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Computer Organization and Architecture : Assignment

   

Added on  2020-05-16

6 Pages778 Words175 Views
Running head: COMPUTER ORGANIZATION AND ARCHITECTUREComputer Organization and ArchitectureName of the StudentName of the UniversityAuthor’s Note
Computer Organization and Architecture : Assignment_1
1COMPUTER ORGANIZATION AND ARCHITECTURE Answer to Question 1:I.The capacity of the drive – Capacity = surfaces * Tracks * Sectors * Bytes= 23 * 512 * 64 * 32= 23 GbII.Rotational Delay – Rotation delay= ½ * [ (60/ RPM) * (1000 / 1 Sec)]= ½ * [(60 / 9600) (1000)]= 6.25 / 2= 3.125 msIII.Access Time – Access time = Seek time + latency= 10 ms + 3.125 ms= 13.125 msAnswer to Question 2:I.Bits required for the opcode is 2 ^8 > 232 Thus, 8 bits of opcodeII.Bits needed for the specification of the register is 2^3 = 8Thus 3 bits are required for specification of the registerIII.Bits left for the address part of the instruction is 32- 8 = 24IV.Maximum allowable size of memory is 2^24
Computer Organization and Architecture : Assignment_2
2COMPUTER ORGANIZATION AND ARCHITECTURE V.Largest unsigned binary number which can be accommodated in one word ofmemory is 2^32-1Answer to Question 3:Total number of address instruction = 2^13= 81925 2-address = 5 * 2^5 * 2^5= 512020 1-address = 20 *2^5= 640Therefore, Number of 0 address instruction still possible = 8192- (5120+640) = 2432Answer to Question 4:Three Address instructionADDR1, X, YR1 M [X] + M [Y]SUBTR2, X, Y R2 M [X] – M [Y]MULR3, X, YR3 M [X] * M[Y]MULA, R1, R2, R4R4 R1 * R2DIVA, R3M [X] R4 / R3Two Address instruction
Computer Organization and Architecture : Assignment_3

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