4 replies to this topic

[Apprentice123]

A program runs in 30s in system A, with CPU A, MEM A and IO A.
The system A is updated a CPU B (twice faster that CPU A) and MEM B (twice faster that MEM A).
After the change the program run in 22s. What percentage of time the program is dedicated to IO operations ?

[dargueta]

Dude...that's homework. We help with homework, but don't do it.

If it weren't for IO time, the execution time would be cut in half. But it isn't; it just reduces by 8 seconds, or -26.7% of the initial time. So we have:

DELTA = (Final - Initial) / Initial

Execution time comprises of instruction time plus memory access time plus IO time, so we have:

DELTA = ((CPUB + MEMB + IO) - (CPUA + MEMA + IO)) / (CPUA + MEMA+IO)

All you need to do is solve for IO and you're good.

[Apprentice123]

dargueta said:

Dude...that's homework. We help with homework, but don't do it.

If it weren't for IO time, the execution time would be cut in half. But it isn't; it just reduces by 8 seconds, or -26.7% of the initial time. So we have:

DELTA = (Final - Initial) / Initial

Execution time comprises of instruction time plus memory access time plus IO time, so we have:

DELTA = ((CPUB + MEMB + IO) - (CPUA + MEMA + IO)) / (CPUA + MEMA+IO)

All you need to do is solve for IO and you're good.

I think

CPU + IO = 30
(CPU/2) + IO = 22

IO = 14s

I correct ?

[dargueta]

Um...yes. My way is much more complicated. As usual. :D

[Apprentice123]

dargueta said:

Um...yes. My way is much more complicated. As usual. :D

Thank you