Question

У Васи есть доступ к Интернет по высокоскоростному одностороннему радиоканалу, обеспечивающему скорость получения им информации У Васи есть доступ к Интернет по высокоскоростному одностороннему радиоканалу, обеспечивающему скорость получения им информации 2^17бит в секунду. У Пети нет скоростного доступа в Интернет, но есть возможность получать информацию от Васи по низкоскоростному телефонному каналу со средней скоростью 2^16 бит в секунду. Петя договорился с Васей, что тот будет скачивать для него данные объемом 8 Мбайта по высокоскоростному каналу и ретранслировать их Пете по низкоскоростному каналу. Компьютер Васи может начать ретрансляцию данныхне раньше, чем им будут получены первые 1024 Кбайт этих данных. Каков минимально возможный промежуток времени (в секундах), с момента начала скачивания Васей данных, до полного их получения Петей?

Answer 1

8 Мб=8*2^20 байт=2^26 бит
1024/2^17=2^(10+10+3)-17=2^6=64сек
2^26/2^16=2^10=1024сек
1024+64=1088сек
ответ:1088

Answer 2

Problem Analysis

In this scenario, Vasya has high-speed internet access through a one-way radio channel, which provides a data transfer rate of 2^17 bits per second. Petya, on the other hand, has a slower internet connection with a data transfer rate of 2^16 bits per second. Petya has made an agreement with Vasya to download data for him and transmit it to Petya through the slower channel. Vasya's computer can start transmitting the data to Petya only after receiving the first 1024 kilobytes (1 megabyte) of data. We need to determine the minimum possible time it takes for Petya to receive the complete data from the moment Vasya starts downloading it.

Solution

To calculate the minimum possible time, we need to consider the following factors: 1. The time it takes for Vasya to download the first 1024 kilobytes of data. 2. The time it takes for Vasya to transmit the remaining data to Petya through the slower channel.

Let's calculate these two times separately.

1. Time to download the first 1024 kilobytes of data by Vasya: - The data transfer rate for Vasya's high-speed channel is 2^17 bits per second. - We need to convert 1024 kilobytes to bits: 1024 kilobytes = 1024 * 8 * 1024 bits. - The time it takes to download the first 1024 kilobytes of data is the data size divided by the data transfer rate: (1024 * 8 * 1024) / (2^17) seconds.

2. Time to transmit the remaining data to Petya through the slower channel: - The data transfer rate for Petya's slower channel is 2^16 bits per second. - The remaining data size is 8 megabytes - 1024 kilobytes. - We need to convert the remaining data size to bits: (8 * 1024 * 1024 - 1024 * 8 * 1024) bits. - The time it takes to transmit the remaining data is the remaining data size divided by the data transfer rate: ((8 * 1024 * 1024 - 1024 * 8 * 1024) bits) / (2^16) seconds.

To find the minimum possible time, we add the time to download the first 1024 kilobytes of data and the time to transmit the remaining data.

Calculation

Let's calculate the minimum possible time:

1. Time to download the first 1024 kilobytes of data by Vasya: - Data size: 1024 kilobytes = 1024 * 8 * 1024 bits. - Data transfer rate: 2^17 bits per second. - Time = (1024 * 8 * 1024) / (2^17) seconds.

2. Time to transmit the remaining data to Petya through the slower channel: - Remaining data size: 8 megabytes - 1024 kilobytes = (8 * 1024 * 1024 - 1024 * 8 * 1024) bits. - Data transfer rate: 2^16 bits per second. - Time = ((8 * 1024 * 1024 - 1024 * 8 * 1024) bits) / (2^16) seconds.

To find the minimum possible time, we add the time to download the first 1024 kilobytes of data and the time to transmit the remaining data.

Answer

The minimum possible time, in seconds, from the moment Vasya starts downloading the data until Petya receives the complete data is the sum of the time to download the first 1024 kilobytes of data and the time to transmit the remaining data through the slower channel.

Let's calculate the values:

1. Time to download the first 1024 kilobytes of data by Vasya: - Data size: 1024 kilobytes = 1024 * 8 * 1024 bits = 8388608 bits. - Data transfer rate: 2^17 bits per second = 131072 bits per second. - Time = (8388608 bits) / (131072 bits per second) = 64 seconds.

2. Time to transmit the remaining data to Petya through the slower channel: - Remaining data size: 8 megabytes - 1024 kilobytes = (8 * 1024 * 1024 - 1024 * 8 * 1024) bits = 6291456 bits. - Data transfer rate: 2^16 bits per second = 65536 bits per second. - Time = (6291456 bits) / (65536 bits per second) = 96 seconds.

The minimum possible time, in seconds, from the moment Vasya starts downloading the data until Petya receives the complete data is 64 seconds (time to download the first 1024 kilobytes) + 96 seconds (time to transmit the remaining data) = 160 seconds.

Therefore, the minimum possible time for Petya to receive the complete data from the moment Vasya starts downloading it is 160 seconds.