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MTU is 200 byte and ip header is 20 byte so In network we send only 200-20=180 byte data

But here Catch is We Can’t Send date packet size is 180 Because It is not divisible by 8 so we take nearest value to 180 which is divisible by 8. SO 176 is divisible by 8 Here We take my packet size 176 byte

-Divisible by 8 Because Offset = 1st byte Number/8

now  Ceil(17076/176) = 967 Packet we send.

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Answer is 97.

Let's see MTU =200B 

So we can send maximum 200B at once. 

Here header size is 20B 

So we can send only 200-20=180B at once but we cannot send data of size not divisible by 8.

We need to send 176 of data + 20 byte in one packet

So number of fragments=17076/176=97.022727

97 fragments. 

But wait, there is some fraction part left (0.022727) 

Let's break it down

If we send 97 fragments with each 176 bytes, in total we are sending =97*176=17072bytes

But total data is 17076bytes i. e we are left with 4byte data. Should we send this with an extra fragment? 

Answer is NO.. because we are sending 196byte data in each fragments but the MTU is 200byte and as it is last byte we can send the remaining data along with this. 

So the last fragment I.e fragment no. 97 will contain total 200 byte of data.

1st fragment 0---175(176B+20B) 

2nd fragment 176---351(176B+20B

..... 

96th  fragment 16720---16895(176B+20B) 

97th fragment 16896---17076(180B+20B) 

If data was given atleast 1 byte extra then we might have to use another fragment for this. 

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