FPinstr. INTinstr. L/Sinstr. Branchinstr. Total Time a. 35s 85s 50s 30s 200s b.
ID: 3616680 • Letter: F
Question
FPinstr. INTinstr. L/Sinstr. Branchinstr. Total Time a. 35s 85s 50s 30s 200s b. 50s 80s 50s 30s 210s a) By how much is the total time reduced if the time for FPoperations is reduced by 20%? b) By how much is the time for INT operations reduced if thetotal time is reduced by 20%? c) Can the total time be reduced by 20% by reducing only thetime for branch instructions? FPinstr. INTinstr. L/Sinstr. Branchinstr. Total Time a. 35s 85s 50s 30s 200s b. 50s 80s 50s 30s 210s a) By how much is the total time reduced if the time for FPoperations is reduced by 20%? b) By how much is the time for INT operations reduced if thetotal time is reduced by 20%? c) Can the total time be reduced by 20% by reducing only thetime for branch instructions?Explanation / Answer
Actual time of FP=35 s
20% of this time=35*20/100=7 sec
Reducing 20% of FP, FP will take=28 S
FP instr
INT instr.
L/S instr
Branch instr
Total Time
28
85
50
30
193
b) if we reduce the total time 20% then it willbe=200-(200*20/100)=200-40=160 Sec
every Instruction will reduce to 193/200 Sec
FP instr
INT instr.
L/S instr
Branch instr
Total Time
35 S
85
50
30
193
New Values will be given below
34 S
82
48
29
193
c) Reducing the total 20% will be 193
if we reduce the branch instruction from 30 to 23 which is 24percent reduction in this command
using same way you can calculate for b.
FP instr
INT instr.
L/S instr
Branch instr
Total Time
28
85
50
30
193
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