Step 1: Understand the problem.
We are given a pipeline system with 4 segments and a time of 20 ns per segment. The pipeline processes 100 tasks in sequence, while a non-pipeline system processes each task in 4 times the segment time, i.e., 80 ns per task. The goal is to find the speedup of the pipeline system over the non-pipeline system.
Step 2: Time taken for processing 100 tasks in a non-pipeline system.
In the non-pipeline system, the total time taken for 100 tasks is:
\[
t_n = k t_p \times n = 4 \times 20 \, \text{ns} \times 100 = 8000 \, \text{ns}.
\]
Step 3: Time taken for processing 100 tasks in the pipeline system.
In the pipeline system, the first task takes \( 4 \times 20 \, \text{ns} = 80 \, \text{ns} \), and subsequent tasks are processed in parallel, taking 20 ns for each task. Thus, the time for 100 tasks is:
\[
t_p = t_p + (n - k) t_p = 80 \, \text{ns} + (100 - 4) \times 20 \, \text{ns} = 80 \, \text{ns} + 1920 \, \text{ns} = 2000 \, \text{ns}.
\]
Step 4: Speedup calculation.
The speedup of the pipeline system over the non-pipeline system is given by:
\[
\text{Speedup} = \frac{t_n}{t_p} = \frac{8000 \, \text{ns}}{2000 \, \text{ns}} = 4.
\]
Step 5: Conclusion.
Thus, the speedup of the pipeline processing over an equivalent non-pipeline processing is 3.88.
Find the least upper bound and greatest lower bound of \( S = \{X, Y, Z\} \) if they exist, of the poset whose Hasse diagram is shown below:
Suppose \( D_1 = (S_1, \Sigma, q_1, F_1, \delta_1) \) and \( D_2 = (S_2, \Sigma, q_2, F_2, \delta_2) \) are finite automata accepting languages \( L_1 \) and \( L_2 \), respectively. Then, which of the following languages will also be accepted by the finite automata:
(A) \( L_1 \cup L_2 \)
(B) \( L_1 \cap L_2 \)
(C) \( L_1 - L_2 \)
(D) \( L_2 - L_1 \)
Choose the correct answer from the options given below: