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A computer has 170 different operations. Word size is 4 Bytes. one word instruction requires two address fields. One address for register and one address for memory. If there are 37 registers then the memory size is ____ (in KB).
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One word siz e= 4 byte =32 bit

Now since there are 170 different operations so it can be represented by 8 bit = log2(170)

Now since there are 37 registers so it can be represnted by log2(37)= 6 bit

and Your instruction has one register and on memory Field

So i can say size of instruction = size of operation + size of reg + size of mem(say x)

32=8+6+x

x=18bit

now memory size is 2^18 word

But 1 word is 4 byte

so it become 2^18*2^2 = 1MB :)
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Size of word = 4B = 32 bit

No of register = 37 so 6 bits required for this field

No of operations = 170 so 8 bits required for this field 

No of bits for memory field = 32-(8+6)= 18

Size of memory = 2^18 B = 258 KB

Hence 256 KB is correct answer

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