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Homework01_key

Homework01_key - CS/ECE 3810 Solutions Homework#1 Fall 2011...

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CS/ECE 3810 Solutions Homework #1 Fall 2011 1. Exercise 1.3: Each part of this problem relies on the formulas for execution time (from lecture and pages 33 and 29): You can use frequency instead of period (one is the inverse of the other): Performance is: 1.3.1: To solve, compute the performance of each processor with known values. Notice the variable for the number of instructions, and notice that the units cancel properly: ( ) ( ) ( ) ( ) ( ) ( )( ) For brevity, I'll omit the intermediate unit computations: ( )( ) ( )( ) ( )( ) Performance is the inverse of time (ignore units): Processor P2 has the highest performance.
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1.3.2: To solve, note that frequency is cycles per second. We have the frequency (from the problem), and we know that the program takes 10 seconds. So, just multiply the cycles/second by 10 seconds to get the total number of cycles. Notice the units work out: ( ) To find the number of instructions, note that we have the number of cycles, and the problem gives cycles per instruction (CPI) for each processor. Divide the total number of cycles by cycles per instruction. Again, start with unit analysis - notice that they again work out: ( ) 1.3.3 (3 points ): To solve, use the equation for execution time, and substitute in the appropriate adjustments. If we want to decrease time by 30%, we must increase CPI by 20%. We will have to adjust the clock frequency by 'k' to make it work: ( ) ( ) ( ) There are many ways to figure out how to adjust the frequency to make this work. I will list two ways. First, you could use the numbers from the previous part. For P1: ( ) ( ) ( ) Solve for k: ( ) ( ) This means for P1, the new clock frequency would need to be: Double check by putting this frequency in and solving for time: ( )
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Interestingly, you don't need to know seconds, instructions, or CPI to determine how to adjust the clock rate. This leads to the second possible approach. Simply note how the equation has changed: ( ) ( ) ( ) In this equation, we have multiplied both sides of the equation by 0.7. (Both sides must remain equal.)
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