Question

6. ) Suppose that a certain communication protocol involves a per-packet overhead of 100 bytes for headers. We send a file of 100 KB (100,000 bytes) from Host A to Host B. There are two links (and one switch) between A and B. Each link has a transmission rate of 1Mbps. Find the total transfer time of moving the file from Host A to Host B for packet size of 1,000, 5,000, and 10,000 bytes, respectively. Ignoring the propagation delay and processing delay at the switch.

Answer 1

Given, overhead per packet = 100 bytes

File size = 10⁵ bytes

Transmission rate = 1 Mbps

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(a) For packet size 1000 bytes

No of packets = 10⁵ / 1000 = 100

For each packet there will be a overhead of 100 bytes.

So, overhead for 100 packets = 100 * 100 = 10000 bytes

Total size of file = 100,000 + 10,000 = 110,000 bytes

Transmission time = Total file size / Transmission rate = 110,000 / 10⁶ = 0.11 seconds

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(b)

For packet size 5000 bytes

No of packets = 10⁵ / 5000 = 20

For each packet there will be a overhead of 100 bytes.

So, overhead for 20 packets = 20 * 100 = 2000 bytes

Total size of file = 100,000 + 2,000 = 102,000 bytes

Transmission time = Total file size / Transmission rate = 102,000 / 10⁶ = 0.102 seconds

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(c) For packet size 10,000 bytes

No of packets = 10⁵ / 10,000 = 10

For each packet there will be a overhead of 100 bytes.

So, overhead for 10 packets = 10 * 100 = 1000 bytes

Total size of file = 100,000 + 1,000 = 101,000 bytes

Transmission time = Total file size / Transmission rate = 101,000 / 10⁶ = 0.101 seconds

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