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Part BCSIR NET December 2024a-step-down-means-no-arrival-during-service

A step down means no arrival during service

Consider an M/G/1 queuing system with arrival rate λ=1\lambda = 1 and independent and identically distributed successive service times having probability density function g(x) = xe⁻ˣ if x > 0, and 0 otherwise. Define, for i=1,2,…,Ii=1i = 1, 2, \dots, I_{i} = 1 if the first transition is from i to i − 1, and Ii=0I_{i} = 0 if the first transition from i is i + 1. Then, Var(I2)Var(I_{2}) equals

  1. A.5/32
  2. B.5/24
  3. C.3/16✓ Correct
  4. D.8/15

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

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50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: An irreducible chain with a stationary distribution converges to it

More on this topic

The chapter behind this: Markov chains: classification and stationary behaviour — free to read

From Limit Theorems and Markov Chains › Markov chains: classification of states, stationary distributions

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