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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 0f 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:

A: How large must the mode field be?

B: How large must the register field be?

C: How large must the address field be?

D: How large is the opcode field?

Respuesta :

ohismike24

Answer:

A. 3 bits required

B. 6 bits are required

c. 8 bits are required

d. 15 bits are required

Explanation:

A. In the address mode selector, we need to specify 1 out of 7 addressing modes. Therefore, there must be 3 bits (2³ = 8)

B. There are 60 address registers. Therefore the bits to represent 60 numbers are 6 bits. (2^6 = 64)

C. As indicated, the memory bit word required is 8 bits.

D. Total Bits = Opcode + Address mode + Register Add + Memory ADD

32 bits = Opcode+ 3 bits + 6 bits + 8 bits

Opcode =  32 bits - 17 bits = 15 bits.

Therefore 15 bits are required for Opcode field.

Cricetus

(a) The mode field is "3 bits" large.

(b) The register field is "6 bits" large.

(c) The address field is "18 bits" large.

(d) The open code is "5 bits" large.

According to the question,

Words,

  • 256 k

bits,

  • 36

registers,

  • 60

(a)

We need to identify the One to seven items.

So,

There must be 3 bits,

  • [tex]23=8[/tex]

(b)

6 bits implies by 60 registers, then

  • [tex]26=64[/tex]

(c)

→ [tex]256 \ K = 28210 =218[/tex]

or,

→ [tex]18 \ bits[/tex]

(d)

→ [tex]32-(3+6+18)[/tex]

→ [tex]32 - 27[/tex]

→ [tex]5 \ bits[/tex]

Thus the above solutions are correct.

Learn more:

https://brainly.com/question/16364860

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