edited by
9,110 views
33 votes
33 votes

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.

edited by

5 Answers

Best answer
46 votes
46 votes

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.

edited by
4 votes
4 votes
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
0 votes
0 votes

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.

Answer:

Related questions