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

1. CSCI 6307 Foundation of Systems Review: Midterm Exam Xiang Lian The University of Texas – Pan American Edinburg, TX 78539 lianx@utpa.edu

2. Review • Chapters A1 ~ A4 in your textbook • Lecture slides • In-class exercises (1) & (2) • Assignments 1 & 2

3. Review • 5 Questions (100 points) • 1 Bonus Question (20 extra points)

4. 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?

5. CPU Clocking • Operation of digital hardware governed by a constant-rate clock Clock period Clock (cycles) Data transferand computation Update state • 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 Chapter 1 — Computer Abstractions and Technology — 5

6. CPU Time • Performance improved by • Reducing number of clock cycles • Increasing clock rate • Hardware designer must often trade off clock rate against cycle count Chapter 1 — Computer Abstractions and Technology — 6

7. 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 Chapter 1 — Computer Abstractions and Technology — 7

8. 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

9. Negative Numbers: 2’s Complement Note: Xn not only represents sign, but has weights (different to the previous sign-magnititude representation). (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 Good: only one zero Easy to add negative and positive numbers Two useful operations: negation and sign-extending Exercise: (11)2 +(01)2=? (11)2 +(10)2=? Chapter 2 — Instruction — 9

10. 2’s Complement: Negation (011)2=3 binary number of -3? (100)2 : complement of (011)2 +(001)2 : plus one =(101)2 : -3 Exercise: find the negation of (110)2 and (100)2 Chapter 2 — Instruction — 10

11. 2’s Complement: Sign Extension Extend (001)2 to 6 bits ?????? (???001) 2 : copy the old bits to the right (000001) 2 : MSBthe remaining bits Exercise: extending (101)2 to 6 bits and verify that they are the same integer. Chapter 2 — Instruction — 11

12. 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. Chapter 2 — Instruction — 12

13. Instruction Operations-Data Transfer • Lw $s1, 30($s2): $s1mem[$s2+30] (load/read word from memory) • Sw $s1, 20($s2): $s1mem[$s2+20] (store/write a word to memory) • Lb $s1, 30($s2): $s1mem[$s2+30] (load/read a byte from memory) Exercise: read a word/byte at $s2+20 in memory to $s2. Chapter 2 — Instruction — 13

14. 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. Chapter 2 — Instruction — 14

15. 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 Chapter 2 — Instruction — 15

16. Instruction Operation-Unconditional Jump • Jump register • Jr $ra • Jump • J 2500 • Jump-and-link instruction • Jal 2500 Chapter 2 — Instruction — 16

17. 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

18. Addition Overflow • Occurs when the result is out the range for a given number of bits. Ex: 1011 (-5) + 1100 (-4) 10111 (7×) 0111 (7) + 0100 (4) 11011 (-5×) 0011 (3) + 0100 (4) 10111 (7) 1111 (-1) + 1100 (-4) 11011 (-5) Chapter 3 — Arithmetic for Computers — 18

19. Checking of Addition Overflow When Two Operands Differ in Sign • Does not occurs Ex: 1011 (-5) + 0100 (4) 11111 (-1) 1111 (-1) + 0100 (4) 10011 (3) Chapter 3 — Arithmetic for Computers — 19

20. Overflow Conditions Chapter 3 — Arithmetic for Computers — 20

21. 1000 × 1001 1000 0000 0000 1000 1001000 Multiplication §3.3 Multiplication • Start with long-multiplication approach multiplicand multiplier product Length of product is the sum of operand lengths Chapter 3 — Arithmetic for Computers — 21

22. IEEE Floating-Point Format single: 8 bitsdouble: 11 bits single: 23 bitsdouble: 52 bits • 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 Chapter 3 — Arithmetic for Computers — 22

23. Converting Floating Point Numbers Ex. Convert 5.1875 to its binary representation (single precision). 5.1875 0.1875×2 =0.375 0 0 0.375×2 =0.75 101. .0011 0.75×2 =1.5 1 101.0011 0.5×2 =1 1 1.010011×22 0 10000001 0100110…0(23 bits) Chapter 3 — Arithmetic for Computers — 23

24. 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

25. Logic Design Basics §4.2 Logic Design Conventions • 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 Chapter 4 — The Processor — 25

26. MIPS Pipeline • Five stages, one step per stage • IF: Instruction fetch from memory • ID: Instruction decode & register read • EX: Execute operation or calculate address • MEM: Access memory operand • WB: Write result back to register Chapter 4 — The Processor — 26

27. Determining Clock Cycle (And) 950 I-Mem: 400ps Add: 100ps Mux: 30ps ALU: 120ps Regs: 200ps D-Mem: 350ps Control: 100ps 950 980 300 700 980 200 1180 1180 600 800 800 950 830 700 Chapter 4 — The Processor — 27

28. Pipelining Analogy §4.5 An Overview of Pipelining • Pipelined laundry: overlapping execution • Parallelism improves performance • Four loads: • Speedup= 8/3.5 = 2.3 • Non-stop: • Speedup= 2n/0.5n + 1.5 ≈ 4= number of stages Chapter 4 — The Processor — 28

29. 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 Chapter 4 — The Processor — 29

30. 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 Chapter 4 — The Processor — 30

31. Good Luck! Q/A