Machine Instruction Practice Questions

The total execution time ($ET$) for an application is given by $ET = (\sum_{i=1}^{n} IC_i \times CPI_i) \times T_c$, where $IC_i$ is the instruction count for type $i$, $CPI_i$ is cycles per instruction for type $i$, and $T_c$ is the clock cycle duration.
Given:
Total instructions = $6.4 \times 10^8$.
Total execution time = 6.3 seconds.
Let's calculate the sum of $(IC_i \times CPI_i)$ for all instruction types:
Branch: $2.25 \times 10^8 \times 2 = 4.5 \times 10^8$
Load: $1.20 \times 10^8 \times 5 = 6.0 \times 10^8$
Store: $1.65 \times 10^8 \times 4 = 6.6 \times 10^8$
Arithmetic: $1.30 \times 10^8 \times 3 = 3.9 \times 10^8$
Sum of $(IC_i \times CPI_i) = (4.5 + 6.0 + 6.6 + 3.9) \times 10^8 = 21.0 \times 10^8 = 2.1 \times 10^9$ cycles.
Now, $T_c = ET / (\sum IC_i \times CPI_i) = 6.3 \, s / (2.1 \times 10^9 \, cycles) = 3 \times 10^{-9} \, s/cycle$.
To convert to nanoseconds: $1 \, ns = 10^{-9} \, s$.
So, $T_c = 3 \, ns/cycle$.
The duration of a clock cycle in nanoseconds is 3.

An Instruction Set Architecture (ISA) defines the interface between software and hardware. It specifies the instructions a processor can execute, including their formats, addressing modes, and the types of operations performed.

Let's examine the options:
(a) The size of the cache memory: Cache memory size is a hardware implementation detail, not part of the ISA. The ISA does not dictate the size of the cache.
(b) The clock frequency of the processor: Clock frequency is a hardware implementation characteristic, part of the microarchitecture, not specified by the ISA.
(c) The number of cache memory levels: Like cache size, the number of cache levels is a microarchitectural detail and not part of the ISA.
(d) The total number of registers: The general-purpose registers (GPRs) available to the programmer are specified in the ISA. The number of registers influences the instruction format and the way data can be manipulated.

Therefore, the total number of registers is part of an Instruction Set Architecture.

The instruction format needs to accommodate the opcode, register operand, and immediate operand within 32 bits.
First, determine bits for the opcode: With 50 distinct instruction types, $\lceil\log_2 50\rceil = 6$ bits are required.
Next, determine bits for the register operand: With 64 general-purpose registers, $\lceil\log_2 64\rceil = 6$ bits are needed to identify a register (like R1).
For the instruction "ADD R1, #25", one register (R1) is specified, requiring 6 bits.
The total bits used for opcode and register are $6 + 6 = 12$ bits.
Since the instruction is 32 bits long, the remaining bits available for the immediate operand are $32 - 12 = 20$ bits.

Given information:

• Instruction format: 32 bits.
• Processor has 50 architectural registers. This means each register field needs $\lceil \log_2 50 \rceil = 6$ bits.
• Instruction set has 150 distinct instructions.
• 1 bit is used in OPCODE to distinguish R-type and I-type instructions.
• The remaining bits in OPCODE indicate the operation.
• R-type format: OPCODE | UNUSED | DST Reg | SRC Reg1 | SRC Reg2.
• I-type format: OPCODE | DST Reg | SRC Reg | Immediate value/address.

Let's find the number of bits for each field:

1. Register fields: Since there are 50 architectural registers, each register field (DST Reg, SRC Reg1, SRC Reg2, SRC Reg) requires 6 bits.
   So, 6 bits per register.

2. OPCODE field (Y):
   Total instructions = 150.
   1 bit is used to distinguish R-type and I-type.
   This means the total number of opcode patterns is used to represent 150 instructions.
   The OPCODE field itself must have enough bits to represent the $2 \times \lceil 150/2 \rceil = 150$ instructions.
   If 1 bit distinguishes types, then we need bits for the operation code.
   Let Y be the total bits for the OPCODE field.
   $2^Y \ge 150$. So $Y \ge \lceil \log_2 150 \rceil = 8$. (since $2^7=128, 2^8=256$).
   The PDF solution uses 9 bits for OPCODE. Let's assume Y=9.
   If OPCODE field is Y bits, and 1 bit distinguishes R/I type, then Y-1 bits are for the actual operation.
   $2^{Y-1} \ge 150/2 = 75$. So $Y-1 \ge \lceil \log_2 75 \rceil = 7$.
   This implies $Y \ge 8$.
   If $Y=9$ bits for OPCODE (as in solution):
   1 bit for R/I type.
   8 bits for specific operation. $2^8 = 256$ operations possible, so 150 instructions can be encoded.
   So, $Y=9$ bits for OPCODE.

3. R-type format: Total 32 bits.
   OPCODE (Y bits) | UNUSED (X bits) | DST Reg (6 bits) | SRC Reg1 (6 bits) | SRC Reg2 (6 bits).
   $Y + X + 6 + 6 + 6 = 32$.
   $9 + X + 18 = 32$.
   $X + 27 = 32$.
   $X = 5$ bits.

4. I-type format: Total 32 bits.
   OPCODE (Y bits) | DST Reg (6 bits) | SRC Reg (6 bits) | Immediate value/address (Z bits).
   $Y + 6 + 6 + Z = 32$.
   $9 + 12 + Z = 32$.
   $21 + Z = 32$.
   $Z = 11$ bits.

We need to calculate $X + 2Y + Z$.
$X = 5$.
$Y = 9$.
$Z = 11$.
$X + 2Y + Z = 5 + 2(9) + 11 = 5 + 18 + 11 = 34$.
The value of $X + 2Y + Z$ is 34.
===END_Q61===

Given instruction format:

• Total bits = 32.
• Processor has 16 general purpose registers. This means register fields require $\lceil \log_2 16 \rceil = 4$ bits.
• Two register operand fields: each 4 bits. So $2 \times 4 = 8$ bits for register operands.
• Addressing mode field: supports 8 addressing modes. This means addressing mode field requires $\lceil \log_2 8 \rceil = 3$ bits.
• 16-bit scalar field: 16 bits.

Instruction format is: Opcode | Addressing Mode | Reg-1 | Reg-2 | Scalar Field.
Total bits used by known fields = $3 + 4 + 4 + 16 = 27$ bits. (Assuming opcode is also known, which it isn't yet.)

Let $O$ be the number of bits for the opcode.
Let $A$ be the number of bits for the addressing mode = 3 bits.
Let $R_1$ be the number of bits for Reg-1 = 4 bits.
Let $R_2$ be the number of bits for Reg-2 = 4 bits.
Let $S$ be the number of bits for scalar field = 16 bits.

The total instruction length is 32 bits.
So, $O + A + R_1 + R_2 + S = 32$.
$O + 3 + 4 + 4 + 16 = 32$.
$O + 27 = 32$.
$O = 32 - 27 = 5$ bits.

The maximum number of unique opcodes possible is $2^O$.
Number of unique opcodes = $2^5 = 32$.
The question states "maximum number of unique opcodes possible for every addressing mode". This wording implies that the 3 bits for addressing modes and 5 bits for opcode are independent, which is correctly captured by assigning specific bit fields to them.
===END_Q57===

To derive the correct operands for the assembly code:

1. U1: The expression $b[i] * 8$ is equivalent to $b[i] << 3$ since $8 = 2^3$. Thus, $U1 = 3$.
2. U2: The register $r3$ points to array $a$. To move to the next element in a 32-bit integer array, the pointer must be incremented by 4 bytes. Thus, $U2 = 4$.
3. U3: Similarly, $r4$ points to array $b$. To move to the next element, $r4$ must be incremented by 4 bytes. Thus, $U3 = 4$.
4. U4: The assembly code requires a loop to repeat the operation until $i=10$. The loop starts at label $L01$. Thus, $U4 = L01$.

Comparing these results $(U1=3, U2=4, U3=4, U4=L01)$ with the given options, we find that they match option B.

The program initializes $R_1$ to $MEMORY[5000] = 10$ and $R_3$ to $3000$.
The loop starts by setting $R_2$ to $MEMORY[R_3]$ (which is $50$) and then adds $R_1$ ($10$) to $R_2$, making $R_2 = 60$.
Inside the loop, $MEMORY[R_3]$ is updated to $R_2$ ($60$), $R_3$ is incremented, and $R_1$ is decremented.
The loop continues as long as $R_1$ is not zero. Since $R_1$ starts at $10$ and decrements, the loop executes $10$ times.
This means memory locations $3000$ through $3009$ will be updated to $60$.
Memory location $3010$ is not accessed or modified by the loop, as $R_3$ goes from $3000$ to $3009$. Therefore, its content remains its initial value of $50$.

To execute the instruction $R0 \leftarrow R1+R2$, we need to perform the following sequence of operations:

1. Instruction Fetch: First, the instruction itself must be fetched from memory. This involves reading the Program Counter (PC) value, sending it to the Memory Address Register (MAR), and then reading the memory content into the Memory Data Register (MDR), which then transfers it to the Instruction Register (IR). This corresponds to step 3 ($PC_r, MAR_w, MEM_r$) followed by step 5 ($MDR_r, IR_w$).

2. Operand Fetch: Next, the operands $R1$ and $R2$ need to be read.

   • $R1$ is read and stored in $TEMP1$: step 2 ($R1_r, TEMP1_w$).
   • $R2$ is read, and the ALU performs the addition with the content of $TEMP1$, storing the result in $TEMP2$: step 1 ($R2_r, TEMP1_r, ALU_{add}, TEMP2_w$). Note that $TEMP1_r$ in step 1 refers to the value of $R1$ loaded into $TEMP1$ in the previous step.

3. Result Write-back: Finally, the result from $TEMP2$ is written back to $R0$: step 4 ($TEMP2_r, R0_w$).

Combining these in order gives: $3 \rightarrow 5 \rightarrow 2 \rightarrow 1 \rightarrow 4$.

The total instruction size is 16 bits, creating a total encoding space of $2^{16}$ unique instructions. This space must be shared among all instruction types. The space an instruction type consumes depends on the number of instructions and the bits needed for operands.

Operand bits for each type:

• Type-1 (3Rs): $3 \times \log_2(16) = 12$ bits
• Type-2 (2Fs): $2 \times \log_2(64) = 12$ bits
• Type-3 (1R+1F): $4 + 6 = 10$ bits
• Type-4 (1F): $6$ bits

The total space used is the sum of (number of instructions) $\times$ $2^{\text{operand bits}}$.
$(4 \times 2^{12}) + (8 \times 2^{12}) + (14 \times 2^{10}) + (N \times 2^6) \le 2^{16}$
$(12 \times 2^{12}) + (14 \times 2^{10}) + (N \times 2^6) \le 2^{16}$
$(3 \times 2^2 \times 2^{12}) + (14 \times 2^{10}) + (N \times 2^6) \le 2^{16}$
$(3 \times 2^{14}) + (14 \times 2^{10}) + (N \times 2^6) \le 2^{16}$
Dividing by $2^6$:
$3 \times 2^8 + 14 \times 2^4 + N \le 2^{10}$
$768 + 224 + N \le 1024$
$992 + N \le 1024 \implies N \le 32$
The maximum value of N is 32.

The design of a RISC (Reduced Instruction Set Computer) processor is based on several key characteristics aimed at simplifying hardware and enabling high performance through pipelining.
I. Register-to-register arithmetic operations only: This is a core concept of the load/store architecture used by RISC. Data must be loaded from memory into registers before arithmetic operations can be performed, and results are stored back to memory.
II. Fixed-length instruction format: This simplifies the instruction decoding process, allowing for faster and simpler control logic.
III. Hardwired control unit: RISC processors use a hardwired control unit, which is faster than the microprogrammed control units typically found in CISC processors.
All three listed characteristics are fundamental design principles of RISC processors.

To determine the memory consumed, we first find the size of a single instruction in bits.
The number of bits for the opcode field is $\lceil \log_2 12 \rceil = 4$ bits.
The number of bits for each register identifier is $\lceil \log_2 64 \rceil = 6$ bits.
The total bits per instruction are:
4 (opcode) + 6 (src reg 1) + 6 (src reg 2) + 6 (dest reg) + 12 (immediate) = 34 bits.
Due to byte-alignment, each instruction must occupy a whole number of bytes. 34 bits requires 5 bytes (40 bits).
For 100 instructions, the total memory is $100 \times 5 = 500$ bytes.