263final_sol_06

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

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 24

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online