Question

How long will it take to send 1.1 million bits using the Stop and Wait ARQ protocol if each packet contains 1000 bits and the only delay is propagation delay

Answer 1

1.111 second

Step-by-step explanation:

We know propagation speed =
`$2 * 10^8$` m/s

=
`$2 * 10^5$` km/s

One packet size = 1000 bit

Distance between sender and the receiver = 1000 m = 1 km

The channel data rate = 1 Mbps =
`$1 * 10^6$` bits per sec

There is no transmission delays of ACKs,

The time to transmit one data packet =
`$T_(trans)+2T_(prop)$`

Here, time to transmit frame =
`$T_(trans)$`

propagation time =
`$T_(prop)$`

Therefore,
`$T_(trans)$` =
`$(bits\ per\ frame)/(transmission \ speed)$`

=
`$(1000)/(1 * 10^6)$` = 0.001 seconds

`$T_(prop)=(distance\ between\ sender\ and\ receiver)/(propagation \ speed)$`

=
`$(1)/(2 * 10^5)$`

= 0.000005 seconds

Therefore, T = 0.001 +2(0.000005)

= 0.00101 seconds

We known, 1.1 million bits= 1100 packets

Therefore to transmit 1 million bits = 1100 x 0.00101

= 1.111 second