Instruction Format, Addressing Capacity, Bus-Width Impact, and Register Bit-Widths (32-bit Hypothetical Microprocessor)

We consider a hypothetical 32-bit microprocessor whose instruction word is 32 bits and has two fields: the first byte (8 bits) holds the opcode, while the remaining 24 bits holds either an immediate operand or an operand address. From this instruction format, the address-related field width is directly $24$ bits, which constrains what can be encoded as a direct memory address. This directly ties into the general principle that the width of the address bus determines the maximum addressable memory locations (typically in bytes).2

Key ideas:

• Opcode
• Immediate operand
• Direct addressing
• Address bus
• Instruction register (IR)

(a) Maximum directly addressable memory capacity (bytes)

Since:

• Instruction size = 32 bits
• Opcode field = first byte = 8 bits
• Remaining field = address/immediate field = $32 - 8 = 24$ bits

For direct addressing, the CPU can encode $2^{24}$ distinct byte addresses. Therefore, the maximum directly addressable memory is: $2^{24}\ \text{bytes} = 16,777,216\ \text{bytes} \approx 16\ \text{MiB}$

This follows the general addressing relationship: if the address contains $n$ bits and memory is byte-addressed, then the number of addressable bytes is $2^n$.2

(b) Impact on system speed for different bus widths

Bus width affects the amount of data moved per transfer and thus the system’s throughput. In general, wider data buses move more bits per memory access, while wider address buses increase the number of accessible locations (addressing capacity) but do not automatically increase per-transfer payload.2

We’ll analyze two hypothetical cases. Let:

• Local address bus width = $A$ bits
• Local data bus width = $D$ bits

Assume a typical RAM operation transfers $D$ bits per memory access. If the CPU needs a 32-bit word but the data bus is narrower, the word will require multiple transfers (and thus more bus cycles). The number of transfers needed for one 32-bit word is approximately: $\left\lceil \frac{32}{D} \right\rceil$

This is consistent with the intuition that a narrower data bus reduces how much data can be moved per transaction, slowing execution.

(i) 32-bit local address bus and 16-bit local data bus

• Addressing capacity: $2^{32}$ locations (if byte-addressed) is addressed-capable, but the instruction’s direct address field (24 bits) still limits direct encoding (part a). The system can still potentially use addressing modes/banking beyond direct field limits, but that’s outside the given instruction format constraint.
• Data movement: $16$-bit data bus means each transfer moves 16 bits $=2$ bytes.

So a 32-bit operand fetch/store generally takes: $\left\lceil \frac{32}{16} \right\rceil = 2\ \text{transfers}$

Impact on speed: compared to an ideal 32-bit-wide data bus, memory accesses that require 32-bit quantities will take roughly about 2× the bus transactions, reducing throughput and slowing instruction execution, especially for instruction fetch and load/store operations.2

(ii) 16-bit local address bus and 16-bit local data bus

• Addressing capacity is now limited by the 16-bit local address bus: $2^{16}\ \text{bytes} = 65,536\ \text{bytes} \ (64\ \text{KiB})$ This directly constrains the maximum directly usable memory space by the local bus.
• Data movement is the same as case (i): 16-bit data bus $\Rightarrow$ again: $\left\lceil \frac{32}{16} \right\rceil = 2\ \text{transfers per 32-bit word}$

Impact on speed: two effects occur:

1. Like (i): narrower data bus causes ~2 transfers per 32-bit word (slower throughput).
2. Additional severe limitation: reduced addressable memory space can force address remapping/banking or prevent programs from fitting; that can increase software complexity and/or cause extra overhead for accessing the needed data/instructions.

Overall, case (ii) is expected to be at least as slow as (i) for transfers and potentially much slower in practice due to reduced accessible memory (and any added management overhead).

Visual intuition (Mermaid)

(c) Bits needed for program counter (PC) and instruction register (IR)

Instruction register (IR)

The IR must hold the entire fetched instruction word. Since instructions are 32-bit, the IR width is: $\text{IR bits} = 32$

This is consistent with the general rule that the instruction register stores the current instruction in full machine-instruction length.

Program counter (PC)

The PC must generate enough address bits to select the next instruction location. In many textbook treatments, the number of PC bits equals the number of bits needed to address all instruction memory locations, which is tied to the address bus width (or instruction-memory address space).

Given the instruction format uses 24 bits for an operand address field, a common interpretation for direct instruction-memory addressing is that instruction addresses also use the same effective address width as “direct addressing.” Under that assumption: $\text{PC bits} = 24$

However, if instead the question intends PC to match the local address bus, then PC bits would be $32$ in case (i). The problem statement in part (c) does not explicitly refer to the bus configuration from (b), so the cleanest inference consistent with the “operand address field” mechanism is to use $24$ bits for direct instruction addressing.

Thus:

• Program counter (PC)
• PC bits: $24$
• IR bits: $32$

(If your course uses PC = address bus width, substitute the relevant bus width accordingly.)