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Computer Organization and Architecture | Document

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Added on  2020-04-01

Computer Organization and Architecture | Document

   Added on 2020-04-01

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Running head: COMPUTER ORGANIZATION AND ARCHITECTURE Computer Organization and ArchitectureName of the Student:Name of the University:Author Note
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1COMPUTER ORGANIZATION AND ARCHITECTUREAnswer to question number 1a.We need 256 Kbytes, i.e., 256 x 1024 x 128 bits.We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.Number of Ram chips required are (256 * 1024 * 128)/(32 * 1024) = 1024b.We need 256 Kbytes, i.e., 256 x 1024 x 8 bits.We have RAM chips of capacity 32 Kbits = 32 x 1024 bits.Number of Ram chips required are (256 * 1024 * 8)/(32 * 1024) = 64c.Number of RAM chips required are 1024 = 2^30.Hence address bits needed for each RAM chip required are 30.d.The number bits present in the RAM = 1024Hence the number of memory banks required for this system are: 256/32 = 8e.Address bits required for all memory = 1024/32 = 32Answer to question number 2a.Number of bits for the opcode = 16.b.Number bits required for the registers = 8Number of bits required for addresses = 16c.The largest unsigned binary number: 2^48 -1.
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2COMPUTER ORGANIZATION AND ARCHITECTUREAnswer to question number 3i)Immediate Value of Add 1000 in immediate addressing is 1000 + 500 = 1500 and the Value is stored in the accumulator.ii)DirectValue of Add 1000 in direct addressing mode is 1400 + 500 =1900.iii)IndirectValue of 1000 is 1400. The value of address 1400 is 1300. Hence 1300 + 500 =1800 is loaded into the accumulator.iv)Indexed1000 + R1 = 1200The value located in 1200 is 1000. Hence the value loaded in the accumulator would be 1000 + 500 = 1500.Answer to question number 42 – addressing machineMOV R1, BADD R1, CMOV R2, D
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