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# CS501 Assignment No. 03 Solution & Discussion Due Date:27-12-2011

 Question No 1 Marks: 5 Let us consider an I/O bus that can transfer 6 bytes of data in one bus cycle. presume that a designer is considering to attach the following two components to this bus:   Hard drive with a transfer rate of 80 M bytes/sec Video card with a transfer rate of 200 M bytes/sec   What will be the implications?(The maximum frequency of the bus is 30 MHz) Question No 2 Marks: 5 If a bus requires 15 nsec for bus requests, 10 nsec for arbitration and the average time to complete an operation is 10 nsec after the access to the bus has been granted, is it possible for such a bus to perform 40 million IOPS?

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### Replies to This Discussion

For complete assignment see the attached file

Attachments:

see pg # 241 of handouts. u can solve both questions very easily.

thnx saadali for this useful information

Maximum frequency -  30 MHz4

Maximum bandwidth of bus - 30 x 4  -120 MB/sec

Demand of bandwidth  = 200+80 = 280 MB/Sec

Which is more than 120 MB/sec

Thus one or both of the components will be operating at reduced bandwidth

QUESTION NO 2

For 40 million IOPS, the average time for each IOP

6

= 1/(40 x 10  )=16 nsec

Sum of three times= 15+10+10=35 nsec i.e

6

bus can perform a maximum of 1(35x 10)= 28.6 million IOPS

Thus , it will not be able to perform 40 million IOPS.

Question:1

Let us consider an I/O bus that can transfer 6 bytes of data in one bus cycle. presume that a designer is considering to attach the following two components to this bus:

Hard drive with a transfer rate of 80 M bytes/sec
Video card with a transfer rate of 200 M bytes/sec

What will be the implications if both of these components need to work simultaneously?
(The maximum frequency of the bus is 30 MHz)

The maximum frequency of the bus is 30 MHz.
Maximum bandwidth of this bus is 30 x 6 = 180 Mbytes/sec.
The demand for bandwidth from Hard drive and Video card is 80 + 200 =280 Mbytes/sec which is more than the 180Mbytes/sec that the bus can provide.
Thus, if the designer uses these two components
With this bus, one or both of these components will be operating at reduced bandwidth.

Question:2

If a bus requires 15 nsec for bus requests, 10 nsec for arbitration and the average time to complete an operation is 10 nsec after the access to the bus has been granted, is it possible for such a bus to perform 40 million IOPS?

For 40 million IOPS, the average time for each IOP is 1 / (40 x ) =16 nsec.
Sum of the three times is 15 + 10 + 10 = 35 nsec
The bus can perform a maximum of 1 / (35 x ) = 28.6 million IOPS.
Thus, it will not be able to perform 40 million IOPS.

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