Introduction to Datapath and its Units

In modern CPUs, the datapath is the hardware that performs operations on data. It consists of various functional units responsible for executing different stages of an instruction. The primary stages in a typical CPU datapath are:

1. Fetch: The instruction fetch unit is responsible for fetching instructions from memory. This involves using the Program Counter (PC) to point to the next instruction to be executed.

2. Decode: The instruction decode unit interprets the fetched instruction. This involves reading the opcode and determining the necessary operations. The control unit generates the required control signals during this stage.

3. Execute: The execution unit performs the required operations on the operands. This could involve arithmetic operations, logical operations, or address calculations.

4. Memory Access: This stage involves accessing memory for load and store instructions. It reads data from memory or writes data to memory.

5. Write Back: The final stage is where the results of the execution or memory access are written back to the register file.

Role of Key Components

• Control Unit: The control unit orchestrates the operation of the CPU by generating the necessary control signals for each stage of the datapath. It ensures that each instruction is executed correctly and in the right order.

• Register File: The register file is a small, fast storage area that holds the data needed by the CPU during execution. It contains multiple registers that can be read from and written to during the instruction cycle.

• Multiport Register File: A multiport register file allows multiple read and write operations simultaneously, improving the performance by enabling concurrent access to the registers.

• Program Counter (PC): The PC holds the address of the next instruction to be fetched. It is incremented after each instruction fetch to point to the subsequent instruction.

Instruction Execution in Single-Cycle Format

In a single-cycle format, each instruction completes its entire execution cycle in one clock cycle. This means that the fetch, decode, execute, memory access, and write-back stages all occur within one clock cycle.

Limitations of Single-Cycle Format

• Clock Speed Limitation: The clock speed is limited by the slowest instruction since each instruction, regardless of its complexity, must complete in one clock cycle. This results in underutilization of the CPU's capabilities and prevents the clock speed from being optimized for simpler instructions.

Instruction Execution in Multi-Cycle Format

In a multi-cycle format, each instruction is broken down into multiple stages, with each stage completing in one clock cycle. Different classes of instructions may require a different number of cycles to complete.

Drawbacks of Multi-Cycle Format

• Complexity: The control unit becomes more complex as it needs to manage different instruction lengths and stages.

• Underutilization: While this allows for a higher clock speed compared to single-cycle format, there are still periods where parts of the datapath are idle, leading to inefficiencies.

Motivation for Pipelining

To better understand the concepts of single-cycle, multi-cycle, and pipelined CPU architectures, let's draw a parallel to a common household task: doing laundry. The stages in the laundry process can be compared to the stages in a CPU datapath. By relating these technical concepts to a familiar process, we can better grasp the motivation behind using pipelining in modern CPUs.

Stages of Laundry and CPU Datapath

Consider the stages of doing laundry:

1. Soaking: Preparing the clothes by soaking them in water.

2. Washing: Washing the clothes to remove dirt.

3. Drying: Removing excess water from the clothes.

4. Ironing: Smoothing out wrinkles in the clothes.

These stages are analogous to the stages in a CPU datapath:

1. Fetch: Fetching the instruction from memory.

2. Decode: Decoding the instruction to understand what needs to be done.

3. Execute: Performing the actual operation (e.g., arithmetic, logic).

4. Memory Access: Accessing memory if needed (e.g., for load/store operations).

5. Write Back: Writing the result back to the register file.

Single-Cycle Laundry

In a single-cycle approach to doing laundry, each piece of clothing must go through all the stages regardless of whether it needs them or not. This is akin to a CPU where each instruction must pass through every stage of the datapath in a single cycle, even if not all stages are required.

Example:

1. Soaking: Every piece of clothing is soaked, even if it's not dirty enough to need it.

2. Washing: Every piece of clothing is washed.

3. Drying: Every piece of clothing is dried.

4. Ironing: Every piece of clothing is ironed, even if it's wrinkle-free.

Limitations of Single-Cycle Laundry

• Inefficiency: Some clothes don't need soaking, drying, or ironing, but they must go through these stages anyway.

• Time-Consuming: The entire process is limited by the slowest stage, making even simple tasks take a long time.

Multi-Cycle Laundry

In a multi-cycle approach, each piece of clothing goes through only the stages it needs, but no other piece of clothing can start until the current one is completely done with all its required stages. This is akin to a CPU where each instruction takes a different number of cycles to complete, but no new instruction can start until the current one finishes.

Example:

1. Soaking: Only soak the clothes that need it. Start soaking the first piece if needed. Other pieces must wait.

2. Washing: Once the first piece is done soaking, start washing it if needed. Other pieces must wait.

3. Drying: Once the first piece is done washing, start drying it if needed. Other pieces must wait.

4. Ironing: Once the first piece is done drying, start ironing it if needed. Other pieces must wait.

Drawbacks of Multi-Cycle Laundry

• Serial Processing: Each piece of clothing must wait for the previous one to finish all its required stages before starting, leading to significant delays.

• Underutilization: Some stages may be idle while waiting for the current piece to move through all its required stages.

Pipelined Laundry

In a pipelined approach, you overlap the stages of the laundry process, allowing multiple pieces of clothing to be in different stages simultaneously. Each piece of clothing only goes through the stages it needs.

Example:

1. Soak the first piece if needed: Start soaking the first piece of clothing.

2. Start washing the first piece while soaking the second piece if needed: Begin washing the first piece as soon as it's done soaking, and simultaneously start soaking the second piece.

3. Start drying the first piece while washing the second piece and soaking the third piece if needed: Move the first piece to drying, start washing the second, and soak the third.

4. Start ironing the first piece if needed while drying the second, washing the third, and soaking the fourth if needed: Iron the first piece while other pieces are in earlier stages.

Advantages of Pipelined Laundry

• Increased Throughput: Multiple pieces of clothing are being processed at the same time, increasing the overall speed of the laundry process.

• Efficient Utilization: Each stage is always busy, reducing idle times and making better use of resources.

• Optimized Performance: The overall time to complete the entire laundry process is reduced significantly.

Motivation for Pipelining in CPUs

Just like pipelining in laundry, pipelining in CPUs allows for multiple instructions to be processed simultaneously, with each instruction being at a different stage of execution. This parallelism leads to:

• Higher Throughput: More instructions are completed in a given time period.

• Improved Efficiency: The CPU resources are utilized more effectively, reducing idle times.

• Optimized Performance: The overall execution time for a program is significantly reduced, allowing for faster processing and improved performance.

Pipelining and Faster Clock Speeds

Pipelining can lead to faster clock speeds and more simultaneous utilization of resources by breaking down the instruction execution process into multiple stages. Each stage performs a part of the instruction processing and passes its results to the next stage at the end of each clock cycle.

What is a Pipeline Stage?

A pipeline stage is a segment of the instruction execution process where a specific subset of operations is performed. For example, in a CPU datapath, common pipeline stages include:

1. Fetch Stage (F): Fetches the instruction from memory.

2. Decode Stage (D): Decodes the fetched instruction and prepares the necessary control signals.

3. Execute Stage (E): Executes the operation specified by the instruction.

4. Memory Access Stage (M): Reads from or writes to memory if required by the instruction.

5. Write Back Stage (W): Writes the result back to the register file.

Each of these stages is designed to complete its assigned operations within one clock cycle. The results of each stage are stored in intermediate registers and passed to the next stage at the beginning of the next clock cycle.

Importance of Uniform Stage Completion Time

To maximize the efficiency of the pipeline, it is crucial that each stage takes approximately the same amount of time to complete its operations. If one stage takes significantly longer than the others, it becomes a bottleneck, reducing the overall clock speed and diminishing the benefits of pipelining. Therefore, the clock speed of a pipelined CPU is determined by the slowest stage.

Breaking Down Longer Stages

If a particular stage takes longer than others, it should be divided into smaller sub-stages to balance the pipeline. This process increases the depth of the pipeline, allowing the CPU to maintain a higher clock speed by ensuring that each stage completes in roughly the same time. For instance:

• If the Execute Stage is significantly longer due to complex arithmetic operations, it can be split into two smaller stages:

  • Execute 1 Stage: Performs part of the arithmetic operation.

  • Execute 2 Stage: Completes the arithmetic operation.

By doing this, each sub-stage fits within the clock cycle duration, preventing any single stage from slowing down the pipeline. This division ensures that the clock speed is not constrained by a single lengthy stage, allowing for a higher overall clock speed.

Simultaneous Utilization of Resources

Pipelining increases the simultaneous utilization of resources by allowing multiple instructions to be in different stages of execution at the same time. For example, while one instruction is in the execute stage, another can be in the decode stage, and yet another can be in the fetch stage. This concurrent processing of multiple instructions keeps all parts of the CPU busy, maximizing resource utilization.

       Cycle
       1  2  3  4  5  6  7  8  9
Inst1  F  D  E  M  W
Inst2     F  D  E  M  W
Inst3        F  D  E  M  W
Inst4           F  D  E  M  W
Inst5              F  D  E  M  W

Performance Equation for Pipelined Execution

To understand the performance of pipelined execution, we need to consider several factors, including the number of pipeline stages, clock frequency, and the number of instructions to be executed. We'll derive a performance equation that takes these factors into account.

Key Variables

• \( n \) : Number of stages in the pipeline.

• \( f \) : Clock frequency.

• CPI: Cycles per instruction for an ideal pipeline (which is \( n \) ).

• \( I \) : Number of instructions to be executed.

Pipelined Execution Breakdown

1. Pipeline Fill Time: The time required to fill the pipeline.

2. Execution Time for Remaining Instructions: The time to run the rest of the program after the pipeline is filled.

Detailed Execution Time

1. Pipeline Fill Time:

   • The time to fill the pipeline is equal to the number of stages \( n \) since each stage must be filled sequentially.

   • This takes \(n -1\) clock cycles.

2. Execution Time for Remaining Instructions:

   • Once the pipeline is filled, each new instruction completes in one clock cycle.

   • This takes \(I\) clock cycles.

Total Execution Time

The total execution time (in clock cycles) for the pipelined processor can be expressed as:

$$\text{Total Cycles} = I + n - 1$$

Simplified Performance Equation

When the number of instructions \( I \) is very high ( \( I \gg n \) ), the pipeline fill time becomes negligible. Therefore, we can approximate the total execution time as follows:

$$\text{Total Cycles} \approx I$$

Execution Time in Seconds

To convert the total execution time into actual time, we consider the clock frequency \( f \) :

$$\text{Execution Time} = \frac{\text{Total Cycles}}{f}$$

For a large number of instructions \( I \) :

$$\text{Execution Time} \approx \frac{I}{f}$$

Performance Equation

Thus, the performance equation for a pipelined CPU can be summarized as:

$$\text{Performance} \approx \frac{f}{I}$$

where:

• \( I \) is the total number of instructions.

• \( f \) is the clock frequency.

Performance Advantage

To illustrate the performance advantage of pipelining, consider the equivalent non-pipelined (single-cycle) version of the CPU. In a non-pipelined CPU:

• CPI (Cycles Per Instruction): Each instruction would take 1 cycle to execute.

• Clock Cycle Time: However, since the entire instruction must be completed in one cycle, the clock cycle time would be \( n \) times longer than that of a pipelined CPU (or equivalently, the clock frequency would be \( \frac{1}{n} \) of the pipelined version).

In a pipelined CPU:

• CPI (Cycles Per Instruction): The effective CPI is still 1, but instructions are divided into \( n \) stages.

• Clock Cycle Time: The clock cycle time is significantly shorter because each stage is shorter.

Speedup Due to Pipelining

The speedup from pipelining can be calculated by comparing the execution time of the non-pipelined version with the pipelined version.

For the non-pipelined version:

• The execution time per instruction is \( n \) times longer.

For the pipelined version:

• The execution time is divided by \( n \) .

Thus, the speedup due to pipelining is approximately \( n \) , which can be expressed as:

$$\text{Speedup} = \frac{\text{Execution Time (Non-Pipelined)}}{\text{Execution Time (Pipelined)}} = n$$

This speedup indicates that a pipelined CPU can theoretically be up to \( n \) times faster than an equivalent non-pipelined CPU, given that all other factors are equal and assuming ideal conditions with no pipeline stalls or hazards.

Real-World Challenges of Pipelining

While pipelining offers significant performance improvements by allowing multiple instructions to be processed simultaneously, it is not without its challenges. In an ideal world, we could infinitely increase the number of pipeline stages to achieve higher performance gains. However, several types of hazards can impede the efficiency of pipelined execution.

Types of Hazards

Structural Hazards

Structural hazards occur when two or more instructions require the same hardware resource simultaneously. This can happen if there are not enough resources (like ALUs, memory ports, or register file ports) to handle all the instructions in the pipeline stages concurrently.

• Example: If a pipeline has only one memory unit, a structural hazard occurs when one instruction needs to read from memory while another needs to write to memory at the same time.

Data Hazards

Data hazards occur when instructions that are close together in the pipeline depend on the same data. There are three main types of data hazards:

1. Read After Write (RAW): Also known as a true dependency, this occurs when an instruction needs to read a value that a previous instruction is writing.

   Example in MIPS ISA:
   ADD R1, R2, R3 # R1 = R2 + R3
   SUB R4, R1, R5 # R4 = R1 - R5
   In this example, the SUB instruction depends on the result of the ADD instruction because it needs to read R1, which is written by the ADD instruction.

    Clock cycle:       1    2    3    4    5    6    7    8    9    
    -------------------------------------------------------------
    ADD R1, R2, R3     F    D    E    M    W
    SUB R4, R1, R5          F    s    s    s    D    E    M    W
   

   Explanation:

   1. The ADD instruction starts at cycle 1 and progresses through the stages without any stalls.

   2. The SUB instruction starts at cycle 2 but stalls for 3 cycles (indicated by 's') after the Fetch (F) stage. This is because the Decode (D) stage of SUB needs to happen after the Write-back (W) stage of ADD to ensure the correct value of R1 is used. After the ADD instruction completes the Write-back stage in cycle 5, the SUB instruction can proceed to the Decode stage in cycle 6 and then continue through the remaining stages.

    Clock cycle:       1    2    3      4    5    6    
    ------------------------------------------------
    ADD R1, R2, R3     F    D    E->    M    W
    SUB R4, R1, R5          F    D    ->E    M    W

Explanation:

1. The ADD instruction starts at cycle 1 and progresses through the stages without any stalls.

2. The SUB instruction starts at cycle 2:

   • During the Decode (D) stage of SUB, the forwarding path allows it to use the result of the ADD instruction as soon as it is produced.

   • The Execute (E*) stage of SUB uses the forwarded result from the ADD instruction, which is available right after the Execute (E) stage of ADD.

   • Thus, SUB can proceed without any stalls and completes normally.

2. Write After Read (WAR): Also known as an anti-dependency, this occurs when an instruction needs to write a value that a previous instruction is reading.

   Example in MIPS ISA:
   LW R1, 0(R2) # Load word into R1 from memory address in R2
   ADD R2, R3, R4 # R2 = R3 + R4
   In this example, the ADD instruction needs to write to R2, but the LW instruction is reading from R2. The ADD instruction must wait until the LW instruction has read R2.

3. Write After Write (WAW): Also known as an output dependency, this occurs when two instructions need to write to the same location.

   Example in MIPS ISA:
   ADD R1, R2, R3 # R1 = R2 + R3
   SUB R1, R4, R5 # R1 = R4 - R5
   In this example, both ADD and SUB instructions are writing to the same register R1. The SUB instruction must wait until the ADD instruction has finished writing to R1.

Control Hazards

Control hazards occur due to the pipeline's handling of branch instructions. When the pipeline encounters a branch instruction, it may not know which path to take (whether the branch will be taken or not) until the branch condition is evaluated. This uncertainty can cause delays and inefficiencies in the pipeline.

Example of a Control Hazard in MIPS ISA

Consider the following MIPS instructions:

LW R10, 0(R9) # Load value into R10 from memory at address in R9
BEQ R1, R2, label # If R1 == R2, branch to label
ADD R3, R4, R5 # Execute this if branch is not taken
label: SUB R6, R7, R8 # Execute this if branch is taken

Explanation

• LW R10, 0(R9): This instruction loads a value from memory into register R10. The value in R10 after this instruction will influence the outcome of the branch instruction.

• BEQ R1, R2, label: The branch instruction checks if R1 is equal to R2. Based on the value loaded in the previous instruction, the branch may or may not be taken.

• ADD R3, R4, R5: This instruction is executed if the branch is not taken (i.e., if R1 is not equal to R2).

• label: SUB R6, R7, R8: This instruction is executed if the branch is taken (i.e., if R1 is equal to R2).

Control Hazard Explanation:

• The control hazard arises because the pipeline does not know whether the branch will be taken until the BEQ instruction is evaluated.

• While the pipeline is fetching and decoding the BEQ instruction, it may have already fetched the next sequential instruction (ADD R3, R4, R5). If the branch is taken, this fetched instruction will need to be discarded, causing a control hazard.

• To mitigate control hazards, techniques such as branch prediction, branch delay slots, or pipeline stalling can be used.

Mitigation with Stalls

       Clock cycle:       1    2    3    4    5    6    7    8    9
       -----------------------------------------------------------------
       LW R10, 0(R9)      F    D    E    M    W
       BEQ R1, R2, label       F    D    s    s    E    M    W
       ADD R3, R4, R5                                   F           
label: SUB R6, R7, R8                                   F

Mitigation with Prediction - branch taken (when prediction is correct)

       Clock cycle:       1    2    3    4    5    6    7    8    9
       -----------------------------------------------------------------
       LW R10, 0(R9)      F    D    E    M    W
       BEQ R1, R2, label       F    D    E    M    W
label: SUB R6, R7, R8               s    F    D    E    M    W

Mitigation with Prediction - branch not taken (when prediction is not correct)

       Clock cycle:       1    2    3    4    5    6    7    8    9    10   11
       ----------------------------------------------------------------------
       LW R10, 0(R9)      F    D    E    M    W
       BEQ R1, R2, label       F    D    E    M    W
       ADD R3, R4, R5                F   D*   FLUSH
label: SUB R6, R7, R8                         F    D    E    M    W

Impact on CPI

Data forwarding helps maintain the CPI by reducing the need for stalls. However, if data forwarding cannot fully resolve the hazard, some stalls might still be necessary, which can increase the CPI for those instructions.

Given that instructions can introduce flushes and stalls, the ideal speedup in a pipelined processor is never fully realized as \( n \) . Flushes and stalls, caused by hazards and mispredictions, introduce delays and idle cycles, preventing the pipeline from operating at maximum efficiency. Consequently, the actual speedup achieved is always less than the theoretical \( n \) times improvement.

Problems

1. Given a pipelined processor with 5 stages, where each stage takes 4 nanoseconds. What is the fastest possible clock speed for this pipeline? Additionally, calculate the Instructions Per Cycle (IPC) and the Instructions Per Second (IPS) for this pipeline, assuming ideal conditions without hazards or stalls.

2. Given a pipelined processor with 5 stages, where each stage takes the following times: Fetch - 2 nanoseconds, Decode - 3 nanoseconds, Execute - 1 nanosecond, Memory Access - 4 nanoseconds, and Write Back - 2 nanoseconds. What is the fastest possible clock speed for this pipeline? Additionally, calculate the Instructions Per Cycle (IPC) and the Instructions Per Second (IPS) for this pipeline, assuming ideal conditions without hazards or stalls.

3. In a real-world pipelined processor, each pipeline stage has a buffer to store results that are pushed to the next stage at the start of the next cycle. Consider a processor with the following stage delays: Fetch - 2 nanoseconds, Decode - 3 nanoseconds, Execute - 1 nanosecond, Memory Access - 4 nanoseconds, and Write Back - 2 nanoseconds. Each stage also includes a constant buffer latch time of 1 nanosecond.

   Given these parameters:

   1. Calculate the fastest possible clock speed for this pipeline.

   2. Determine the Instructions Per Cycle (IPC).

   3. Calculate the Instructions Per Second (IPS) for this pipeline, assuming ideal conditions without hazards or stalls.

4. Consider a single-cycle processor where each instruction takes one cycle to execute, and the clock speed is 1 GHz (1 nanosecond per cycle). This processor is then pipelined into a 4-stage pipeline, and the clock rate is increased by a factor of 4 to 4 GHz (0.25 nanoseconds per cycle). Calculate the time needed to execute 10 instructions, 100 instructions, and 1000 instructions for both the single-cycle and pipelined processors. For the pipelined processor, include the pipeline fill time. Show how, when the instruction count increases and ( I >> n ), the time to fill the pipeline becomes insignificant.

5. Consider a CPU with a 5-stage pipeline operating at a clock frequency of 2 GHz. The instruction mix consists of 50% arithmetic, 30% control, and 20% memory instructions. Assume 20% of arithmetic instructions introduce a 1-cycle stall, 10% of control instructions introduce a 3-cycle flush, and 25% of memory instructions introduce a 2-cycle stall. Calculate the overall performance of the CPU, taking into account the stalls and flushes introduced by these instructions. Determine the effective CPI and the Instructions Per Second (IPS) for the given instruction mix and pipeline characteristics. Assume there are 1 billion instructions in the program.

Solutions

Problem 1

1. Fastest Possible Clock Speed:

   The fastest possible clock speed for a pipelined processor is determined by the longest stage delay. Given each stage takes 4 nanoseconds, the clock period \(T\) is 4 nanoseconds.

   To find the clock speed (frequency), we use the formula:

   $$\text{Clock Speed} = \frac{1}{\text{Clock Period}}$$

   Since the clock period \(T = 4 \, \text{nanoseconds} = 4 \times 10^{-9} \, \text{seconds}\):

   $$\text{Clock Speed} = \frac{1}{4 \times 10^{-9}} = 2.5 \times 10^8 \, \text{Hz} = 250 \, \text{MHz}$$

   Therefore, the fastest possible clock speed for this pipeline is 250 MHz.

2. Instructions Per Cycle (IPC):

   In an ideal pipelined processor without hazards or stalls, one instruction completes every clock cycle after the pipeline is filled.

   $$\text{IPC} = 1$$

3. Instructions Per Second (IPS):

   The Instructions Per Second (IPS) is calculated by multiplying the Instructions Per Cycle (IPC) by the clock speed:

   $$\text{IPS} = \text{IPC} \times \text{Clock Speed}$$

   Substituting the values:

   $$\text{IPS} = 1 \times 250 \times 10^6 = 250 \times 10^6 \, \text{instructions per second}$$

   $$\text{IPS} = 250 \, \text{million instructions per second}$$

Final Answers

• Fastest Possible Clock Speed: 250 MHz

• Instructions Per Cycle (IPC): 1

• Instructions Per Second (IPS): 250 million instructions per second

Problem 2

1. Fastest Possible Clock Speed:

   The fastest possible clock speed for a pipelined processor is determined by the longest stage delay. The time taken by each stage is as follows:

   • Fetch: 2 nanoseconds

   • Decode: 3 nanoseconds

   • Execute: 1 nanosecond

   • Memory Access: 4 nanoseconds

   • Write Back: 2 nanoseconds

The slowest stage is the Memory Access stage, which takes 4 nanoseconds. Therefore, the clock period (T) is 4 nanoseconds.

To find the clock speed (frequency), we use the formula:

$$\text{Clock Speed} = \frac{1}{\text{Clock Period}}$$

Since the clock period \((T = 4 \text{nanoseconds} = 4 \times 10^{-9} \text{seconds})\):

$$\text{Clock Speed} = \frac{1}{4 \times 10^{-9}} = 2.5 \times 10^8 \, \text{Hz} = 250 \, \text{MHz}$$

Therefore, the fastest possible clock speed for this pipeline is 250 MHz.

2. Instructions Per Cycle (IPC):

   In an ideal pipelined processor without hazards or stalls, one instruction completes every clock cycle after the pipeline is filled.

   $$\text{IPC} = 1$$

3. Instructions Per Second (IPS):

   The Instructions Per Second (IPS) is calculated by multiplying the Instructions Per Cycle (IPC) by the clock speed:

   $$\text{IPS} = \text{IPC} \times \text{Clock Speed}$$

   Substituting the values:

   $$\text{IPS} = 1 \times 250 \times 10^6 = 250 \times 10^6 \, \text{instructions per second}$$

   $$\text{IPS} = 250 \, \text{million instructions per second}$$

Final Answers

• Fastest Possible Clock Speed: 250 MHz

• Instructions Per Cycle (IPC): 1

• Instructions Per Second (IPS): 250 million instructions per second

Problem 3

1. Fastest Possible Clock Speed:

   To determine the fastest possible clock speed for the pipeline, we need to consider both the stage delays and the buffer latch time. The total delay for each stage is the stage delay plus the buffer latch time.

   The stage delays and buffer latch times are:

   • Fetch: 2 nanoseconds + 1 nanosecond = 3 nanoseconds

   • Decode: 3 nanoseconds + 1 nanosecond = 4 nanoseconds

   • Execute: 1 nanosecond + 1 nanosecond = 2 nanoseconds

   • Memory Access: 4 nanoseconds + 1 nanosecond = 5 nanoseconds

   • Write Back: 2 nanoseconds + 1 nanosecond = 3 nanoseconds

The slowest (longest) stage determines the clock period for the pipeline. The slowest stage is Memory Access with a total delay of 5 nanoseconds.

The clock period (T) is 5 nanoseconds.

To find the clock speed (frequency), we use the formula:

$$\text{Clock Speed} = \frac{1}{\text{Clock Period}}$$

Since the clock period \((T = 5 \text{nanoseconds} = 5 \times 10^{-9} \text{seconds})\):

$$\text{Clock Speed} = \frac{1}{5 \times 10^{-9}} = 2 \times 10^8 \, \text{Hz} = 200 \, \text{MHz}$$

Therefore, the fastest possible clock speed for this pipeline is 200 MHz.

2. Instructions Per Cycle (IPC):

   In an ideal pipelined processor without hazards or stalls, one instruction completes every clock cycle after the pipeline is filled.

   $$\text{IPC} = 1$$

3. Instructions Per Second (IPS):

   The Instructions Per Second (IPS) is calculated by multiplying the Instructions Per Cycle (IPC) by the clock speed:

   $$\text{IPS} = \text{IPC} \times \text{Clock Speed}$$

   Substituting the values:

   $$\text{IPS} = 1 \times 200 \times 10^6 = 200 \times 10^6 \, \text{instructions per second}$$

   $$\text{IPS} = 200 \, \text{million instructions per second}$$

Final Answers

• Fastest Possible Clock Speed: 200 MHz

• Instructions Per Cycle (IPC): 1

• Instructions Per Second (IPS): 200 million instructions per second

Problem 4

To compare the execution time for a single-cycle processor and a pipelined processor, we will calculate the time needed for 10, 100, and 1000 instructions.

1. Single-Cycle Processor

For a single-cycle processor with a clock speed of 1 GHz (1 nanosecond per cycle), each instruction takes exactly one cycle to execute.

• Time per instruction: 1 nanosecond

For \(I\) instructions, the total execution time \(T\) is:

$$T_{\text{single-cycle}} = I \times 1 \, \text{nanosecond}$$

• 10 instructions:

$$T_{\text{single-cycle}} = 10 \times 1 = 10 \, \text{nanoseconds}$$

• 100 instructions:

$$T_{\text{single-cycle}} = 100 \times 1 = 100 \, \text{nanoseconds}$$

• 1000 instructions:

$$T_{\text{single-cycle}} = 1000 \times 1 = 1000 \, \text{nanoseconds}$$

2. Pipelined Processor

For a pipelined processor with 4 stages and a clock speed of 4 GHz (0.25 nanoseconds per cycle), the time to execute an instruction after the pipeline is filled is 1 cycle (0.25 nanoseconds per instruction).

However, the pipeline takes time to fill. The formula to calculate the number of cycles needed to execute \(I\) instructions in a pipeline with \(n\) stages is:

$$\text{Cycles} = I + n - 1$$

Given that \(n = 4\) stages:

$$\text{Cycles} = I + 4 - 1 = I + 3$$

The total execution time \(T\) for the pipelined processor is:

$$T_{\text{pipelined}} = \text{Cycles} \times \text{Clock Period}$$

The clock period for the pipelined processor is 0.25 nanoseconds:

$$T_{\text{pipelined}} = (I + 3) \times 0.25 \, \text{nanoseconds}$$

• 10 instructions:

$$T_{\text{pipelined}} = (10 + 3) \times 0.25 = 13 \times 0.25 = 3.25 \, \text{nanoseconds}$$

• 100 instructions:

$$T_{\text{pipelined}} = (100 + 3) \times 0.25 = 103 \times 0.25 = 25.75 \, \text{nanoseconds}$$

• 1000 instructions:

$$T_{\text{pipelined}} = (1000 + 3) \times 0.25 = 1003 \times 0.25 = 250.75 \, \text{nanoseconds}$$

3. Observation: Impact of Pipeline Fill Time

As the number of instructions \(I\) increases:

• For small \(I\) (e.g., 10 instructions), the pipeline fill time is a significant part of the total execution time.

• For larger \(I\) (e.g., 1000 instructions), the time to fill the pipeline becomes negligible compared to the total execution time.

Thus, as \(I \gg n\) (where \(n\) is the number of pipeline stages), the pipeline fill time becomes insignificant, demonstrating the advantage of pipelining for large numbers of instructions.

Final Results

• Single-Cycle Processor:

  • 10 instructions: 10 nanoseconds

  • 100 instructions: 100 nanoseconds

  • 1000 instructions: 1000 nanoseconds

• Pipelined Processor:

  • 10 instructions: 3.25 nanoseconds

  • 100 instructions: 25.75 nanoseconds

  • 1000 instructions: 250.75 nanoseconds

Problem 5

To calculate the overall performance of the CPU, we need to determine the effective CPI (Cycles Per Instruction) considering the stalls and flushes caused by different instruction types. We then use this effective CPI to compute the Instructions Per Second (IPS).

1. Given Information

• Clock frequency: 2 GHz (2 billion cycles per second)

• Instruction mix:

  • Arithmetic Instructions: 50%

  • Control Instructions: 30%

  • Memory Instructions: 20%

• Stalls and Flushes:

  • 20% of Arithmetic Instructions cause a 1-cycle stall.

  • 10% of Control Instructions cause a 3-cycle flush.

  • 25% of Memory Instructions cause a 2-cycle stall.

• Total Instructions: 1 billion ((10^9) instructions)

2. Calculate the Effective CPI

The effective CPI (Cycles Per Instruction) is calculated by taking into account the base CPI of the pipeline and adding the additional cycles due to stalls and flushes.

$$\text{Effective CPI} = \text{Base CPI} + \text{Stalls/Flushes per Instruction}$$

The base CPI for an ideal pipeline (without stalls) is 1. However, we must add the additional cycles introduced by the instruction mix.

Calculate additional cycles for each instruction type:

1. Arithmetic Instructions:

   • 50% of total instructions

   • 20% cause a 1-cycle stall

$$\text{Additional Cycles (Arithmetic)} = 0.5 \times 0.2 \times 1 = 0.1$$

2. Control Instructions:

   • 30% of total instructions

   • 10% cause a 3-cycle flush

$$\text{Additional Cycles (Control)} = 0.3 \times 0.1 \times 3 = 0.09$$

3. Memory Instructions:

   • 20% of total instructions

   • 25% cause a 2-cycle stall

$$\text{Additional Cycles (Memory)} = 0.2 \times 0.25 \times 2 = 0.1$$

Total Additional Cycles per Instruction:

$$\text{Total Additional Cycles} = 0.1 + 0.09 + 0.1 = 0.29$$

Effective CPI Calculation:

$$\text{Effective CPI} = 1 + 0.29 = 1.29$$

3. Calculate Instructions Per Second (IPS)

Instructions Per Second (IPS) can be calculated using the clock frequency and the effective CPI:

$$\text{IPS} = \frac{\text{Clock Frequency}}{\text{Effective CPI}}$$

Given the clock frequency is 2 GHz ((2 \times 10^9) cycles per second):

$$\text{IPS} = \frac{2 \times 10^9}{1.29} \approx 1.55 \times 10^9 \, \text{instructions per second}$$

4. Calculate Total Execution Time for 1 Billion Instructions

The total execution time (T) for 1 billion instructions can be calculated using the effective CPI:

$$T = \frac{\text{Total Instructions} \times \text{Effective CPI}}{\text{Clock Frequency}}$$

$$T = \frac{10^9 \times 1.29}{2 \times 10^9} = \frac{1.29}{2} \, \text{seconds} = 0.645 \, \text{seconds}$$

Final Answers

• Effective CPI: 1.29

• Instructions Per Second (IPS): 1.55 billion instructions per second

• Total Execution Time for 1 Billion Instructions: 0.645 seconds