CH - site.iugaza.edu.ps

Lecture # 2

CH.1

Computer Networks and the Internet

Islamic University of Gaza

Faculty of Engineering

Computer Engineering Department

Networks Discussion ECOM 4021

By

Eng. Wafaa Audah

Feb. 2013

Networks Discussion Eng. Wafaa Audah

(Theoretical material: page 2-7, Review questions and problems :8-end)

1.1 What is the Internet

End systems (hosts): computing devices that are interconnected by

the network like PC, server, cell phones, laptops.

End Systems are connected together by communication links and

packet switches.

Packet switches come in many shapes, but the two most famous types

in today's Internet are routers and link-layer switches.

Bandwidth mean transmission rate of data, measured by bit/sec.

End systems access the Internet through Internet Service Providers

(ISPs).

End systems and packet switches run protocols that control the

sending and receiving of information within the Internet. The Transmission Control Protocol (TCP) and the Internet Protocol (IP) are

two of the most important protocols in the Internet.

Protocol:

Computer network protocol

1.2 Network Edge

Network edges mean the applications and hosts.

Access network is the physical links that connect an end system to

the first router (also known as the "edge router") on a path from the

end system to any other distant end system.

A protocol defines the format and the order for

messages exchanged

between two or more

communicating entities, as well as the actions

taken on the transmission

and/or receipt of the

message.

2


Network core includes interconnected routers and network of

networks.

Information about the lasting of this section which include:

Internet access networks and physical media, are available at Dr.

slides :lecture 1 _slides 13-28 and textbook

1.3 Network Core

There are two fundamental approaches for moving data through a

network of links and switches: circuit switching and packet switching.

Circuit Switching:

- Dedicated link that represents end-end resources reserved for

"call" and dedicated resources which mean "no share".

- Network resources (e.g., bandwidth) divided into "pieces" *****

- Every user has a bandwidth as the following equation show:

Bandwidth one user = Total bandwidth / Number of users

- Represent guaranteed transmission (ensure: data arrival, correct

arrival and ordered data).

- Call setup required

- Resource piece idle if not used by owning call ……. Disadvantage.

3


***** Dividing link bandwidth into "pieces":

- Frequency division: Time is taken completely and frequency is

divided between users ( Every user take a part of frequency all the

time).

- Time division: Frequency is taken completely and time is divided into

slots that every user take a slot that is repeated periodically ( Every

user take the total frequency periodically)

Packet Switching:

- Data stream is divided into packets.

- No dedicated link that represents No end-end resources reserved.

- No dedicated resources which means "sharing" (resources used as needed).

- Network resources (bandwidth) are not divided into "pieces", each packet uses full link bandwidth.

- No guarantee of transmission (No ensuring for : data arrival, correct

arrival and ordered data).

4


1.4 Delay Loss and Throughput in Packet-Switched Networks

Dealing with packet switching means: probability of many types of delays

which are:

- Nodal processing: Time taken by the node itself in order to (e.g.) check bit errors or determine output link.

- Queuing delay: When packets arrival rate to the link exceeds

output link capacity, packets queue in router buffers period of

time, wait for their transmission chance.

- Transmission delay: The time needed to load all the packet into

the link, which is L/R (R=link bandwidth (bps), L=packet length

(bits).

- Propagation delay: The time needed for a packet to across the

link, which is d/s (d = length of physical link, s = propagation speed in medium.

- Store and forward delay: The time needed for the node (router)

to receive complete packet before forwarding it to another node.

Nodal Delay (End-to-End Delay)

Note:

When packets arrival rate to the link exceeds output link capacity: queuing

happenes, but if the queuing buffer is full, then packet loss will happen.

5


Throughput

Rate (bits/time) at which bits transferred between sender/receiver and it

measured by getting the Minimum **** of links rates between sender and

receiver.

Note: Minimum link rate will get the control of all transmission links

because it can allow transmission just at its rate even if there are higher

rated links, that means getting the throughput as its rate.

- The following figure describe the meaning of how minimum link rate

describe the throughput, suppose that every shape is a link and its area

describe its rate……… as a figure shows, smaller shape control whole system

because data are passed through it just according to its area which

represent rate.

1.5 Protocol layers, service models

- Internet protocol stack model provides reliable byte-stream

service between client and service processes.

link 1 link 2 link 3

Data

Data

Data

6


- ISO/OSI reference model

Note:

According to Internet protocol stack model

- Switch works at link and physical layers.

- Router works at network, link and physical layers.

For sections: 1.6, 1.7 see Dr. slides (Lecture 3_11-25) and textbook

7


Review Questions and Problems

Review Questions

R.11 What advantage does a circuit-switched network have over a

packet-switched network? What advantages does TDM have over FDM in a

circuit-switched network?

- A circuit-switched network can guarantee a certain amount of end-to-

end bandwidth for the duration of a call, but packet-switched

networks cannot make any end-to-end guarantees for bandwidth.

- In TDM each signal uses all of the bandwidth some of the time, TDM

provides greater flexibility and efficiency, by dynamically allocating

more time periods to the signals that need more of the bandwidth,

while reducing the time periods to those signals that do not need it.

FDM lacks this type of flexibility, as it cannot dynamically change the

width of the allocated frequency.

R.12 Why is it said that packet switching employs statistical

multiplexing? Contrast statistical multiplexing with the multiplexing that

takes place in TDM.

In a packet switched network, the packets from different sources flowing on

a link do not follow any fixed, pre-defined pattern. In TDM circuit switching,

each host gets the same slot in a revolving TDM frame.

R.15 Suppose users share a 2 Mbps link. Also suppose each user

transmits continuously at I Mbps when transmitting, but each user

transmits only 20 percent of the time.

a. When circuit switching is used, how many users can be supported?

b. For the remainder of this problem, suppose packet switching is used.

Why will there be essentially no queuing delay before the link if two or

fewer users transmit at the same time? Why will there be a queuing delay

if three users transmit at the same time?

c. Find the probability that a given user is transmitting.

8


R.19 Suppose Host A wants to send a large file to Host B. The path from

Host A to Host B has three links, of rates RI =500 kbps, R2 =2 Mbps, and

R) = I Mbps.

a. Assuming no other traffic in the network, what is the throughput for

the file transfer.

b. Suppose the file is 4 million bytes. Dividing the file size by the through

put, roughly how long will it take to transfer the file to Host B?

c. Repeat (a) and (b), but now with R2 reduced to 100 kbps.

processing delay, transmission delay, propagation delay and queuing delay.

All of these delays are fixed, except for the queuing delays, which are

variable according to the pressure of the packets on link.

R.16 Consider sending a packet from a source host to a destination host

over a fixed route. List the delay components in the end-to-end delay.

Which of these delays are constant and which are variable?

a) according to equation:

Bandwidth one user = Total bandwidth / Number of users

I M= 2 M/N N = 2 users can be supported.

b) Each user requires 1Mbps when transmitting, if two or fewer users

transmit simultaneously, a maximum of 2Mbps will be required. Since the

available bandwidth of the shared link is 2Mbps, there will be no queuing

delay before the link. Whereas, if three users transmit simultaneously, the

bandwidth required will be 3Mbps which is more than the available

bandwidth of the shared link. In this case, there will be queuing delay before

the link.

c) Probability that a given user is transmitting = 0.2 (each user transmits

only 20 percent of the time)

9


Problems

P.3 Consider the circuit-switched network in Figure 1.12. Recall that

there are n circuits on each link.

a. What is the maximum number of simultaneous connections that can

be in progress at any one time in this network?

b. Suppose that all connections are between the switch in the upper-

left-hand corner and the switch in the lower-right-hand corner. What is

the maximum number of simultaneous connections that can be in

progress?

a) throughput according to the previous information above_ page 6:

min (500k,2M,1M) bit/sec =500k bit/sec.

b) 500*10^3 sec

8*4*10^6 ? : 64 second

c) a-100k bit/sec (the same "min" approach), b- 100*10^3 sec

8*4*10^6 ? : 320 second

10


P.4 Review the car-caravan analogy in Section l.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?

a) We can have n connections between each of the four pairs of adjacent

switches. This gives a maximum of 4n connections.

b) We can n connections passing through the switch in the upper-right-hand

corner and another n connections passing through the switch in the lower-

left-hand corner, giving a total of 2n connections.

.

. .

.

.

. .

.

n

n

n n

n

n

n

n

2 n

2 n

11


Tollbooths are 75 km apart, and the cars propagate at 100km/hr. A tollbooth

services a car at a rate of one car every 12 seconds.

a) There are ten cars. It takes 120 seconds, or 2 minutes, for the first

tollbooth to service the 10 cars. Each of these cars has a propagation

delay of 45 minutes (travel 75 km) before arriving at the second

tollbooth. Thus, all the cars are lined up before the second tollbooth

after 47 minutes. The whole process repeats itself for traveling

between the second and third tollbooths. It also takes 2 minutes for

the third tollbooth to service the 10 cars. Thus the total delay is 96

minutes.

Explanation:

- Time _ start -stage1:

tr_10 cars loaded to link 1 (10*12sec=120sec=2min)+ tp

(75k/100k/hr=45min)= 47min.

- Time _ stage1-stage 2 (10 cars loaded to link 2): the same of the

previous work =47min

- Time_ stage2-end tr_10 cars passed tollbooth and loaded out of

it: 10*12sec=120sec=2min

- Total end to end delay = 47+47+2 (min) = 96 min

The following figure explain the work

Example

at

section

1.4

12


.

•1

75 •2

.

•3

P.6 In this problem we consider sending real-time voice from Host A to

Host B over a packet-switched network (VoIP). Host A converts analog

voice to a digital 64 kbps bit stream on the fl y. Host A then groups the

bits into 56-byte packets. There is one link between Host A and B; its

transmission rate is 2 Mbps and its propagation delay is 10 msec. As soon

as Host A gathers a packet, it sends it to Host B. As soon as Host B

receives an entire packet, it converts the packet's bits to an analog

signal. How much time elapses from the time a bit is created (from the

original analog signal at Host A) until the bit is decoded (as part of the

analog signal at Host B)?

Link 1 Link 2

start stage 1 stage 2 end

tp tp

tr _10cars tr _10cars

tr _10cars

caravan

End-to-end delay


The arriving packet must first wait for the link to transmit 6,750 bytes or

54,000 bits. Since these bits are transmitted at 2 Mbps, the queuing delay is

27 msec. Generally, the queuing delay is

- Waiting link to be free : (L-x)/R.

- Waiting n-queue packets: nL/R

- Total delay = (nL + (L - x))/R

P.11 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?

Consider the first bit in a packet. Before this bit can be transmitted, all of

the bits in the packet must be generated. This requires

1. The time required to make analog- to digital conversion

56*8/(64*10^3)sec=7msec.

2. The time required to transmit the packet is

56*8/(2*10^6)sec=224μsec.

3. Propagation delay = 10 msec.

The delay until decoding is

7msec +μ224sec + 10msec = 17.224msec

A similar analysis shows that all bits experience a delay of 17.224 msec.

L n packets in

queue

Queu

packet X L-x

A B

New packet

Link


P.24 Suppose two hosts, A and B, are separated by 20,000 kilometers

and are connected by a direct link of R = 2 Mbps. Suppose the

propagation speed over the link is 2.5 . 10H meters/sec.

a. Calculate the bandwidth-delay product, R . d prop '

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose

the file is sent continuously as one large message. What is the maximum

number of bits that will be in the link at any given time?

c. Provide an interpretation of the bandwidth-delay product.

d. What is the width (in meters) of a bit in the link? Is it longer than a

football field?

e. Derive a general expression for the width of a bit in terms of the

propagation speed s, the transmission rate R, and the length of the link

m.

The queuing delay is 0 for the first transmitted packet, L/R for the second

transmitted packet, and generally, (n-1)L/R for the nth transmitted packet.

Thus, the average delay for the N packets is

(L/R + 2L/R + ....... + (N-1)L/R)/N

= L/(RN) * (1 + 2 + ..... + (N-1))

= L/(RN) * N(N-1)/2

= LN(N-1)/(2RN)

= (N-1)L/(2R)

Note that here we used the well-known fact that

1 + 2 + ....... + N = N(N+1)/2

P.12 Suppose N packets arrive simultaneously to a link at which no

packets are currently being transmitted or queued. Each packet is of

length L and the link has transmission rate R. What is the average

queuing delay for the N packets?


a) 80,000,000 bits

b) 800,000 bits, this is because that the maximum number of bits that will

be in the link at any given time = min(bandwidth delay product, packet size)

= 800,000 bits. (the same concept of throughput measurement)

c) .25 meters

P.26 Consider problem P24 but now with a link of R =1 Gbps.

a. Calculate the bandwidth-delay product, R . d prop

b. Consider sending a file of 800,000 bits from Host A to Host B. Suppose

the file is sent continuously as one big message. What is the maximum

number of bits that will be in the link at any given time?

c. What is the width (in meters) of a bit in the link?

a) R* (D/V) = 2M *20,000 k / 2.5*10^8 = 160,000 bits

b) 160,000 bits because the link can only has a number of bits equals

bandwidth delay product.

c) The bandwidth-delay 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 / bandwidth-delay product, so 1 bit is

125 meters long.

e) s/R


P.30 In modern packet-switched networks, 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 1.28

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 1.28. 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?

b. Now suppose that the message is segmented into 4,000 packets, with

each packet being 2,000 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?

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.

d. Discuss the drawbacks of message segmentation.


See You at lecture 3

Best Wishes

a) Time to send message from source host to first packet switch =

(8*10^6)/(2*10^6)=4sec . With store-and-forward switching, the total time

to move message from source host to destination host = 3*4 sec=12 sec.

(three links has the same rate so the same time will be needed)

b) Time to send 1st packet from source host to first packet switch =

(2*10^3)/(2*10^6) =1msec

- Time at which 2nd packet is received at the first switch = time at which

1st packet is received at the second switch: 2*1msec=2msec

c) Time at which 1st packet is received at the destination host = 1msec*3=3

msec. After this, every 1msec one packet will be received; thus time at

which last (4000th) packet is received = 3msec+3999*1msec=4.002sec. It can

be seen that delay in using message segmentation is significantly less

(almost 1/3rd).

Detailed explanation ****

d) Drawbacks: i. Packets have to be put in sequence at the destination.

ii. Message segmentation results in many smaller packets. Since header size

is usually the same for all packets regardless of their size, with message

segmentation the total amount of header bytes is more.

**** Assume that we have 3-packets (small number instead of 4000)

As it clear, 2nd packet need just 1msec after the arrival of 1st packet

(4msec-3msec), third packet has the same thing. Meaning for 4000 packets:

1st packet needs 3msec to reach destination and all other packets need just

1msec because every packet needs 1msec after its previous packet reach

1msec

2msec

3msec

4msec

5msec

1st packet

2nd packet

3rd packet

S sw1 sw2 d

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