A station A will have to transfer a message to another station B. The size of th

Question

A station A will have to transfer a message to another station B. The size of the message received by the IP entity is 7200 bytes. The IP datagram is identified with 0000H. The) datagram (s) carrying the message will have to cross 2 networks (RES1, RES2) Each network is configured with a specific MTU (for example, the RES1 MTU is 1200 bytes: MTU is the acronym for Maximum Transmission Unit, which defines the maximum size that the network frame can carry, i.e. the maximum size of the IP datagram). The MTU of the network RES2 is 900 bytes.

a) For each network, give the number of IP datagrams needed to carry the message in cases where the datagram is fragmented, give the value of the fields following of each fragment: fragment offset (fragment offset) and bit (More bit).

(please don't copy other incorrect answers, and please show all work)

Explanation / Answer

Before entering the first network, we have IP datagram of size 7200(data) + 20( Ip Header)=7220B

Tansimission on first network:

        First network MTU is 1200Bytes.So we can send atmost 1176 bytes fragments. Because 1200-20 (IP header)=1180, Fragment size should be multiple of 8 so we have to send maximum of 1176Bytes each time.

Fragment1: IP header of 20 bytes, and data of 1176 bytes. Offset is 0, More bit=1.
Fragment2: IP header of 20 bytes, and data of 1176 bytes. Offset is 1176/8 = 147, More bit=1.
Fragment3: IP header of 20 bytes, and data of 1176 bytes. Offset is 2352/8 = 294, More bit=1.
Fragment4: IP header of 20 bytes, and data of 1176 bytes. Offset is 3528/8 = 441, More bit=1.
Fragment5: IP header of 20 bytes, and data of 1176 bytes. Offset is 4704/8 = 588, More bit=1.
Fragment6: IP header of 20 bytes, and data of 1176 bytes. Offset is 5880/8 = 735, More bit=1.
Fragment7: IP header of 20 bytes, and data of 164 bytes. Offset is 7056/8 = 882, More bit=0.

Tansimission on Second network:

        First network MTU is 900Bytes.So we can send atmost 880B fragments. Because 900-20 (IP header)=880.

Fragment 1 goes through second network:
         It need to split the fragment into 2 fragments.
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment8: IP header of 20 bytes, and data of 880 bytes. Offset is 0, More bit=1.
Fragment9: IP header of 20 bytes, and data of 296 bytes. Offset is 880/8 = 110, More bit=0.

Fragment 2 goes through second network:
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment10: IP header of 20 bytes, and data of 880 bytes. Offset is 1176/8 = 147, More bit=1.
Fragment11: IP header of 20 bytes, and data of 296 bytes. Offset is 2056/8 = 257, More bit=0.

Fragment 3 goes through second network:
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment12: IP header of 20 bytes, and data of 880 bytes. Offset is 2352/8 = 294, More bit=1.
Fragment13: IP header of 20 bytes, and data of 296 bytes. Offset is 3232/8 = 404, More bit=0.

Fragment 4 goes through second network:
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment14: IP header of 20 bytes, and data of 880 bytes. Offset is 3528/8 = 441, More bit=1.
Fragment15: IP header of 20 bytes, and data of 296 bytes. Offset is 4408/8 = 551, More bit=0.

Fragment 5 goes through second network:
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment16: IP header of 20 bytes, and data of 880 bytes. Offset is 4704/8 = 588, More bit=1.
Fragment17: IP header of 20 bytes, and data of 296 bytes. Offset is 5584/8 = 698, More bit=0.

Fragment 6 goes through second network:
         Now total data in fragment 1 is 1176 now split fragment into 2 parts 880B + 296B

Fragment18: IP header of 20 bytes, and data of 880 bytes. Offset is 5880/8 = 735, More bit=1.
Fragment19: IP header of 20 bytes, and data of 296 bytes. Offset is 6760/8 = 845, More bit=0.

Fragment 7 goes through second network:
         Now total data in fragment 1 is 164 Bytes, so no need to split fragment.

Fragment20: IP header of 20 bytes, and data of 164 bytes. Offset is 7056/8 = 882, More bit=0.

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