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 = 2 32 Calculating maximum “L” value:

Question

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Answer 1

Expert Solution & Answer

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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Answer 2

Expert Solution & Answer

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 $= 232 + 528857934$

$= 4$

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 3

Expert Solution & Answer

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 $= 232 + 528857934$

$= 4$

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

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 $= 232 + 528857934$

$= 4$

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

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 $= 232 + 528857934$

$= 4$

Answer 8

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:

Answer 9

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:

Answer 10

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 $= 232 + 528857934$

$= 4$

Answer 11

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 $= 232 + 528857934$

$= 4$

Answer 12

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Check out a sample textbook solution

See solution

Answer 13

Ch. 3 - Prob. R1RQ Ch. 3 - Prob. R2RQ Ch. 3 - Consider a TCP connection between Host A and Host...Ch. 3 - Prob. R4RQ Ch. 3 - Prob. R5RQ Ch. 3 - Prob. R6RQ Ch. 3 - Suppose a process in Host C has a UDP socket with...Ch. 3 - Prob. R8RQ Ch. 3 - Prob. R9RQ Ch. 3 - In our rdt protocols, why did we need to introduce...

Solution Summary: The author calculates the time taken for transmitting data through the 155 Mbps link.

Explanation of Solution

Explanation of Solution

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Chapter 3 Solutions