COA NUmerical Gate Application

Question

Answer 1

size of memory unit = 256k words

Since it is not mention, we will consider word addressable

i)

address part = log 256k = 18

register code = log 64 = 6

indirect bit = 1

so operation code = 32-18-6-1 = 7

ii)

Indirect bit	operation code	register code	address bits
1	7	6	18

iii)

bits for address input = log 256k = 18

bits for data = 1 word = 32 bits

Comments

As I know unless otherwise specified we use byte addressing.How can you assume its word addressable?

Answer 2

256K word = 2¹⁸ ie. 18 bits for the address is required.
1 bit for indeirect access.
64 register required 6 bits for the access.
total size of the memory reference is 32 Bit.

so bit required for OPCODE = 32 - 1 - 18 - 6 = 7 bits.

B) So we have.

31 - (30----24) - (23----18) - (17------0)
I      Opcode     Register     Address.

C) total Address bit is 18 bit and memroy refernce is 32 bit.