Homework07-sol - 3/26/03 5:44 AM ECSE-2660 Computer...

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Unformatted text preview: 3/26/03 5:44 AM ECSE-2660 Computer Architecture, Networks, & Operating Systems, Spring 2002 Page 1 of 4 Homework #07 (Solution) 3/26/2003 5:44 AM 1. (15 points) Problem 6.23 from the text (the worst ordering will have two stalls). lw $3, 0($5) add $7, $7, $3 #requires stall on #3 lw $4, 4($5) add $8 $8, $4 #requires stall on $4 add $10, $7, $8 sw $6, 0($5) beq $10, $11, Loop Grading: 3pts for trying, +6 pts for rearranging for each stall. 2. (15 points) Problem 6.24 from the text. 36 beqd $1, $3, 8 40 sub $10 $4, $8 #delay slot, always executed Grading: 5pts for trying, +10pts for sub after beqd, +7pts for any code that works using beqd. 3. (20 points) Consider a computer system that is organized as shown below: CPU Cache Main Memory (DRAM) Bus The following table summarizes some relevant parameters relating to the memory system: Calculate the average memory access time assuming that the main memory (DRAM) can be operated at its maximum possible speed, i.e. 80ns per word 60% of the time, and at 165ns per word 40% of the time. Hint: Make a tree diagram. Cache (SRAM) Access Time 20ns DRAM Access Time – best case (see comment below) 80ns DRAM Cycle Time – worst case (see comment below) 165ns Cache Hit Rate 98% Cache Block Size 4 words 3/26/03 5:44 AM...
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This note was uploaded on 04/25/2010 for the course ECSE ecse-2340 taught by Professor Wozny during the Spring '09 term at Rensselaer Polytechnic Institute.

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Homework07-sol - 3/26/03 5:44 AM ECSE-2660 Computer...

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