Computer Network Throughput

Question

Suppose Alice wants to send a 200 kB file to Bob as a single message. The path from Alice to Bob

has three links of transmission rates: R1 = 2 Mbps, R2 = 500 kbps, and R = 1 Mbps.

(a) Assuming no other traffic in the network and ignoring processing/queuing/propagation delays,

what is the throughput for the file transfer?

(b) Assuming no other traffic in the network and ignoring processing/queuing/propagation delays,

how much time will it take for Bob to receive the message completely?

(c) Now, assume that Alice is transmitting the file in two packets of 100 kB each. Assuming there

is no other traffic on the network, will there be any queuing delays? If so, where will the queuing

occur?

Answer 1

First we need to convert the 200 kB (kiloByte since you wrote a capital letter B) to Mb (megabits - note the lowercase b) so that we can then do the math for each of the 3 hops in the network. Since a Byte is 8 times larger than a bit, the math is as follows...

1. 200 kB = 1600 kb
2. 1600 kb is 1.6 Mb

Now we can figure out the time it takes for each hop, expressed as millseconds (ms)...

1. R1 = 800ms (1.6 / 2 Mbps)
2. R2 = 3200ms (1.6 / .5 Mbps since 500 Kbps is .5 Mbps)
3. R3 = 1600ms (1.6 / 1 Mbps)

So the total transmission time is 5600ms or 5.6 seconds to complete the 3 hops.

(a) Approx. 291 Kbps (291 x 5.6 = approx. 1600). This is the same as approx. 36.4 kB (kiloBytes since 8 bits = 1 Byte) per second

(b) It will take about 5.6 seconds to receive the message.