CDA3101-F11-Quiz4-KEY

CDA3101-F11-Quiz4-KEY - Shift Only 0 0 0 0 0 0 0 1 1 0 1 1...

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CDA3101 Fall 2011 Quiz 4 KEY , p. 1 of 2 Problems: Solution : Multiplicand 0 0 1 1 0 1 1 1 1 1 223 (hereafter called “Mcand”) Multiplier x 1 1 1 0 0 0 0 0 1 0 -126 Y 0 0 -1 0 0 0 0 1 -1 0 recoded multiplier (showing 0 -> 1 changes) Operation Accumulated Result ___________ Shift Only 0 0 0 0 0 0 0 0 0 0 0 Add -Mcand + 1 1 0 0 1 0 0 0 0 1 1 1 0 0 1 0 0 0 0 1 0 Shift 1 1 1 0 0 1 0 0 0 0 1 0 Add Mcand + 0 0 1 1 0 1 1 1 1 1 0 0 0 1 1 0 1 1 1 1 1 0 Shift 0 0 0 0 1 1 0 1 1 1 1 1 0 Shift Only 0 0 0 0 0 1 1 0 1 1 1 1 1 0 Shift Only 0 0 0 0 0 0 1 1 0 1 1 1 1 1 0
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Unformatted text preview: Shift Only 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 0 Shift Only 0 0 0 0 0 0 0 0 1 1 0 1 1 1 1 1 0 Add -Mcand + 1 1 0 0 1 0 0 0 0 1 1 1 0 0 1 0 0 1 0 0 0 1 1 1 1 1 0 Shift 1 1 1 0 0 1 0 0 1 0 0 0 1 1 1 1 1 0 Shift Only 1 1 1 1 0 0 1 0 0 1 0 0 0 1 1 1 1 1 0 Shift Only 1 1 1 1 1 0 0 1 0 0 1 0 0 0 1 1 1 1 1 0 -28098 CDA3101 Fall 2011 Quiz 4 KEY , p. 2 of 2 Solution: (cont d) Analysis : In the preceding algorithm instance, there are 10 shifts and 3 additions. Since Adds/Subs take 2 cycles and Shifts take 1 cycle, we have: Total No. Cycles = 10 shifts x 1 cycle/shift + 3 adds x 1 cycle/add = 10 + 3 = 13 cycles...
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CDA3101-F11-Quiz4-KEY - Shift Only 0 0 0 0 0 0 0 1 1 0 1 1...

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