Xiang Lian The University of Texas – Pan American Edinburg, TX 78539 lianx@utpa

CSCI 6307 Foundation of Systems CSCI 6307 Foundation of Systems Review: Midterm ExamReview: Midterm Exam

Xiang LianThe University of Texas – Pan American

Edinburg, TX [email protected]

Review

• Chapters A1 ~ A4 in your textbook

• Lecture slides

• In-class exercises (1) & (2)

• Assignments 1 & 2

2

Review

• 5 Questions (100 points)

• 1 Bonus Question (20 extra points)

3

Chapter 1 Computer Abstractions and Technology

• Classes of computers

• Evolution of programming languages

• Performance evaluation of computers– What are the criteria of the evaluation?– How to compare the performance of two

computers?

4

Chapter 1 — Computer Abstractions and Technology — 5

CPU Clocking• Operation of digital hardware governed by a

constant-rate clock

Clock (cycles)

Data transferand computation

Update state

Clock period

Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250×10–12s

Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0×109Hz


CPU Time

• Performance improved by– Reducing number of clock cycles– Increasing clock rate– Hardware designer must often trade off clock

rate against cycle count

Rate Clock

Cycles Clock CPU

Cycles Clock CPUTime CPU

PeriodClock


Instruction Count and CPI

• Instruction Count for a program– Determined by program, ISA and compiler

• Average cycles per instruction– Determined by CPU hardware– If different instructions have different CPI

• Average CPI affected by instruction mix

Rate Clock

CPICount nInstructio

Period ClockCPICount nInstructioTime CPU

nInstructio per CyclesCount nInstructioCycles Clock

Chapter 2 Instructions• Binary representation

– 2's complement– Positive/negative integers binary numbers– Addition, negation, sign extension

• Instructions– Arithmetic: add, sub, addi– Data transfer: lw, sw, lb– Logical: and, or, andi, sll, srl– Conditional branch: beq, bne, slt, sltu, slti– Jump: jr, j, jal

• Please understand the meanings of these instructions, and use them to write simple assembly programs.

• Given a set of instructions, write down the output of the code

8

Negative Numbers: 2’s Complement

(xnxn-1…x0)2=-xn×2n+xn-1×2n-1+…+x0×20

Ex:

(00)2= =-0×21+0×20=0

(01)2= =-0×21+1×20=1

(10)2= =-1×21+0×20=-2

(11)2= =-1×21+1×20=-1

9

Good: 1.only one zero2.Easy to add negative and positive numbers3.Two useful operations: negation and sign-extending

Exercise: (11)2 +(01)2=?

(11)2 +(10)2=?

Note: Xn not only represents sign, but has weights (different to the previous sign-magnititude representation).

Chapter 2 — Instruction — 9

2’s Complement: Negation

(011)2=3 binary number of -3?

(100)2 : complement of (011)2

+(001)2 : plus one

=(101)2 : -3

10

Exercise: find the negation of (110)2 and (100)2

Chapter 2 — Instruction — 10

2’s Complement: Sign Extension

Extend (001)2 to 6 bits ??????

(???001) 2 : copy the old bits to the right

(000001) 2 : MSBthe remaining bits

Exercise: extending (101)2 to 6 bits and verify that they are the same integer.

11Chapter 2 — Instruction — 11

Instruction Operations-Arithmetic

• Add $s1,$s2,$s3: $s1=$s2+$s3• Sub $s1,$s2,$s3: $s1=$s2-$s3• Addi $s1,$s2, 20: $s1=$s2+20

Exericse:• For $s1=1,$s2=2,$s3=3, write the results

after executing each instructions.• Write an instruction so that $s1=$s1-1.


Instruction Operations-Data Transfer

• Lw $s1, 30($s2): $s1mem[$s2+30] (load/read word from memory)

• Sw $s1, 20($s2): $s1mem[$s2+20] (store/write a word to memory)

• Lb $s1, 30($s2): $s1mem[$s2+30] (load/read a byte from memory)

Exercise: read a word/byte at $s2+20 in memory to $s2.


Instruction Operation-Logical

• And– And $s1,$s2,$s3

• Or– Or $s1,$s2,$s3

• And immediate– Andi $s1,$s2,20

• Shift left logic– Sll $s1, $s2,10

• Etc.


Instruction Operation-Conditional Branch

• Branch if equal– Beq $s1, $s2, 25

• Branch if not equal– Bne $s1, $s2, 25

• Set on less than– Slt $s1,$s2,$s3

• Set on less than unsigned– Sltu $s1,$s2,$s3

• Set on less than immediate– $slti $s1,$s2,20


Instruction Operation-Unconditional Jump

• Jump register– Jr $ra

• Jump– J 2500

• Jump-and-link instruction– Jal 2500


Chapter 3 Arithmetic

• Overflow of addition operator– How to detect it?

• Multiplication– Two approaches

• Floating point representation– Scientific notation– Conversion from decimal number to binary number– Conversion from binary number to floating point

number– Conversion from floating point number to IEEE

format

17

Addition Overflow

• Occurs when the result is out the range for a given number of bits.

Ex:

18

1011 (-5)+ 1100 (-4) 10111 (7×)

0111 (7)+ 0100 (4) 11011 (-5×)

1111 (-1)+ 1100 (-4) 11011 (-5)

0011 (3)+ 0100 (4) 10111 (7)

Chapter 3 — Arithmetic for Computers — 18

Checking of Addition Overflow When Two Operands Differ in Sign• Does not occurs

Ex:

19

1011 (-5)+ 0100 (4) 11111 (-1)

1111 (-1)+ 0100 (4) 10011 (3)


Overflow Conditions

20

Operation Oprand A Operand B Result indicating overlfow

A+B >=0 >=0 <0

A+B <0 <0 >=0

A-B >=0 <0 <0

A-B <0 >=0 >=0



Multiplication• Start with long-multiplication approach

1000× 1001 1000 0000 0000 1000 1001000

Length of product is the sum of operand lengths

multiplicand

multiplier

product

§3.3 Multiplication


IEEE Floating-Point Format

• S: sign bit (0 non-negative, 1 negative)• Normalize significand: 1.0 ≤ |significand| < 2.0

– Always has a leading pre-binary-point 1 bit, so no need to represent it explicitly (hidden bit)

– Significand is Fraction with the “1.” restored• Exponent: excess representation: actual exponent + Bias

– Ensures exponent is unsigned– Single: Bias = 127; Double: Bias = 1203

S Exponent Fraction

single: 8 bitsdouble: 11 bits

single: 23 bitsdouble: 52 bits

Bias)(ExponentS 2Fraction)(11)(x

Converting Floating Point Numbers

23

5.1875

101. .0011

0.1875×2 =0.375 0

0.375×2 =0.75 0

0.75×2 =1.5 1

0.5×2 =1 1101.0011

1.010011×22 0 10000001 0100110…0(23 bits)

Ex. Convert 5.1875 to its binary representation (single precision).


Chapter 4 Processor

• CPU overview• Logical design

– Combinational element– State element– Their differences

• Datapath of instruction execution– 5 stages– Calculation of the clock cycle

• Pipeline– Speedup computation– 3 hazards of the pipeline

24

Chapter 4 — The Processor — 25

Logic Design Basics

§4.2 Logic Design C

onventions

• Information encoded in binary– Low voltage = 0, High voltage = 1– One wire per bit– Multi-bit data encoded on multi-wire buses

• Combinational element– Operate on data– Output is a function of input

• State (sequential) elements– Store information


MIPS Pipeline

• Five stages, one step per stage1. IF: Instruction fetch from memory

2. ID: Instruction decode & register read

3. EX: Execute operation or calculate address

4. MEM: Access memory operand

5. WB: Write result back to register

Determining Clock Cycle (And)


I-Mem: 400psAdd: 100psMux: 30psALU: 120psRegs: 200psD-Mem: 350psControl: 100ps

300

200

700

600

700

800

800

830950

950

950

980

980

11801180


Pipelining Analogy• Pipelined laundry: overlapping execution

– Parallelism improves performance

§4.5 An O

verview of P

ipelining Four loads: Speedup

= 8/3.5 = 2.3 Non-stop:

Speedup= 2n/0.5n + 1.5 ≈ 4= number of stages


Pipeline Speedup

• If all stages are balanced– i.e., all take the same time

– Time between instructionspipelined

= Time between instructionsnonpipelined

Number of stages

• If not balanced, speedup is less

• Speedup due to increased throughput– Latency (time for each instruction) does not

decrease


Hazards

• Situations that prevent starting the next instruction in the next cycle

• Structure hazards– A required resource is busy

• Data hazard– Need to wait for previous instruction to

complete its data read/write

• Control hazard– Deciding on control action depends on previous

instruction

Good Luck!Good Luck!

Q/AQ/A

Xiang Lian The University of Texas – Pan American Edinburg, TX 78539 lianx@utpa

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