The memory unit of a computer has 4M words of 32 bits each. The computer has an
ID: 3820859 • Letter: T
Question
The memory unit of a computer has 4M words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field a Mode field to specify 1 of 6 addressing modes a Register Address field to specify one of 30 registers a Memory Address field Assume an instruction is 32 bits long. Show your reasoning for your answers to the following: a. How large must the Mode field be? b. How large must the Register Address field be? c. How large must the Memory Address field be? d. How large is the opcode field?Explanation / Answer
The operation field of an instruction specifies the operation to be performed. This operation must be executed on some data stored in computer register as memory words.
Operands are chosen depends on the type of operand address.
The addressing mode specifies a rule for interpreting or modifying the address field of the instruction between the operand is activity referenced.
How large must mode field will be?
For the given question we need to have 1 to 7 data’s, so it should contain 3 bits (2^3 = 8)
- Will get program verstality.
- To decrease the number of bits in the addressing fields of the instruction.
The reference register contains a memory address, and the address field
contains a displacement from that address.
The register reference may be explicit or implicit. In some implementation, a single segment/base register is employed and is used implicitly.
How large must the register address filed be?
This register addressing is same to direct addressing. Here address filed refer to register.
It requires only less memory.
The disadvantage of register addressing is that the address space is very limited.
For the given question register address must be
60 registers implies 6 bits (2^6 = 64)
The address field references a main memory address, and the reference
register contains a positive displacement from that address. In this case also the
register reference is sometimes explicit and sometimes implicit.
How large must the memory address field be?
To make the computer processed data in binary principle of bits in the data should not be greater than the capacity of the register in the microprocessor. Capacity registry microprocessor is usually in accordance with size of the data bus to a computer, although there are versions where the capacity of multiple register data bus size, and is then filled into two bus clock readings.
256K = 2 8 2 10 = 2 18 , or 18 bits
How large is the opcode field?
An opcode is a single instruction that can be executed by the CPU. In machine language it is a binary loaded into the instruction register.
32 - (3 + 6 + 18) = 5 bits
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