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1.10

給分別的指令數、CPI

1.10.1

求指令數

a:

processor instruction

(1) 2560 + 1280 + 256 = 4096

(2) 1280 + 640 + 128 = 2048

(4) 640 + 320 + 640 = 1024

(8) 320 + 160 + 32 = 512

Total instruction

(1) 4096*1 = 4096

(2) 2048*2 = 4096

(4) 1024*4 = 4096

(8) 512*8 = 4096

b:

processor instruction

(1) 2560 + 1280 + 256 = 4096

(2) 1350 + 800 + 128 = 2278

(4) 800 + 600 + 64 = 1464

(5) 600 + 500 + 32 = 1132

Total instruction

(1) 4096*1 = 4096

(2) 2278*2 = 4556

(4) 1464*4 = 5856

(8) 1132*8 = 9056

1.10.2

找Execution time (clock rate = 2GHz)

a:

(1)

2560*1 + 1280*4 + 256*2

 = 2560 + 5120 + 512

 = 8192

8192/2GHz = 4096 ns

(2)

1280*1 + 640*4 + 128*2

 = 1280 + 2560 + 256

 =4096

4096 / 2GHz = 2048 ns

(4)

640*1 + 320*4 + 64*2

 = 640 + 1280 + 238

 = 2048

2048 / 2GHz = 1024 ns

(8)

320*1 + 160*4 + 32*2

 = 320 + 640 +64

 = 1024

1024 / 2GHz = 512 ns

b:

(1)

2560*1 + 1280*4 + 256*2

 = 2560 + 5120 + 512

 = 8192

8192 / 2GHz = 4096 ns

(2)

1350*1 + 800*6 + 128*2

 = 1350 + 4800 + 256

 = 6406

6406 / 2GHz = 3203 ns

(4) 800*1 + 600*9 + 64*2

 = 800 + 5400 + 128

 = 6328

6328 / 2GHz = 3164ns

(8) 600*1 + 500 *13 + 32*2

 = 600 + 6500 + 64

 = 7164

7164 / 2GHz = 6582ns

1.10.3

Arithmetic 的指令數變為兩倍,對於Execution time的影響

a:

(1)

2560*2*1 + 1280*4 + 256*2

 = 5120 + 5120 + 512

 = 10752

10752 / 2GHz = 5376ns

(2)

1280*2*1 + 640*4 + 128*2

 = 2560 + 2560 + 256

 = 5376

5376 / 2GHz = 2688ns

(4)

640*2*1 + 320*4 + 64*2

= 1280 + 1280 + 128

= 2688

2688 / 2GHz = 1344ns

(8)

320*2*1 + 160*4 + 32*2

 = 640 + 640 + 64

 = 1344

1344 / 2GHz = 372ns

b:

(1)

2560*2*1 + 1280*4 + 256*2

= 5120 + 5120 + 512

= 10752

10752 / 2GHz = 5376ns

(2)

1350*2*1 + 800*6 + 128*2

 = 2700 + 4800 + 256

 = 7756

7756 / 2GHz = 3878ns

(4)

800*2*1 + 600*9 + 64*2

 = 1600 + 5400 + 128

 = 7128

7128 / 2GHz = 3564ns

(8)

600*2*1 + 500*13 + 32*2

 = 1200 + 6500 + 64 

 = 7764

7764 / 2GHz = 3882ns

<我是Exercise 1.10的分隔線>

也是給a、b

分別有 Cores per processor、Instructions per core、Averge CPI

<我是Exercise 1.10的分隔線>

1.10.4

假設 3GHz clock frequency

求execution time

a:

(1)

( 1*10^10*1.2 ) / 3GHz

 = 4s

(2)

( 5*10^9*1.3 ) / 3GHz

 = 2.167s

(4)

( 2.5*10^9*1.5 ) / 3GHz

 = 1.25s

(8)

( 1.25*10^9*1.8 ) / 3GHz

= 0.75s

b:

(1)

( 1*10^10*1.2 ) / 3GHz

= 4s

(2)

(5*10^9*1.2) / 3GHz

= 2s

(4)

( 2.5*10^9*1.2 ) / 3GHz

= 1s

(5)

( 1.25*10^9*1.2 ) / 3GHz

 = 0.5s

1.10.5

3GHz Clock:

Voltage

= (2/5)+0.4 = 1

(1)

Power 

= (5*10^(-3))*(1.0^2)*(3*10^3)

= 15W

(2)

15*2 = 30

(4)

15*4 = 60

(8)

15*8 = 120

500MHz Clock:

Voltage

 = (0.5/5)+0.4

 = 0.5

(1)

Power

 = (5*10^(-3))*(0.5^2)*(5*10^2)

 = 0.625W

(2)

0.625*2 = 1.25W

(4)

0.625*4 = 2.5W

(8)

0.625*8 = 5W

1.10.6

3Gz Clock

a.

(1)

4*15 = 60

(2)

30*2.167 = 65.01

(4)

60*1.25 = 75

(8)

120*0.75 = 90

b.

(1)

4*15 = 60

(2)

2*30 = 60

(4)

1*60 = 60

(8)

0.5*120 = 60