Question

Difficulty: MediumSequential Processes and State Transitions

An automated server optimization pipeline adjusts cache memory CC (in megabytes) and latency LL (in milliseconds) across three sequential cycles (k=1,2,3k = 1, 2, 3). The initial values at cycle 00 are C0=100C_0 = 100 MB and L0=50L_0 = 50 ms. The state transitions occur according to the following rules:

- Cycle 1 (k=1k = 1): C1=C0+20C_1 = C_0 + 20, and L1=L00.2×C0L_1 = L_0 - 0.2 \times C_0.
- Cycle 2 (k=2k = 2): If L135L_1 \le 35, then C2=C130C_2 = C_1 - 30 and L2=L1+10L_2 = L_1 + 10; otherwise, C2=C1+10C_2 = C_1 + 10 and L2=L15L_2 = L_1 - 5.
- Cycle 3 (k=3k = 3): If C2<100C_2 < 100, then C3=1.5×C2C_3 = 1.5 \times C_2 and L3=L215L_3 = L_2 - 15; otherwise, C3=C2+20C_3 = C_2 + 20 and L3=L2+5L_3 = L_2 + 5.

Which of the following statements regarding the system metrics across these cycles are true? Select all that apply.

  1. At the end of Cycle 1, the latency metric L1L_1 is equal to 3030 ms.Answer
  2. B
    At the end of Cycle 2, the cache memory C2C_2 is equal to 130130 MB.
  3. At the end of Cycle 3, the cache memory C3C_3 is equal to 135135 MB.Answer
  4. D
    At the end of Cycle 3, the latency metric L3L_3 is equal to 3535 ms.
  5. The net change in cache memory from Cycle 0 to Cycle 3 (C3C0C_3 - C_0) is 3535 MB.Answer

Answer

The statements asserting that L1=30L_1 = 30 ms at the end of Cycle 1, C3=135C_3 = 135 MB at the end of Cycle 3, and the net change in cache memory C3C0=35C_3 - C_0 = 35 MB are correct.
The calculation reveals the complete sequence of state pairs (Ck,Lk)(C_k, L_k): (100,50)(120,30)(90,40)(135,25)(100, 50) \rightarrow (120, 30) \rightarrow (90, 40) \rightarrow (135, 25). Therefore, at Cycle 1, L1=30L_1 = 30 ms is true. At Cycle 3, C3=135C_3 = 135 MB is true. Finally, the net difference C3C0=135100=35C_3 - C_0 = 135 - 100 = 35 MB is also true.

Step-by-Step Solution

1
Calculate state values at Cycle 1 (k=1k=1)
C1=100+20=120C_1 = 100 + 20 = 120 MB, and L1=500.2×100=30L_1 = 50 - 0.2 \times 100 = 30 ms.
Direct evaluation of the Cycle 1 transition equations starting from C0=100C_0 = 100 and L0=50L_0 = 50.
2
Evaluate condition and calculate state values at Cycle 2 (k=2k=2)
Since L1=3035L_1 = 30 \le 35 is true, C2=12030=90C_2 = 120 - 30 = 90 MB and L2=30+10=40L_2 = 30 + 10 = 40 ms.
The condition L135L_1 \le 35 is satisfied, requiring the application of the primary branch.
3
Evaluate condition and calculate state values at Cycle 3 (k=3k=3)
Since C2=90<100C_2 = 90 < 100 is true, C3=1.5×90=135C_3 = 1.5 \times 90 = 135 MB and L3=4015=25L_3 = 40 - 15 = 25 ms.
The condition C2<100C_2 < 100 is satisfied, triggering the multiplicative increase in cache memory.
4
Verify overall net change and evaluate all option statements
C3C0=135100=35C_3 - C_0 = 135 - 100 = 35 MB. Statements matching L1=30L_1 = 30 ms, C3=135C_3 = 135 MB, and net change =35= 35 MB are true.
Compares calculated metrics against each provided statement.

Key Concept

Multi-stage conditional state transitions and sequential numerical processing
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