GATE CSE 2016 Set 2 | Question: 53

Question

A network has a data transmission bandwidth of $20 \times 10^{6}$ bits per second. It uses CSMA/CD in the MAC layer. The maximum signal propagation time from one node to another node is $40$ microseconds. The minimum size of a frame in the network is __________ bytes.

Comments

https://gateoverflow.in/3834/gate2005-it-71?show=312877#a312877

Check answer here for all similar questions.

Answer 1

Since, CSMA/CD
Transmission Delay = RTT

Hence,
$L=B  \times \text{ RTT}$
$\implies L=B  \times 2 \times T_{\text{propagation delay}}$
$\implies L=(20 \times 10^6) \times 2 \times 40 \times 10^{-6}$
$\qquad =20 \times 2\times 40$
$\qquad =1600 \text{ bits}$
$\qquad = 200\text{ bytes}$

Hence, 200 Bytes is the answer.

Comments

How we can know the answer is in bit or byte

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Read question carefully it is given in the question that answer should be in bytes

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because bandwidth is 20*10^6 bits

L=2* propagation delay * bw

L = 2* (40/10^6) * 20*10^6

L = 1600 bits(because of bandwidth in bits)

L=1600/8

L=200 Bytes

Answer 2

For minimum size of packet to detect the collison in CSMA/CD  --
Tt ≥ 2*Tp

L / B ≥ 2*Tp

L ≥ 2*Tp*B

L ≥ 2 * (40*10^-6) * (20*10^6)  ≥ 1600 bits ≥ 200 bytes

Answer 3

 

The following image explains the answer

Answer 4

Collision detection condition of CSMA/CD protocol is 

Transmission Time >= 2 x Propagation Time + Transmission Time of Jam signal

Here, we are not considering Jam signal. So, it’s transmission time will be considered as 0.

This collision detection condition puts a restriction on the size of frame being transmitted: 

Minimum Frame size  = 2 x Propagation delay x Bandwidth

Minimum Frame size  = 2*(40*10^-6)*(20*10^6) = 1600 bits = 200 Bytes.