Consider an implementation of the instruction set architecture P1 has a clock rate of 4.8 GHz and CPIs (Cycles Per Instruction) of 1 for arithmetic, 4 for load/store and 8 for branch. A program execution consists of 1 million instructions, including 50% arithmetic, 30% load/store, and 20% branch instructions. In this case, the average CPI is .
For a 10% improvement in performance of P1, a load/store instruction will take cycles on average if other classes are NOT improved at all.

a) 3.3

b) 3

Explanation:

1)

As we know ,

CPI = sum of ( instructions× clock cycle)

     = 50%(1) + 30%(4) + 20%(8)

    = 50% + 120% + 160%

    = 330%

    = = 3.3

⇒CPI = 3.3

2)

Let a = number of cycles on average

As given,

a 10% improvement in performance of P1,

⇒New CPI = 3

⇒3 = 50% + 30% (x) + 20%(8)

⇒3 = 50% + 30% (x) + 160%

⇒3 = 210% + 30%(x)

⇒3 = 2.1 + 0.3x

⇒3 - 2.1 = 0.3x

⇒0.9 = 0.3x

⇒x = = 3

⇒x = 3

∴ we get

Number of cycles on average = 3

power left de cliffs or inoonsw

explanation:

d. long term memory