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Some more definitions:
MIPS millions of instructions per second
MOPS millions of operations per second
MFLOPS millions of floating point operations per second
MIOPS millions of integer operations per second
(Classical, old, terms. Today would be billions.)
Amdahl's Law (many forms, understand the concept)
the part of time improved
new time = ------------------------- + the part of time not improved
factor improved by
old time = the part of time improved + the part of time not improved
old time
speedup = --------
new time
Given: on some program, the CPU takes 9 sec and the disk I/O takes 1 sec
What is the speedup using a CPU 9 times faster?
9 sec
Answer: new time = ----- + 1 sec = 2 sec
9
old time = 9 + 1 = 10 sec
speedup = 10 / 2 = 5 a pure number
------------------------------------------------------------------------------
Amdahl's Law (many forms, understand the concept)
new performance
speedup = ---------------
old performance
Given: Performance of M1 is 100 MFLOPS and 200 MIOPS
Performance of M2 is 50 MFLOPS and 250 MIOPS
On a program using 10% floating point and 90% integer
Which is faster? What is the speedup?
Answer; .1 * 100 + .9 * 200 = 190 MIPS
.1 * 50 + .9 * 250 = 230 MIPS (M2 is faster)
speedup = 230/190 = 1.21
------------------------------------------------------------------------------
old performance
new performance = -----------------------------------------------------
fraction of old improved
------------------------ + fraction of old unimproved
improvement factor
Given: half of a 100 MIPS machine is speeded up by a factor of 3
what is the speedup relative to the original machine?
1 1
Answer: new performance = --------- * 100 MIPS = ---- * 100 MIPS = 150 MIPS
0.5 .666
--- + 0.5
3
1
speedup = 150 / 100 = 1.5 (same as -------------------------------)
fraction improved
------------------ + fraction
improvement factor unimproved
------------------------------------------------------------------------------
SPEC Benchmarks
------------------------------------------------------------------------------
CPI is average Clocks Per Instruction. units: clock/inst
MHz is frequency, we use millions of clocks per second. units: clock/sec
MIPS is millions of instruction per second. units: inst/sec
Note: MIPS=MHz/CPI because (clock/sec) / (clock/inst) = 10^6 inst/sec
( 5/4 of people do not understand fractions. )
Computing average CPI, Clocks Per Instruction
-------given--------------- ----------compute------------
type clocks %use product
RR 3 25% 3 * 25 = 75
RM 4 50% 4 * 50 = 200
MM 5 25% 5 * 25 = 125
______ ____
100% 400 sum
400/100 = 4 average CPI
-------given--------------- ----------compute------------
type clocks instructions product
RR 3 25,000 3 * 25,000 = 75,000
RM 4 50,000 4 * 50,000 = 200,000
MM 5 25,000 5 * 25,000 = 125,000
_______ _______
100,000 400,000 sum
400,000/100,000 = 4 average CPI
Find the faster sequence of instructions Prog1 vs Prog2
-------given---------------------
type clocks
A 1
B 2
C 3
instruction counts for A B C
Prog1 2 1 2
Prog2 4 1 1
----------compute------------------------------
Prog1
A 1 2 1 * 2 = 2
B 2 1 2 * 1 = 2
C 3 2 3 * 2 = 6
__ sum
10 clocks
Prog2
A 1 4 1 * 4 = 4
B 2 1 2 * 1 = 2
C 3 1 3 * 1 = 3
__ sum
9 clocks more instructions yet faster
speedup = 10 clocks / 9 clocks = 1.111 a number (no units)
CS411_opcodes.txt different from Computer Organization and Design 8/29/06
rd is register destination, the result, general register 1 through 31
rs is the first register, A, source, general register 0 through 31
rt is the second register, B, source, general register 0 through 31
--val---- generally a 16 bit number that gets sign extended
--adr---- a 16 bit address, gets sign extended and added to (rx)
"i" is generally immediate, operand value is in the instruction
Opcode Operands Machine code format
6 5 5 5 5 6 number of bits in field
nop RR 00 0 0 0 0 00
add rd,rs,rt RR 00 rs rt rd 0 32
sub rd,rs,rt RR 00 rs rt rd 0 34
mul rd,rs,rt RR 00 rs rt rd 0 24
and rd,rs,rt RR 00 rs rt rd 0 04
srl rd,rt,shf RR 00 0 rt rd shf 03 rt>>shf
sll rd,rt,shf RR 00 0 rt rd shf 02 rt<<shf
cmpl rd,rt RR 00 0 rt rd 0 5
j jadr J 02 ------jadr--------
addi rd,rs,val M 07 rs rd ---val----
beq rs,rt,adr M 11 rs rt ---adr----
lw rd,adr(rx) M 35 rx rd ---adr----
sw rt,adr(rx) M 43 rx rt ---adr----
instruction bits (binary of 6 5 5 5 5 6 format above)
3 3 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 1
1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0 9 8 7 6 5 4 3 2 1 0
| | | | |
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 nop
0 0 0 0 0 0 a a a a a b b b b b r r r r r -ignored- 1 0 0 0 0 0 add r,a,b
0 0 0 0 0 0 a a a a a b b b b b r r r r r -ignored- 1 0 0 0 1 0 sub r,a,b
0 0 0 0 0 0 a a a a a b b b b b r r r r r -ignored- 0 1 1 0 0 0 mul r,a,b
0 0 0 0 0 0 a a a a a b b b b b r r r r r -ignored- 0 0 0 1 0 0 and r,a,b
0 0 0 0 0 0 0 0 0 0 0 b b b b b r r r r r s s s s s 0 0 0 0 1 0 sll r,b by s
0 0 0 0 0 0 0 0 0 0 0 b b b b b r r r r r s s s s s 0 0 0 0 1 1 srl r,b by s
0 0 0 0 0 0 0 0 0 0 0 b b b b b r r r r r -ignored- 0 0 0 1 0 1 cmpl r,b
0 0 0 0 1 0 -----address to bits (27:2) of PC------------------ j adr
0 0 0 1 1 1 x x x x x r r r r r ---2's complement value-------- addi r,val(x)
0 0 1 0 1 1 a a a a a b b b b b ---2's complement address------ beq a,b,adr
1 0 0 0 1 1 x x x x x r r r r r ---2's complement address------ lw r,adr(x)
1 0 1 0 1 1 x x x x x b b b b b ---2's complement address------ sw b,adr(x)
Definitions:
nop no operation, no programmer visible registers or memory
are changed, except PC <= PC+4
j adr bits 0 through 25 of the instruction are inserted into PC(27:2)
probably should zero bits PC(1:0) but should be zero already
lw r,adr(x) load word into register r from memory location (register x plus
sign extended adr field)
sw b,adr(x) store word from register b into memory location (register x plus
sign extended adr field)
beq a,b,adr branch on equal, if the contents of register a are equal
to the contents of register b, add the, shifted by two,
sign extended adr to the PC (The PC will have 4 added by then)
addi r,val(x) add immediate, the contents of register x is added to the
sign extended value and the result put into register r
add r,a,b add register a to register b and put result into register r
sub r,a,b subtract register b from register a and put result into register r
mul r,a,b multiply register a by register b and put result into register r
and r,a,b and register a to register b and put result into register r
sll r,b by s shift the contents of register b by s places left and put
result in register r
srl r,b by s shift the contents of register b by s places right and put
result in register r
cmpl r,b one's complement of register b goes into register r
Also: no instructions are to have side effects or additional "features"
last updated 8/27/2006 (slight difference in opcodes from previous semesters)
General register list (applies to MIPS ISA and project)
(note: project op codes may differ from MIPS/SGI)
Register notes
0 $0 zero value, not writable
1 $1
2 $2 $v0 return values (convention, not constrained by hardware)
3 $3 $v1
4 $4 $a0 arguments (convention, not constrained by hardware)
5 $5 $a1
6 $6 $a2
7 $7 $a3
8 $8 $t0 temporaries(not saved by software convention over calls)
9 $9 $t1
10 $10 $t2
11 $11 $t3
12 $12 $t4
13 $13 $t5
14 $14 $t6
15 $15 $t7
16 $16 $s0 saved by software convention over calls
17 $17 $s1
18 $18 $s2
19 $19 $s3
20 $20 $s4
21 $21 $s5
22 $22 $s6
23 $23 $s7
24 $24 $t8 more temporaries
25 $25 $t9
26 $26
27 $27
28 $28 $gp global pointer ( more designations by software convention)
29 $29 $sp stack pointer
30 $30 $fp frame pointer
31 $31 $ra return address
Remember: From a hardware view registers 1 through 31 are general purpose
and identical. The above table is just software conventions.
Register zero is always zero!
Basic digital logic
IA-64 Itanium
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