[COA] Memory

Question

A digital computer has a memory unit with 24 bits per word. The instruction set consists of 150 different operations. All instructions have a operation code part (opcode) and an address part ( allowing for only one address). Each instruction is stored in one word of memory.

a. How many bits are needed for the opcode?
b. How many bits are left for the address part of the instruction?
c. What is the maximum allowable size for memory?
d. What is the largest unsigned binary number that can be accommodated in one word of memory?

Comments

a) 8 bits

b) 16 bits

c) 16 MB

d) 2^24 -1

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Yeah,I got the same.

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can u please explain the reason for each of these values

the explanation please

Answer 1

opcode size = 8 bits

address size = 24 - 8 =16 bits

maximum allowable memory = (2 ^16)*24 bits since 2^16 addresses and each location can hold 24 bits.

maximum unsigned binary = (2^24)-1

Answer 2

i)For representing 150 operations we need 8 bits. So 8 bits for opcode.

ii)leftover bits for address of instruction is 24-8=16bits

iii)since memory unit is 24 bits, size of memory 2^24 i.e 16MB

iv)largest unsigned int 2^24 - 1

Comments

For option c ,how is it 2^24B?

24 is the data width

As address part in instruction is 16 bits,it means 2^16 memory locations can be addressed and hence 64 KB memory.

I also got 16 MB earlier,now i am having some doubt.Can you re check?