# GATE | GATE-CS-2005 | Question 73

• Difficulty Level : Hard
• Last Updated : 28 Jun, 2021

In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 bytes and each packet contains a header of 3 bytes, then the optimum packet size is:
(A) 4
(B) 6
(C) 7
(D) 9

Explanation: Dividing a message into packets may decrease the transmission time due to parallelism as shown in the following figure.

But after a certain limit reducing the packet size may increase the transmission time also.

Following figure shows the situation given in question.

Let transmission time to transfer 1 byte for all nodes be t. The first packet will take time = (packet size)*3*t. After the first packet reaches the destination, remaining packets will take time equal to (packet size)*t due to parallelism.

```If we use 4 bytes as packet size, there will be 24 packets
Total Transmission time = Time taken by first packet +
Time taken by remaining packets
= 3*4*t + 23*4*t = 104t

If we use 6 bytes as packet size, there will be 8 packets
Total Transmission time = 3*6*t + 7*6*t = 60t

If we use 7 bytes as packet size, there will be 6 packets
Total Transmission time = 3*7*t + 5*7*t = 56t

If we use 9 bytes as packet size, there will be 4 packets
Total Transmission time = 3*9*t + 3*9*t = 54t
```

Source: question 2 of https://www.geeksforgeeks.org/computer-networks-set-6/

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