MTU with header

Question

If suppose MTU =620 (including header)

and 1380 bytes has to be sent (header is 20B)

s data to be sent is 1380 -20 =1360

MTU = 600 -20 =600

so max capacity to be sent as fragments is 600 + header or 596 + header?

Comments

Data size of fragment should be a multiple of 8..So if MTU is 620 B [ MTU takes the header into account as well ] , so data size that can be routed  =  600 B (620 - 20) ..

As 600 is a multiple of 8 hence no need of subtracting a few bytes from it further as it is already a multiple of 8 which is needed for fragmentation as 1 offset of a fragment references to 8 bytes of data in a fragment(This is the convention)..

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fragment send will be $620(600+20),620(600+20),180(160+20),$

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Suppose MTU with header is 630

then data part = 630 -20 =610

now 610 is not divisible by 8

then now how will we round off? to the higer value of 8 that is divisible or to the lower value?

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The nearest lower value we will have to take into account..As higher value means packet size > MTU of router which is not possible