Computer Architecture-Secondary Memory

Question

Susheel is setting up a website. He bought a fancy new hard disk which advertises:
• an 8 ms average seek time.
• 10000 RPM or roughly 6 ms per rotation.
• a2 ms overhead for each disk operation.
• a transfer speed of 10,000,000 bytes per second
Susheel had enough money left for a 10,000,000 bytes per second network connection. His system bus has a maximum baqndwidth of 133 mega bytes per second and his HTML files have an average size of 8000 bytes. How much time will it take on an average to read a random HTML file from the disk?

Comments

17.01 ms

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@Tesla! can you post the solution?

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Total time in loading is seek + rotational + transfer

Transfer =$\frac{800}{10*10^{6}}$=0.08 ms

So 8+3+0.08=11.08ms

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@Tesla! disk transfer speed 10 mega byte per sec is less than the maximum bandwidth 133 mega byte per sec. shouldn't we use the speed of secondary device as it will limit the transfer speed?

please clear my doubt. and also provide some good resource for secondary memory concept. I always stuck in these type of question.

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@ananya jaiswal ,yes i too think that disk transfer speed is the bottleneck and it is considered while the calculation of answer.

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@satendra @Ananya Jaiswal 1 Both of you are correct I made a stupid mistake, the bottleneck is disk transfer rate.

The basic concept is given in all standard book is to apply some logic to get 

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@Tesla

transfer time will be .8 microsec

not milisec

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14.06 ms

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8 ms(seek time) + 3 ms(rotational time) + 0.8 ms(transfer time) + 2ms(overhead) =13.8 ms

Answer 1

Answer is 13.8 ms

as 8ms ( seek time ) + 2 ms ( disk overhead) +  0.8 ( for transfer bottle neck bandwidth will be used which is 10Mbps not 133 Mbps )