Lecture 6 7 8

Lecture 6 7 8 - 1.4 Network Models 1.4 Network Models Now...

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1.4 Network Models usc-ee-den-ee555-200901-silvester section 1.4 1 1.4 Network Models Now suppose we have a communication network of nodes. Each node is a router and has an output buffer for each link leaving that node. Example: 4 node network 1 3 2 4 1 3 2 4 1,2 2,1 4,3 3,4 4,1 1,4 3,1 2,3 1.4 Network Models usc-ee-den-ee555-200901-silvester section 1.4 2 1.4.1 Traffic Matrix Define traffic matrix of flow between source s and destination d : {} sd γ = Γ in packets per second. It is often convenient to represent this with a scale factor outside the matrix so that the elements of the matrix represent the fraction of flow between s and d . For our example, let us assume that the traffic pattern is uniform, i.e. the same amount of traffic flows between any pair of nodes (not allowing a node to talk to itself, even though they sometimes do!) Then, = Γ 0 12 / 1 12 / 1 12 / 1 12 / 1 0 12 / 1 12 / 1 12 / 1 12 / 1 0 12 / 1 12 / 1 12 / 1 12 / 1 0 and the total network traffic or load is just γ . 1.4.2 Flow Matrix Based on the routing algorithm (or table or matrix) we can compute the flows (in packets per second) on each link, F . Define sd ij f as the flow on link ( i,j ) due to [ s,d ]
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1.4 Network Models usc-ee-den-ee555-200901-silvester section 1.4 3 traffic. Then we may write: sd ij sd sd ij f δ γ = where sd ij is the fraction of [ s,d ] traffic that uses the link ( i,j ). For single path routing this will be either 1 or 0 depending on whether the [ s,d ] traffic uses the link or not. The total flow on link ( i,j ) is given by ∑∑ = s d sd ij sd ij f For our example, assume that the traffic from 1 to 3 flows via node 2; the traffic between 2 and 4 flows via node 3 and that routing is symmetrical. = + + + + + + + + = 0 2 1 2 0 3 3 0 2 1 2 0 12 0 0 0 0 42 43 41 24 34 42 31 32 13 24 23 31 21 14 13 12 F And the total network flow f = = = 3 4 12 16 ij f packets per second. 1.4 Network Models usc-ee-den-ee555-200901-silvester section 1.4 4 The links may have different capacities so we would also have a capacity matrix, { } ij c C = in bits per second; define the average packet length to be L bits. 1.4.3 Link Utilisation We could then find the link utilizations ij ij ij c Lf = ρ . For our example, let us assume that all C c ij = are equal = = Ρ 0 6 / 1 12 / 1 6 / 1 0 4 / 1 4 / 1 0 6 / 1 12 / 1 6 / 1 0 } { C L ij The maximum traffic that the network can sustain (for the specified traffic pattern) can by increasing γ until one of the links saturates; i.e. / ij ij f cL . In our example the most heavily loaded link is (2,3) or (3,2) and for that link we have: /4 1 4 / LC CL ≤⇒
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1.4 Network Models usc-ee-den-ee555-200901-silvester section 1.4 5 Note that this pattern is rather non-uniform. We could smooth it out by splitting the 2 hop traffic over the (two) available paths, i.e. 1 to 3 could use (1,2,3) half of the time and (1,4,3) half of the time, resulting in: 12 13 42 14 13 24 21 31 24 23 24 13 32 31 42 34 31 24 41 31 42 43 42 13 0 / 2/ 2 / 2 /2 0 0 0 02 2 202 12 22 0 F γ γγ ++ = −+ + + + ⎡⎤ ⎢⎥ = ⎢−
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This note was uploaded on 12/22/2010 for the course EE 555 taught by Professor Silvester during the Fall '08 term at USC.

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Lecture 6 7 8 - 1.4 Network Models 1.4 Network Models Now...

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