Computer Architecture Assignment

Added on - 28 May 2020

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Running Head: COMPUTER ARCHITECTURE1Computer ArchitectureInstitutionDateName
COMPUTER ARCHITECTURE2Question 3Length of instruction = 13 bitsAddress field size = 5 bits.Two address instructions = 5One address instructions = 20Maximum instruction code represented utilizing 13 bits = 2^13 = 8192And the address field size is 5 bits therefore, each address will contain 2^5 Possible value= 322-Address instruction will require 32*32 address space. But we have 5 such instructions.Thus 5*32*32 = 5120The 1-Address instruction will require 32 address space but we have 20 instructions.Thus 20*32 = 640The number of 0-address instructions to accommodate therefore is:The total – (2 Address instr + 1 Address instr) (Harris & Harris, 2015)8192- (5120+640)8192-5760= 2432
COMPUTER ARCHITECTURE3Question 4Take:P=X+YQ=Y-ZR=X*YThen,A=P×QR2 Machine instruction: -Two address machineholdsone source operand and destinationoperandusing the register.MOV T1, XADD T1, YMOV P, T1MOV T2, YSUB T2, ZMOV Q, T2MOV T3, XMUL T3, YMOV R, T3
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