t It follows that a leaky bucket regulator has envelope + t Is this envelope causal? Is it sub-additive?

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For any arrival process A, the departure process of the regulator satisfies D = A*E Thus E is the service curve of the regulator Proof: Note down from the board

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A process A with envelope E 1 is passed through a leaky bucket regulator with envelope E 2 Show that the resulting output has envelope E 1 * E 2 Lets work this out! Consider the output of a voice coder Emits bits at the rate R A has envelope E = Rt Regulate this process by a leaky bucket regulator with envelope E(t) = + t The resulting process will have an envelope min(Rt, + t)

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What do we do if we know the envelopes? We can upper bound the delay of an arrival process if its envelope and the minimum service curve of the network element is known. Let an arrival process A be transmitted through a network element with minimum service curve S(t) D is the departure process d max = inf {d: E* d S}

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Here, d max is the least amount the envelope E(t) be shifted so that it falls below he service curve S. The delay produced by the network element is upper bounded in terms of d max D A * dmax Proof: D A * S A * (E* dmax ) = (A * E) * dmax A * dmax

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Example computations of d max E(t) = min(Rt, + t) S(t) = ct d max = (R-c)/c(R - ) Lets work it out!

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Consider an arbitrary envelope E(t) and S(t) = ct It follows that E(t- d max ) ct Not quite rigorous! E(t) c(t+ d max ) d max (E(t)-ct)/c More rigorously, d max = sup t 0 (E(t)-ct)/c If delay upper bound must be less than some T, then the minimim required service rate c min is such that sup t 0 (E(t)- c min t)/ c min = T It follows that c min = sup t 0 (E(t)/ T + t)

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