Bits, Bytes, transmission rates

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SUMMARY

The discussion focuses on calculating the storage and transmission requirements for a 100,000-word English book and 100 images, each 1 MB, using a transmission rate of 1 Mbit/s. It is established that the book can be stored in 600 kB of memory, and it takes approximately 4.8 seconds to transmit the text. Additionally, transmitting 100 images at the same rate requires nearly 15 minutes, confirming the significant impact of transmission rates on data transfer times.

PREREQUISITES
  • Understanding of data storage units (bytes, kilobytes, megabytes)
  • Knowledge of transmission rates (Mbit/s)
  • Basic arithmetic for calculating data transfer times
  • Familiarity with character encoding (1 byte per character)
NEXT STEPS
  • Research data transmission protocols and their efficiencies
  • Learn about data compression techniques to reduce file sizes
  • Explore the impact of network latency on data transfer times
  • Investigate different storage formats and their space requirements
USEFUL FOR

Students in computer science, network engineers, and anyone involved in data transmission and storage optimization.

Timiop2008
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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.
 
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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.
 

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