A digital computer has a memory unit with 32 bits per word. The instruction set

Question

A digital computer has a memory unit with 32 bits per word. The instruction set consists of 110 different operations. All instructions have an operation code part (opcode) and two address fields: one for a memory address and one for a register address. This particular system includes eight general-purpose, user-addressable registers. Registers may be loaded directly from memory, and memory may be updated directly from the registers. Direct memory-to-memory data movement operations are not supported. Each instruction stored in one word of memory.

a)      How many bits are needed for the opcode?

7 bits are needed for opcode

b)      How many bits are needed to specify the register?

32-7/4 = 6 bits

c)      How many bits are left for the memory address part of the instruction?

25-6 =19 bits

d)     What is the maximum allowable size for memory?

e)      What is the largest unsigned binary number that can be accommodated in one word of memory?

I think I have A, B, & C.  I need help with D & E. (Also if you see and errors in a-c, please note)

Explanation / Answer

b) log 8 to base 2 = 3 bits

c) 32- 3- 7 = 22

d) 2^22

e) 2^32 -1

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