Test by Bikram | Mock GATE | Test 4 | Question: 33

Question

A router receives an $IP$ packet containing $400$ data bytes and forwards that packet in a network where maximum transmission unit is $100$ bytes. If the $\text{IP}$ header is $15$ bytes long, then the number of fragments required is ______.

Comments

i am getting  fragments viz;

80|20  80|20  80|20  80| 20  20|20 is it correct i am not getting the solution provided below 

please check it @Mk Utkarsh @abhishekmehta4u

Answer 1

hope this helps!!!!

Comments

@akash sir, the data unit need to be a multiple of 8. Right? It should be 80 and not 85.

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400 B includes header or not ? Although ans would be same.

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@akash.dinkar12 shouldn't the size of a fragment be divisible by 8? how did you take size as 85 in the selected answer? :/

Answer 2

$( 100 - 15) = 85 /8 = 10.5 = 10$
that means $10 * 8 B + 15 B header = 95 B$
so number of fragments require $( 100 - 95) = 5$

Comments

Hi Bikram sir, Shouldn't the IP header be 15x4=60(4 is scaling factor), as the minimum IP header length is 20 Bytes.

The answer should be then 9 packets, If there is any error in approach please tell me.

Answer 3

If the IP header is 15 bytes long

The size of IPv4 header is between 20 and 60 Bytes. But it is given 15 bytes, so let's play along.

(We only should use scaling, when just the number is given. Here, the unit "bytes" is explicitly mentioned)

 

Only two things you must keep in mind:

1. The payload must be a multiple of 8 in all but the last fragment. $^{[1]}$
2. Sum of fragmented payloads = original payload (not original packet size).

So, keeping these two golden rules in mind, we'll fragment

15

400

into

 

15

80

15

80

15

80

15

80

15

80

 

So, 5 fragments.

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$^{[1]}$ The reason goes a little deep. Read the answer by Ryan here: https://stackoverflow.com/questions/7846442/why-the-ip-fragments-must-be-in-multiples-of-8-bytes