Chapter 1, Problem P31P

a.

Explanation of Solution

Given,

The length of the message is 8 × 10⁶ bits

The transmission rate of the link is 2 Mbps

$\text{Transmission}\text{delay}\text{=}\frac{\text{Length}}{\text{Transmission}\text{rate}}$

Note: $\text{1}\text{Mbps}\text{=}{\text{10}}^{\text{6}}\text{bps}$

Substitute, “8 × 10⁶” for “Length” and “2 × 10⁶” for “Transmission rate” is given below:

The time required to send the message from source host to first packet switch is

$\frac{\text{8}\text{×}}{}$

b.

Explanation of Solution

$\text{Transmission}\text{delay}\text{=}\frac{\text{Length}}{\text{Transmission}\text{rate}}$

Note: $\text{1}\text{Mbps}\text{=}{\text{10}}^{\text{6}}\text{bps}$

Substitute, “1 × 10⁴” for “Length” and “2 × 10⁶” for “Transmission rate” is given below:

The time required to move the first packet from source host to the first switch is,

$\frac{\text{1}\text{×}{\text{10}}^{\text{4}}}{\text{2}}$

c.

Explanation of Solution

$\text{Transmission}\text{delay}\text{=}\frac{\text{Length}}{\text{Transmission}\text{rate}}$

Note: $\text{1}\text{Mbps}\text{=}{\text{10}}^{\text{6}}\text{bps}$

Substitute, “1 × 10⁴” for “Length” and “2 × 10⁶” for “Transmission rate” is given below:

The time required to move the first packet from source host to the first switch is,

$\frac{\text{1}\text{×}{\text{10}}^{\text{4}}}{\text{2}\text{×}{\text{10}}^{\text{6}}}$

d.

Explanation of Solution

Reasons to use message segmentation:

• In case of bit errors and single bit errors, the entire message must be retransmitted. Hence, message segmentation is used to eliminate retransmission...

e.

Explanation of Solution

Drawbacks of message segmentation:

• Packets present in the link must be kept in a specified sequence at the destination...