P5. Review the car-caravan analogy in Section 1.4. Assume a propagation speed

of 100 km/hour.

a. Suppose the caravan travels 150 km, beginning in front of one tollbooth,

passing through a second tollbooth, and finishing just after a third tollbooth.

What is the end-to-end delay?

Total Travel 150 Kms.

Speed 100 Km/Hr.

Total Transmission Delay 150/100 = 1.5 Hrs.

Time to service a single car 12 Sec.

Time to service 10 cars 120 Sec. = 2 Min.

Time to service 10 cars X (3 Toll booths) 360 Sec. = 6 Min.

End-to-End Delay 1.5 Hrs. + 6 Min = 1 Hr. 36 Min.

b. Repeat (a), now assuming that there are eight cars in the caravan instead of

ten.

Total Travel 150 Kms.

Speed 100 Km/Hr.

Total Transmission Delay 150/100 = 1.5 Hrs.

Time to service a single car 12 Sec.

Time to service 8 cars 96 Sec. = 1.6 Min.

Time to service 8 cars X (3 Toll booths) 288 Sec. = 4.8 Min.

End-to-End Delay 1.5 Hrs. + 4.8 Min = 1 Hr. 34 Min. 48 Sec.

P9. Consider the discussion in Section 1.3 of packet switching versus circuit

switching in which an example is provided with a 1 Mbps link. Users are

generating data at a rate of 100 kbps when busy, but are busy generating data

only with probability p = 0.1. Suppose that the 1 Mbps link is replaced by a 1

Gbps link.

a. What is N, the maximum number of users that can be supported

simultaneously under circuit switching?

Transmission Rate 1 Gbps.

Data Generation Rate (per user) 100 Kbps.

Maximum Number of Possible Users 1 Gbps. / 100 Kbps. = 10000 Users

b. Now consider packet switching and a user population of M users. Give a

formula (in terms of p, M, N) for the probability that more than N users are

sending data.

Complete Population of Users = M Transmitting Users = N Data Generation Probability of each of the Users = P = 0.1 The probability of exactly K users are using the network at a given time is:

KMK pPK

M

1

Hence, if - at least (N+1) users are sending at the same time, the probability will be:

M

Ni

iMi pPi

M

1

1

P12. A packet switch receives a packet and determines the outbound link to

which the packet should be forwarded. When the packet arrives, one other

packet is halfway done being transmitted on this outbound link and four other

packets are waiting to be transmitted. Packets are transmitted in order of

arrival. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is

the queuing delay for the packet? More generally, what is the queuing delay

when all packets have length L, the transmission rate is R, x bits of the currently-

being-transmitted packet have been transmitted, and n packets are already in

the queue?

The arriving packet has to wait for the link that is transmitting 4.5 packets = 6,750 Bytes = 54,000 Bits. Transmission Speed = 2 Mbps

Thus, Queuing Delay = 54,000

2= 27 .

As a rule, we can say that the General Queuing Delay = R

xLnL )(

P31. In modern packet-switched networks, including the Internet, the source host segments long, application-layer messages (for example, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into the original message. We refer to this process as message segmentation. Figure 1.27 illustrates the end-to-end transport of a message with and without message segmentation. Consider a message that is 8 106 bits long that is to be sent from source to destination in Figure 1.27. Suppose each link in the figure is 2 Mbps. Ignore propagation, queuing, and processing delays. a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from the source host to the first packet switch? Keeping in mind that each switch uses store-and-forward packet switching, what is the total time to move the message from source host to destination host? Message Length = 8 106 Bits Transmission Rate = 2 Mbps = 2 106 Bits/Sec.

Time for the message to be sent from source to the 1st switch = 8106

2106= 4 .

Since we have a Store-and-Forward packet switching, and 3 links. Therefore: The total time to move message from source host to destination host = 4 Sec x 3 Links = 12 Sec.

b. Now suppose that the message is segmented into 800 packets, with each packet being 10,000 bits long. How long does it take to move the first packet from source host to the first switch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the first switch. At what time will the second packet be fully received at the first switch?

Time to send 1st packet from source host to 1st packet switch = 1104

2106= 5 .

The 2nd packet will take the same time as the 1st one, so it would be = 2 5 = 10 .

c. How long does it take to move the file from source host to destination host when message segmentation is used? Compare this result with your answer in part (a) and comment. Time for 1st packet to be received at the destination host = 5 . 3 = 15. From there, each packet will take 5 mSec. to be received. Hence, the time for the 800th packet to be received is = 15 + (5 799) = 4.01 . It is obvious as we can observe that the delay of the message using segmentation has been significantly reduced by about 66.6 % one third of the time for sending without message segmentation.

d. In addition to reducing delay, what are reasons to use message segmentation? Couple of reasons can be found:

- If any failure of delivery message, only a small portion has to be retransmit instead of having the entire message to be resent.

- Segmentation comes of a benefit when the packet(s) encounters any congestions on its way.

- Without Message Segmentation the network will be facing some severe delays (depends on the sizes of transmitted messages). Each message will have to wait for the other to get through. Let us consider sending Videos vs. Text!

e. Discuss the drawbacks of message segmentation. - The total amount of packets' headers bytes is relatively large and more than the ones

we find in packets without segmentation. - If one segmented packet is missing, then the overall message cannot be read. - Packets has to be organized in sequence at the destination. - Consumes more bandwidth (Overhead).