TEST 1 – Tuesday March 3 Lectures 1 - 8, Ch 1,2 HW Due Feb 24 –1.4.1 p.60 –1.4.4 p.60 –1.4.6 p.60 –1.5.2 p.60-61 –1.5.4 p.61 –1.5.5 p.61

Computer Architecture CSE 3322Lecture 8

TEST 1 – Tuesday March 3

Lectures 1 - 8, Ch 1,2

Web Sitecrystal.uta.edu/~cse3322

TEST 1 – Tuesday March 3Lectures 1 - 8, Ch 1,2

• HW Due Feb 24– 1.4.1 p.60– 1.4.4 p.60– 1.4.6 p.60– 1.5.2 p.60-61– 1.5.4 p.61– 1.5.5 p.61

CPI

Invest Resources where time is Spent!

CPI = Clock Cycles / Instruction Count = (CPU Time * Clock Rate) / Instruction Count

“Average clock cycles per instruction”

CPU Time = Instruction Count x CPI / Clock Rate = Instruction Count x CPI x Clock Cycle Time

Average CPI = SUM of CPI (i) * I(i) for i=1, n Instruction Count

Average CPI = SUM of CPI(i) * F(i) for i = 1, nF(i) is the Instruction Frequency

Example (RISC processor)

Typical Mix

Base Machine (Reg / Reg)

Op Freq Cycles F(i)CPI(i) % Time ALU 50% 1

Load 20% 5

Store 10% 3

Branch 20% 2


Typical Mix


Op Freq Cycles F(i)CPI(i) % Time

ALU 50% 1 .5

Load 20% 5 1.0

Store 10% 3 .3

Branch 20% 2 .4

2.2 = CPI ave


Typical Mix



ALU 50% 1 .5

Load 20% 5 1.0

Store 10% 3 .3

Branch 20% 2 .4

2.2 = CPI ave

CPU Time(i) = Instr Cnt(i) * CPI(i) * Clk Cycle TimeCPU Time Inst Cnt * CPI ave * Clk Cycle Time

% Time = F(i) * CPI(i) / CPI ave


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2 = CPI ave

CPU Time(i) = Instr Cnt(i) * CPI(i) * Clk Cycle TimeCPU Time Inst Cnt * CPI ave * Clk Cycle Time

% Time = F(i) * CPI(i) / CPI ave


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2 = CPI ave

How much faster would the machine be if a better data cachereduced the average load time to 2 cycles?


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 (2) 1.0 (.4) 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2 (1.6)

How much faster would the machine be if a better data cachereduced the average load time to 2 cycles?

2.2/1.6 = 1.375

CPU Time = Inst Cnt * CPI ave * Clk Cycle Time


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2

How much faster would the machine be if a better data cachereduced the average load time to 2 cycles? CPI = 1.6

How does this compare with using branch prediction to shave a cycle off the branch time?


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 (1) .4 (.2) 18%

2.2 (2.0)


How does this compare with using branch prediction to shave a cycle off the branch time? CPI = 2.0


Typical Mix



ALU 50% 1 .5 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2



What if two ALU instructions could be executed at once?


Typical Mix



ALU 50% 1 (.5) .5 (.25) 23%

Load 20% 5 1.0 45%

Store 10% 3 .3 14%

Branch 20% 2 .4 18%

2.2 (1.95)



What if two ALU instructions could be executed at once? CPI=1.95

A compiler designer is trying to decide between two code sequences for a particular machine. Based on the hardware implementation, there are three different classes of instructions:

Class A has 1 cycle Class B has 2 cycles Class C has 3 cycles

The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C

The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C.

Which sequence will be faster? How much?What is the CPI for each sequence?



The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second

sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be faster? How much?

What is the CPI for each sequence?



The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be

faster? How much? 10 / 9 = 1.11What is the CPI for each sequence?



The first code sequence has 5 instructions: 2 of A, 1 of B, and 2 of C 2*1+1*2+2*3 = 10 The second sequence has 6 instructions: 4 of A, 1 of B, and 1 of C. 4*1+1*2+1*3 = 9 Which sequence will be faster?

How much? 10 / 9 = 1.11What is the CPI for each sequence? 10/5 = 2

9/6 = 1.5

MIPS = Instruction Count

Execution time x 10 6

A popular performance metric is MIPS, the numberof millions of instructions per second.

For a given program,





1. Cannot compare if instruction set is different





1. Cannot compare if instruction set is different2. Highly dependent on the program





1. Cannot compare if instruction set is different2. Highly dependent on the program3. Can be inversely proportional to performance

Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions: Class A has 1 cycle,Class B has 2 cycles, Class C has 3 cycles

Instruction counts ( billions)Code from A B CCompiler 1 5 1 1Compiler 2 10 1 1

• Which sequence will be faster according to MIPS?• Which sequence will be faster according to execution

time?

MIPS example

Two different compilers are being tested for a 100 MHz. machine with three different classes of instructions:

Class A Class B Class C

CPI 1 2 3 Instruction counts ( billions)

Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 5+1x2+1x3=10 billionCompiler 2

MIPS example




Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billionCompiler 2 15 billion

MIPS example




Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billion 1010x10-8=100Compiler 2 15 billion

MIPS example




Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billion 100 secCompiler 2 15 billion 150 sec

MIPS example




Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billion 100 sec 7x103/100Compiler 2 15 billion 150 sec

MIPS example


Class A Class B Class CCPI 1 2 3

Instruction counts ( billions)Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billion 100 sec 70Compiler 2 15 billion 150 sec 12x103/150

MIPS example




Code from A B C TotalCompiler 1 5 1 1 7Compiler 2 10 1 1 12 CPU cycles Exec Time MIPSCompiler 1 10 billion 100 sec 70Compiler 2 15 billion 150 sec 80

MIPS example

• Performance best determined by running a real application– Use programs typical of expected workload– Or, typical of expected class of applications

e.g., compilers/editors, scientific applications, graphics, etc.

Benchmarks

• Performance best determined by running a real application– Use programs typical of expected workload– Or, typical of expected class of applications

e.g., compilers/editors, scientific applications, graphics, etc.

• Small benchmarks– nice for architects and designers– easy to standardize– can be abused

Benchmarks

SPEC CPU 2006

• SPEC - System Performance Evaluation Cooperative• 12 Integer Benchmarks• 17 Floating Point Benchmarks• Fig 1.20 P.49

Execution Time After Improvement =

Execution Time Unaffected +

( Execution Time Affected / Amount of Improvement )

Amdahl's Law


Execution Time Unaffected + ( Execution Time Affected / Amount of Improvement )

• Example:Suppose a program runs in 100 seconds on a

machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?

Amdahl's Law



• Example:Suppose a program runs in 100 seconds on a machine,

with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? Improved Time = (100 – 80) + 80/n = 100/4

Amdahl's Law



• Example:Suppose a program runs in 100 seconds on a machine, with

multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster? Improved Time = (100 – 80) + 80/n = 100/4

20 + 80/n = 25 80 = 5n ; n = 16

Amdahl's Law



• Example:Suppose a program runs in 100 seconds on a

machine, with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?

How about making it 5 times faster?

Amdahl's Law



• Example:Suppose a program runs in 100 seconds on a machine,

with multiply responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?

How about making it 5 times faster?Improved Time = (100 –80) + 80/n = 100/5

Amdahl's Law



• Example:Suppose a program runs in 100 seconds on a machine, with multiply

responsible for 80 seconds of this time. How much do we have to improve the speed of multiplication if we want the program to run 4 times faster?

How about making it 5 times faster?Improved Time = (100 –80) + 80/n = 100/5 20 + 80/n = 20

80/n = 0 Impossible!

Amdahl's Law

• Performance is specific to a particular program/s

– Total execution time is a consistent summary of performance

Remember

• Performance is specific to a particular program/s– Total execution time is a consistent summary of

performance

• For a given architecture performance increases come

from:– increases in clock rate (without adverse CPI affects)

Remember

• Performance is specific to a particular program/s– Total execution time is a consistent summary of

performance

• For a given architecture performance increases come from:– increases in clock rate (without adverse CPI affects)– improvements in processor organization that lower CPI

Remember

• Performance is specific to a particular program/s– Total execution time is a consistent summary of performance

• For a given architecture performance increases come from:– increases in clock rate (without adverse CPI affects)– improvements in processor organization that lower CPI– compiler enhancements that lower CPI and/or instruction

count

Remember

Documents

TEST 1 – Tuesday March 3 Lectures 1 - 8, Ch 1,2 HW Due Feb 24 –1.4.1 p.60 –1.4.4 p.60 –1.4.6 p.60 –1.5.2 p.60-61 –1.5.4 p.61 –1.5.5 p.61