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# System prepended metadata

title: 1.5)

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                                Homework 1
# 1.5)

A.

P1= 3.0Ghz & CPI - 1.5

P1: 3.0GHz / 1.5 = 2 * 10^9 instructions per second

P2= 2.5Ghz & CPI - 1

P2: 2.5GHz / 1 = 2.5 * 10^9 instructions per second

P3= 4.0Ghz & CPI - 2.2

P3: 4.0 GHz / 2.2 = 1.81 * 10^9 instructions per second

( P2 is the highest one.)

B.

Cycles:

P1: 3GHz * 10 = 3 * 10^10 cycles
P2: 2.5GHz * 10 = 2.5 * 10^10 cycles
P3: 4GHz * 10 = 4 * 10^10 cycles

Num of instructions:
P1: 3GHz * 10 / 1.5 = 2 * 10^10 instructions
P2: 2.5GHz * 10 / 1.0 = 2.5 * 10^10 instructions
P3: 4GHz * 10 / 2.2 = 1.82 * 10^10 instructions

C.

Execution time * 0.7 = (Num of instructions * CPI * 1.2) / New Clock rate
New Clock rates:

P1: 3GHz * 1.71 = 5.13 GHz

P2: 2.5GHz * 1.71 = 4.27 GHz

P3: 4GHz * 1.71 = 6.84 GHz

# 1.6)

A.

Class A : 10^5

Class B : 2 X 10^5

Class C : 5 × 10^5

Class D : 2 × 10^5


Time = No. instr. × CPI/clock rate
Total time P1 = (10^5 + 2 * 10^5 * 2 + 5 * 10^5 * 3 + 2* 10^5 * 3) /(2.5 * 10^9) = 10.4 ×10^−4s
Total time P2 = (10^5 * 2 + 2 * 10^5 * 2 + 5 * 10^5 * 2 + 2 * 10^5 * 2) / (3 * 10^9) = 6.66*10^−4s
CPI(P1) = 10.4 * 10^−4 × 2.5 * 10^9/10^6 = 2.6
CPI(P2) = 6.66 * 10^−4 × 3 * 10^9/10^6 = 2.0

B.

clock cycles (P1) = 10^5 * 1 + 2 * 10^5 * 2 + 5 * 10^5 * 3 + 2 * 10^5 * 3 = 26 * 10^5
clock cycles (P2) = 10^5 * 2 + 2 * 10^5 * 2 + 5 * 10^5 * 2 + 2 * 10^5 * 2 = 20 * 10^5

# 1.7)

A. 

Compiler A CPI = 1.1 / (1.0E9 × 1.0E-9) = 1.1 
Compiler B CPI = 1.5 / (1.2E9 × 1.0E-9) = 1.25

B.

1.5s/1.1s = 1.36 
B’s processor is 1.36 times faster than A. 

C.

Cycles = CPI * instructions = 1.1 X 6.0 X 108 = 6.6 X 108 
Time = 6.6 X 108 X 1ns = 0.66s 

Speed up for A: 1.1s / 0.66s = 1.67 
Speed up for B: 1.5s / 0.66s = 2.27
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# 1.11)

1.
CPI(bzip2)=0.94

2.
SPECratio(bzip2)=9650/750=12.86 

3.
10% increase.

4.
CPUtime(after)/CPU time(before)= 1.1*1.05=1.155. Thus, CPU time is increased by 15.5%. 

5.
SPECratio(after)/SPECratio(before)=CPU time(before)/CPU time(after)=1/1.1555 =   0.86. 
 The SPECratio is decreased by 14%. 

6.
CPI= 1.37. 

7.
Clock rate ratio = 4GHz/3GHz=  1.33 
CPI at 4GHz=1.37,CPI at 3GHz=0.94, ratio=  1.45 

the number of instructions are reduced by 15%.

8.
700/750=0.933. CPU time reduction is 6.7%.

9.
Number of instruments= 2146*10^9.
 
10.

Clock rate new= No.instr * CPI/0.9 * CPU time = 1/0.9 clock rate old= 3.33GHz
 
11.
 
Clock rate new= No. instr * 0.85 * CPI/0.80 * CPU time = 0.85/0.80 clock rate old= 3.18GHz.

# 1.14)

A.

clock cycles = 512*10^6.
T CPU= 0.256s.

CPI improved fp=(256−462)/50<0, thus not possible 

B.
CPI improved l/s=(256−198)/80=0.725

C.

CPI int= 0.6*1=0.6
CPI fp= 0.6*1=0.6
CPI l/s= 0.7*4=2.8 
CPI branch= 1.4
TCPU before improvement=0.256s 
TCPU after improvement=0.171s
                               