Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

A pitfall cited in section 1.10 of the text book is expecting improve the overal

ID: 3877203 • Letter: A

Question

A pitfall cited in section 1.10 of the text book is expecting improve the overall performance of a computer by improving only one aspect of the computer. Consider a computer running a program that requires 200s, with 50s spent executing FP instructions, 65s executed Load/Store instructions, and 40s spent executing branch instructions, and the rest time for INT instructions.

By how much (percentage) is the total time reduced if the time for FP operations is reduced by 20%?

By how much (percentage) should the time for INT operations be reduced in order to reduce the total time by 15%?

Can the total time be reduced by 25% by reducing only the time for branch instructions? Why?

Explanation / Answer

Given that CPU total time = 200s
Execution time for Floating point instructio = 50s
Execution time for L/S instruction = 65s

Execution time for branch instruction = 40s


1. As per given FP operations is reduced by 20%

So that, reduced time for FP instructions would be= 50s*(100-20)/100 = 40s

Then the total predicted time would be = 40s+(200 -50) = 190s
Then, the total reduced time = CPU total time – Total predicted time = 200 –190 =10s

Time reduced percentage = Total reduced time / CPU total time

= (10s/200s) *100

= 5%


2. As per given if total time is reduced by 15%,
Total Time for INT operation = 200-50-65-40 = 45sec

Percentagereduced = [(Total_Timeold - Total_Timenew) / Total_Timeold] * 100%
==> 15% = (200 - Total_Timenew ) / 200
==> 0.15= (200 - Total_Timenew ) / 200
==> Total_Timenew = (200 -30) = 170s

Total_Timenew would be equal to = Total_Timeold - (x% of Time_TakenINT_op)
==> 170s = 200s - (x% of 45s)
==> (x/100) * 45s = 200s - 170s
==> x = (30/45) * 100%
==> x = 66.666 %

3)
Supose if the time for branch instruction is reduced, then sould be reduced total time by 25%

The total time after 25% reduction = CPU total time – CPU total time * 25/100

= 200 – 200*0.25

= 150
it means that must be substracted from it will be 50sec After reduction.

But we know that, branch instructions are having atmost 40s, so we cannot reduce 25% of total time

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote