Consider sending a 3,000-byte datagram into a link that has an MTU of 1500 bytes

Question

Consider sending a 3,000-byte datagram into a link that has an MTU of 1500 bytes, and then into a link that has an MTU of 500 bytes. Suppose the original datagram is stamped with the identification number 466.

(a) (0.5 point) How many fragments are generated?

(b) (1.5 points) What is the identification number, size, flag, and offset for each? Fill in the following table:

fragment

Identification number

length

Frag flag

Offset

1

2

3

4

5

6

7

8

9

fragment

Identification number

length

Frag flag

Offset

1

2

3

4

5

6

7

8

9

Explanation / Answer

The given datagram size is 3,000.MTU=1500

so,first fragment packet will be identifier will be =1,R=0,DF=0.MF=1,flag offset=0

Heade size[20]+data gram size[1480]=1500.

Flag offset is calculated as sum of datagram size divisible by 8. i.e. 1480/8 = 185.

Identification number is a a unique identifier is assigned to each message being fragmented.

This value is placed in the Identification field in the IP header of each fragment sent.

second fragment will be id=1,DF=0,MF=0,FO=185

i.e 1480/4=370 and similarly for third fragment i.e 1480/2=555.

so,totally four fragments are generated.

A three-bit field follows and is used to control or identify fragments. They are in order,high order,low order

i.e bit 0:reserved must be zero.

bit 1:dont fragment(DF).

bit2 :more fragment(MF).

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