GATE IT 2008 | Question: 39

Question

Consider a CPU where all the instructions require $7$ clock cycles to complete execution. There are $140$ instructions in the instruction set. It is found that $125$ control signals are needed to be generated by the control unit. While designing the horizontal microprogrammed control unit, single address field format is used for branch control logic. What is the minimum size of the control word and control address register?

• A. $125, 7$
• B. $125, 10$
• C. $135, 9$
• D. $135, 10$

Comments

single address field format is used for branch control logic                    

What is the significance of this line in the question?

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@sachinmittal1 sir in this question if we consider microinstruction and control word same then answer is 135 ,10

and if we consider different then answer is 125 ,10 what should we consider

Answer 1

Given, 1 instruction takes 7 clock cycles to complete execution.

we know, that each micro-operation takes 1 clock cycle to execute, so, indirectly the question is trying to imply that one instruction has 7 micro-operations.

Now, given, there are 140 instructions in an instruction set, so, the total number of micro-operations is 140*7 = 980 micro-operations.

we know, that each micro-operation is equivalent to a control word in the control memory, so there is a total of 980 control words in the control memory.

So, address size of the control memory = log(980) base2 = 10 bits

according to the format of Control word, number of bits in Control word = number of bits of control signal + number of bits of address size = 126(given data) + 10 = 135 bits 

therefore, answer is Option D

Answer 2

Ans  is  B.

since it  uses horizontal microprogrammed  that  requires i bit /control sognal  .  so  for  125  control signal we need 125  bit.

and   total no  of  microoperation intructions  are  140 * 7  =  980   that  requires   10  bit.

Comments

why 140*7 ???
is it total no of microoperation or total no of clock cycle ??

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it is microoperation ..

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i think control word size is 125+10=135. so now answer will be (D)

Answer 3

 In horizontal microprogrammed approach 1 bit is taken for each control signal therefore 125 control signals are there and each instruction require 7 cycle so 140 * 7 = 980 cycles are needed for  the instructions.Hence, log₂980 = 10 bits for Control Address register and 125 + 10 = 135 bits for control word.

Comments

Why are we adding 10 bits to Control word? Why are we accomodating 980 cycles in bits? Could you please explain?

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"single address field format is used for branch control logic"

What does it mean??

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"single address field format is used for branch control logic"

They are talking about in control word their is next address field for fetching the next control instruction from control memory.

Answer 4

Ans is B
A control word is a word whose individual bits represent the individual control signals. Since there are 125 control signals so minimum size of control word is 125.
There are 140 instructions and each instruction completes in 7 clock cycles, so there are 140*7=980 mictroinstructions that are stored in the control memory. So the control address register requires 10 bits.((2^9=512)<980<=(2^10=1024))