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WOOD 492 MODELLING FOR DECISION SUPPORT. Lecture 25 Simulation. Review. Simulation: used to imitate the real system using computer software, helpful when system is too complex or has many stochastic elements

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review
Review
  • Simulation: used to imitate the real system using computer software, helpful when system is too complex or has many stochastic elements
  • Discrete event simulation: if the state of the system changes at random points in time as a result of various events
  • Different probability distributions are used for different purposes

Wood 492 - Saba Vahid

distributions
Distributions
  • Various probability distributions are used for different random events
  • Poisson : distribution of number of arrivals per unit of time
  • Exponential : distribution of time between successive events (arrivals, serving customers,…)
  • Uniform: for random number generation
  • Normal : for some physical phenomenon's, normally used to represent the distributions of the means of observations from other distributions
  • Binomial: coin flip

Wood 492 - Saba Vahid

cumulative distribution function cdf
Cumulative Distribution Function (CDF)

Highlighted area: P(x<=t)

  • CDF is calculated using the area under the probability density graph (PDF):
  • Assume x is a random variable and t is a possible value for x. If we show the PDF of x with f(x) and the CDF with F(x):

=(the area under f(x) up to point t)

f(x)

t

x

F(x)

1.0

P(x<=t)

t

x

Wood 492 - Saba Vahid

example 16 a discrete event simulation
Example 16 – A discrete event simulation
  • Simulate a queuing system :
    • One server
    • Customers arrive according to a Poisson distribution (mean arrival rate λ = 3 per hour)
    • Service rate changes according to a Poisson distribution (mean service rate μ = 5 customers per hour)

Wood 492 - Saba Vahid

probability reminder
Probability reminder
    • When the arrival rate α(number of arrivals per unit of time t) follows a Poissondistribution with the mean of αt, it means that inter-arrival times (the time between each consecutive pair of arrival) follow an exponential distribution with the mean of 1/α
    • If x belongs to an Exponential distribution with the mean 1/α:
  • Therefore, if customers arrive with the mean rate of 3 per hour, the inter-arrival time has an exponential distribution with the mean of 1/3 hour (on average one arrival happens every 1/3 hour)

So, for example, the probability of an arrival happening in the first hour (time of event, x, is less than or equal to 1 hour, t)

Wood 492 - Saba Vahid

example 16 queuing system
Example 16 – Queuing system
  • State of the system at each time t
    • N(t) = number of customers in the queue at time t
  • Random events in the simulation:
    • Arrival of customers (mean inter arrival times are 1/3 hour)
    • Serving the customers (mean service times are 1/5 hour)
  • System transition formula:
    • Arrival: reset N(t) to N(t)+1
    • Serve customer: reset N(t) to N(t)-1
  • How to change the simulation clock (2 ways):
    • Fixed-time increment
    • Next-event increment

Wood 492 - Saba Vahid

fixed time increment for example 16
Fixed-time increment for Example 16
  • Two steps:
    • Advance the clock by a small fixed amount (e.g: 0.1 hour)
    • Update N(t) based on the events that have occurred (arrivals and serving customers)

Example: let’s move the clock from t=0 to t=0.1 hr

N(0)=0

Probability of an arrival happening in the first 0.1 hr is:

Probability of a departure happening in the first 0.1 hr is:

How to use these probabilities?

Wood 492 - Saba Vahid

using random numbers to generate events
Using random numbers to generate events
  • To see if the events should occur or not, we use a random number generator to generate a uniform random number between [0,1] (e.g. in Excel there is a RAND() function that does this)
  • If the random number is less than the calculated probability (in previous slide) we accept the event, if not we reject it.
  • let’s assume we’ve generated a random number for the arrival of customers with Rand() function, random_A=0.1351

Random_A < 0.259 so we accept the arrival

We must generate a new random number for each case, so let’s assume random_D=0.5622

Random_D >= 0.393 so we reject the departure

N(1) = N(0)+ 1 (arrival) – 0 (departure) = 0+1=1

Example 16

Wood 492 - Saba Vahid

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