sp96 midterm1 solution

# Computer Organization and Design: The Hardware/Software Interface

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Name: 1 Feb 21, 1996 University of California College of Engineering Computer Science Division -EECS Sp 1996 D.E. Culler CS 152 Midterm I Your Name:________SOLUTION_______________ ID Number:_______________________________________________________ Discussion Section:__________________________________________________ You may bring two pages of notes and you may use a calculator, but no book or computer. Please print you name clearly on the cover sheet and on every page. The point value of each question is indicated in brackets. There are a total of 120 points. You have 170 min- utes. Show your work. Write neatly and be well organized. It never hurts to make it easy to grade. Good luck. Problem Possible Score 1 40 2 20 3 20 4 20 5 20 Total 120

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Name: 2 Feb 21, 1996 Problem 1 (40 points) 1a [3] State the five major components of a computer. Processor datapath Processor Control Memory Input Output 1b [5] State five major distinct issues that must be addressed in an instruction set architec- ture. programmable storage data types and encodings set of operations instruction formats number of operands where besides memory can operands be located how memory operands are speified (addressing modes) 1c [2] Define Little Endian. word is addressed by the byte address of the least significant byte (least significant byte is at lowest address in the word.) 1d[3] Decode the following MIPS instruction using the opcode encoding table at the end of the exam (Fig A.18) 10001100111001110000000000000111. Give its RTL (register transfer language) meaning. lw \$7, 7(\$7) R[7] <– mem(R[7] + 7) 1e [3] What is the value of 1000 1100 1110 0111 0000 0000 0000 0111 as a 32bit 2s com- plement number? number is negative so, comp+1=> 0111 0011 0001 1000 1111 1111 1111 1001 -(2^30 + 2^29 + 2^28 + 2^25 + 2^24 + 2^20 + 2^19 + 2^16 - 7) = 1931018233 1f [3] What is the value of 10001100111001110000000000000111 as a single-precision IEEE floating-point number?
Name: 3 Feb 21, 1996 -1.11001110000000000000111 x 2^(-102) or about -3.56 x 10^(-31) 1g. [2] DRAM memory chips increase in capacity by a factor of 16 every how many years? 4x per 3 years => 16x per 6 years 1h. [3] Under what conditions is CPI a valid metric of performance comparison? Time = Instruction Count x CPI x Cycle Time, so Same program, same instruction set, and same cycle time 1i. [4] State three different methods for evaluating branch conditions. Explain the advan- tages and disadvantages of each. conditions codes. condition is set implicitly when doing normal operations, e.g. arithmetic, so some explicit comparisons can be avoided. Introduces implicit dependences between instructions. condition registers & compare instructions. Simple to implement, but tends to increase instruction count. compare&branch instructions. Reduce the number of instructions, but difficult to implement and may impact cycle time.

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