AinShams University
Faculty of Engineering
CSE 351s: Computer
Networks
Section 1
Eng. Noha Wahdan
Class textbook:
ANDREW S. TANENBAUM, DAVID J.
WETHERALL, “Computer Networks”, 5th
edition, PRENTICE HALL, 2011.
Sheet 1: Question 2
An alternative to a LAN is simply a big timesharing system with terminals for
all users. Give two advantages of a client-server system using a LAN.
Sheet 1: Question 2
An alternative to a LAN is simply a big timesharing system with terminals for
all users. Give two advantages of a client-server system using a LAN.
Advantages:
➢ The LAN model can be grown incrementally.
➢ No single point of failure (if servers are replicated).
Internet structure: a “network of networks”
▪ hosts connect to Internet via access
Internet Service Providers (ISPs)
▪ access ISPs in turn must be
interconnected
mobile network
national or global ISP
• so that any two hosts (anywhere!)
can send packets to each other
▪ resulting network of networks is
very complex
• evolution driven by economics,
national policies
local or
regional ISP
home network
content
provider
network
datacenter
network
enterprise
network
Let’s take a stepwise approach to describe current Internet structure
Internet structure: a “network of networks”
Question: given millions of access ISPs, how to connect them together?
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
Introduction: 1-5
Internet structure: a “network of networks”
Question: given millions of access ISPs, how to connect them together?
access
net
access
net
access
net
access
net
access
net
access
net
access
net
connecting each access ISP to
each other directly doesn’t scale:
O(N2) connections.
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
Introduction: 1-6
Internet structure: a “network of networks”
Option: connect each access ISP to one global transit ISP?
Customer and provider ISPs have economic agreement.
access
net
access
net
access
net
access
net
access
net
access
net
access
net
global
ISP
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
access
net
Introduction: 1-7
Internet structure: a “network of networks”
But if one global ISP is viable business, there will be competitors ….
access
net
access
net
access
net
access
net
access
net
access
net
access
net
ISP A
ISP B
access
net
access
net
access
net
ISP C
access
net
access
net
access
net
access
net
access
net
access
net
Introduction: 1-8
Internet structure: a “network of networks”
But if one global ISP is viable business, there will be competitors …. who will
want to be connected
Internet exchange point
access
net
access
net
access
net
access
net
access
net
IXP
access
net
ISP A
access
net
ISP B
IXP
access
net
access
net
access
net
ISP C
access
net
peering link
access
net
access
net
access
net
access
net
access
net
Introduction: 1-9
Internet structure: a “network of networks”
… and regional networks may arise to connect access nets to ISPs
access
net
access
net
access
net
access
net
access
net
IXP
access
net
ISP A
access
net
ISP B
IXP
access
net
access
net
access
net
ISP C
access
net
access
net
regional ISP
access
net
access
net
access
net
access
net
Introduction: 1-10
Internet structure: a “network of networks”
… and content provider networks (e.g., Google, Microsoft, Akamai) may
run their own network, to bring services, content close to end users
access
net
access
net
access
net
access
net
access
net
IXP
access
net
ISP A
Content provider network
IXP
ISP B
access
net
access
net
access
net
access
net
ISP C
access
net
access
net
regional ISP
access
net
access
net
access
net
access
net
Introduction: 1-11
What’s a protocol?
mobile network
Protocols define the format, order of
messages sent and received among
network entities, and actions taken
on message transmission, receipt
4G
national or global ISP
Streaming
video
IP
Skype
local or
regional ISP
▪ protocols are everywhere
• control sending, receiving of
messages
• e.g., HTTP (Web), streaming video,
Skype, TCP, IP, WiFi, 4G, Ethernet
home network
content
provider
network
HTTP
datacenter
network
Ethernet
TCP
enterprise
network
WiFi
What’s a protocol?
A human protocol and a computer network protocol:
Hi
TCP connection
request
Hi
TCP connection
response
Got the
time?
GET http://gaia.cs.umass.edu/kurose_ross
2:00
<file>
time
Q: other human protocols?
Introduction: 1-13
Layered Internet protocol stack
▪ application: supporting network applications
• HTTP, IMAP, SMTP, DNS
▪ transport: process-process data transfer
• TCP, UDP
▪ network: routing of datagrams from source to
destination
• IP, routing protocols
▪ link: data transfer between neighboring
network elements
• Ethernet, 802.11 (WiFi), PPP
application
application
transport
transport
network
link
physical
▪ physical: bits “on the wire”
Introduction: 1-14
Sheet 1: Question 17
What is the main difference between TCP and UDP?
TCP is connection oriented with reliable transport between sending and receiving
process, whereas UDP is a connectionless service that does not provide reliability or
flow control.
Sheet 1: Question 12
Two networks each provide reliable connection-oriented service. One of
them offers a reliable byte stream and the other offers a reliable message
stream. Are these identical? If so, why is the distinction made? If not, give an
example of how they differ.
Message and byte streams are different. In a message stream, the network keeps
track of message boundaries. In a byte stream, it does not. For example, suppose a
process writes 1024 bytes to a connection and then a little later writes another 1024
bytes. The receiver then does a read for 2048 bytes. With a message stream, the
receiver will get two messages of 1024 bytes each. With a byte stream, the message
boundaries do not count, and the receiver will get the full 2048 bytes as a single unit.
The fact that there were originally two distinct messages is lost.
Encapsulation: an
end-end view
source
message
segment
M
Ht
M
datagram Hn Ht
M
frame
M
Hl Hn Ht
application
transport
network
link
physical
link
physical
switch
M
Ht
M
Hn Ht
M
Hl Hn Ht
M
destination
Hn Ht
M
application
transport
network
link
physical
Hl Hn Ht
M
network
link
physical
Hn Ht
M
router
Introduction: 1-17
Sheet 1: Question 10
What are two reasons for using layered protocols? What is one possible
disadvantage of using layered protocols?
Advantages:
➢ Layering makes it easier to design and manage each module (structured way to
discuss system components).
➢ Layering eases maintenance and updating of system.
Sheet 1: Question 10
What are two reasons for using layered protocols? What is one possible
disadvantage of using layered protocols?
Disadvantages:
➢ The performance of a layered system is likely to be worse than the performance of
a monolithic system, as one layer may duplicate lower-layer functionality. For
example, many protocol stacks provide error recovery on both a per-link basis
(data-link layer) and an end-to-end basis (Transport layer).
➢ Another drawback is that functionality at one layer may need information (for
example, a timestamp value) that is present only in another layer.
Sheet 1: Question 16
A system has an n-layer protocol hierarchy. Applications generate messages
of length M bytes. At each of the layers, an h-byte header is added. What
fraction of the network bandwidth is filled with headers?
Sheet 1: Question 16
A system has an n-layer protocol hierarchy. Applications generate messages
of length M bytes. At each of the layers, an h-byte header is added. What
fraction of the network bandwidth is filled with headers?
➢ Total header size = n × h = nh
➢ Total size of message with headers = M + nh
➢ Fraction of wasted bandwidth with headers = nh/(nh+M)
Sheet 1: Question 27
Suppose the algorithms used to implement the operations at layer k is
changed. How does this impact operations at layers k − 1 and k + 1?
It has no impact on the operations at layers k-1 or k+1.
Packet delay: four sources
A
____________
B
dproc
Processing
delay
dqueue
queueing
delay
▪ check bit errors
▪ determine o/p link
▪ typically < microsecs
▪ time waiting at output link
for transmission
▪ depends on congestion
level of router
dtrans
transmission
delay
dprop
propagation
delay
▪ d: length of physical link
▪ L: packet length (bits)
▪ R: link transmission rate ▪ s: propagation speed
(~2x108 m/sec)
(bps)
▪ dprop = d/s
▪ dtrans = L/R
dtotal = dproc + dqueue + dtrans + dprop
Sheet 1: Question 23
An image is 1600 × 1200 pixels with 3 bytes/pixel. Assume the image is
uncompressed. How long does it take to transmit it over a 56-kbps modem
channel? Over a 1-Mbps cable modem? Over a 10-Mbps Ethernet? Over
100-Mbps Ethernet? Over gigabit Ethernet?
Image size L = 1600 × 1200 × 3 = 5.76 × 106 bytes = 46.08 × 106 bits
Transmission delay = L / R
If R = 56× 103 bit/sec: delay = (46.08 × 106 ) / (56× 103) = 822.857 sec
If R = 1× 106 bit/sec: delay = (46.08 × 106 ) / (1× 106) = 46.08 sec
If R = 10× 106 bit/sec: delay = (46.08 × 106 ) / (10× 106) = 4.608 sec
If R = 100× 106 bit/sec: delay = (46.08 × 106 ) / (100× 106) = 0.4608 sec
If R = 1× 109 bit/sec: delay = (46.08 × 106 ) / (1× 109) = 0.04608 sec = 46.08 msec
Sheet 1: Question 25
List two advantages and two disadvantages of having international standards
for network protocols.
Advantages:
➢ Makes communication easier.
➢ The widespread use of any standard will give it economies of scale, as with VLSI
chips.
Sheet 1: Question 25
List two advantages and two disadvantages of having international standards
for network protocols.
Disadvantages:
➢ Once a standard has been widely adopted, it is difficult to change even if new and
better techniques or methods are discovered.
➢ By the time a standard has been accepted, it may be obsolete.
➢ The political compromises necessary to achieve standardization may lead to poor
standards.
Sheet 1: Question 20
When a file is transferred between two computers, two acknowledgement
strategies are possible. In the first one, the file is chopped up into packets,
which are individually acknowledged by the receiver, but the file transfer as a
whole is not acknowledged. In the second one, the packets are not
acknowledged individually, but the entire file is acknowledged when it
arrives. Discuss these two approaches.
Acknowledgement per packet
Acknowledgement per file
Higher number of acknowledgment messages. Saves network bandwidth by using a smaller
number of acknowledgment messages.
An efficient approach for unreliable
transmission lines as only the lost packet
would have to be re-transmitted.
An efficient approach for reliable transmission
lines, and inefficient for high-loss transmission
lines as it requires the entire file to be
retransmitted if even a single packet is lost.
Sheet 1: Question 21
Mobile phone network operators need to know where their subscribers’
mobile phones (hence their users) are located. Explain why this is bad for
users. Now give reasons why this is good for users.
Having mobile phone operators know the location of users lets the operators learn
much personal information about users, such as where they sleep, work, travel, and
shop.
This information might be sold to others or stolen. It could also let the government
monitor citizens.
On the other hand, knowing the location of the user lets the operator send help to
the right place in an emergency.
Introduction: 1-29
AND
OR
Introduc
tion: 130
Sheet 1: Question 9
A disadvantage of a broadcast subnet is the capacity wasted when multiple
hosts attempt to access the channel at the same time. As a simplistic example,
suppose that time is divided into discrete slots, with each of the n hosts
attempting to use the channel with probability p during each slot. What
fraction of the slots will be wasted due to collisions?
Probability of collision in a given slot
Node 1
1
1
Node 2
2
2
Node 3
3
C
1
1
2
3
E
C
S
E
C
3
E
S
S
C: collision
S: success
E: empty
Sheet 1: Question 9
A disadvantage of a broadcast subnet is the capacity wasted when
multiple hosts attempt to access the channel at the same time. As a
simplistic example, suppose that time is divided into discrete slots, with
each of the n hosts attempting to use the channel with probability p
during each slot. What fraction of the slots will be wasted due to
collisions?
For each time slot: σ 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑒𝑣𝑒𝑛𝑡𝑠 = 1
Probability of a collision = 1 – (Probability of successful transmission +
Probability of idle slot)
Probability of idle slot = Probability (Node_1 not transmitting and
Node_2 not transmitting …. and Node_n not transmitting)
Probability of idle slot = (1 – p) (1 – p ) …. (1 – p) for n nodes = (1 – p )n
Sheet 1: Question 9 (Continued)
• Probability of successful transmission for node_1
= Probability that node_1 will transmit and all other nodes not
transmitting = Probability (Node_1 transmitting and Node_2 not
transmitting and Node_3 not transmitting …. and Node_n not
transmitting) = p(1 – p)n-1
• Probability of successful transmission from any node = Probability
(successful transmission from Node_1 OR successful transmission
from Node_2 OR ….. OR successful transmission from Node_N)
= p(1 – p)n-1 + p(1 – p)n-1+ ………… + p(1 – p)n-1
= n×p (1-p)n-1 = np(1 – p)n-1
Sheet 1: Question 9 (Continued)
• Wasted fraction of slots
= Probability of a collision
= 1 – (Probability of successful transmission + Probability of idle slot)
= 1 – np(1 – p)n-1 – (1 – p )n
Notes if we have N users, and each of them can
transmit at any time with probability p:
• The probability that all users transmit at the same time =
Probability (User_1 transmitting and User_2 transmitting and User_3 transmitting ….
and User_N transmitting) = pN
• The probability that all users will not transmit at a given time =
Probability (User_1 not transmitting and User_2 not transmitting and User_3 not
transmitting …. and User_N not transmitting) = (1-p)N
• The probability that a specific user (e.g. 1st user) is transmitting while others are not =
Probability (User_1 transmitting and User_2 not transmitting and User_3 not
transmitting …. and User_N not transmitting) = p(1 – p)N-1
• The probability that any user is transmitting while others not =
Probability (User_1 only transmits OR User_2 only transmits OR ….. OR User_N only
transmits) = Probability (User_1 transmitting and all other users not transmitting) +
Probability (User_2 transmitting and all other users not transmitting) + …. + Probability
(User_N transmitting and all other users not transmitting) = N×p (1-p)N-1
Sheet 1: Question 9
A disadvantage of a broadcast subnet is the capacity wasted when multiple
hosts attempt to access the channel at the same time. As a simplistic example,
suppose that time is divided into discrete slots, with each of the n hosts
attempting to use the channel with probability p during each slot. What
fraction of the slots will be wasted due to collisions?
Probability of collision in a given slot
Node 1
1
1
Node 2
2
2
Node 3
3
C
1
1
2
3
E
C
S
E
C
3
E
Another Solution???
S
S
C: collision
S: success
E: empty
Sheet 1: Question 9 (Alternative solution)
A disadvantage of a broadcast subnet is the capacity wasted when
multiple hosts attempt to access the channel at the same time. As a
simplistic example, suppose that time is divided into discrete slots, with
each of the n hosts attempting to use the channel with probability p
during each slot. What fraction of the slots will be wasted due to
collisions?
Probability Collision
= P Any 2 nodes transmitting + P Any 3 nodes transmitting
+ P Any 4 nodes transmitting + ⋯ + P(N nodes transmitting)
𝑛
n
=෍
× 𝑝𝑖 × (1 − 𝑝)n−i
𝑖
𝑖=2
Thanks