Unformatted text preview: Problem 2
a) A circuitswitched network would be well suited to the application described, because the application involves long sessions with predictable smooth bandwidth requirements. Since the transmission rate is known and not bursty, bandwidth can be reserved for each application session circuit with no significant waste. In addition, we need not worry greatly about the overhead costs of setting up and tearing down a circuit connection, which are amortized over the lengthy duration of a typical application session. b) Given such generous link capacities, the network needs no congestion control mechanism. In the worst (most potentially congested) case, all the applications simultaneously transmit over one or more particular network links. However, since each link offers sufficient bandwidth to handle the sum of all of the applications' data rates, no congestion (very little queuing) will occur. Problem 8
a) 10,000 M Mn M −n b) p (1 − p ) n n = N +1 Problem 9
Consider a packet of length L which 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 di, si, and Ri denote the length, propagation speed, and the transmission rate of link i for i = 1, 2, 3. The packet switch delays each packet by dproc. Assuming no queuing delays, in terms of di, si, Ri, and L, what is the total endtoend delay for the packet? Suppose the packet is 1500 bytes, propagation speed is 2.5e108 m/s, transmission rates of all three links are 2Mbps, the packet switch processing delay is 3msec, the length of the three links is 5000 km, 4000km and 1000 km. What is the endtoend delay? ! ! ! "#!$ ! % !&!&' % Problem 10
! # ( !$ " ) * + ! "# ! $ ! % ' % Problem 23
% ! + ,%# ' %# , $#$" , %# , $#$" ,  . / $## ,$# 0'&"##### ( ) 23 4$## ' &1 " $ ," ! & ! # ! ' () * Problem 24
% + 1 1 " $ 3 # 6 4 5 " ! $ " " $ % ) ' " ! " " / # 2 ) '% ! !" ! ./ 0 ! " a) 160,000 bits b) 160,000 bits c) The bandwidthdelay product of a link is the maximum number of bits that can be in the link. d) the width of a bit = length of link / bandwidthdelay product, so 1 bit is 125 meters long, which is longer than a football field e) s/R
5$ $ 5 /, 0 , 0 8 .6/ !7 5" 7 9 5 :5 ;* <( ...
View
Full Document
 Spring '10
 Cetti
 Bit rate, Orders of magnitude, Byte, packet switch delays

Click to edit the document details