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)

1. Determine the clock cycle time ($t_{\text{clock}}$): The stage delays are $180, 250, 150, 170,$ and $250\text{ ns}$.

   $$\text{Maximum stage delay} = \max(180, 250, 150, 170, 250) = 250\text{ ns}$$

   $$\text{Latch delay} = 10\text{ ns}$$

   $$t_{\text{clock}} = 250 + 10 = 260\text{ ns} = 0.26\,\mu\text{s}$$

2. Determine the total number of clock cycles: For a $k$-stage pipeline executing $n$ instructions with no stalls:

   $$k = 5, \quad n = 1000$$

   $$\text{Total cycles} = k + (n - 1) = 5 + 999 = 1004\text{ cycles}$$

3. Calculate total execution time:

   $$T = 1004 \times 260\text{ ns} = 261040\text{ ns} = 261.04\,\mu\text{s}$$

Thus, the time taken is 261.04 microseconds (acceptable range: 260.20 to 261.20).