263final_sol_06

263final_sol_06 - EE263 Prof S Boyd Dec 8–9 or Dec 9–10...

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Unformatted text preview: EE263 Prof. S. Boyd Dec. 8–9 or Dec. 9–10, 2006. Final exam solutions 1. Analysis and optimization of a communication network. A communication network is modeled as a set of m directed links connecting nodes. There are n routes in the network. A route is a path, along one or more links in the network, from a source node to a destination node . In this problem, the routes are fixed, and are described by an m × n route-link matrix A , defined as A ij = braceleftBigg 1 route j passes through link i 0 otherwise. Over each route we have a nonnegative flow , measured in (say) bits per second. We denote the flow along route j as f j , and we call f ∈ R n the flow vector . The traffic on a link i , denoted t i , is the sum of the flows on all routes passing through link i . The vector t ∈ R m is called the traffic vector . Each link has an associated nonnegative delay , measured in (say) seconds. We denote the delay for link i as d i , and refer to d ∈ R m as the link delay vector . The latency on a route j , denoted l j , is the sum of the delays along each link constituting the route, i.e. , the time it takes for bits entering the source to emerge at the destination. The vector l ∈ R n is the route latency vector . The total number of bits in the network at an instant in time is given by B = f T l = t T d . (a) Worst-case flows and delays. Suppose the flows and link delays satisfy (1 /n ) n summationdisplay j =1 f 2 j ≤ F 2 , (1 /m ) m summationdisplay i =1 d 2 i ≤ D 2 , where F and D are given. What is the maximum possible number of bits in the network? What values of f and d achieve this maximum value? (For this problem you can ignore the constraint that the flows and delays must be nonnegative. It turns out, however, that the worst-case flows and delays can always be chosen to be nonnegative.) (b) Utility maximization. For a flow f j , the network operator derives income at a rate p j f j , where p j is the price per unit flow on route j . The network operator’s total rate of income is thus ∑ n j =1 p j f j . (The route prices are known and positive.) The network operator is charged at a rate c i t i for having traffic t i on link i , where c i is the cost per unit of traffic on link i . The total charge rate for link traffic 1 is ∑ m i =1 t i c i . (The link costs are known and positive.) The net income rate (or utility) to the network operator is therefore U net = n summationdisplay j =1 p j f j − m summationdisplay i =1 c i t i . Find the flow vector f that maximizes the operator’s net income rate, subject to the constraint that each f j is between 0 and F max , where F max is a given positive maximum flow value. Solution: The traffic vector t is given by t = Af , and the route latency vector l is given by l = A T d ....
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263final_sol_06 - EE263 Prof S Boyd Dec 8–9 or Dec 9–10...

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