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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.

### 1 comment

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

Check answer here for all similar questions.

## 5 Answers

Best answer

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.

### 3 Comments

How we can know the answer is in bit or byte
Read question carefully it is given in the question that answer should be in bytes
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
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

The following image explains the answer

by

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.

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