A digital computer has a memory unit with 24 bits per word. Theinstruction set c

Question

A digital computer has a memory unit with 24 bits per word. Theinstruction 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 the opcode?
b) How many bits are left for the address part of theinstruction?
c) What is the maximum allowable size for memory?
d) What is the largest unsigned binary number that can beaccommodated in one word memory?

Explanation / Answer

Dear User, a) The instruction set consists of 150 operations     28 = 256 which is larger than150, so take the lower bit that is 27 = 128. 150operations between 128 and 256.     therefore opcode needs 8 bits. b) 24-8 = 16. The 16 bits are left for theaddress part of the instruction c) The maximum allowable size = 2^16 =65536     The maximum size is 65536-1 = 65535 d) The largest unsigned binary number that can beaccommodated in one word memory                2^15 -1= 32768-1=32767 I hope this will helps toyou

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