Alice and Bob locate at two cities 2000 Km apart. Alice needs to send a picture

Question

Alice and Bob locate at two cities 2000 Km apart. Alice needs to send a picture of 4M Bytes to Bob. Suppose that each packet carries 1000 bits of data, the channel data rate is 1 Mbps, and the propagation speed is 2 times 10^8 m/s. Ignore transmission delays of ACKs, waiting and processing delays. Also, we assume that no data or control frame is lost or damaged. Ignore the overhead due to the header and trailer. It a system uses the Go-Back-N ARQ protocol with a window size of 7, how long does it take? What is the link utilization?

Explanation / Answer

RTT = 2PD (since no transmission delays)

PD = 2000km/(2 * 10^8) = 10 ms

So, RTT = 2 * 10ms = 20ms.

In RTT, in Go-Back-N protocol, N frames are transmitted. N = 7 here.

So, time to transmit 7 frames = 20 ms.

Each frame of of size 1000.

So, time to transmit 7 * 1000 = 7000 bits = 20 ms.

Time taken to send 32M (2^25) bits = 20 ms * 2^25/7*10^3 = 95869ms=95.8 sec~96 sec.

Link Utilisation:

Utilisation=(1-P)/(1+2a)

a=propogation Time/Transmission Time

Transmission time=framelenght/datarate=7000(7-Kbit) bits/1Mbps =7 ms

a=10ms/7ms=1.42

1+2a=1+2.84=3.84~4

p= probability of single frame error=0

Link Utilisation=1/4=25%

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