Consider IP datagram as illustrated in Figure 6 of RFC 791, except that the tota

Question

Consider IP datagram as illustrated in Figure 6 of RFC 791, except that the total length field is 1200. If this datagram were to traverse a local network that had a maximum transmission unit size of 500 bytes it would have to be fragmented. Assume it is fragmented into three smaller datagrams according to the algorithm presented Section 3.2 (see subsection titled ``An Example Fragmentation Procedure'') of RFC 791. With this technique, each fragment (except the last) is made the maximum allowable size. What are the values of the header length, total length, identification, flags, and fragment offset fields for all three fragment headers?

Assume that the datagram fragments from the previous problem finally reach the destination host which has to reassemble them into a single datagram before passing them “up” to the TCP module. How does the IP module on the destination host know which fragments belong to the original IP datagram and how long it is (Hint: the “more fragments” bit plays a role in this)?

Explanation / Answer

Given, MTU = 500 Bytes Total length = 1200 For the first and second fragments which has maximum allowable size, Old Internet Header Length OIHL = Internet Header Length = 5 Old total length OTL = Total length, TL = 1200 Old Fragment Offset OFL = Fragment Offset FO = 0 Old More Fragments flag OMF = More Fragments flag, MF = 1 TL = (IHL x 4 ) + (NFB x 8) NFB = (MTU – IHL x 4)/8 = (500 – 5 x 4)/8 =(500 – 20) / 8 = 480 / 8 = 60 TL = (IHL x 4 ) + (NFB x 8) = (5x4) + (60 x 8) = 20 + 480 = 500 So for the 1st and 2nd fragments, IHL = 5, TL = 500, MF = 1, FO = 0, identification = 111, flag = 0 For the last fragment, IHL

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