6 replies to this topic

[Apprentice123]

A 2Ghz CPU executes 2 instructions/clock. An operation I/O that demands the execution of 100,000 instructions by the operating system and 100,000 by the running application (transfer of 64 KB). Set the throughput required for the bus system so that this bus can support the maximum traffic allowed by the CPU

[WingedPanther]

I'm not sure what your question about the problem is.

[dargueta]

I smell homework...what's your question?

[JCoder]

The question is clear: what is the throughput?
It is easy:

1 IO op. = 200 000 CPU cycles
Thus, a 2GHz CPU can do 10000 IO operations per second.

10000 /s * 64 kB = 640 000 kB / s = 625 MB / s = 5000 Mbit / s ~ 5 Gbit / s

[Apprentice123]

JCoder said:

The question is clear: what is the throughput?
It is easy:

1 IO op. = 200 000 CPU cycles
Thus, a 2GHz CPU can do 10000 IO operations per second.

10000 /s * 64 kB = 640 000 kB / s = 625 MB / s = 5000 Mbit / s ~ 5 Gbit / s

Sorry, I did not understand your calculation.
We have:
CPU 2 Ghz that run 2 instructions/clock
IO run 100.000 instructions of operation system and run 100.000 instructions by the sofware in execution.
Transference 64KB

The question is what the transfer rate for the bus ? (The bus need support the maximun traffic in CPU)

[JCoder]

Oops, I missed the statement saying there are 2 instructions per cycle executed. So the final transfer should be twice as high as in my previous calculations - 10 Mbit/s.

Which part you don't understand?
The idea is simple - you just calculate how many I/O operations per second your CPU can handle. And if you know this and the size of the I/O operation (64 kB), just by multiplying you get the transfer rate.

[Apprentice123]

JCoder said:

Oops, I missed the statement saying there are 2 instructions per cycle executed. So the final transfer should be twice as high as in my previous calculations - 10 Mbit/s.

Which part you don't understand?
The idea is simple - you just calculate how many I/O operations per second your CPU can handle. And if you know this and the size of the I/O operation (64 kB), just by multiplying you get the transfer rate.

I do not understand the steps of your calculation. The problem is CPU 2Ghz processes 2 instructions per click of clock. 100.000 instructions of operation system (SO) 100.000 instructions of software.

IO op = 200.000 / 2 instruction per second = 100.000 ?
Transfer rate = 100.000 * 64 Kb = 6400000 Kb/s = 6400 Mb/s = 6,4 Gb/s ??

Edited by Apprentice123, 07 December 2009 - 03:47 AM.