Transcript Slide 1
A study guide: Please do not take this document as strict guideline! 1) Expect short answer questions regarding fundamental definitions from lecture slides/book such as: Amdahl’s law, what it is, how is it important or what is underflow? Remember this is an open book exam, all you need to know is where that information is. 2) There will be performance related questions more on the lines of comparing two models using CPU execution time formulas, CPI, MIPS or Amdahl’s law. Refer to slides, assignments, and examples from the book. 3) Other topics: MIPS ISA Features, characteristics, definitions Addressing Modes Need to understand a given assembly language Need to be able to convert a simple C instruction into assembly Need to know basic MIPS instructions Arithmetic Addition, Subtraction ALU Design: slt, bne, beq , how to implement them at gate level, go through the lecture slides, how did we insert slt instruction to the ALU? What if I give you another instruction? CLA Design, it’s significance? Multiplication, Division Algorithms (refer to assignments and book examples) IEEE Floating Point Arithmetic Conversion Key Issues !!! • Limit your answers to the space provided. If you use more space than is provided, you are probably doing something wrong. • Manage your time carefully. If you do not understand something in a question, ask the instructor for help. • You may refer to ONLY ECE369 textbook “Computer Organization and Design”, by Hennessy and Patterson during the exam. NO other source is allowed for reference. • Show your work in order to receive credit $L6: • • Consider a system with a CPU, connected to separate instruction and data memories. System is executing the following MIPS code. The code is a loop, which you can assume iterates many times. Also assume that the beq instruction is taken 10% of the time. Suppose the CPU can execute instructions infinitely fast, but, of course, it must fetch instructions from the (instruction) memory, and fetch and store data from and to the (data) memory. In case you have forgotten, the instruction size is 4 bytes, and the word size is also 4 bytes. $L4: lw $2,0($4) beq $2,$0,$L4 lw $2,0($5) lw $3,0($6) addu $2,$2,$3 sw $2,0($4) addu $4,$4,4 addu $6,$6,4 addu $5,$5,4 addu $7,$7,1 slt $2,$7,10000 bne $2,$0,$L6 • What is the average number of instructions per iteration? 90% of the time we execute 12 instrs. and make 4 data memory refs. 10% of the time we execute 8 instrs. and make 1 data ref. Therefore, avg. number of instrs. per iteration = 0.9x12 + 0.1x8 = 11.6, • What is the average number of data references per iteration? the avg. number of data refs per iteration = 0.9x4 + 0.1x1 = 3.7. • In a floating-point number system, larger bases improve the accuracy at the cost of sacrificing the range. • T. • More powerful instructions lead to higher performance since the total number of instructions executed is smaller for a given task with more powerful instructions. • F • Programs written in assembly language are inherently machine-specific and must be totally rewritten to run on another computer architecture. • T • We want to design the instruction set of a RISC processor, with the following characteristics: 128 General Purpose registers; 96 Instructions; 32bit instruction word length; 3-operand arithmetic and logical operations (ex. ADD R1, R2, R3); 2 register operands and 1 immediate constant arithmetic and logical instructions (ex. ANDI R1, R2, 5). The instructions have a single opcode, in a fixed position inside the instruction word, which is also used to specify the size of the operands (byte, half-word, word). • A) Draw a simple illustration showing the typical 3-operand arithmetic instruction for this processor, clearly indicating the number of bits used for each field of the instruction (op-code, operands, etc.) • Solution: For 96 instruction, we need 7 bit (27 = 128 > 96) of Opcode, and for 128 register, we need 7 bit for each operand. • B) How many bits can be used for describing the constant used in immediate arithmetic and logical instructions? (e.g. ADDI) • Solution: 11bit