Network_Design_Issues yrgfw wdhf ewwdghghf e

CSE 306
Presented by: Dr. Amandeep Singh

Network Layer Design Issues
• Store-and-Forward Packet Switching
• Services Provided to the Transport Layer
• Implementation of Connectionless Service
• Implementation of Connection-Oriented Service
• Comparison of Virtual-Circuit and Datagram Subnets

Store-and-Forward Packet Switching
The environment of the network layer protocols.
fig 5-1

Services provided to Transport layer
The services need to be carefully designed with the following
goals in mind:
1. The services should be independent of the router
technology.
2. The transport layer should be shielded from the number,
type, and topology of the routers present.
3. The network addresses made available to the transport
layer should use a uniform numbering plan, even across
LANs and WANs.

Packet Switching
• Datagram Approach: Connectionless Service
• Virtual Circuit Approach: Connection Oriented
Service

POLL 1
• Datagram approach is
a) Connection oriented service
b) Connectionless service

Implementation of Connectionless Service
Routing within a diagram subnet.

Implementation of Connection-Oriented Service
Routing within a virtual-circuit subnet.

POLL 2
• Connection identifier is linked with
a) Connectionless service
b) Connection oriented Service
c) None

Comparison of Virtual-Circuit and Datagram
Subnets
5-4

POLL 3
• Which one of the following deals with transaction
processing system
• Datagram Approach
• Virtual Connection Approach

Network Layer Services
• Packetizing: encapsulating the payload(data received from upper
layer) at source and decapsulating at the destination.
• Routing: To find the best path from source to destination using
routing protocols.
• Forwarding: Action applied by each router when packet arrives at
one of its interface using routing or forwarding table.
• Routing and Forwarding are related to each other.

NETWORK-LAYER PERFORMANCE
The performance of a network can be measured in
terms of:
• Delay
• Throughput
• Packet loss

POLL 4
• Is Delay a parameter for network Performance
a) Yes
b) No
c) Don’t Know
d) Didn’t understood the question

Delay
The delays in a network can be divided into four
types:
• Transmission delay
• Propagation delay
• Processing delay
• Queuing delay.

• Transmission Delay
Delaytr = (Packet length) / (Transmission rate).
• Propagation Delay
Delaypg = (Distance) / (Propagation speed).
• Processing Delay
Delaypr = Time required to process a packet in a router or a destination host
• Queuing Delay
Delayqu = The time a packet waits in input and output queues in a router
• Total Delay
Total delay = (n + 1) (Delaytr + Delaypg + Delaypr) + (n) (Delayqu)

Throughput
• Throughput = minimum {TR1, TR2, . . . TRn}.

Packet Loss
• When a router receives a packet while processing another packet, the
received packet needs to be stored in the input buffer waiting for its
turn.
• A router, however, has an input buffer with a limited size. A time
may come when the buffer is full and the next packet needs to be
dropped.
• The effect of packet loss on the Internet network layer is that the
packet needs to be resent, which in turn may create overflow and
cause more packet loss.

An IPv4 address is 32 bits long.
Note

The IPv4 addresses are unique and universal.
Note

The address space of IPv4 is
232 or 4,294,967,296.

Figure Dotted-decimal notation and binary notation for an IPv4address

Change the following IPv4 addresses from binary notation to
dotted-decimal notation.
Solution
We replace each group of 8 bits with its equivalent decimal
number (see Appendix B) and add dots for separation.

Change the following IPv4 addresses from dotted-decimal
notation to binary notation.
Solution
We replace each decimal number with its binary equivalent
(see Appendix B).

Find the error, if any, in the following IPv4
addresses.

In classful addressing, the address space is divided into five classes: A, B, C,
D, and E.
Note

Figure Finding the classes in binary and dotted-decimalnotation

Find the class of each address.
a. 00000001 00001011 00001011 11101111
b. 11000001 10000011 00011011 11111111
c. 14.23.120.8
d. 252.5.15.111

Table Number of blocks and block size in classful IPv4addressing

In classful addressing, a large part of the available addresses were wasted.
Note

Table Default masks for classfuladdressing

Classful addressing, which is almost obsolete, is replaced with classless addressing.
Note

Figure A block of 16 addresses granted to a smallorganization

In IPv4 addressing, a block of addresses can be defined
as
x.y.z.t /n
in which x.y.z.t defines one of the addresses and the /n
defines the mask.

The first address in the block can be found by setting the rightmost 32 − n
bits to 0s.
Note

A block of addresses is granted to a small organization. We know that
one of the addresses is 205.16.37.39/28. What is the first address in
the block?
Solution
The binary representation of the given address is 11001101
00010000 00100101 00100111
If we set 32−28 rightmost bits to 0, we get
11001101 00010000 00100101 0010000
or 205.16.37.32.

The last address in the block can be found by setting the rightmost 32 − n
bits to 1s.
Note
19.3
8

Find the last address for the block in Last
Example
19.3
9
Solution
The binary representation of the given address is
11001101 00010000 00100101 00100111
If we set 32 − 28 rightmost bits to 1, we get 11001101
00010000 00100101 00101111
or 205.16.37.47

The number of addresses in the block can be found by using the formula 232−n.
Note

Find the number of addresses in
last example
Solution
The value of n is 28, which mean the
number of addresses is 2 32−28 or 16.

Another way to find the first address, the last
address, and
the number of addresses is to represent the mask as a 32-bit binary (or
8-digit hexadecimal) number. This is particularly useful when we are
writing a program to find these pieces of information. In Example the
/28 can be represented as
11111111 11111111 11111111 11110000
(twenty-eight 1s and four 0s).
Find
a.The first address
b.The last address
c.The number of addresses.

Solution
a. The first address can be found by ANDing the given
addresses with the mask. ANDing here is done bit by
bit. The result of ANDing 2 bits is 1 if both bits are 1s;
the result is 0 otherwise.

b. The last address can be found by ORing the given
addresses with the complement of the mask. ORing here
is done bit by bit. The result of ORing 2 bits is 0 if both
bits are 0s; the result is 1 otherwise. The complement
of a number is found by changing each 1 to 0 and each
0 to 1.

The number of addresses can be found by
decimal
complementing the mask, interpreting it as a
number, and adding 1 to it.

Figure A network configuration for the block
205.16.37.32/28

The first address in a block is normally not assigned to any
device;
it is used as the network address that represents the
organization to the rest of the world.

Figure Two levels of hierarchy in an IPv4 address

Figure A frame in a character-orientedprotocol

Each address in the block can be considered as a two-level hierarchical
structure: the leftmost n bits (prefix) define the network; the rightmost
32 − n bits define the host.

Figure Three-level hierarchy in an IPv4 address

• An ISP is granted a block of addresses starting with
190.100.0.0/16 (65,536 addresses). The ISP needs to
distribute these addresses to three groups of
customers as follows:
a. The first group has 64 customers; each needs 256
addresses.
b. The second group has 128 customers;
each needs 128 addresses.
c. The third group has 128customers; each needs 64
addresses.
Design the subblocks and find out how many addresses are
still available after these allocations.
Example

Example
Group 1
For this group, each customer needs 256 addresses. This
means that 8 (log2 256) bits are needed to define each host.
The prefix length is then 32 − 8 = 24. The addresses are

Example
Group2
For this group, each customer needs 128 addresses. This means that 7
(log2 128) bits are needed to define each host. The prefix length is
then 32 − 7 = 25. The addresses are

Example
For this group, each customer needs 64 addresses. This means that 6
(log264) bits are needed to each host. The prefix length is then 32 − 6 =
26. The addresses are
Number of granted addresses to the ISP: 65,536 Number of
allocated addresses by the ISP: 40,960 Number of available
addresses: 24,576

Figure An example of address allocation and distribution by anISP

Network_Design_Issues yrgfw wdhf ewwdghghf e

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