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csse232_hw3_soln

csse232_hw3_soln - Fall 2009-2010 CSSE 232 CSSE 232...

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Fall 2009-2010 CSSE 232 CSSE 232 – Computer Architecture I Rose-Hulman Institute of Technology Computer Science and Software Engineering Department Homework 3 Solutions 1. (10 points) F 2. (20 points) The problem statement tells us how much the cycle time increases and that the CPI does not change. We must figure out how many load instructions can be eliminated. 35% of instructions are memory access and 2/3 of those are loads. This leads us to: ExecTime old ExecTime new = CycleTime old × CPI old × # Inst old 1 . 1 CycleTime old × CPI old × (1 - (2 / 3)(0 . 35) x ) # Inst old = 1 . 0 and x = 0 . 3896 . If 38.96% of the loads can take advantage of addm then the new machine will have equivalent performance. Note: In practice the average CPI would also change and would need to be recalculated. 3. (20 points) a = b + c ; b = a + c ; d = a - b ; Accumulator Instruction Code Data load AddressB 3 4 add AddressC 3 4 store AddressA 3 4 add AddressC 3 4 store AddressB 3 4 neg 1 0 add AddressA 3 4 store AddressD 3 4 Total 22 28 20 October 2009 Page 1
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Fall 2009-2010 CSSE 232 Code size is 22 bytes, and memory bandwidth is 22 + 28 = 50 bytes.
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