GATE CSE 2004 | Question: 67

Question

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?

• A. $10, 3, 1024$
• B. $8, 5, 256$
• C. $5, 8, 2048$
• D. $10, 3, 512$

Comments

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

Answer 1

$x + y + 13 = 26  \rightarrow (1)$
$y = 3$  $(y)$ 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: $A$

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?

Answer 2

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