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 2

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

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? 4

5. Chapter 1 — Computer Abstractions and Technology — 5 CPU Clocking Operation of digital hardware governed by a constant-rate clock Clock (cycles) Data transfer and computation Update state Clock period Clock period: duration of a clock cycle e.g., 250ps = 0.25ns = 250×10 –12 s Clock frequency (rate): cycles per second e.g., 4.0GHz = 4000MHz = 4.0×10 9 Hz

6. Chapter 1 — Computer Abstractions and Technology — 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

7. Chapter 1 — Computer Abstractions and Technology — 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

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 8

9. Negative Numbers: 2’s Complement (x n x n-1 …x 0 ) 2 =-x n ×2 n +x n-1 ×2 n-1 +…+x 0 ×2 0 Ex: (00) 2 = =-0×2 1 +0×2 0 =0 (01) 2 = =-0×2 1 +1×2 0 =1 (10) 2 = =-1×2 1 +0×2 0 =-2 (11) 2 = =-1×2 1 +1×2 0 =-1 9 Good: 1.only one zero 2.Easy to add negative and positive numbers 3.Two useful operations: negation and sign-extending Exercise: (11) 2 +(01) 2 =? (11) 2 +(10) 2 =? Note: X n not only represents sign, but has weights (different to the previous sign-magnititude representation). 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 10 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. 11 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. 12 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. 13 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. 14 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 15 Chapter 2 — Instruction — 15

16. Instruction Operation-Unconditional Jump Jump register –Jr $ra Jump –J 2500 Jump-and-link instruction –Jal 2500 16 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 17

18. 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

19. 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) Chapter 3 — Arithmetic for Computers — 19

20. Overflow Conditions 20 OperationOprand AOperand BResult indicating overlfow A+B>=0 <0 A+B<0 >=0 A-B>=0<0 A-B<0>=0 Chapter 3 — Arithmetic for Computers — 20

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

22. Chapter 3 — Arithmetic for Computers — 22 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 SExponentFraction single: 8 bits double: 11 bits single: 23 bits double: 52 bits

23. Converting Floating Point Numbers 23 5.1875 101..0011 0.1875×2 =0.3750 0.375×2 =0.75 0 0.75×2 =1.51 0.5×2 =11 101.0011 1.010011×2 2 0 10000001 0100110…0(23 bits) Ex. Convert 5.1875 to its binary representation (single precision). 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 24

25. Chapter 4 — The Processor — 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

26. Chapter 4 — The Processor — 26 MIPS Pipeline Five stages, one step per stage 1.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

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

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

29. Chapter 4 — The Processor — 29 Pipeline Speedup If all stages are balanced –i.e., all take the same time –Time between instructions pipelined = Time between instructions nonpipelined Number of stages If not balanced, speedup is less Speedup due to increased throughput –Latency (time for each instruction) does not decrease

30. Chapter 4 — The Processor — 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

31. Good Luck! Q/A