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Computer Networking : A Top-Down Approach

Computer Networking : A Top-down Approach

Authors:
James F. Kurose, Keith W. Ross
Exercise:
Problems
Chapter:
Computer Networks And The Internet
Edition:
6
ISBN:
9780132856201
Question:
20

 

Question

Consider the throughput example corresponding to Figure 1.20(b). Now suppose that there are M client-server pairs rather than 10. Denote Rs, Rc, and R for the rates of the server links, client links, and network link. Assume all other links have abundant capacity and that there is no other traffic in the network besides the traffic generated by the M client-server pairs. Derive a general expression for throughput in terms of Rs, Rc, R, and M.

Answer

Given data:

Rs  = Server link rate 

Rc = Client link rate

R  = Network link rate

M = Client-server pair

Instantaneous throughput and average throughput are two types of throughput.  The server throughput Rc faster than Rs.

Networks always depends on client-server links(M). The min is a simple two link network links on the network.

Therefore, general expression for throughput in terms of Rs, Rc, R, and M is min {Rs ,Rc , R/M}.

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