A maximum-length IP datagram carrying 65,536 bytes can be transported over a WAN

Question

A maximum-length IP datagram carrying 65,536 bytes can be transported over a WAN with a maximum packet length of 292 bytes, which includes the 20-byte IP header. Determine

a. How many fragments are required.

b. The offset values in binary, hex, decimal for the first four and last two fragments’and the number of bytes in the last fragment.

a. How many fragments are required.

b. The offset values in binary, hex, decimal for the first four and last two fragments’and the number of bytes in the last fragment.

a. How many fragments are required.

b. The offset values in binary, hex, decimal for the first four and last two fragments’and the number of bytes in the last fragment

Explanation / Answer

total number of bytes = 65,536

maximum packet length = 292 bytes

header 20-byte IP header

a) number of bytes transmitted per packet= 292-20 = 272 bytes

number of fragments required - 241 packets

b) offset values for first four =0000/8 [fragment offset = position of the fragment in orginal message * 8]

offset values for last two fragments =2,800/8=350

number of bytes in last fragment =879

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