The rnairr rnemory of a computer has 256 K words of 32 bits each. The instructbfisetof the CPU of the computer consists of 150 dift'erent operations. All instructiorff h6vean operation code and two (02) equal operand fields. The length of each instruction inthe instruction set is equal to a memory word. Draw the instruction format and specifuthe number of bits in each field

The given computer has a main memory with a capacity of 256 K words, where each word is 32 bits long. The CPU's instruction set consists of 150 different operations. Each instruction in the set has an operation code and two equal operand fields. The length of each instruction is equal to a memory word.

To draw the instruction format, we need to specify the number of bits in each field.

Let's break it down:

1. Operation code field: This field represents the operation being performed by the CPU. Since there are 150 different operations, we need enough bits to represent all of them. To determine the number of bits required, we can use the formula: log2(n), where n is the number of operations. In this case, log2(150) is approximately 7.23. Since we cannot have fractional bits, we round it up to 8 bits.

2. Operand fields: There are two equal operand fields in each instruction. Since each operand field is equal in size, we can divide the remaining bits equally between them. To calculate the number of bits in each operand field, we subtract the number of bits used for the operation code field from the total number of bits in a memory word. In this case, the total number of bits in a memory word is 32, and the number of bits used for the operation code field is 8. So, we have 32 - 8 = 24 bits remaining. Dividing this by 2 gives us 12 bits for each operand field.

Therefore, the instruction format can be represented as follows:

| Operation Code (8 bits) | Operand Field 1 (12 bits) | Operand Field 2 (12 bits) |

In summary, the instruction format consists of an 8-bit operation code field and two 12-bit operand fields.