Question

Question 7 
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a) Assume that packet size is 1KB, the round-trip-time is 25ms, and the bandwidth of the channel is 1500KB/s. What is the throughput of stop-and-wait?

b) What is the minimum number of logical channels in the concurrent-logical channels protocol that are needed to ensure we maximize the throughput

c) What if instead of using the number of logical channels you gave in b), we use twice that. What will happen to the throughput? Briefly argue why.

Answer 1

7) (a) Packet size = 1 KB

RTT = 25 ms

Bandwidth = 1500 KB/s

Throughput for stop and wait = 1 KB / 25 ms = 1000 B / 25 * 10^-3 s = 40 KB/s

(b) The minimum number of concurrent logical channels protocols that are needed = Bandwidth * delay of the line

1500 KB/s * 25*10^-3 s = 1500*10³ B/s * 25*10^-3 s = 37500 B = 37.5 KB / 1 KB = 37.5 =~ 38

(c) We know that throughput = logical channel size / RTT

If the logical channel size is doubled then the value will be (37.5 * 2) = 75 / Logical channel = 75 packets

1 packet = 1 KB

Throughput = 75 KB / 25 ms = 75 * 10³ B / 25 * 10^-3 s = 3 * 10⁶ B/s = 3 MB/s

So, the throughput will increase.

Hope this helps.