Chapter 9, Problem R1RQ

Explanation of Solution

Assume:

$\text{1 kbyte = 8000 bits}$

Calculate for 100 Kbytes:

$\begin{array}{c}\text{=}\frac{\text{(8000×100}\text{Kbytes)}}{\text{1Kbytes}}\\ \text{=800000}\text{bits}\end{array}$

Therefore, bit rate for 20 seconds is 800000 bits.

Calculate byte transferred for victor video:

Bit rate of Victor video is 4Mbps.

To calculate number of bytes transferred in 67 minutes:

$\begin{array}{c}\text{= 4}\text{Mbps}\text{×67}\text{min ×60}\text{sec}\\ \text{=}\text{16080}\text{Mbps ÷}\text{8}\\ \text{=}\text{2010}\text{Mbytes}\\ \text{=2 Gbytes}\end{array}$

Therefore, number of bytes transferred in 67 minutes is 2Gbytes.

Calculate byte transferred for Facebook Franck:

Bit rate for Facebook Franck is 40 Kbps:

To calculate number of bytes transferred in 67 minutes:

$\begin{array}{c}\text{= 40 Kbps}\text{×67 min ×60 sec}\\ \text{=}\text{160800}\text{kbps}\\ \text{=}\text{20100}\text{kbytes}\\ \text{=20}\text{.1}\text{Mbytes}\end{array}$

Therefore, number of byte transferred in 67 minutes is 20.1Mbytes.

Calculate byte transferred for Maratha music:

Bit rate of Maratha music is 200 kbps.

To calculate number of bytes transferred in 67 minutes:

$\begin{array}{c}\text{= 200 Kbps}\text{×67}\text{min ×60}\text{sec}\\ \text{=}\text{804000}\text{kbps}\\ \text{=}\text{100500}\text{kbytes}\\ \text{=100}\text{Mbytes}\end{array}$

Therefore, numbers of bytes transferred in 67 minutes are 100Mbytes.

Reconstruct of the table:

	Bit rate	Bytes transferred in 67 min
Facebook Frank	40 kbps	20 Mbytes
Maratha music	200 kbps	100 Mbytes
Victor video	4 kbps	2 GBytes