Bits, Bytes, transmission rates

[Timiop2008]

Problem :
a) A good sized English book has about 100000 words. If an average English word is six characters long and each character needs 1 byte, show that the book can be stored in 600 kbyte of memory and show that the text takes nearly 5 seconds to be transmitted at 1 Mbit s-¹. Show that it takes nearly a quarter of an hour to transmit 100 images, each of 1Mbyte on a channel with a transmission rate of 1 Mbit s-¹.

Attempt:
600000 X 8 =4800000 bits - text
600 X 1024 X 8 =4915200 bits - available

1Mbit = 1,000,000 bits = 125000 bytes
600000/125000 = 4.8 seconds

100Mbytes = 100,000,000 bytes / 1,000,000 bytes = 100?

These attempts are total guesses. Can somebody please try and explain the proper solutions.

 

[LowlyPion]

Almost.

The first one is a little ambiguous in that it says 6 characters per word. You might inquire of your instructor why it isn't 7 since every word needs a delimiter like a space or a period. But let's presume that 6 is the round number they want you to consider. So yes, you have more bits available than required.

And yes, too on transmitting them in 4.8 seconds.

For the last one simply use your transmission rate from 2).
1*10⁶ X 100/12.5*10⁴ = 8*10² sec

Divide by 60 and you should be getting close.

 

[thomate1]

Sure, I can provide a more detailed explanation for you.

First, let's define some terms. A bit is the smallest unit of data in a computer, and it can have a value of either 0 or 1. A byte is a unit of measurement for data storage, and it is equal to 8 bits. Transmission rate refers to the speed at which data can be transmitted over a network or communication channel, and it is typically measured in bits per second (bps) or multiples of bits per second, such as kilobits per second (Kbps) or megabits per second (Mbps).

Now, let's look at the first part of the problem, which asks us to show that a good sized English book can be stored in 600 kilobytes (kbyte) of memory. We are given that the book has 100000 words, and we are told that an average English word is six characters long. So, we can calculate the total number of characters in the book by multiplying 100000 words by 6 characters per word, which gives us 600000 characters. Since each character needs 1 byte of storage, the total storage needed for the book is 600000 bytes. However, we are asked to show this in kilobytes, so we divide 600000 bytes by 1024 (the number of bytes in a kilobyte) to get 585.94 kilobytes. We can round this up to 600 kbytes, which is the amount of memory needed to store the book.

Now, let's look at the second part of the problem, which asks us to show that the text takes nearly 5 seconds to be transmitted at 1 megabit per second (1 Mbps). To do this, we need to convert the storage size (600000 bytes) into bits, since transmission rates are measured in bits per second. Since 1 byte is equal to 8 bits, we can multiply 600000 bytes by 8 to get 4800000 bits. Next, we can divide this by the transmission rate of 1 Mbps to get the time it takes to transmit the text. This gives us 4800000 bits / 1000000 bits per second = 4.8 seconds. So, it would take nearly 5 seconds to transmit the text at 1 Mbps.

Finally, let's look at the last part of the problem, which asks us to show that it