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

Its answer should be $(D)$ because $140$ instructions, each requiring $7$ cycles means $980$ cycles which will take $10$ bits.

Since it is horizontal for control word, $125 \text{ control signals} + 10$ bits $=135$ bits will be required.

Comments

 While designing the horizontal microprogrammed control unit, single address field format is used for branch control logic.

KIndly elaborate this part also. it is confusing why explicitly it is mentioned that single address field is used for branch control logic. is there mutli adress field format  also possible?? if it is can you provide some link to read theory about it.

 

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There are 140 instructions and 140*7 microinstructions.

 I think u are confused between these terms - instruction and microinstructions.

@Pratush Priyam Kaun

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Atleast the first micro-instruction is common to all 140 instructions (as this is instruction fetch). So, number of micro-instructions must be atmost 140 * 6 + 1 = 841 right?

Answer 2

Ans  is  D Part.

Explanation -->

140 (No. of instructions)

7 (No. of cycles required per instruction) 

We need 980 (140*7) memory locations to store control signal corresponding to different instructions. 980 location could be addressed by 10 bits ($\lceil{\log_2(980)}\rceil$) (= Minimum Size Control Address Register).

Horizontal Micro-programmed Control Unit (means no encoding)

Single Address Field (means in each address of control memory contains control signals + address).

125 (No. of different control signal) 

Minimum size of the control word = 10 + 125 = 135 (Width of Control Memory)

Comments

sir, i am not getting how 140 instructions * 7 clock cycles gives 980 microoperations/memory locations. plz help.

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every clock cycle requires a series of micro operations and hence for every clock cycle we need a control word or micro instruction in the control memory.

One micro instruction can initiate various microperations.

hence 140*7 control words are needed.

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Thanks Chotu

Answer 3

control address register will contain address of control word..control word will is a micro operation (cond flags+control signals+ next control word address)..since 980 cycles are needed..in each cycle 1 micro operation can be performed so 980 micro operations are there in control memory so 10 bits are required for addressing purpose
and control word size=125(control signal bits) +10=135

Comments

@Anurag_s sir , i think we can perform more than one micro operations if those are independent, means using different set of components.  Ex- Incrementing PC + transfering content from MBR to IR.

Answer 4

each microintruction in an instruction require one cycle, so thats why 140 is multiplied with 7 to get total no of microinstructions which are  980

these 980 micro instruction will be stored in control memory (rom or ram) so to uniquely identify each of them we require 10 bits,

microinstruction is divided into 3 parts - >   flag      microoperation     nextAddress

since nothing mentioned about flags take it 0 - these flags tells about branch and non branch microinstruction

microoperation - > 125 contol signal mentioned are nothing but microoperation and since we are using horizontal
                                   microinstruction these 125 bits will not be encoded so take it 125

next address -> 10 bits from those 140 instructions we calculated are next address bits

total size of control word = 0+125+10 = 135

now control address register size is equal to next address field bits = 10 bits
so answer us -- >   135, 10

Comments

min. size of control word should be 125.

ans is B

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Good Explanation!

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Thanks!