Hw 1 sol - 1.2N R = 1.5N C Then, N C /N R = 1.2/1.5 = 0.8...

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EEL 3801 - Computer Organization Spring, 2010 HW #1 Solution 1.3-(a) Load A, R0 Load B, R1 Add R0, R1 Store R1, C 1.3-(b) Yes, Move B, C Add A, C 1.4-(a) Non-overlapped time for Program i is 19 time units composed as: Overlapped time is composed as:
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Time between successive program completions in the overlapped case is 15 time units, while in the non-overlapped case it is 19 time units. Therefore, the ratio is 15/19. 1.5-(a) Let T R = (N R * S R ) / R R and T C = (N C * S C ) / R C be execution times on the RISC and CISC processors, respectively. Equating execution times and clock rates, we have
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Unformatted text preview: 1.2N R = 1.5N C Then, N C /N R = 1.2/1.5 = 0.8 Therefore, the largest allowable value for N C is 80% of N R . 1.6-(a) Let cache access time be 1 and main memory access time be 20. Every instruction that is executed must be fetched from the cache, and an additional fetch from the main memory must be performed for 4% of these cache accesses. Therefore, 1 . 11 ) 20 04 . ( ) 1 . 1 ( 20 . 1 = × + × × = factor Speedup...
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This note was uploaded on 04/02/2010 for the course EEL 3801 taught by Professor Froosh during the Spring '08 term at University of Central Florida.

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Hw 1 sol - 1.2N R = 1.5N C Then, N C /N R = 1.2/1.5 = 0.8...

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