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Study on Computer Organization and Architecture

   

Added on  2020-05-16

6 Pages806 Words217 Views
Running head: COMPUTER ORGANIZATION AND ARCHITECTUREITC544 Computer Organization and ArchitectureName of the StudentName of the UniversityAuthor’s Note
Study on Computer Organization and Architecture_1
1COMPUTER ORGANIZATION AND ARCHITECTURE Answer 1:i.Capacity of the drive can be calculated using the formulaCapacity C = Surfaces x Tracks x Sectors x Bytes= 23 x 512 x 64 x 32= 24117248= 23 Gbii.The formula used for the calculation of the rotational delay is given belowRotation delay= ½ x [(60 /RPM) x (1000 /1 Sec)]= ½ x [(60 /9600) (1000)]= 6.25 /2= 3.125 millisecondsiii.The Access Time is calculated using the formulaAccess time = Seek time + latency= (10 + 3.125) milliseconds= 13.125 millisecondsAnswer 2:i.The calculation for finding the Bits needed for the opcode is given below:= 28 > 232 Therefore, 8 bits needed for the opcodeii.The calculation for the number of Bits needed to specify the register is givenbelow:=> 8 = 23Therefore, 3 bits needed to specify the register
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2COMPUTER ORGANIZATION AND ARCHITECTURE iii.The calculation for finding the number of bits left for the address part is givenbelow:=> Memory unit – opcode=> 32- 8 = 24iv.The calculation for finding the maximum allowable size of memory is givenbelow: => 224v.The calculation for finding the largest unsigned binary number which can beaccommodated in one word of memory is => 232-1Answer 3:The calculation for number of 0 – address instruction possible for the accommodationof instruction architecture is given below: Instruction length = 13 bitsTherefore, total number of address instruction = 2^13= 8192FIVE 2-address instruction = 5 * 2^5 * 2^5= 5120TWENTY 1-address instruction= 20 *2^5= 640Therefore, 0 - address instruction left= 8192- (5120 + 640)= 2432
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