When studying a computer science course, one has to solve problems of finding the amount of information stored on a carrier or transmitted over a communication channel for a certain time. The units for measuring the amount of information are bit, nibble, byte, word, double word and their derivatives.
Instructions
Step 1
When calculating, keep in mind that one nibble is four bits, a byte is eight bits, a word is sixteen, and a double word is thirty-two. A kilobyte is equal to 1024 bytes, megabytes - 1024 kilobytes, gigabytes - 1024 megabytes, terabytes - 1024 gigabytes. Similarly, kilobits, megabits, gigabits and terabits are translated into each other. Bits are designated by a lowercase letter "b", bytes - by an uppercase letter "B".
Step 2
To find out the amount of information stored on a medium, add together the volumes of all files stored on it. If they are all the same, just multiply the volume of one of them by their number. Note that on some file systems, the length of all files is automatically rounded up to a predetermined value. It is usually 4096 bytes. For example, if the disk contains four files of 30, 50, 58749 and 14358 bytes, then their total size is 4096 + 4096 + 61440 + 20480 (the last two values are obtained by multiplying the number 4096, respectively, by 15 and 5), or 90112 bytes.
Step 3
Calculate the amount of information transmitted via the communication channel for a given period of time as follows. Since the data transfer rate is indicated in bits per second and their derivatives, first convert it to bytes per second or their derivatives by dividing by 8. For example, 56 kbps (kilobits per second) = 7 kbps (kilobytes per second). Then multiply this speed by the time expressed per second. For example, in 10 seconds at the above speed, 70 KB (kilobytes) will be transmitted over the channel. If data is transmitted via GPRS and the tariff is not unlimited, the result should always be rounded up to the threshold specified by the provider. So, if 1 kilobyte is transmitted over such a channel, and the threshold is 10 kilobytes, the cost of transferring such a volume of data will be the same as for 10 kilobytes.
Step 4
If the length of the text in characters is specified in the conditions of the problem, keep in mind that in different encodings one character corresponds to a different number of bits. In the Baudot code, there are 5 bits per character, in the ASCII code - 7 (but the peculiarities of data storage in computing devices lead to the fact that 8 bits are spent on its storage), in the encodings 866, KOI-8P, KOI-8U, 1251 and similar - 8 bits, and in Unicode - 16 bits (except for characters from the ASCII table, which occupy 8 bits in Unicode).
