Homework 2 Solutions - EEL 4768 Homework #2 Solutions 1a....

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1a. 52% of instructions are not loads, stores or conditional branches, and hence are always 16 bits long. 36% are loads and stores and 12% are conditional branches. Using the data from the table: (.52)(16 bits) + (.17)(.36)(16 bits) + (.43)(.36)(24 bits) + (.98)(.12)(24 bits) + (.40)(.36)(32 bits) + (.02)(.12)(32 bits) = 20.52 bits 1b. (.52)(24 bits) + (.6)(.36)(24 bits) + (.4)(.36)(48 bits) + (.98)(.12)(24 bits) + (.02)(.12)(48 bits) = 27.51 bits 2. ICorig * CCorig * CPIorig = ICnew * CCnew * CPInew for equal performance. As stated, CCnew = 1.1 * CCorig and CPIorig = CPInew so ICorig/ICnew = 1.1 Assuming arbitrarily that ICorig has 100 instructions, ICnew = 1.1/100 = 90.9 instructions. The table for gcc in B.27 shows loads are 25.1% of the original mix. So in our 90.9 instructions for ICnew, 74.9 are still non-load instructions. This leaves 16 load instructions. We has 25.1 in the original mix and 16 in the new mix, so the new mix has 16/25.1 = 63.78% of the loads of the original, so 36.21% of the loads have been removed.
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This note was uploaded on 10/09/2010 for the course EEE 423 taught by Professor Mr.mackey during the Spring '09 term at The School of the Art Institute of Chicago.

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Homework 2 Solutions - EEL 4768 Homework #2 Solutions 1a....

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