Question

a computer uses a memory unit with 256k words of 32 bits each.A binary instruction code is stored in one word of memory.The instruction has four parts: an indirect bit, an operation code, a register code part to specify one of 64 registers, and an address par.

a) how many bits are there in the operation code, the register code part, and the address part?

Answer 1

a.) 7 bits
6 bits
18 bits respectively

Answer 2

• Memory Unit:

        |    0      |       1      |      2       | ....... |   256   |

          32  bits    32 bits    32 bits             32 bits

• Instruction word format :

        Indirect bit | OP-code |  Register  |  Address

             1 bit            7 bits         6 bits            18 bits

a)  Indirect bit has  1 bit

   Address: 2^8 ( 256 kB ) *  2^10 ( 1024 bytes/kB )

                       = 2^18 resulting in 18 bits

   Register: 64 registers

                       = 2^6 resulting in 6 bits

   OP-code: 32 - 1 - 18 - 6 bits = 7 bits