PROBLEMS

Module 05: Performance Metrics

RTT (Round Trip Time)

1. Calculate the total time required transfer a 1.5 MB file in the following cases, assuming

an RTT of 80 ms, a packet size of 1-KB data and an initial 2xRTT of handshaking before

data is sent.

a) The bandwidth is 10 Mbps, and data packets can be sent continuously.

b) The bandwidth is 10Mbps, but after we finish sending each data packet we must wait

one RTT before sending the next.

c) The link allows infinitely fast transmit, but limits bandwidth such that only 20 packets

can be sent per RTT.

Solution:

a) 1.5 MB = 12,582,912 bits. 2 initial RTT’s (160 ms) + 12,582,912/10,000,000 bps

(transmit) + RTT/2 (propagation) ≈ 1.458 seconds.

b) Number of packets required = 1.5 MB/1 KB = 1,536. To the above we add the time

for 1,535 RTTs (the number of RTTs between when packet 1 arrives and packet

1,536 arrives), for a total of 1.458+122.8 = 124.258 seconds.

c) Dividing the 1,536 packets by 20 gives 76.8. This will take 76.5 RTTs (half an RTT

for the first batch to arrive, plus 76 RTTs between the first batch and the 77th partial

batch), plus the initial 2 RTTs, for 6.28 seconds.

2. Consider a point to point link 55 km in length. At what bandwidth would propagation

delay equal transmit delay for 100 byte packets?

Solution: Propagation delay is 55 × 103 m/ (2 × 108 m/sec) = 275 µs

800 bits/275 µs is 2.91 Mbits/sec.

3. Calculate the delay x bandwidth product for a 45-Mbps link with a delay 50 ms.

Solution: delay x bandwidth = 50 x 10 -3 sec x 45 x 106 bits/sec

= 2.25 x 10 6 bits

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4. Calculate the minimum RTT and delay x bandwidth product for a 128-Kbps point-to-

point link that is set up between two nodes A and B. The distance from A to B is 55 Gm

and data travels over the link at the speed of light- 3x108 m/sec.

Solution:

Propagation delay on the link is (55 × 109)/(3 × 108) = 184 seconds.

Thus the RTT is 368 seconds.

Module 11: Error Detection

CRC

1. For the given message M=101010. Check whether the transmitted message is error free

or not using CRC algorithm.

Step1: Compute M(x) * xk

Step2: Divide T(x) by C(x)

Step3: Find remainder T(x) / C(x) = R(x)

Step4: subtract T(x) – R(x) = D(x) [D(x) is exactly divisible by C(x)]

Step5: Transmit D(x) M(x) = 101010

C(x) = x3 + x1 (1010)

Generator

10001

1010 101010000

1010

1000

1010

00100

-Message

-Remainder

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Message transmitted is:

101010100 is transmitted

101010100 is exactly divisible by 1010. (Error Free).

Module 26: Bridges and Switches

Delay diagram for Datagram circuits.

P1

P1

P2

Packet

P2

Source P2 Destination

For example we divide the message into 3 packets, so k=3 then the equation becomes

=Lt+LP+(k-1)P

=3t+3P+(3-1)P here number of hops =3, 2 switches

=3t+3P+2P between source and destination

=3t+5P

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T

Source

1 2 3

Switch1

1 2 3

Switch2

1 2 3

Destination

t P t P t P P P = 3t+5P

Module 27: Switches and Connectionless

Find the shortest path in the network Using Spanning tree algorithm

A

EB3

B B4 B5D F

B2

C G

B1

H

Fig: Extended LAN with Loops

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Focusing on the node B3:

B3 receives (B2,0,B2)

Since 2>3 , B3 accepts B2 as root

B3 adds 1 to the distance advertised by B2 (0) and sends (B2,1,B3) towards B5

B2 accepts B1 as root and sends (B1,1,B2) towards B3

B5 accepts B1 as root and sends (B1,1,B5) towards B3

stops forwardingB3 accepts B1 as root and notes that B2 and B5 are closer to root. Thus B3

Hence the above figure is modified as follows:

A

EB3

B

B22

B4

D B5 F

C G

B1

H

Fig: Spanning tree with some ports not selected

Module31: Congestion Control

Leaky Bucket Algorithm

Calculate the maximum output burst using token bucket algorithm for the following data C = 250 Kb, M= 25 Mbps,r = 2 Mbps .

Using token bucket algorithm we know that,Maximum output burst = C + rS =MS

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And S=C/M-rS – Burst lengthM – Maximum output rateS – Maximum byte in lengthsr– Token arrival rateC – Capacity of token bucket in byte

So, S=C/M-r=250/ (25-2)

S=11mlMaximum output burst = C + rS =MS

=25*11

Maximum output burst =275

Module 32: Routing Algorithms

Dijktra’s Algorithm

Consider the network,

1 2 3

5 2 1

63 4

1 3 2

52 4

For source =1

Iteration N 2 3 4 5 6

Initial {1} 3 2 5 ∞ ∞

1. {1,3} 3 2 4 ∞ 3

2. {1,2,3} 3 2 4 7 3

3. {1,2,3,6} 3 2 4 5 3

4. {1,2,3,4,6} 3 2 4 5 3

5 {1,2,3,4,5,6} 3 2 4 5 3

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The spanning tree for this network is

1 2 32 1

63 4

2

2 5

Distance Vector Algorithm

Consider the network,

1 2 3

5 2 1

63 4

1 3 2

2 54

For destination=6

Iteration 1 2 3 4 5

Initial (-1, ∞) (-1, ∞) (-1, ∞) (-1, ∞) (-1, ∞)

Indicates initially destination n node thinks that all other nodes are at ‘∞’ distance from me

1. (-1, ∞) (-1, ∞) (6, 1) (-1, ∞) (6, 2)

2. (3, 3) (5, 6) (6, 1) (3, 3) (6, 2)

3. (3, 3) (4,4 ) (6, 1) (3, 3) (6, 2)

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4. (3, 3) (4,4 ) (6, 1) (3, 3) (6, 2)

The spanning tree for this network is

1 2 3

2 1

64

2

12 5

Module 35: BGP

IP-Address

A block of address es is granted to a small organization. We know that one of the addresses is205.16.37.39/28. What is the first address, last address, and the number of addresses.

The binary representation of the given address is

11001101 00010000 00100101 00100111

i. Mask is 28, so 32-28=4

We should make 4 right most bits to 0, we get

11001101 00010000 00100101 00100000

So the starting block address is 205.16.37.32

ii. Mask is 28, so 32-28=4

We should make 4 right most bits to 1, we get

11001101 00010000 00100101 00101111

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So the last block address is 205.16.37.47

iii. We know that , 232-n

232-28 =24 = 16 addresses

Module 43: Network Security

RSA Algorithm

Encrypt the following using RSA algorithm

i) P=5,q=11,e=7,P=18

n=p*q

=5*11

n =55

φ = (p-1)*(q-1)

=4*10

φ =40

Given encryption key e=7

Compute‘d’ such that e*d=1mod φ

7*d=1 mod 40

7*23=1mod40 (7*23=161 if we divide it by 40 we will get remainder as 1)

d=23

We know that, Encryption can be done using

C=Memod n

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=187 mod 55

= (184 (182 (18 mod 55)))

=36*49*18 mod 55

C=17

We know that, Decryption can be done using

M=Cd mod n

=1723 mod 55

= (1716 (174 (172 (17 mod 55)))

M=18

ii) a=3,q=11,x=3,M=18

n=p*q

=3*11

n =33

φ = (p-1)*(q-1)

=2*10

φ =20

Given encryption key e3

Compute ‘d’ such that e*d=1mod φ

3*d=1 mod 20

3*7=1mod20 (3*7=21 if we divide it by 20 we will get remainder as 1)

d=7

We know that, Encryption can be done using

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C=Memod n

=93 mod 33

= (92 (9 mod 33))

C=9

We know that, Decryption can be done using

M=Cd mod n

=97 mod 33

= (94 (92 (9 mod 33)))

M=9

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