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Process-to-Process Delivery-Data Communication Systems-Assignment Solution, Exercises of Data Communication Systems and Computer Networks

Birla Institute of Technology and ScienceData Communication Systems and Computer Networks

This file contains solution to problems related Data Communication Systems. Mr. Prajin Ahuja assigned task at Birla Institute of Technology and Science. Its main points are: Models, Physical, Datalink, Network Transport, Applicaiton, Framing, Receiver, Damaged, Lost, Frames, Establishing, Presentation

Typology: Exercises

2011/2012

Uploaded on 07/26/2012

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CHAPTER 23
Process-to-Process Delivery:
Solutions to Odd-Numbered Review Questions and Exercises
Review Questions
1. Reliability is not of primary importance in applications such as echo, daytime,
BOOTP, TFTP and SNMP. In custom software, reliability can be built into the cli-
ent/server applications to provide a more reliable, low overhead service.
3. Port addresses do not need to be universally unique as long as each IP address/port
address pair uniquely identify a particular process running on a particular host. A
good example would be a network consisting of 50 hosts, each running echo server
software. Each server uses the well known port number 7, but the IP address,
together with the port number of 7, uniquely identify a particular server program
on a particular host. Port addresses are shorter than IP addresses because their
domain, a single system, is smaller than the domain of IP addresses, all systems on
the Internet.
5. The minimum size of a UDP datagram is 8 bytes at the transport layer and 28 bytes
at the IP layer. This size datagram would contain no data–only an IP header with
no options and a UDP header. The implementation may require padding.
7. The smallest amount of process data that can be encapsulated in a UDP datagram
is 0 bytes.
9. See Table 23.1.
Tab le 23. 1 Answer to the Question 9.
Fields in UDP Fields in TCP Explanation
Source Port Address Source Port Address
Destination Port Address Destination Port Address
Tot al L en gth There is no need for total length.
Checksum Checksum
Sequence Number UDP has no flow and error control.
Acknowledge Number UDP has no flow and error control.
Header Length UDP has no flow and error control.
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1

CHAPTER 23

Process-to-Process Delivery:

Solutions to Odd-Numbered Review Questions and Exercises

Review Questions

  1. Reliability is not of primary importance in applications such as echo, daytime, BOOTP, TFTP and SNMP. In custom software, reliability can be built into the cli- ent/server applications to provide a more reliable, low overhead service.
  2. Port addresses do not need to be universally unique as long as each IP address/port address pair uniquely identify a particular process running on a particular host. A good example would be a network consisting of 50 hosts, each running echo server software. Each server uses the well known port number 7, but the IP address, together with the port number of 7, uniquely identify a particular server program on a particular host. Port addresses are shorter than IP addresses because their domain, a single system, is smaller than the domain of IP addresses, all systems on the Internet.
  3. The minimum size of a UDP datagram is 8 bytes at the transport layer and 28 bytes at the IP layer. This size datagram would contain no data–only an IP header with no options and a UDP header. The implementation may require padding.
  4. The smallest amount of process data that can be encapsulated in a UDP datagram is 0 bytes.
  5. See Table 23.1.

Table 23.1 Answer to the Question 9.

Fields in UDP Fields in TCP Explanation Source Port Address Source Port Address Destination Port Address Destination Port Address Total Length There is no need for total length. Checksum Checksum Sequence Number UDP has no flow and error control. Acknowledge Number UDP has no flow and error control. Header Length UDP has no flow and error control.

a. None of the control bits are set. The segment is part of a data transmission with- out piggybacked acknowledgment. b. The FIN bit is set. This is a FIN segment request to terminate the connection. c. The ACK and the FIN bits are set. This is a FIN+ACK in response to a received FIN segment.

Exercises

  1. See Figure 23.1.
  2. The server would use the IP address 130.45.12.7 , combined with the well-known port number 69 for its source socket address and the IP address 14.90.90.33 , com- bined with an ephemeral port number as the destination socket address.
  3. 16 bytes of data / 24 bytes of total length = 0.
  4. 16 bytes of data / 72 byte minimum frame size = 0.
  5. It looks as if both the destination IP address and the destination port number are corrupted. TCP calculates the checksum and drops the segment.
  6. See Figure 23.2.
  7. Every second the counter is incremented by 64,000 × 2 = 128,000. The sequence number field is 32 bits long and can hold only 2^32 −1. So it takes (2 32 −1)/(128,000) seconds or 33,554 seconds.
  8. See Figure 23.3.
  9. The largest number in the sequence number field is 2 32 −1. If we start at 7000, it takes [(2^32 − 1) −7000] / 1,000,000 = 4295 seconds.

Reserved UDP has no flow and error control. Control UDP has no flow and error control. Window Size UDP has no flow and error control. Urgent Pointer UDP cannot handle urgent data. Options and Padding UDP uses no options.

Figure 23.1 Solution to Exercise 13

Table 23.1 Answer to the Question 9.

Fields in UDP Fields in TCP Explanation

69

0

52010

48

Data (40 bytes)

Figure 23.4 Solution to Exercise 31

Figure 23.5 Solution to Exercise 33

Before

After receiving ACK

After sending 1000 bytes

2001 3001 5000

2500 3001 6499

2500 4001 6499

OutOfOrder

Received

winSize lastACK

cumTSN cumTSN

(^87654321) To process

5 4

1800

18 17 16 14131211

16,

11,

2019