Question

Many computers use one byte (8 bits) of data to represent each letter of the alphabet. Suppose there are 25 million words in a book that we want to store in an area of just 1 mm²and the average word contains 6 letters.

1. What is the byte density of this storage (bytes/in²)?
2. What is the bit density of this storage (bits/in²)?
3. What area is needed to record one byte? one bit?
4. Is this storage superior or inferior to common information storage today, e.g. tens to hundreds of megabytes/in² for CD-ROMs and DVDs?

Answer 1

Note: Done accordingly. Please comment for any problem. Please Uprate. Thanks.

Total letters = 25 * 10⁶ * 6 = 150 * 10⁶ letters

Total Bytes = 150 megabytes

Total bits = 150 * 10⁶ * 8 bits = 1200 megabits

1 mm² = 0.00155 in²

a) What is the byte density of this storage (bytes/in²)?

(150 * 10⁶)/( 0.00155) = 96774.19 megabytes/in²

b) What is the bit density of this storage (bits/in²)?

(1200* 10⁶)/( 0.00155) = 774193.548387 megabits/in²

c) What area is needed to record one byte? one bit?

Area for 1 byte = 1 byte/ density in bytes = 1/96774.19 megabytes/in²

= 0.00001033333 * 10^-6 in² = 0.01 * 10^-9 in²

Area for 1 bits = 1 bit/ density in bits = 1/774193.548387 megabits/in²

= 0.00000129 * 10^-6 in² = 0.0013 * 10^-9 in²

d) Yes it is far superior then CD-ROMs and DVDs as it can store thousands of megabytes per square inch.