midterm_fall03 - CS/ECE 552 Introduction to Computer...

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Unformatted text preview: CS/ECE 552 Introduction to Computer Architecture Midterm Exam Monday, October 20, 2003 7:15-9:15 p.m Name:_________________________________________________________________________ Limit your answers to the space provided. Unnecessarily long answers will be penalized. If you use more space than is provided, you are probably doing something wrong. Use the back of each page for any scratch work. Problem 1. Problem 2. Problem 3. Problem 4. Problem 5. Problem 6. Problem 7. Total ______________ (out of 20 points) ______________ (out of 12 points) ______________ (out of 14 points) ______________ (out of 11 points) ______________ (out of 19 points) ______________ (out of 8 points) ______________ (out of 16 points) ______________ (out of 100 points) Write the logic equations for the following signals, in the space provided. (The equations should be in terms of the inputs to the logic block for which they represent the outputs.) (14 points) GI = 0 PI = 3 P III = 3 C 16 = C 11 = C 54 = (a) (1) Let Xi , Ci and Si , with i = 0, 1, ..., 63, represent the individual bits of the number to be incremented, the carries, and the sums, respectively. Let G I and P I , i = 0,....,63 represent the first level generates and propagates. i i Let G II and P II , j = 0,...,15 represent the second level generates and propagates, and let G III and P III , k = 0,..,3 j j k k represent the third level generates and propagates. S 39 = The figure below shows the general structure of a 64-bit carry lookahead adder with three levels of lookahead. Some signals are shown in the figure to give you an idea of the structure of the adder; however not all relevant signals are shown (for example the carries from the first level lookahead to the single-bit adders). You are to use this structure (i.e., single-bit units with 3 levels of lookahead) to design a 64-bit incrementer. As discussed in class, and as you did in the homework, we can get an incrementer from an adder by making the Y input zero, and C 0 = 1. Incrementer Design (20 points) 16-bit CLA 16-bit CLA 16-bit CLA Third Level Lookahead Second Level Lookahead G III 0 Single Bit Adders P III 0 Page 2 of 10 G II P II 0 0 C0 First level Lookahead GI PI 0 0 Signal G III 1 C 48 S 19 Time Ready Comments (vi) (v) (iv) (iii) (ii) (i) (b) (2) Since an add operation has 3 operands (2 input and 1 output), add instructions must be 3-address instructions. Pipelining improves performance by decreasing the latency of each instruction. More powerful instructions lead to higher performance since the total number of instructions executed is smaller for a given task with more powerful instructions. The sign extension unit in the datapath you have studied is sequential logic. In a load/store architecture, the only instructions that access memory are load and store. A barrel shifter to shift a 16-bit input number right by up to 15 bit positions can be built with 4 stages of 2-1 multiplexors. Assume that each gate delay is time units, and all gates are available with up to 4 inputs. Further assume that all Xi and C 0 are ready at time 0. In the table below, show at what time the chosen signal value is ready. Use the comment column for any comments (comments will be used to determine partial credit in case of an incorrect answer). (6 points) For each of the following questions, answer TRUE(T) if the statement is true, and FALSE if the statement is false. You will be credited 2 points for each correct answer and penalized 1 point for each incorrect answer. It is possible that you may perceive a question to be ambiguous. If that happens, please state any assumptions that you make in your answer. The minumum score for this question will be zero. TRUE and FALSE questions. (12 points) Page 3 of 10 Page 4 of 10 (3) (i) Short Questions (14 points) What is the primary implementation advantage of fixed-length instructions. (3 points) (ii) What are the 5 components of a modern computer system? (Hint: Two of these components can be combined and called the processor.) (2 points) (iii) What is the formula for the total execution time of a program? (2 points) (iv) In a system executing jobs, when is: (a) throughput = 1/latency (2 points) (b) throughput > 1/latency (2 points) (v) What is a stored program computer? (3 points) Page 5 of 10 (4) (i) More Short Questions (11 points) What are structural hazards in a pipeline? (3 points) (ii) What are branch hazards in a pipeline? (3 points) (iii) In the MIPS instruction set, the top 6 bits (31..26) are used for an opcode. With 6 bits, we can have 64 different combinations. Does this imply that the total number of instructions in the MIPS architecture is less than 64? Explain. No credit without an explanation. (5 points) Cycles 1 2 3 4 5 6 7 8 9 10 11 12 13 14 IF ADD1 ADD2 R1 <-- R2 + R3 R4 <-- R1 + R5 R5 <-- Mem[R4] R6 <-- Mem[R5] R7 <-- R5 + R4 ADD1 ID EX ADD1 ADD2 LOAD1 LOAD2 ADD3 MEM WB (b) (a) (5) THIS QUESTION CONTINUES ON THE NEXT PAGE For the following instruction sequence, show the pipeline timing in the table. Part of the timing for the first instruction is shown. (12 points) What purpose does the bypass serve? (3 points) Pipelining (19 points) Following is a simple pipeline; with a bypass from the output of the MEM stage (i.e., the MEM/WB latch) to the EX stage. (Note: There is no bypass from the EX stage, i.e., the EX/MEM latch, back to the EX stage.) Assume that the register file is written in the first half and read in the second half of the clock cycle. Hence, a register value can be written and read in the same cycle. Stall Memory Inst IF insert 0s ID Hazard Detect Reg. File ALU EX bypass Mem Data MEM WB Reg. File Page 6 of 10 Page 7 of 10 (iii) Suggest technique(s) to eliminate the bubbles in part (ii). Which bubbles would your proposed technique(s) eliminate? (4 points) (6) Machine Performance ( 8 points) Consider two machines, machine A and machine B. Machine B runs floating-point instructions 6 times faster than machine A. (i) Consider a program that takes 100 seconds to run on machine A, and spends half its time in floating-point instructions. How much faster is machine B (over machine A) at executing this program? (3 points) (ii) Suppose we want a benchmark program to run 3 times faster on machine B than it runs on machine A. What fraction of its time does it spend executing floating point instructions on machine A? What fraction of its time does the same program spend executing floating-point instructions on machine B? (5 points) Page 8 of 10 (7) Datapath (16 points) In this problem you will extend the multi-cycle datapath as presented in the course text in Figures 5.33 and 5.34. These figures have been provided on the last page of this exam. It is possible to decrease the execution time of a program by combining two or more instructions (that would normally take two or more instruction times to execute) into a single instruction that takes only one instruction time (or a fraction more than one) to execute. These hybrid instructions are chosen because the instructions they replace appear often as a group in programs, and because they can be implemented without extensive modification to the datapath. Modify the datapath as necessary to allow the implementation of each instruction. Be sure to include complete information about any new signals that appear in the modified datapath. After the datapath is modified, fill in the list of control signals that must appears at each clock cycle in order implement the instruction. To reduce the number of signals that you have to write, assume that the initial value all signals is zero. Remember that if you assert a signal, you must de-assert it when necessary. The number clock cycles needed for each instruction may vary. Use as many as you need, but make the implementation efficient as possible. The signals for the first clock cycle already appear in the table. You should not have modify them. to of of as to ALL OF YOUR WORK FOR EACH INSTRUCTION SHOULD APPEAR ONLY ON THE PAGE FOR THAT INSTRUCTION. Note: In figure 5.34, ALU stands for arithmetic logic unit, MDR for memory data register, PC for program counter, and IR for instruction register. (i) On the next page, modify the datapath(and control signals, as necessary), and fill in the list of control signal to implement a decrement and branch if non-zero (DCRBNEQ) instruction. This instruction has a 6 bit opcode (bits 26-31), a 5-bit Rs register field (bits 21-25), a 5-bit Rt register field (bits 16-20), and a 16-bit displacement field (bits 0-15). The semantics of the instruction are as follows: the Rs register is decremented by 1 and the new value is stored in the Rs register. If the new value is non-zero, the branch is taken, else it is not. The branch target address is calculated by adding the 16-bit displacement, sign extended and shifted left by 2 bits, to PC+4 (i.e, the PC of the next instruction). The Rt register field of the instruction is not used. (Note: you may not need all the clock cycles provided in the table on the next page.). Points will be taken off for solutions that require more hardware, or take more time steps, than necessary. (16 points) Clock Cycle 8 7 6 5 4 3 2 1 Instruction -> IR and PC=PC+4 Functional Description MemRd=1, ALUSrcA=0, IorD=0, IRWrite=1, ALUSrcB=01 ALUOp=00, PCWrite=1, PCSource=00 Signals Page 9 of 10 ...
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This test prep was uploaded on 03/29/2008 for the course ECE 552 taught by Professor Wood during the Spring '05 term at University of Wisconsin.

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