TKI 3206 – Operations Research 204 – Exponential Distribution and Poisson Process

Sinta R Sul istyo

sinta .sul istyo@ugm.ac. id

Last Lectureo Unconditional Probability

o Classification of States

o Long Run Behavior of DTMC

Outlineo Cost Calculation

o Exponential Distribution

o Poisson Process

Cost Calculation

State Based Costs𝜋𝑗 = long run (L-R) average fraction of time DTMC spends in state 𝑗.

State based costs

𝑐(𝑖) = expected cost per visit to state 𝑖

= expected cost per time unit spent in state 𝑖

Long run average (state based) costs per unit time:

𝑖∈𝑆

𝑐 𝑖 𝜋𝑖

Transition Based Costs𝜋𝑗 = long run (L-R) average fraction of time DTMC spends in state 𝑗.

Transition based costs

𝑐(𝑖, 𝑗) = expected cost per transition from state 𝑖 to state 𝑗

Long run average (transition based) costs per unit time:

𝑖∈𝑆

𝑗∈S

𝑐 𝑖, 𝑗 𝑃𝑖𝑗 𝜋𝑖

Example𝑃 =

01

0.7 0.30.4 0.6

𝜋0 = 4/7 𝜋1 = 3/7

𝑋𝑛 = 01

𝑆 = 0,1

Assume 𝑋𝑛 is a DTMC with transition probability matrix 𝑃

When up, machine makes nothing w.p 0.1, 1 item w.p 0.3, and 2items w.p 0.6. Each item can be sold for $ 200. Operating cost per day $50 only if the machine is up. When down, repair costs $ 100per day. Expected cost per repair is $ 80.

How much is the L-R average daily profit?

if machine down on day 𝑛if machine up on day 𝑛

Exponential Distribution

Exponential Distribution𝑋~𝐸𝑥𝑝(𝜆) 𝑋~𝐸𝑥𝑝𝑜(𝜃)

𝑋 is a continuous RV

Probability Density Function (PDF)

𝑓𝑋 𝑥 = 𝜆𝑒−𝜆𝑥

0

Cumulative Density Function (CDF)

𝐹𝑋 𝑥 = 𝑃 𝑋 ≤ 𝑥 = 1 − 𝑒−𝜆𝑥

∞

𝐸 𝑋 = 1/𝜆 𝑉𝑎𝑟 𝑋 = 1/𝜆2

𝑥 ≥ 0𝑥 < 0

𝑥 ≥ 0otherwise

Properties of ExponentialProperty 1: memoryless property

𝑋 ~ 𝐸𝑥𝑝 (𝜆)

∀ 𝑠, 𝑡 ≥ 0

𝑃 𝑋 > 𝑠 + 𝑡 𝑋 > 𝑡} =𝑃 𝑋 > 𝑠 + 𝑡, 𝑋 > 𝑡

𝑃 𝑋 > 𝑡= ⋯ = 𝑃 𝑋 > 𝑠

ExampleThe amount of time a person spends in a bank is exponentially distributed with mean 10 minutes.

a. What is the probability that a customer will spend more than 15 minutes in the bank?

b. What is the probability that a customer will spend more than 15 minutes in the bank given that he is still in the bank after 10 minutes?

Properties of ExponentialProperty 2: minimum of some exponential RVs

𝑋1, 𝑋2, …𝑋𝑛 are independent

𝑋𝑖~𝐸𝑥𝑝 (𝜆𝑖) 𝑖 = 1,2, … , 𝑛

𝑌 = min𝑖≤𝑖≤𝑛

𝑋𝑖

𝑌 ~ 𝐸𝑥𝑝 ( 𝑖=1𝑛 𝜆𝑖)

Properties of ExponentialProperty 3: sum of some exponential RVs

𝑋1, 𝑋2, …𝑋𝑛 independent

𝑋𝑖~𝐸𝑥𝑝 (𝜆) 𝑖 = 1,2, … , 𝑛

𝑍 = 𝑖=1𝑛 𝑋𝑖

𝑍 ~ 𝐸𝑟𝑙𝑎𝑛𝑔 𝑛, 𝜆 ∀𝑛

Erlang Distribution𝑋 ~ Erlang (𝑛, 𝜆)

𝑓 𝑥 = 𝜆 𝜆𝑥 𝑛−1𝑒−𝜆𝑥

𝑛 − 1 !0

𝐹 𝑥 = 1 −

𝑖=0

𝑛−1𝜆𝑡 𝑖

𝑖!𝑒−𝜆𝑡

𝐸 𝑋 = 𝑛/𝜆

𝑉𝑎𝑟 𝑋 = 𝑛/𝜆2

𝑥 ≥ 0

𝑥 < 0

𝑡 ≥ 0

ExampleThe lifetime of an electronic tube is exponentially distributed with mean 1,000 hours. There are three of these tubes (original and 2 spares). When one burns out, it is replaced by a spare.

What is the expected amount of time Y until the last spare burns out?

Properties of ExponentialProperty 4: between two exponential RVs

𝑋 ~ 𝐸𝑥𝑝 𝜆

𝑌 ~ 𝐸𝑥𝑝 (𝜇)

𝑃 𝑋 < 𝑌 =𝜆

𝜆 + 𝜇

𝑃 𝑋 = 𝑌 = 0

𝑃 𝑋 > 𝑌 =𝜇

𝜆 + 𝜇

ExampleSuppose a stereo system consists of two main parts, a radio and a speaker. If the lifetime of the radio is exponential with mean 1000 hours and the lifetime of the speaker is exponential with mean 500 hours, independent of the radio’s lifetime, then

What is the probability that the system’s failure (when it occurs) will be caused by the radio failing?

Counting Process

Counting Process{𝑁 𝑡 } is a counting process if 𝑁(𝑡) is the number of events by time 𝑡

𝑡 > 𝑠 𝑁 𝑡 − 𝑁 𝑠 = increment from 𝑠 to 𝑡

= number of events in (𝑠, 𝑡]

The Poisson Process (PP)

The Poisson Process (PP)A PP with rate 𝜆 > 0 is a counting process with

i. 𝑁 0 = 0

ii. 𝑁 𝑡 has independent increments

iii. The increment over any period of length 𝑡 is Poisson distributed with parameter 𝜆𝑡

𝑁 𝑡 ~ 𝑁 𝑡 + 𝑠 − 𝑁 𝑠 ~ 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 (𝜆𝑡)

Poisson DistributionA Poisson (𝜆𝑡) RV is discrete with PMF

𝑃 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆𝑡 = 𝑛 = 𝜆𝑡 𝑛

𝑛!𝑒−𝜆𝑡

0

𝐸 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆 = 𝜆

𝑉𝑎𝑟 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆 = 𝜆

𝑛 ∈ 0,1,2, …

otherwise

ExampleThe arrival of customers at a café is a Poisson process with rate 4 per minute. Find the probability that there is no less than 3 arrivals between time (0,2] (the first two minutes)!

Inter-arrival Time 𝑁(𝑡) = number of events by time 𝑡

𝑇1 = time when first event occur

𝑇𝑛 = time between events 𝑛 and 𝑛 − 1 for 𝑛 ≥ 2

𝑇1, 𝑇2, … are all 𝐸𝑥𝑝 (𝜆) distribution independent

Hence,

Time between events in a PP(𝜆) are iid 𝐸𝑥𝑝 (𝜆) RV’s

In a PP, we cannot have two events take place at same time.

𝑆0 = 0

𝑆𝑛 = time when event 𝑛 takes place (𝑛 ≥ 1)

𝑆1 = 𝑇1

𝑆2 = 𝑇1 + 𝑇2

𝑆𝑛 = 𝑖=1𝑛 𝑇𝑖

𝑆𝑛 ~ 𝐸𝑟𝑙𝑎𝑛𝑔 (𝑛, 𝜆)

Note that,

𝑆𝑛 ≤ 𝑡 = 𝑁 𝑡 ≥ 𝑛

𝑃 𝑆𝑛 ≤ 𝑡 = 𝑃 𝑁 𝑡 ≥ 𝑛 = 1 − 𝑃 𝑁 𝑡 < 𝑛 = 1 − 𝑖=0𝑛−1𝑃{𝑁 𝑡 = 𝑖}

= 1 − 𝑖=0𝑛−1𝑃 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝜆𝑡 = 𝑖 = 1 − 𝑖=0

𝑛−1 𝜆𝑡 𝑖

𝑖!𝑒−𝜆𝑡 𝑡 ≥ 0

ExampleSuppose that people immigrate into a territory at a Poisson rate λ=1 per day.

a. What is the expected time until the tenth immigrant arrives?

b. What is the probability that the elapsed time between the tenth and the eleventh arrival exceeds two days?

Next Lectureo a little bit more on Poisson Process

o Semi Markov Processes

o Continuous Time Markov Chain

Question?

thankyouAcknowledgement:

Prof. Sigrún Andradóttir