Transfer rate =2 *10^8 m/s

Time to transfer =(4*10^3)/(2*10^8)=2*10^-5 sec

Data rate = 10^7 bps

Packet size = 2*10^-5*10^7 bits = 200 bits

so its option D ....

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+17 votes

A $2$ $km$ long broadcast LAN has $10^7$ bps bandwidth and uses CSMA/CD. The signal travels along the wire at $2 \times 10^8$ m/s. What is the minimum packet size that can be used on this network?

- $50$ $\text{bytes}$
- $100$ $\text{bytes}$
- $200$ $\text{bytes}$
- None of the above

+28 votes

Best answer

In CSMA/CD, to detect a collision the transmission time (which depends on the

packet size) must be greater than twice the propagation delay.

Propagation delay here = $\dfrac{2 km}{ 2 \times 10^8 m/s}=10 \text{ microseconds}$

Now, transmission time for $x$ bytes = $\dfrac{x \times 8}{10^7} = 0.8x\text{ microseconds}$

So, $0.8x > 2 \times 10 \implies x > 25 \text{ bytes}$

So, None of these.

packet size) must be greater than twice the propagation delay.

Propagation delay here = $\dfrac{2 km}{ 2 \times 10^8 m/s}=10 \text{ microseconds}$

Now, transmission time for $x$ bytes = $\dfrac{x \times 8}{10^7} = 0.8x\text{ microseconds}$

So, $0.8x > 2 \times 10 \implies x > 25 \text{ bytes}$

So, None of these.

+3

Total distance for RTT = 4 Km

Transfer rate =2 *10^8 m/s

Time to transfer =(4*10^3)/(2*10^8)=2*10^-5 sec

Data rate = 10^7 bps

Packet size = 2*10^-5*10^7 bits = 200 bits

so its option D ....

Transfer rate =2 *10^8 m/s

Time to transfer =(4*10^3)/(2*10^8)=2*10^-5 sec

Data rate = 10^7 bps

Packet size = 2*10^-5*10^7 bits = 200 bits

so its option D ....

0

(doubt) yes the packet size ahould be 25 bytes, but question says what could be minimum size .

a packet should be equal to or more than 25bytes so can't we say 50 bytes(option a) is fulfilling the question criteria?

a packet should be equal to or more than 25bytes so can't we say 50 bytes(option a) is fulfilling the question criteria?

+7 votes

For CSMA/CD Protocol formula of minimum packet size

L >= 2*T_{p}*B

where B is the bandwidth and T_{p} is the propagation delay which is D/V D is the distance and V is the velocity that signal travels through the wire so by putting all these values we get that L>= 200 bits which is L>= 25 bytes

Ans D

+5 votes

RTT = 2 * Propagation Time

Transmission Time = Length of packet / Bandwidth

RTT = 2 (d/v) = 2(2000/2×10^{8})

Therefore to find minimum size of the packet,

RTT <= Length of packet / Bandwidth

Length of packet >= RTT x Bandwidth

= 2(2000/2×10^{8}) x 10^{7} = 200bits = 25bytes

Therefore, minimum size of the packet = 25bytes

–5 votes

Lets explore the answer

As we know:

**TT = framesize/Bandwidth;**

and minimum time taken to acquire the line is = 2*pt

TT = 2*pt

min frame size = 2*pt*b.w;

**=200 bits(C)**

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