This document contains solutions to problems related to computer networking concepts. For problem 1, the solutions discuss the advantages of datagram versus virtual circuit networks when routers frequently fail or fixed bandwidth is required between nodes. For subsequent problems, solutions provide forwarding tables for datagram and virtual circuit networks, address prefix matching, subnet masking, and looking up university IP address locations.

1. Name:Ahmed Osama Osman
ID number :11010214464
Date: November 19th
2014
Problem # 5 Solutions
Q1.
In this question, we consider some of the pros and
cons of virtual-circuit and datagram networks.
a.) Suppose that routers were subjected to conditions
that might cause them to fail fairly often. Would this
argue in favor of a VC or datagram architecture?
Why?
b.) Suppose that a source node and a destination
require that a fixed amount of capacity always be
available at all routers on the path between the source
and destination node, for the exclusive use of traffic
flowing between this source and destination node.
Would this argue in favor of a VC or datagram
architecture? Why?
c.) Suppose that the links and routers in the network
never fail and that routing paths used between all
source/destination pairs remains constant. In this

2. scenario, does a VC or datagram architecture have
more control traffic overhead? Why?
Solution:
a)
Datagram: In the case of connection-oriented network,
every router failure will involvethe routing of that
connection. At a minimum, this will require the router that
is upstreamfrom the failed router to establish a new
downstream part of the path to the destinationnode, with
all of the requisite signaling involved in setting up a path.
Moreover, all of therouters on the initial path that are
downstream from the failed node must take down thefailed
connection, with all of the requisite signaling involved to do
this. Whereas in thecase of connectionless datagram
network, no signaling is required to either set up a
newdownstream path or take down the old downstream
path.
b)
VC: In this case in order for a router to maintain an
available fixed amount of capacity onthe path between the
source and destination node for that source-destination
pair, it wouldneed to know the characteristics of the traffic
from all sessions passing through that link.That is, the
router must have per-session state in the router. This is
possible in aconnection-oriented network, but not with a
connectionless network. Thus, a connection-oriented VC
network would be preferable.

3. c)
In this scenario, datagram architecture has more control
traffic overhead. This is due tothe various packet headers
needed to route the datagrams through the network. But in
VC
Reference :
http://www.scribd.com/doc/221350919/Assignment-Chap-4#scribd
Q2.
Consider the network below.
a.) Suppose that this network is a datagram network.
Show the forwarding table in router A, such that all
traffic destined to host H3 is forwarded through
interface 3.
b.) Suppose that this network is
a11001000 10010001 01010001 01010101
11100001 01000000 11000011 00111100
11100001 10000000 00010001 01110111
datagram network. Can you write down a
forwarding table in router A, such that all traffic from
H1 destined to host H3 is forwarded through interface
3, while all traffic from H2 destined to host H3 is
forwarded through interface 4? (Hint: this is a trick
question.)
c.) Now suppose that this network is a virtual circuit
network and that there is one ongoing call between H1

4. and H3, and another ongoing call between H2 and H3.
Write down a forwarding table in router A, such that
all traffic from H1 destined to host H3 is forwarded
through interface 3, while all traffic from H2 destined
to host H3 is forwarded through interface 4.
d.) Assuming the same scenario as (c), write down the
forwarding tables in nodes B, C, and D.
Solution:
a. Destination Address H3
Link Interfac 3
b. No, because forwarding rule is only based on
destination address.
c.Incoming interface Incoming VC Outgoing Interface
1 12 3
2 63 4
outgoing vc 22 18

5. refrences:
file:///C:/Users/A/Downloads/CS132_EECS148_ProblemSet3_Solution.pdf
Q3.
Consider a datagram network using 32-bit host
addresses. Suppose a router has four links, numbered
0 through 3, and packets are to be forwarded to the
link interfaces as follows:
Destination Address Range Link
Interface
11100000 00000000 00000000 00000000
Through 0
11100000 00111111 11111111 11111111
11100000 01000000 00000000 00000000
Through 1
11100000 01000000 11111111 11111111
11100000 01000001 00000000 00000000
Through 2
11100001 01111111 11111111 11111111
Otherwise 3

6. a.) Provide a forwarding table that has five entries,
uses longest prefix matching, and forwards
packets to the correct link interfaces.
Prefix Match Link Interface
11100000 00 0
11100000 01000000 1
11100001 2
11100000 01000001 2
Otherwise 3
Reference :
http://www.scribd.com/doc/221350919/Assignment-Chap-4#scribd
b.) Describe how your forwarding table determines the
appropriate link interface for datagram’s with
destination addresses:
11001000 10010001 01010001 01010101
11100001 01000000 11000011 00111100
11100001 10000000 00010001 01110111
Solution:
a)
0
1
2
3
b)

7. Address Short
Answer
11001000 10010001 01010001
01010101
11100001 01000000 11000011
00111100
11100001 10000000 00010001
01110111
Q4.
Consider a router that interconnects three subnets:
Subnet 1, Subnet 2, and Subnet 3. Suppose all of the
interfaces in each of these three subnets are required
to have the prefix 223.1.17/24. Also suppose that
Subnet 1 is required to support at least 60 interfaces,
Subnet 2 is to support at least 90 interfaces, and
Subnet 3 is to support at least 12 interfaces. Provide
three network addresses (of the form a.b.c.d/x) that
satisfy these constraints.
Solution:
223.1.17/24 (Network ID) 11011111 00000001 00010001 ????????
Subnet 1 11011111 00000001 00010001 0???????
Subnet 2 11011111 00000001 00010001 10??????
Subnet 3 11011111 00000001 00010001 11??????
All 223.1.17/24
Subnet 1 223.1.17.0/25

8. Subnet 2 223.1.17.128/26
Subnet 3 223.1.17.192/26
Reference:
https://uniteng.com/wiki/lib/exe/fetch.php?media=classlog:computernetwork:hw4_report.pdf
Q5.
Consider a subnet with prefix 128.119.40.128/26. Give
an example of one IP address (of form
xxx.xxx.xxx.xxx) that can be assigned to this network.
Suppose an ISP owns the block of addresses of the
form 128.119.40.64/26. Suppose it wants to create four
subnets from this block, with each block having the
same number of IP addresses. What are the prefixes
(of form a.b.c.d/x) for the four subnets?
Solution:
IP address range: 128.119.40.128 ‐ 128.119.40.191
Note: 128.119.40.128(last 8 bits: 10000000, reserved
address), 128.119.40.191(last 8 bits:
10111111, reserved address, broadcast)
Four subnets:
128.119.40.64/28
128.119.40.80/28
128.119.40.96/28
128.119.40.112/28

9. Reference:
https://uniteng.com/wiki/lib/exe/fetch.php?media=classlog:computernetwork:hw4_report.pdf
Q6.
Use the whois service at the American Registry for
Internet Numbers(http://www.arin.net/whois) to
determine the IP address blocks for three universities.
Can the whois services be used to determine with
certainty the geographical location of a specific IP
address? Use www.maxmind.com to determine the
locations of the Web servers at each of these
universities.
Solution:
Geographic locations of where these IP addresses are not
tracked and provided by arin’s whois service since this service
only give contact and registration information, autonomous
system numbers (ASN), organizations or customers that are
associated with these resources, and related Points of Contact
(POC).
1. Arizona State University.
2. University of Arizona.
3. Franklin W. Olin College of Engineering
Reference:
https://uniteng.com/wiki/lib/exe/fetch.php?media=classlog:computernetwork:hw4_report.pdf