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A vertical microprogrammed control unit supports 512 instructions. The system is using 8 conditional flags and contains 31 control signals. Each instruction on an average required 1 μ operation. The approximate size of control memory is ________ in bytes.
in CO and Architecture by Loyal | 396 views
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Ans. 1088 Bytes.
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Same type of question:-   https://gateoverflow.in/181814/control-memory

But i didn't understood how they calculate the size of control memory.

Shouldn't it be :- Size of control memory = #Instruction * size of control word.

In above question size of control word = 17 Bit.

So answer is :- Size = 512 * 17 / 8 = 1088 B. (By Made easy solution)

 

But in above link the last Step is different please Clear my doubt which approach is good.

 

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Anybody Help me here. Little guidance will be appreciated :)
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Yes, there will be 512 µ operations or 512 µ Instructions(Each instruction has 1 microoperation)Since the question is saying about vertical µ programming, So we need not generate 31 control signals at the same time, So we will use a decoder  of 5 * 32 size which needs 5 bits for control signal field of a control word and since there are 8 flag conditions are used for condition checking, so we need MUX 8*1 using 3 select lines, we need 3 bits for storing this information in control word, Apart from this, since there are 512 micro Instructions present in control memory , so we need 9 bits for the next address field of a control word.

So the size of a control word = 5 + 3 + 9 = 17 bits

Size of control memory = Number of control words(Micro Instructions) * size of each control word

                                      = 512 * 17 bits

                                      = 1088 Bytes

2 Answers

+3 votes

1088 B 

 

by Boss
0 votes
Yes . 512 nos instructions( 2^9) correspond to 9 bits.

8 nos FLAGs ( 2^3) correspond to 3 bits and 31 Control Signals ( 2^5) corresponds to 5 bits.

Hence total nos of bits =9+3+5= 17 bits.

Hence the size of Control memory= (512*17)/8 byte =1088 bytes.
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