Full reward for whoever can explain how to solve this: The memory unit of a comp
ID: 3561989 • Letter: F
Question
Full reward for whoever can explain how to solve this:
The memory unit of a computer has 256K words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify 1 of 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:
How large must the mode field be? _____
How large must the register field be? ____
How large must the address field be? ____.
How large is the opcode field? _____
Explanation / Answer
For memory of 256K words we need 18 bits as 256 = 2^18. Inorder to adjust 7 addressing modes we need 3 bits since 2^3 = 8 which is the least higher number of combinations.To addjust 60 registers we need atleast 6 bits as 64= 2^6 which is the least higher number of combinations.The remaining bits in the 32-bit word can be used for (opcodes), 32 -( 18 + 3 + 6) = 5. So there can be 32= 2^5 opcodes.
The instruction format can be in below sizes
---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
mode:24 to 26
reg_address :18 to 23
mem_address: 1 to 17
opcode: 27 to 31
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