a. How long does it take computer A to receive acknowledgement for a packet? Ass

Question

a.How long does it take computer A to receive acknowledgement for a packet? Assume the speed of light for signal propagation and neglect computer B processing time.

b.How long does it take for computer A to send out a packet of size 1000 bytes if the throughput is 10,000 Kbps?

c.How much time is computer A idle during a one packet cycle assuming error free communication?

d.Calculate the utilization of the media between A and B assuming error free communication?

e.Repeat part (d) for system A and B that uses a sliding window ARQ with a window size of 7.

Explanation / Answer

1.

time taken for computer A to receive acknowledgement for a packet = time taken for the packet to reach B + time taken for ack to reach from B to A = 2*(propagation delay)

=2*(propagation delay)----propagation delay Tp=distance/speed

=2*(distance /speed)

=2*600 km/3x10^8m/sec

=2*600000m/3x10^8m/sec

=2*2x0^-3 sec

=4x10^-3 sec

2.

Transmission time(Tx)

Tx=length of the frame/bit rate ------(bit rate is bandwidth)

=1000Bytes/10000kbps

=1000 x 8 bits / 10000x10^3 bps

=8x10^-4 sec

3

idel time = 2Tp (as in stop and wait sender will wait until he recieves the ack from the reciever and for this it require 2Tp time)

=4x10^-3

4.

effecienty = Tx/(Tx+2Tp) (total time required is = tranmission time+2*(propagation time) in which only for a time of Tx computer will be working and for the remaining time the computer will sit idle )

=8x10^-4 /(8x10^-4 + 2*2x10^-3)

=0.1666

5.

efficency = min( W*Tx/(Tx+2Tp) ,1)

----W*Tx/(Tx+2Tp)

=7*8x10^-4 /(8x10^-4 + 2*2x10^-3)

=7*0.1666

=1.16

so efficiency = min(1.16 ,1)
=1

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