A digital computer has a memory unit with 24 bits perword. The instruction set c
ID: 3610629 • Letter: A
Question
A digital computer has a memory unit with 24 bits perword. The instruction set consists of 150 different operations. Allinstructions have an operation code part (opcode) and an addresspart (allowing for only one address). Each instruction is stored inone word of memory. a. How many bits are needed for theopcode? b. How many bits are left for the addresspart of the instruction? c. What is the maximum allowable size formemory? d. What is the largest unsigned binary numberthat can be accommodated in one word of memory? A digital computer has a memory unit with 24 bits perword. The instruction set consists of 150 different operations. Allinstructions have an operation code part (opcode) and an addresspart (allowing for only one address). Each instruction is stored inone word of memory. a. How many bits are needed for theopcode? b. How many bits are left for the addresspart of the instruction? c. What is the maximum allowable size formemory? d. What is the largest unsigned binary numberthat can be accommodated in one word of memory?Explanation / Answer
Dear, (a) The instruction set consists of 150 differentoperations,we take 28 =156 which is larger than 15027 willgive only 128 instructions,So we take 8 bits for theopcode.
(b) 24 - 8 = 16.
There are 16 bitsavailable for the address part of the instruction
(c) 216 = 65536. Maxnumber is 65536-1 = 65535.
The maximum allowablesize for memory is 65535.
(d) 215 = 32768-1 =32767. 32767 is the largest unsigned binary number that can beaccommodated in one word of memory I hopethis will helpful foryou............. 32767 is the largest unsigned binary number that can beaccommodated in one word of memory I hopethis will helpful foryou............. I hopethis will helpful foryou.............
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