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

Copy my answer below into overleaf to see the mathematical symbols properly

### (a) Throughput Calculation:

Throughput (\(T\)) is given by the minimum of the transmission rates of the links along the path.

\[ T = \min(R1, R2, R) \]

Substitute the given values:

\[ T = \min(2 \, \text{Mbps}, 500 \, \text{kbps}, 1 \, \text{Mbps}) \]

Let's convert the units to Mbps:

\[ T = \min(2 \, \text{Mbps}, 0.5 \, \text{Mbps}, 1 \, \text{Mbps}) \]

\[ T = 0.5 \, \text{Mbps} \]

### (b) Time Calculation:

Time (\(t\)) for Bob to receive the message completely is given by:

\[ t = \frac{\text{File Size}}{\text{Throughput}} \]

Let's assume the file size is 200 kB:

\[ t = \frac{200 \, \text{kB}}{0.5 \, \text{Mbps}} \]

Convert units to seconds:

\[ t = \frac{200 \times 8 \times 10^3 \, \text{bits}}{0.5 \times 10^6 \, \text{bps}} \]

\[ t = 3.2 \, \text{seconds} \]

### (c) Queuing Delays:

Since Alice is transmitting the file in two packets of 100 kB each, queuing delays may occur if the rate at which packets are transmitted exceeds the rate at which they can be processed.

Calculate the time to transmit each packet:

\[ t_{\text{packet}} = \frac{100 \, \text{kB}}{0.5 \, \text{Mbps}} \]

\[ t_{\text{packet}} = \frac{100 \times 8 \times 10^3 \, \text{bits}}{0.5 \times 10^6 \, \text{bps}} \]

\[ t_{\text{packet}} = 1.6 \, \text{seconds} \]

Compare \(t_{\text{packet}}\) to the transmission time of the link with the smallest rate (0.5 Mbps). Since \(t_{\text{packet}} < t_{\text{link}}\), there won't be queuing delays, and the packets can be transmitted without waiting.