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A register to register machine supports 2-address. 1-address and zero-address instructions instruction register size is 24 bits and register set size is 480. If there are 48 2-address instructions and 2048 Zero - address instructions then what is the maximum possible number of 1 - address instruction?
 

closed as a duplicate of: Co doubt
asked in CO and Architecture by Junior (615 points)
closed by | 147 views
0
8188..?
0
512?
0
0
oh shit :/ i reversed the formula ... it should be 8188 :p
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Is there any formula to calculate it?

1 Answer

+1 vote
This Computer supports 2 Address, 1-Address ,Zero Instruction instructions.

Instruction register size =24 bit

Register Set size = 480 = 2^9 = 9 bits  needed to address these registers

Address 1 = 9 bit Address 2 = 9bit

Opcode = 24-(9+9) =6 =2^6 =64 instructions are possible.

48 2- Address Instruction then 64-48 = 16, then 1 Address be 16(2^9).

Now for Zero address instructions, let x be a number of unused opcode = 2^9*x = 2048, x =4

Number of Opcodes used in 1 Address instruction = 16 *2^9 -4 =8188
answered by Active (1.7k points)
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0
In the question they mentioned 2048 zero address instruction.. how u took it as x and equated to 2018?

can't understand the logic.. is there any formulae..

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