edited by
13,426 views
39 votes
39 votes

The microinstructions stored in the control memory of a processor have a width of $26$ bits. Each microinstruction is divided into three fields: a micro-operation field of $13$ bits, a next address field $(X),$ and a MUX select field $(Y).$ There are $8$ status bits in the input of the MUX.

How many bits are there in the $X$ and $Y$ fields, and what is the size of the control memory in number of words?

  1. $10, 3, 1024$
  2. $8, 5, 256$
  3. $5, 8, 2048$
  4. $10, 3, 512$
edited by

2 Answers

Best answer
55 votes
55 votes
$x + y + 13 = 26  \rightarrow (1)$
$y = 3$  $(y)$ is no of bits used to represent 8 different states of multiplexer $ \rightarrow (2)$
$x$ is no of bits required represent size of control memory
$x = 10$ from $(1)$ and $(2)$

$\therefore$ Size of control memory $= 2^x = 2^{10}= 1024$

Correct Answer: $A$
edited by
2 votes
2 votes
The number of bits in Control memory =26.
From the given data each instruction divided into op field (13)+X(next address field)+Y(MUX)
8(23) status bits in the inputs of the MUX then three bits in the MUX select field.
No. of bits in control memory next address field=26-13-3 =10
size of the control memory in number of words is 210=1024 words
Answer:

Related questions

59 votes
59 votes
4 answers
1
go_editor asked Apr 24, 2016
19,515 views
Consider the following program segment for a hypothetical CPU having three user registers $R_1, R_2$ and $R_3.$\begin{array}{|l|l|c|} \hline \text {Instruction} & \text...
35 votes
35 votes
3 answers
2
52 votes
52 votes
10 answers
3
17 votes
17 votes
3 answers
4