When studying a course of informatics it is necessary to solve problems on finding of volume of the information which is stored on the carrier or transferred of a communication channel for defined time. Units of measure of volume of information are the bit, half-byte, byte, the word, the double word and their derivatives.
The sponsor of placement P&G Articles on the subject "How to Find Information Volume" For what is necessary defragmentation of disks How to find information volume How to transfer megabytes to megabits
At calculations consider that one half-byte is equal to four bits, byte - to eight bits, the word - sixteen, and the double word - to thirty two. The kilobyte is equal to 1024 bytes, megabyte - to 1024 kilobytes, gigabyte - to 1024 megabytes, terabyte - to 1024 gigabytes. Are similarly translated each other and kilobits, megabits, gigabits and terabits. Bits "would" be designated by a lower case letter, bytes - a capital letter of "B".
To learn the volume of information which is stored on the carrier, put among themselves volumes of all of the files which are stored on it. If all of them identical, simply increase the volume of one of them by their quantity. Consider that in some file systems length of all files is automatically rounded towards increase to some in advance preset value. Usually it is equal 4096 bytes. For example, if on a disk there are four files with a capacity of 30, 50, 58749 and 14358 bytes, their total volume is equal 4096+4096+61440+20480 (the last two values are received by multiplication of number 4096, respectively, on 15 and on 5), or 90112 bytes.
Count the volume of information transferred on a communication channel for the set period so. As the speed of data transmission is specified in bits a second and their derivatives, transfer it to bytes per second or their derivatives in the beginning, having divided on 8. For example, 56 kb / from (kilobits per second) = 7 kb / about (kilobyte per second). Then increase this speed by time expressed in a second. For example, in 10 seconds at a speed stated above on the channel 70 kb (kilobyte) will be transferred. If data are transferred on the GPRS channel, and a tariff not unlimited, the result should be rounded always towards increase to the threshold specified by provider. So, if on such channel 1 kilobyte is transferred, and the threshold is equal the 10th kilobyte, the cost of transfer of such volume of data will be same, as for the 10th kilobyte.
If in statements of the problem text length in symbols is specified, consider that in various codings of one symbols there corresponds various number of bits. In Baudot's code 5 bits, in the ASCII code - 7 (but features of data storage in computers lead to that is spent for its storage of 8 bits), in codings 866, KOI-8R, KOI-8U, 1251 and similar - 8 bits, and in the coding Unicode - 16 bits are the share of one symbol (except symbols from the table ASCII which occupy 8 bits and in Unicode).