Consider a set of mobile computing clients in a certain town who each
need to be connected to one of several possible base stations. We’ll
suppose there are n clients, with the position of each client specified
by its (x, y) coordinates in the plane. There are also k base stations; the
position of each of these is specified by (x, y) coordinates as well.
For each client, we wish to connect it to exactly one of the base
stations. Our choice of connections is constrained in the following ways.
There is a range parameter r—a client can only be connected to a base
station that is within distance r. There is also a load parameter L—no
more than L clients can be connected to any single base station.
Your goal is to design a polynomial-time algorithm for the following
problem. Given the positions of a set of clients and a set of base stations,
as well as the range and load parameters, decide whether every client can
be connected simultaneously to a base station, subject to the range and
load conditions in the previous paragraph

answer given in the explanation

Explanation:

We have n clients and k-base stations, say each client has to be connected to a base station that is located at a distance say 'r'. now the base stations doesn't have allocation for more than L clients.

To begin, let us produce a network which consists of edges and vertex

Network (N) = (V,E)

where V = [S, cl-l, - - - -  cl-n, bs-l - - - - - - bs-k, t]

given that cl-l, - - - - - cl-n represents nodes for the clients

also we have that bs-l, - - - - - bs-k represents the nodes for base station

Also

E = [ (s, cl-i), (cl-i,bs-j), (bs-j,t)]

(s, cl-i) = have capacity for all cl-i (clients)

(cl-i,bs-j) = have capacity for all cl-i  clients & bs-j (stations)

⇒ using Fond Fulkorson algorithm we  find the max flow in N

⇒ connecting cl-i clients to  bs-j stations

      like (cl-i, bs-j) = 1

   if f(cl-i, bs-j)  = 0

⇒ say any connection were to produce a valid flow, then

if cl-i (clients) connected                f(s,cl-i) = 1 (o otherwise)

if cl-i (clients) connected  to bs-j(stations)   f(cl-i,bs-j) = 1 (o otherwise)

   f(bs-j,t) = no of clients  (cl-i)  connected to bs-j

⇒ on each node, the max flow value (f) is longer than the no of clients that can be connected.

⇒ create the connection between the client and base station i.e. cl-l to base bs-j iff    f(cl-i, bs-j) = 1

⇒ when considering the capacity, we see that any client cannot directly connect to the base stations, and also the base stations cannot handle more than L clients, that is because the load allocated to the base statsion is L.

from this, we say f is the max no of clients (cl-i) that can be connected if we find the max flow, we can thus connect the client to the base stations easily.

cheers i hope this helps

ok

explanation:

k

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