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A digital computer has memory unit with $24$ bits word.The instruction set consists of $150$ different operations. All instructions have an operation code part and an address part. Each instruction is stored in one word of memory.

$Q1$ How many bits are needed for the OP-CODE and how many bit are left for the address of the instruction

1. $4,8$
2. $8,16$
3. $8, 64$
4. $16,64$

$Q2$ What is the maximum available size for memory and the largest unsigned binary number that can be accommodated in one word of  memory

1. $2^{16}, 2^{24} + 1$
2. $2^{16}, 2^{24}$
3. $2^{16},2^{24}-1$
4. $\textrm{None of these}$

edited | 139 views
+1
1 ka b

2 ka c
0
Can you please explain how ?

## 2 Answers

+4 votes

.........

by Boss (35.4k points)
+2 votes

24 bits word

word contains opcode bits + address bits.

Instructions set contain 150 different operations so to identify each uniquely would require at least 8 bits (2^8=256).

so 16 bits left for address. So for question no 1 Ans is B

16-bit for address so total memory that can be addressable is 2^16.

word size is 24 bit so largest unsigned binary number is 2^24 - 1.

So for question no 2 Ans is C

by (147 points)

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