EBK COMPUTER NETWORKING
EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Chapter 1, Problem P20P

Explanation of Solution

Let Rs be server link, Rc be client link, R be network link and M number of client-server pairs present in the links.

The general expression for throughput in terms of Rs,Rc,R and M is as follows,

Throughput=min{Rs,Rc,RM}

Therefore, th

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Students have asked these similar questions
Consider a router buffer preceding an outbound link. In this problem, you will use Little’s formula, a famous formula from queuing theory. Let N denote the average number of packets in the buffer plus the packet being transmitted. Let a denote the rate of packets arriving at the link. Let d denote the average total delay (i.e., the queuing delay plus the transmission delay) experienced by a packet. Little’s formula is N=a⋅d . Suppose that on average, the buffer contains 10 packets, and the average packet queuing delay is 10 msec. The link’s transmission rate is 100 packets/sec. Using Little’s formula, what is the average packet arrival rate, assuming there is no packet loss?
Consider Figure 2.12, for which there is an institutional network connected to the Internet. Suppose that the average object size is 1,000,000 bits and that the average request rate from the institution’s browsers to the origin servers is 16 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is three seconds on average (see Section 2.2.5). Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use ∆/(1 - ∆b), where ∆ is the average time required to send an object over the access link and b is the arrival rate of objects to the access link. Find the total average response time. I get ∆=1000000/15000000= 1/15, then ∆/(1 - ∆b)= (1/15)/(1-(1/15)*16)=-1, so the finally answer is : 3+(-1)=2s or 3+0=3s?
Consider the throughput example. Now suppose that there are 4 client-server pairs. Denote Rs1, Rs2, Rs3,Rs4 ,Rc1 ,Rc2 ,Rc3, Rc4 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 4 client-server pairs.  Question: Assume that for i = {1,2,3,4}, Rsi = Rci = R , assume that all routers have infinite buffercapacity, meaning that packets are never dropped and are eventually forwarded by a router. Whatis the throughput for the Client Server pair 1? What is the throughput for Client Server pair 3?

Chapter 1 Solutions

EBK COMPUTER NETWORKING

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