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 = $\text{4 × 8}$

     $\text{= 32 bits}$

Therefore, the sequence numbers of bits = ${\text{2}}^{\text{32}}$

Calculating maximum “L” value:

b.

Explanation of Solution

Given data:

MSS = 536 bytes.

Therefore, the segments data = $2^{32}/536$

$=8012999$

Given that the total header files = 66 bytes.

Calculate the time taken for transmitting data:

Sum of bytes through the 155 Mbps link $\text{= 8012999×66 bytes}$

      $\text{=528857934 bytes}$

Therefore, transmitted data ${\text{= 2}}^{\text{32}}\text{ + 528857934}$

         $\text{= 4}{}^{}$