Problem1 A. Give one advantage and one disadvantage of circuit--switched network

Question

Problem1

A. Give one advantage and one disadvantage of circuit--switched networks over packet--switched networks.

B. Give three examples of access networks. State their typical speeds and whether they provide shared or dedicated access.

C. Give two examples of guided physical media and two examples of unguided physical media, and state their typical speeds.

D. List four factors that contribute to the end--to--end delay in a packet--switched network. Which of these are constant and which of these depend on the load in the network? Explain.

E. Suppose you would like to urgently deliver S=40 terabytes data from Boston to Los Angeles. You have available a C=100Mbps dedicated link for data transfer. Would you prefer to transmit the data via this link or instead use Fedex overnight delivery?

Problem 2:

How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5×108 m/s, and transmission rate 2 Mbps?

Problem 3:

More generally, given a packet of length L bits, a link of distance d km, propagation speed s m/s and transmission rate R bps, what do you think are the quantities that are most important in determining delay… and why? (there is no single correct answer, the intent is to have you think about the “why” part)

Problem 4:

Consider a packet of length L which begins at end system A and travels over three links to a destination end system. These 3 links are connected by 2 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.

a. Assuming no queuing delay, and writing your answer in terms of di, Si, Ri and L, what is the total end to end delay for the packet (qualitatively, or symbolically)?

b. Suppose now the packet is 1,500 bytes, the propagation speed on all 3 links is 2.5*108 m/s, the transmission rates on all 3 links are 2 Mbps, the packet switch 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?

Explanation / Answer

Problem1:

A.

Advantages of Circuit switched networks over packet switched networks:

1. The dedicated link between sender and receiver provides a guarnateed data rate.

2. There is no delay in circuit switched networks when the path is established. In packet switched networks there takes transmission delays.

3. This is used for long continous transmission.

Disadvantages of Circuit Switching:

1. Even if the channel is free it doesn't allow any other data since its has dedicated conncection.

2. Efficiency of network is low. Requies more bandwidth and takes lot of time to establish physical link between sender and reciever.

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B.

Access networks is divided into three categories:

1. Residential access networks is connecting a home end system into the network. modem speeds allow dialup access at rates up to 56 Kbps.

It is point-to-point access.

2. Institutional access networks is connecting an end system in a business or educational institution into the network.It is shared or dedicated cable connects end system and router. The speed is 10 Mbs, 100Mbps Gigabit Ethernet.

3. Mobile access networks is connecting a mobile end system into the network. It is a shared wireless access network connects end system to router. The speed of Lucent Wavelan is 10 Mbps

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C.

i. Examples of guided pysical media are Copper , fibres, coaxial cables and twisted pairs. Category 3: traditional phone wires has 10 Mbps ethernet. Category 5 TP: 100Mbps ethernet

ii. Examples of guided physical media are microwave(up to 45 Mbps channels), LAN (e.g., waveLAN has 2Mbps, 11Mbps), wide-area (e.g., cellular CDPD has 10’s Kbps) satellite(up to 50Mbps channel).

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D.

1. Packets experience delay on end-to-end path. The four factors are transmission, propagation, nodal processing, queuing.

Constant: transmission delay, propagation delay

Depend on the load: nodal processing, queueing delay

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E.

40terabytes =40*10power(12)*8bits.

The dedicated link may take 400 *10 power(12)*8 /(10*10power(9))= 320000seconds =3.7days.

With FedEx overnight delivery, you can guarantee the data arrives in one day, and it only costs you no more than $100.

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Problem2:

Time(propagate) = distance/propagation speed = (2500*10power(3))/(2.5x10power(8)) = 10.00 msec

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Problem3:

This delay is not depend on packet length or on transmission rate.

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Problem4:

Given: N(links)=3, L= 1,500 bytes

si= 2.5*10power(8) m/s on all 3 links

Ri= 2 Mbps, dproc= 3 msec

Formula for Delay is given below:

Delay(end-to-end)= N(dproc+(L/Ri)+ (di meters/si sec))

Substitute the values for i = 1 in the formula and computing the value:

=N(dproc+(L/R1)+ (d1 meters/s1 m/sec))=3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (5,000,000 m/2.5*10power(8) m/sec))=0.087 seconds

For i=2,

=N(dproc+(L/R2)+(d2 meters/s2 m/sec))= 3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (4,000,000 m/2.5*10power(8) m/sec))=0.075 seconds

For i=3,

=N(dproc+(L/R3)+(d3 meters/s3 m/sec))= 3(0.003 sec+(1,500*8 bits/2,000,000 bits/sec)+ (1,000,000 m/2.5*10power(8) m/sec))=0.039 seconds

The total delay on three links can be computed by adding each case values i.e 0.087 + 0.075 + 0.039 = 0.201 sec

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