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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. $10, 3, 512$
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I am not  getting how the size of control memory$=1024$?

I am ok with calculating $X$ and $Y$.But i think

control memory size=number of different microinstruction * size of each micro instruction 

                               =$1024 *26 \,\,bits=26624bits$=$3328 Bytes$=$3328 words$ (if i consider word size = 1 Byte)

If the solution is correct then this solution is wrong

https://gateoverflow.in/112421/vertical-microprogrammed

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I think the word size is considered to be 26bits.
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 @sourav, Control memory is nothing but the collection of control words, in this question next address field is 10 which means there are 1024 control words are there.

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Notice in this question Load and increment both are connected to MUX output. But at a time only one will be active.
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@chhotu ... so does is affect the width of micro-instruction ?? I  also have the same doubt like @ Sourav X = 10 ,so their are 2^10 microinstructions are present so control memory size should be 1024 * 26 bit.
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what is the purpose of mux here?what functionality do the status bits provide?
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Read last line of question carefully. We just need to say size in words, that is how many words will be there, question is not asking for absolute size i.e. in bits or bytes.

Hope this is clear.
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This question was asked in BARC 2019 (with the exact same values and options)

I couldn't add a tag to the question, I think it's because I don't have the question-editing privilege
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2 Answers

53 votes
53 votes
Best answer
$x + y + 13 = 26  \rightarrow (1)$
$y = 3$  $(y)$ 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: $A$
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4 Comments

Both the video are same, ....please verify that only one video cover the topic completely
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any significance of the figure given?
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Can you provide a link to the full course of these lectures?
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1 vote
1 vote
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:

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