EBK COMPUTER NETWORKING
EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
Expert Solution & Answer
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Chapter 1, Problem P23P

a.

Explanation of Solution

Let the pair of packets be A and B.

Given,

  • The rate of the first link from server to client is Rsbps.
  • The length of the each packet is L bits.
  • Packet B starts the transmission only after the completion of transmission of the packet A

b.

Explanation of Solution

When both the packets are sent back to back, the second packet B must cross the input queue of second link before the completion of the transmission of the first packet A.

Then, LRs+LRs+dprop<LRs+dprop+LRc

In the above inequality expression, the left hand side represents the time required by the packet B to cross the input queue of second link before the completion of the transmission of the packet A whereas the right hand side represents the time required by the packet A to complete the transmission on the second link.

If packet B is sent with the delay of time T seconds then,

LRs+LRs+dprop+

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Students have asked these similar questions
Consider figure 2. Assume that we know the bottleneck link along the path from the server to the client is the first link with rate Rs bits/sec. Suppose we send a pair of packets back to back from the server to the client, and there is no other traffic on this path. Assume each packet of size L bits, and both links have the same propagation delay dprop. (a). What is the packet inter-arrival time at the destination? That is, how much time elapses from when the last bit of the first packet arrives until the last bit of the second packet arrives? (b). Now assume that the second link is the bottleneck link (i.e., Re < Rs). Is it possible that the second packet queues at the input queue of the second link? Explain. Now suppose that the server sends the second packet T seconds after sending the first packet. How large must T be to ensure no queuing before the second link? Explain.
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 a packet of length L that begins at end system A and travels over three links to a destination end system. These three links are connected by two packet switches. Let d, s, and R denotes the length, propagation speed, and the transmission rate of link i, for i=1,2,3 . The packet switch delays each packet by d . Assuming no queuing delays, in terms of d, s , R, (i=1,2,3),  and L, what is the total end-to-end delay for the packet? Suppose now the packet is 1,500 bytes, the propagation speed on all three links is the transmission rates of all three links are 2 Mbps, the packet switch processing delay is 3 msec, the length of the first link is 5,000 km, the length of the second link is 4,000 km, and the length of the last link is 1,000 km. For these values, what is the end-to-end delay?  In the above problem, suppose R1=R2=R3=R and dproc=0. Further, suppose the packet switch does not store-and-forward packets but instead immediately transmits each bit it receives before waiting…

Chapter 1 Solutions

EBK COMPUTER NETWORKING

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