Nt1310 Unit 7 Data Communication And Computer Networks

DATA COMMUNICATION AND COMPUTER NETWORKS

HOMEWORK – 2

Chapter 2

Problem-10:

Consider a short, 10-meter link, over which a sender can transmit at a rate of 150 bits/sec in both directions. Suppose that packets containing data are 100,000 bits long, and packets containing only control (e.g., ACK or handshaking) are 200 bits long. Assume that N parallel connections each get 1/N of the link bandwidth. Now consider the HTTP protocol, and suppose that each downloaded object is 100 Kbits long, and that the initial downloaded object contains 10 referenced objects from the same sender. Would parallel downloads via parallel instances of non-persistent HTTP make sense in this case? Now consider persistent HTTP. Do you expect significant gains over the

non-persistent case? Justify and explain your answer.

Sol:

Given length of link =10 meters.

Also, 150 bits/second can be transmitted to and fro.

This is the case where each downloaded object can be completely put into one data packet.

Let P denote the one-way propagation delay between two communication party’s sender and receiver.

In the first case,

This is the case of Parallel downloads via non-persistent connections.

Parallel download would allow 10 connections share the 150 bits/sec bandwidth.

Therefore, each gets just 15 bits/sec.

Thus, the total time needed to receive all objects is given by:

(200/150+P + 200/150 +P + 200/150+P + 100,000/150+ P) + (200/ (150/10) +P + 200/ (150/10) + P + 200/ (150/10) +P + 100,000/ (150/10) +P)

= 7377 + 8*P (seconds)

In the second case,

This is the case of persistent HTTP connection case. The total time needed is given by:

(200/150+P + 200/150 +P + 200/150+P + 100,000/150+ P)+ 10*(200/150+P + 100,000/150+ P)

=7351 + 24*P (seconds)

We have speed of light = 3*106^8m/sec,

So, P = 10/ (3*10^8) = 0.03 microsecond.

But P is negligible compared with transmission delay T. Therefore, with parallel downloads, we can observe that the persistent HTTP does not have significant gain (less than 1 percent) over the non-persistent case.

Problem-18:

a. What is a whois database?
Sol:
A whois database is a querying database which is a query and response protocol that mainly stores as well as displays registered users and assignees on internet that publicly displays domain names, IP addresses. Also, it is the database which more clearly displays address information along with creation and expiry dates.

b. Use various whois databases on the Internet to obtain the names of two DNS servers. Indicate which whois databases you used.
Sol:
Various whois databases I found were
ARIN, RIPE, APNIC, LACNIC, INTERNIC
There are few DNS servers associated with above:
1. ICANN
2. Google
3. Bing
4. Yahoo

c. Use nslookup on your local host to send DNS queries to three DNS servers: your local DNS server and the two DNS servers you found in part (b). Try querying for Type A, NS, and MX reports. Summarize your findings.
Sol:
Using nslookup to send DNS queries,

DNS Servers obtained in part (b) are, ICANN, Google, Bing and Yahoo.

Querying on local DNS Server, iupui.edu

When I tried querying Type A, NS and MX reports the following results are obtained.
For Type A, the server name and its IP address is obtained for iupui.edu.
For NS – Name server, the name servers,
dns2.iu.edu, dns1.iu.edu, dns1.illinois.edu
For MX reports, we obtained the mail exchanger preference which is a value and mail-exchanger like exchanger-relay.iupui.edu

d.

Use nslookup to find a Web server that has multiple IP addresses. Does the Web server of your institution (school or company) have multiple IP addresses?

Sol:

From the above part (b), Yahoo.com and google.com are the web servers that have multiple IP addresses

Also, the web server of my school has multiple IP Addresses (two addresses) as shown below,

e. Use the ARIN whois database to determine the IP address range used by your university.

Sol:

The address range used by our university IUPUI is,

1. IUPUI Net: 134.68.0.0-134.68.255.255

2. IUPUI Net2: 149.166.0.0-149.166.255.255

f. Describe how an attacker can use whois databases and the nslookup tool to perform reconnaissance on an institution before launching an attack.

Sol:

An attacker using whois database and nslookup can find out each and every IP Address and also the target IP Addresses the company is using by port mapping during or before his attack with their associated DNS servers.

G. Discuss why whois databases should be publicly available.

Sol:
Whois databases should be publicly available because they are the main reference to find and store records of registered users and assignees on internet resources like domain names, IP addresses, autonomous systems and also finding creation and expiration dates of those IP address blocks.

Problem-26:
Suppose Bob joins a Bit Torrent, but he doesn’t want to upload any data to any other peers (so called free-riding).
a. Bob claims that he can receive a complete copy of the file that is shared by the swarm. Is Bob’s claim possible? Why or why not?

Sol:

Peer-to-peer system: It is the system which consists of number of peers and at any point of time any peer can act as a centralized server depending upon the target resource they are holding.

Therefore it is true that Bob can receive a complete copy of file that is shared by swarm because a peer-to-peer connection depends upon how many peers present and as long as there are enough number of peers, Bob can receive data.

b. Bob further claims that he can further make his “free-riding” more efficient by using a collection of multiple computers (with distinct IP addresses) in the computer lab in his department. How can he do that?

Sol:
The peer-to-peer connection consists of multiple computers with distinct IP addresses which act as peers and each PC can act as a client or server at any point of time depending upon the resource to be shared.
So, Bob can run a client on each machine, and allow each computer to “free-ride” the torrent so that he can get the required files.

Chapter 3

Problem-14:

Consider a reliable data transfer protocol that uses only negative acknowledgments.
Suppose the sender sends data only infrequently. Would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot of data to send and the end-to-end connection experiences few losses. In this second case, would a NAK-only protocol be preferable to a protocol that uses ACKs? Why?

Sol:

For the first case where data is sent infrequently, a NAK-only protocol would not be preferable to a protocol that uses ACK’s.

NAK and ACK are negative and positive acknowledgments sent in communication protocols to indicate the status of message delivery. NAK is sent when there is loss of a packet due to error or delay of message or packet.
Consider a packet p sent by sender to receiver. In a NAK-only protocol, the loss of packet p cannot be realized until packet p+1 is correctly received. The loss of packet p can be realized when packets p-1 and p+1 are received at the receiver’s side correctly. So when there is loss of packet x due to delay or some error, a NAK is sent for retransmission where it takes lot of time to recover packet p.

The second case where the data is sent too often, a NAK-only protocol would be preferable.

As the data is sent frequently, the interval between two packets is small. Also, the error occurrence in delivery of packets is minor for such a case. Therefore, a NAK-only protocol best suits this case because it saves lot of time avoiding sending ACK’s for each and every frequently received packets.

Problem-23:

Consider the GBN and SR protocols. Suppose the sequence number space is of size k. What is the largest allowable sender window that will avoid the occurrence of problems such as that in Figure 3.27 for each of these protocols?

Solution:

In this question, we are supposed to sort out the sequence number space, such that it should be large enough to fit the entire receiver and the sender window without any overlapping condition. The condition is that from figure 3.27 we have to avoid the leading edge of the receiver’s window wrapped around the sequence number space overlap with the trailing edge of the sender’s window.

Here,
Leading edge: One with “highest” sequence number.
Trailing edge: One with “lowest” sequence number.
W = Size of window.

So, we should determine the large range of sequence number that can be covered by both sender and receiver window.

Assuming that receiver is waiting for packet T which has the lowest sequence number. In this case,
Window range = [T, T+w-1].
Before this, the receiver has received packets T-1 and w-1 and if the acknowledgments have not yet been received by the sender, then the values [m-w, m-1] may still be extending back. In this case, if no ACK’s have been received by the sender,
Senders window = [T-w, T-1].
Therefore, the trailing and leading edges of the sender’s and receiver’s window = T-w and T+w-1.

So, to avoid the overlapping condition of leading edge and trailing edges of sender’s and receiver’s window, the sequence number space must be large enough to accommodate sequence numbers which is twice the size of w (2w).

Problem-33:

In Section 3.5.3, we discussed TCP’s estimation of RTT. Why do you think TCP avoids measuring the SampleRTT for retransmitted segments?

Sol:

If TCP doesn’t avoid measuring the SampleRTT for retransmitted packets, an incorrect value of SampleRTT would be obtained. This is due to miss assumption of the acknowledgement received on the sender side.
Consider a case when a sender sends packet M and this is delayed due to some error. The source again sends the copy of same file with other packet P. Now packet p is the retransmitted packet and source calculates SampleRTT for packet P. After this, if an acknowledgement for the first packet M arrives, the source will mistakenly assumes that it is ACK of packet P and starts calculating the value of SampleRTT which is the incorrect value.
Therefore, it is better to avoid considering the retransmitted packets for calculating sampleRTT.