University of Windsor, Ontario Canada

88-556 Assignment No2 Computer Networks – Dr. Kemal Tepe

Carlos Eduardo Palacio 2/26/2008

88-556 Assignment No2 February 26, 2008

Table of Contents Figures .................................................................................................................................... 2 Tables ...................................................................................................................................... 2 Assignment Questionnaire ..................................................................................................... 3 Answers .................................................................................................................................. 4

Figures Figure 1 Four Way Handshake ................................................................................................ 7 Figure 2 Four Way Handshake optimized Host B ready ......................................................... 7

Tables Table 1 Packet Segmentation IP Header Data Length, Segment Offset, More Filed ............. 4 Table 2 IP 137.207.1.17/24 in Binary Octet............................................................................ 4 Table 3 Subnet 1 with 125 Hosts ............................................................................................ 4 Table 4 Subnet 2 with 60 Hosts .............................................................................................. 5 Table 5 Subnet 3 with 60 Hosts .............................................................................................. 5 Table 6 TCP Slow Start Tahoe with Threshold 16 ................................................................... 6 Table 7 TCP Slow Tahoe with no Threshold ........................................................................... 6 Table 8 Estimated RTT and Possible Time Out base on Sampled RTT.................................... 7

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88-556 Assignment No2 February 26, 2008

Assignment Questionnaire 1. Consider a router needs to send 3000 byte datagram into a link whose MTU is 500 Bytes. If the original datagram’s ID is 422, please find data length, segment offset and more field of IP header of each datagram after fragmentation? 2. Consider a router that interconnects three subnets. The administrator of the network has 137.207.1.17/24 as network address. If the first subnet has 125 hosts, and second and third have 60 hosts. Provide network addresses ranges of these three sub networks and their subnet masks? 3. Consider sending a large file from host A to host B. Assume the maximum segment size (MSS) is 1454 Bytes. a. What is the maximum value of the file that TCP sequence numbers are not exhausted? b. How long does it take to send this file over 1 mega bit per second (Mbps) link? (Do not consider about flow control problems). 4. If there are 100 segments that need to be transferred from host A to host B, and TCP employs slow-start Tahoe congestion control with threshold of 16 segments. Assuming there are no missed ACKs, how long does it take to transfer the segments from host A to host B if: a. RTT is 2 sec, and segment transmission time is 0.25 seconds? b. If there is no threshold in the congestion scheme? c. Instead of dynamic congestion window, TCP employs static congestion window of 2? 5. If the sampled RTTs in the host are (1.6, 2.0, 2.2., 1.5, 1.8) milliseconds (ms). Find the estimated RTT and possible timeout periods for α=0.875 6. Explain how four-way handshake works in the TCP connection termination?

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88-556 Assignment No2 February 26, 2008

Answers 1. The original datagram = 3000 Bytes, this means that there is already one header but is not include n the datagram. Is important that we will have to include headers for each segment that means that the capacity of data will be reduce by 20 octet o 20 Bytes. The MTU (maximum transmission unit) = 500 Bytes.

Table 1 Packet Segmentation IP Header Data Length, Segment Offset, More Filed

Packet Number 1 2 3 4 5 6 7

Data Length (octets)

Header Length

480 octets 480 octets 480 octets 480 octets 480 octets 480 octets 120 octets

Segment Offset

20 octets 20 octets 20 octets 20 octets 20 octets 20 octets 20 octets

More Field

0 60 *64 bit 120*64 bit 180*64 bit 240*64 bit 300*64 bit 360*64 bit

Bytes of the Datagram Left

1 1 1 1 1 1 0

2520 2040 1560 1080 600 120 0

2. Router interconnects three subnets and the Administrator have a network address IP 137.207.1.17/24. First subnet will have 125 hosts, Second and Third 60 hosts. Table 2 IP 137.207.1.17/24 in Binary Octet

.

.

137

.

207

1

0

0

0 1

0 0

1

1

2

3

4 5

6 7

8

.

1 9

1 1 0

0 1 1

0 1 2

1

1 1 3

1 1 4

1 1 5

1 1 6

.

0 1 7

0 1 8

0 1 9

0 2 0

17 0 2 1

Network ID

Number of Hosts IP Range for the Host Subnet Mask Gateway Broadcast

0 2 3

1 2 4

.

0

0

0

1

0

Host ID

Table 3 Subnet 1 with 125 Hosts

Subnet

0 2 2

1 125 137.207.1.1 137.207.1.126 255.255.255.128 137.207.1.0 137.207.1.127

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0

0

1

88-556 Assignment No2 February 26, 2008 Table 4 Subnet 2 with 60 Hosts

Subnet

2

Number of Hosts

60

IP Range for the Host Subnet Mask

137.207.1.129 137.207.1.190 255.255.255.192

Gateway

137.207.1.128

Broadcast

137.207.1.191

Table 5 Subnet 3 with 60 Hosts

Subnet Number of Hosts IP Range for the Host Subnet Mask

2 60 137.207.1.193 137.207.1.254 255.255.255.128

Gateway

137.207.1.192

Broadcast

137.207.1.255

3. Consider sending a large file and the Maximum segment size (MSS) is 1454 Bytes. a. Maximum Value of File that TCP sequence numbers are not exhausted, is related with the sequence number capacity in the TCP header that is 2 . That in terms of how many numbers can generate starting from 0 is 2  42,94,967,296 and how big should be the file if we want to use all the sequence numbers will be 2  1454   6,244,882,448,384  b. Link of 1 Mbps (mega per second) = 1024 bits per second, and the MSS in bits is 1454  8  11632  for 1 MSS we have   11632    11,36     1024 /  But we need to transfer 2 MSS that will be the same if we say that we need to 6,244,882,448,384  42,94,967,296 !""  11,36    48,790,828,482.56    1547.14 years 4. 100 segments to transfer from ) *  TCP Slow start using Tahoe with Threshold 16 segments. a. RTT is 2 seconds and Segment transmission time is 0.25 seconds.

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88-556 Assignment No2 February 26, 2008 Table 6 TCP Slow Start Tahoe with Threshold 16

Window

Packets Send

Packets to Send

2 4 8 16 17 18 19 20

2 6 14 30 47 65 84 100

Comment

98 94 86 70 Threshold 53 35 16 0 Window is 20 but only 16 used

Time RTT

Segment Transmission

Total Time

2 2 2 2 2 2 2 2

0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25

2.25 4.5 6.75 9 11.25 13.5 15.75 18

Time RTT

Segment Transmission

Total Time

2 2 2 2 2 2

0.25 0.25 0.25 0.25 0.25 0.25

2.25 4.5 6.75 9 11.25 13.5

b. No threshold Table 7 TCP Slow Tahoe with no Threshold

Window

Packets Send

Packets to Send

2 4 8 16 32 64

2 6 14 30 62 100

Comment

98 94 86 70 38 0 Window is 64 but only 38 used

c. Static Window = 2 This means 100 packets and you send 2 each time = 50 transmissions +50 ,-.  +2   / 0.25  00  112.5   5. If the Sampled RTT in the host are (1.6, 2.0, 2.2, 1.5, 1.8) α= 0.875 1.233456  7α  Estimated>??@AB C / ++1 D α0  Sample>?? 0

HIJKL  M+1 D N 0  HI233O P / +N  Q".R233 D 1.233 Q0 ,. S -I  1.233 / +4  HI233 0

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88-556 Assignment No2 February 26, 2008 Table 8 Estimated RTT and Possible Time Out base on Sampled RTT

Sample 1 2 3 4 5

T

U

0.875 0.875 0.875 0.875 0.875

0.25 0.25 0.25 0.25 0.25

Sample RTT

Estimated RTT Old

1.6 2 2.2 1.5 1.8

0.0000 0.2000 0.4250 0.6469 0.7535

Estimated RTT New 0.2000 0.4250 0.6469 0.7535 0.8843

Dev(K) Dev(K+1) 0.0000 0.3500 0.6563 0.8805 0.8470

0.3500 0.6563 0.8805 0.8470 0.8641

Time out Interval 1.6000 3.0500 4.1688 4.1414 4.3409

6. Four way Handshake algorithm • Host A Send Termination i request to Host B. • Host B Send ACK for the i request to Host A. • Host B Send Termination j request to Host A. • Host A Send ACK for the j request to Host B.

Figure 1 Four Way Handshake

Figure 2 Four Way Handshake optimized Host B ready

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