Question

Документ объемом 4 Мб можно передать с одного компьютера на другой двумя способами: а) сжать, передать, распаковать. б) передать по каналу связи без архивации Какой способ быстрее и на сколько если средняя скорость передачи по каналу связи составляет 2^18 бит в секунду, объем сжатого архиватором документа равны 50% исходного. Время требуемая на сжатие документа 20 сек., а на распаковку 5 сек. Решение!

Answer 1

4 Мб = 4 * 1024 * 1024 = 2^22 бит.
а) Сжимаем (20 секунд, объём документа становится 2^21 бит), передаём (2^21 / 2^18 = 8 секунд), распаковываем (5 секунд) - итого получаем 33 секунды
б) передаём без архивации: 2^22 / 2^18 = 16 секунд.
Ответ: быстрее без архивации, на 33 - 16 = 17 секунд.

Answer 2

Comparison of Two Methods for Transferring a 4MB Document

To determine which method is faster for transferring a 4MB document between two computers, we will compare two approaches:

a) Compress, transfer, and decompress the document. b) Transfer the document over a communication channel without compression.

Let's calculate the time required for each method and compare the results.

Method a: Compress, Transfer, and Decompress

Given: - Document size: 4MB - Compression ratio: 50% of the original size - Compression time: 20 seconds - Decompression time: 5 seconds

To calculate the time required for this method, we need to consider the following steps:

1. Compression: The document needs to be compressed before transferring it. Since the compressed size is 50% of the original size, the compressed document will be 2MB. 2. Transfer: The compressed document, which is 2MB in size, needs to be transferred over the communication channel. 3. Decompression: Once the document is received on the other computer, it needs to be decompressed before it can be used.

Let's calculate the time required for each step:

1. Compression time: 20 seconds 2. Transfer time: The document size is 2MB, and the average transfer speed over the communication channel is 2^18 bits per second. To convert the document size to bits, we multiply it by 8 (1 byte = 8 bits). Therefore, the document size in bits is 2MB * 8 = 16Mb. The transfer time can be calculated using the formula: transfer time = document size / transfer speed. Substituting the values, we get: transfer time = 16Mb / 2^18 bps = 16Mb / 262144 bps = 0.061 seconds. 3. Decompression time: 5 seconds

Adding up the times for each step, we get: Total time =