In computer science the set of the most various units of measure of information is used. From the bit which is considered as the smallest unit to the terabytes capable to contain the whole libraries and hundreds of movies. Each such unit has the scope. So, for example, information transfer speed is traditionally measured in kilobits per second (recently more often – in Megabits). Volumes of files were usually measured in kilobytes earlier. Now – in Megabytes and Gigabytes. However at the solution of practical tasks different units quite often should be given to "an identical denominator".
It is required to you
Instruction
1. To transfer numerical value of volume of information from kilobits to kilobytes, divide the number of kilobits into eight. That is: kb = kb / 8. So, for example, if the volume of the file is 800 kilobits, then its size in kilobytes will be 800/8 – the 100th kilobyte.
2. To transfer from kilobits per second to kilobytes per second information transfer speed, also divide the number of kilobits per second into 8. That is: kb / with = (kbps) / 8. This formula it is useful at assessment of time necessary for "downloading" files. However considering that in one megabyte 1024 kilobytes, and in one megabit – 1024 kilobits, are more practical to use more modern option: Mbps = (Mbps) / 8.
3. To calculate, what is the time it will be required for downloading of the file, divide the speed of your modem into 8 (it is specified usually in Mbps if it is, of course, not Dial-Up the modem). Now divide into the received modem speed (expressed already in Mbps) the size of your file in megabytes. As a result time (number of seconds) necessary for loading of information will turn out. If the number turned out too large, then divide it into 60 – the number of minutes will turn out. Having divided this result into 60 once again, it is possible to receive a number of hours, etc.
4. Let, for example, there is a "standard" 3G modem with a speed of reception of these 3.6 Mbps. On the server the video adjusted by the size to a usual DVD disk, that is 4700 MB is placed. Means to loading of this movie on the local computer it will be required: 4700 / (3.6 / 8) = 4700 / 0.45 = 10444 seconds ili10444/60 = 174 minutes ili174/60 = 2.9 hours.