A 5-stage instruction pipeline has stage delays of 180, 250, 150, 170, and 250, respectively, in nanoseconds. The delay of an inter-stage latch is 10 nanoseconds. Assume that there are no pipeline stalls due to branches and other hazards. The time taken to process 1000 instructions in microseconds is ______. (Rounded off to two decimal places)

The correct answer is 261.04.

Step-by-Step Explanation:

1. Determine the Clock Cycle Time ($T_p$) of the Pipeline:
In a pipelined processor, the clock cycle time is determined by the delay of the slowest stage plus any overhead, such as the delay introduced by the inter-stage latches (registers). This is expressed by the formula:

$$T_p=max\left(\text{Stage Delays}\right)+\text{Latch Delay}$$

Given the stage delays are 180, 250, 150, 170, and 250 nanoseconds, and the latch delay is 10 nanoseconds, we have:

$$T_p=max(180,250,150,170,250)+10=250+10=260\text{ns}$$

2. Calculate the Total Time to Process the Instructions:
For a pipeline with $k$ stages executing $n$ instructions with no stalls, the total number of clock cycles required is given by:
- The first instruction takes $k$ clock cycles to exit the pipeline (to fill the pipeline).
- The remaining $n-1$ instructions each take 1 clock cycle to complete.
Thus, the total cycles required is $k+n-1$.

The total execution time is calculated as:

$$\text{Total Time}=(k+n-1)\times T_p$$

Substituting the given values, where the number of stages $k=5$ and the number of instructions $n=1000$:

$$\text{Total Time}=(5+1000-1)\times 260\text{ns}$$

$$\text{Total Time}=1004\times 260\text{ns}=261040\text{ns}$$

3. Convert the Time to Microseconds ($\text{μs}$):
Since $1\text{μs}=1000\text{ns}$, we divide the result by 1000:

$$\text{Total Time}=\frac{261040}{1000}\text{μs}=261.04\text{μs}$$

Therefore, the time taken to process 1000 instructions is 261.04 microseconds.