ISRO2009-34

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 $\text{(X)},$ and a MUX select field $\text{(Y)}.$ There are $8$ status bits in the inputs of the MUX. How many bits are there in the $\text{X} $ and $\text{Y}$ 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. $10, 3, 512$

Answer 1

MUX has 8 states bits as input lines so we require 3 select inputs to select and input lines.

No. of bits in control memory next address field=26-13-3 =10

10 bit addressing .we have 2^10=1024 memory size

So X,Y size=10,3

So (A) is correct option.

Comments

@Pranabesh Ghosh ,

it sounds odd but my doubt is  "there is 8 status bits in the input of MUX" means there is 8 states can possible which requires 3 bits ???  please explain first line of your ans..

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It means that we have 8 inputs to the MUX so it is 8:1 MUX and therefore it require 3 select inputs. Which makes Y = 3. Now remaining bits = 10 = X. Number of words in control memory is 2¹⁰ = 1024. So option A) is answer.

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Thanks Rahul Jain25..

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