Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN: 9780133594140
Author: James Kurose, Keith Ross
Publisher: PEARSON
Expert Solution & Answer
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Chapter 3, Problem P26P

a.

Explanation of Solution

Given:

The TCP sequence number field= 4 bytes.

In order to represent the size of sequence number in bits, then convert the bytes value into its corresponding bits as follows:

One byte= 8 bits.

The bits value for 4 bytes = 4 × 8

     = 32 bits

Therefore, the sequence numbers of bits = 232

Calculating maximum “L” value:

b.

Explanation of Solution

Given data:

MSS = 536 bytes.

Therefore, the segments data = 232/536

=8012999

Given that the total header files = 66 bytes.

Calculate the time taken for transmitting data:

Sum of bytes through the 155 Mbps link = 8012999×66 bytes

      =528857934 bytes

Therefore, transmitted data = 232 + 528857934

         = 4

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