B.N.Bandodkar College of Science, Thane. Subject : Computer Simulation and Modeling.

B.N.Bandodkar College of Science, Thane Subject : Computer Simulation and Modeling. Simulation is a powerful technique for solving a wide variety of problems. To simulate is to copy the behaviors of a system or phenomenon under study. The simulation technique is used when the problem cannot be epressed in the Mathematical model. We will be dealing with only numerical sequential simulation hence we have to study data values. Though simulation is indeed versatility tool, It provides only satisfied estimates rather than eact result and it only compares the alternatives rather than generating an optimal one. Hence when all fails go for Simulation. In simulation we have to generate data values. Hence we have to study some statistical models or distributions of data values. Statistical Models/ distributions The value of the characteristic under study is known as a variable. Eg demand, arrival time, service time, marks scored I).Discrete variable : If the number of possible values of the variable is finite or countably infinite then the variable is called as a discrete variable. e.g Number of arrivals at the airport. Number of Jobs completed per day. Let X be a discrete random variable. The possible values of X are denoted by the sample space S X. Then probability function P( i )= P(X= i ) i S X i=,, 3. is known as probability mass function (pmf). This function satisfy the following two conditions ) P( i ) 0 for all i ) P( i )=

Cumulative probability distribution function (cdf) is defined as F() = P(X ) R F() = P ( i) Properties of cdf: ) F(- ) = 0 and F( ) = ) F is non decreasing function of 3) F is continuous at right. 4) P (a < X < b) = F(b) F(a) Mean, Median, mode and variance are the characteristics of the distribution. If M is the median then F(M)= ½ Mode is that value of X at which P() is maimum Mean =E(X) = i *P( i ) Variance = V(X) = E(X-E(X)) = E(X ) (E(X)) e.g. Tossing a die. A die is loaded in such a way that P(i) is proportional to i. P(i)=ki P(i)= k+k+3k+4k+5k+6k= =» k=/ X: 3 4 5 6 P() / / 3/ 4/ 5/ 6/ E(X)= */ +*/+.+6*6/= 9/=4.3333 E(X)= */+ */..+6 *6/= 44/= V(X)= - (4.3333) =.5 Mode = 6

Graph of p.d.f. 6/ / 0 3 4 5 6 Cdf of X F() = 0 < = / < = 3/ < 3 =6/ 3 < 4 = 0/ 4 < 5 = 5/ 5 < 6 = 6 P(<X< 5)= F(5)-F() =(5/) (3/)=(/) P(< X< 5) = P(< X< 5)= F(5)-F() =5/ -/=4/=/3 P(X > 4) = -P(X< 4) =-0/=/

Graph of cdf 6/ 5/ 4/ 3/ / 0 3 4 5 6 Standard discrete distributions Bernoulli distribution : This distribution best describes all situations where a "trial" is made resulting in either "success" or "failure," such as when tossing a coin, or when modeling the success or failure of a surgical procedure. The Bernoulli distribution is defined by pmf as: P()= p if = =(-p) if =0 = 0 otherwise This can be written as P() = p (-p) -, for = 0, and 0<p<

= 0 otherwise Where p is the probability that a particular event will occur Mean = E(X)= *p+0*(-p)=p E(X )= *p-0 *(-p) =p Variance =E(X )-(E(X)) = p-p = p(-p)=pq Mean=p Variance=pq Binomial distribution : The binomial distribution is useful for describing distributions of binomial events, The random variable X that denotes the number of success in n Bernoulli trials has a binominal distribution given by p (), where P() = n C p q n- for = 0,,,...,n 0<p< where p is the probability that the respective event will occur q is equal to -p n is the maimum number of independent trials. Mean = np variance = npq (n, p) are parameters of the binomial distribution. 3 Poisson distribution The Poisson distribution is also sometimes referred to as the distribution of rare events. Eamples of Poisson distributed variables are number of accidents on the particular road on the particular day, number of printing mistakes per page, or the number of defects found in a production process. It is defined by pmf as: P() = (λ e -λ )/! for = 0,,,..., λ>0

According to mathematical derivation Poisson distribution is the limiting case of binomial distribution such that number of trials can be very large probability of success of a trial is (n ), average number of success are finite [np = finite constant =λ] Mean = λ and variance = λ Note : If X, X,..Xn are n independent Poisson variates with parameter λ, then the distribution of X i is also Poisson with parameter n λ. 4 Geometric Distribution If independent Bernoulli trials are made until a "success" occurs, then the total number of trials required is a geometric random variable. The geometric distribution is defined as: P() = pq -, for =,,...,.. 0<p< p is the probability of a success at a given trial, q=-p E (X) = p q and V(X) = p 5 Negative binomial distribution If independent Bernoulli trials are made until a "k successes" occurs, then the total number of trials required is a negative binomial random variable. Its pmf is given by P()= - C k- p k q -k for =k, k+,k+.. 0<p<, p is the probability of a success at a given trial and q=-p E (X) = p k kq and V(X) = p

Summary : Name of the distribution Probability mass function Mean Variance Eample Bernoulli 0<p< P=P(success) Binomial 0<p< P+q= Poisson λ>0 Geometric 0<p< P+q= Negative binomial 0<p< P+q= P() = p (-p) -, for = 0, = 0 otherwise P() = n C p q n- = 0 o.w. for = 0,,.,n P() = (λ e -λ )/! for = 0,,,..., =0 o.w P() = pq -, for =,,..., =0 o.w. P()= - C k- p k q -k for =k, k+, k+,.. =0 o.w. p P(-p) Eperiment with two possible outcomes. Success & failure np npq Number of successes in n Bernoulli trials λ λ Number of events occurring on a unit interval. λ is mean rate per unit time /p q/p Number of trials to obtain the first success in a sequence of Bernoulli trials k/p kq/p Number of trials to obtain the k th success in a sequence of Bernoulli trials

Eamples: ) On an average % defectives items are produced. A lot of size 50 is selected if it contains more than defectives then lot is rejected. Compute the probability that lot is rejected. ) 40% of the printers are rejected at the inspection centre. Find the probability that i) first accepted printer is the third one inspected. ii) five printers are inspected in order to accept three. iii) out of seven printers inspected four are accepted. 3) The calls due to the failure of a computer occur in accordance with Poisson distribution with a mean of per day. Find the probability that i) there are three calls for computer failure on the net day. ii) Two or more calls on the net day. iii) At least one calls. 4) Of the orders a shop receives 5% are welding jobs and 75% are machining jobs. What is the probability that i) out of net five jobs will be machining jobs? ii) Net four jobs will be welding jobs? 5) Students arrival at the library follows Poisson with mean 0 per hr. Determine the probability that i) there are 50 arrivals in the net one hr. ii) there are no arrivals in the net one hr. iii) there are 75 arrivals in the net hrs. 6) Mr. A receives four calls in a day (Poisson distribution).what is the probability that on the net day, the number of calls received will eceed the average by more than one standard deviation. 7) In the production of ball bearing bubbles or depressions occur, then this bearing is unfit for sale. It has been noted that on an average one in every 800 of the ball bearing has a defect. What is the probability that a random sample of 4000 will yield fewer than three defective ball bearings? 8) A random variable X has pmf P()=/(n+) over the range{0,, n}. Find mean and variance of this distribution.

9) An industrial chemical that will retard the spread of fire in paint has been developed. The local sales representative has determined, from past eperience that 48% of the sales calls will result in an order. a) What is the probability that the first order will come on the fourth sales call of the day? b) What is the probability that the third order will come on the sith sales call of the day? c) If eight sales calls are made in a day, what is the probability of receiving eactly si orders? d) If four sales calls are made before lunch, what is the probability that one or less results in an order? 0) Player A is currently winning 0.55 of his games. There are 5 games in the net two weeks. What is the probability that he will win more games than he will lose? ) If r.v. is geometrically distributed over.. with P () = 3P(3) Find P( = odd ) ) A boy is throwing stones at a target, if the probability of hitting the target is /5. i)what is the probability that target it hit on the 5 th attempt? ii) find the probability that si trials were required to hit target thrice. 3) State and prove Memory- less property of geometric distribution. i. P(X > s + t / X>t) = P( X>s) ii. P(X = s + t / X>t) = P( X=s)

II) Continuous Distributions Continuous random variables can be used to describe random phenomena in which the variable of interest can take on may value in some interval - for eample, the time to failure or the length of a rod. Let the random variable X be a continuous r.v. P(-d/ X +d/)= f()d Then the function f() is said to be the probability density function of X. The properties of p.d.f. ) f() 0 for all X S ) f ( ) d Cumulative probability distribution function (cdf) is defined as F() = P(X ) R F() = f ( ) d Properties of cdf: ) F(- ) = 0 and F( ) = ) F is non decreasing function of 3) F is a continuous function of X 4) P( < X < ) =F( ) F( ) If M is the median then F(M)= ½ Mode is that value of X at which f() is maimum. df ( ) d Mode is obtained by solving the equation 0 Mean =E(X) = f ( ) d S E(X ) = f ( ) d S Variance = V(X) = E(X-E(X)) = E(X ) (E(X))

Standard continuous distributions ) Uniform distribution. A random variable X is uniformly distributed on the interval (a, b) if its pdf is given by f() = b a, 0, otherwise The cdf is given by 0, b, a a F () =, < a a < b b Note that P ( < X < ) = F ( ) F ( ) = b a is proportional to the length of the interval, for all,and satisfying a < b. The mean and variance of the distribution are given by E (X) = a b and V (X) = ( b a) The pdf and cdf when a = and b = 6 are shown in Figure f () 0. 0. 0 3 4 5 6

F ().0 0.8 0.6 0.4 0. 0 3 4 5 6 The uniform distribution plays a vital role in simulation. Random numbers, uniformly distributed between zero and, provide the means to generate random events. Eample: The demand (in Kg)for cakes is uniformly distributed over the range (000, 500). Find the probability that on a given day demand is i) less than or equal to 500 kg. ii) demand is between (00, 000 ) iii) demand is more than 000 kg. Hear a= 000, b= 500 i) P( X<500)= F( 500) =(500-000)/(500-000)=500/500=/3 ii) P(00<X<000)= F(000)-F(00)=(000/500)-(00/500)=8/5 iii) P(X>000) =-P(X<000) =-(000/500)=/3 ) Triangular distribution. A random variable X has a triangular distribution if its pdf is given by

( a) ( b a)( c a), ( c ) ( c b)( c a) f () =, 0, a b b< c elsewhere The cdf for the triangular distribution is given by F () = 0, ( a) ( b a)( c, a), ( c ) ( c b)( c a), a a < b Height = / (c - a) a b c Graph of pdf E (X) = a b 3 c and Mode= b Variance= a b c ab 8 ac bc Mode= 3Mean-(a+c) The mode is used more often than the mean to characterize the triangular distribution. As shown in above fig, its height is / (c - a) above the ais. The variance, V (X), of the triangular distribution is little used.

Eample The central processing requirements, for programs that will eecute, have a triangular distribution with a = 0.05 second, b =. seconds, and c = 6.5 seconds. Determine the probability that the CPU requirement for a random program is.5 seconds or less. The required probability is P(X<.5)= F (.5) Since b(=.) <.5 < c (=6.5) (6.5.5) (6.5 0.05)(6.5 F (.5) = - 0. 54 Thus, the probability is 0.54 that the CPU requirement is.5 seconds or less. Problems:. The current reading on Sag Reva s gas mileage indicator is an average of 5.3 miles per gallon. Assume that gas mileage on Sag s car follows a triangular distribution with a minimum value of zero and a maimum value of 50 miles per gallon. What is the value of the median?. The consumption of raw material for a fabrication firm follows triangular distribution with minimum 00units, maimum 75 units and mean is 40 units. Obtain its median. 3. Time required to assemble a component follows triangular distribution with a=0seconds, c= 5 seconds, the median is 5, compute mode. 3) Eponential distribution. A random variable X is said to be eponentially distributed with parameter > 0 if its pdf is given by e f () =, 0 0, elsewhere.) F() = t t e e dt [ ] 0 e 0 e Therefore F()= 0 <0 =- e -λ >0

The eponential distribution has mean and variance given by E (X) = and V (X) = The pdf and cdf are shown in the following Figures f() F() 0 0 Thus, the mean and standard deviation are equal Cdf of eponential pdf of eponential distribution. One of the most important properties of the eponential distribution is that it is memory less, which means that for all s 0 and t 0,

P (X > s + t X > t) = P (X > s) (A) Let X represent the life of a component (a battery, light bulb, computer chip, laser ray, etc.) and assume that X is eponentially distributed. Equation (A) states that the probability that the component lives for at least s + t hours, given that it has survived t hours, is the same as the initial probability that it lives for at least s hours. If the component is alive at time then the distribution of the remaining amount of time s (if X > t), that it survives, namely X > t, is the same as the original distribution of a new component. That is, the component does not remember that it has already been in use for a time t. A used component is as good as new. That Equation (A) holds is shown by eamining the conditional probability P (X > s + t X > s) = P( X P( X s t) s) F( s t) F( s). P (X > s + t X > s) = e e ( s t) s e t = P (X > t) E Suppose that the life of an industrial lamp, in thousands of hours, is eponentially distributed with failure rate = /3 (one failure every 3000 hours, on the average),find the probability that i) it will last longer than its mean life. ii) lamp will last between 000 and 3000 hours iii) lamp will last for another 000 hours, given that it is operating after 500 hours. Solution : life (000hr) average failure rate = /3 ( out 3). i)the probability that the lamp will last longer than its mean life of 3000 hours is given by P (X > 3) = P = (X 3) = F (3). P (X > 3) = ( e -3/3 ) = e - = 0.368 Regardless of the value of, this result will always be the same That is, the probability that the random variable is greater than its mean is 0.368, for any value of.

ii) The probability that the industrial lamp will last between 000 and 3000 hours is determined by P ( X 3) = F (3) - F () F (3) - F () = ( e -3/3 ) - ( e -/3 ) = -0.368 + 0.53 = 0.45 iii) The probability that the industrial lamp will last for another 000 hours, given that it is operating after 500 hours. Problems: P (X > 3.5 X >.5) = P (X > ) = e -/3 = 0.77 ) For an eponentially distributed random variable X, find the value of that satisfies the following relationship: P (X ) = 0.9 P (X 3) )The time to service customers at a bank s teller counter is eponentially distributed with mean 50 secs. What is the probability that the two customers in front of an arriving customer will each take less than 30 secs to complete their transactions? 4 )Gamma distribution. A function used in defining the gamma distribution is the gamma function, which is defined for all > 0 as ( ) 0 e d where ( ) = ( - ) ( - ) If is an integer, then using () = we can show ( ) = ( - )! A random variable X is gamma distributed with parameters and if its pdf is given by

f() f () = 0, ( ( ) ) e, > 0 otherwise The parameter is called the shape parameter and is called the scale parameter. Several gamma distributions for = and various values of are shown in the following Figure The pdf of several Gamma dist when scale parameter =. 0.8 0.6 0.4 0. B= B= B=3 0 0 0.4 0.8..6.4.8 The mean and variance of the gamma distribution are given by E (X) = and V (X) = Mode = (β ) / (θ) The cdf of X is given by

F () = 0, ( ( t) ) When is an integer, the gamma distribution is related to the eponential distribution in the following manner: If the random variable, X, is the sum of independent, eponentially distributed random variables, each with parameter, then X has a gamma distribution with parameters and. Thus, if where the pdf of X j is given by X = X + X +. +X e t dt, > 0 g ( j ) = ( ) e 0,, 0 and the X j are mutually independent. 5)The Erlang distribution is a special case of the Gamma distribution where the shape parameter β is an integer. In the Gamma distribution, this parameter is not restricted to the integers. The p.d.f is given by f () = ( 0, ( ) )! e, >0 Mean= E (X) = and V (X) =, Mode = (β ) / (θ) 6)Weibull distribution. The random variable X has a Weibull distribution if its pdf has the form f () = ( ) 0, ep ( ) υ, ( otherwise

The three parameters of the Weibull distribution are υ (- < υ < ), which is the location parameter; ( > 0), which is the scale parameter; and ( > 0), which is the shape parameter. E (X) = υ + and V (X) = The cdf of the Weibull distribution is given by 0, F () =, ep >v v When υ = 0, the Weibull pdf becomes f () = ( ) ep 0 0, when υ = 0 and =. Letting, = the Weibull distribution is reduced to f () = / e 0,, 0 otherwise which is an eponential distribution with parameter = /. Eample :The time it takes for an aircraft to land and clear the runway at a major international airport has a Weibull distribution with υ =.34 minutes, = 0.5, and = 0.04 minute. Determine the probability that an incoming airplane will take more than.5 minutes to land and clear the runway. In this case P (X >.5) is determined as follows: P (X >.5) = -F (.5) = -( ep.5.34 0.04 0.5 ) =-( e - )= 0.35

Therefore, the probability that an aircraft will require more than.5 minutes to land and clear the runway is 0.35 Eample The time to failure for a flat panel screen is known to have a Weibull distribution with υ = 0, = /3, and = 00 hours. Find the mean time to failure. Also evaluate the probability that a unit fails before 000 hours. The mean time to failure is given E (X) = υ + =0+00 (3 + )= 00 (3 + ) = 00 (3!) = 00 hours The probability that a unit fails before 000 hours is determined by F(000) F (000) = ep 000 00 / 3 = e 0 3 = e -.5 = 0.884 7) Normal distribution. A random variable X with mean (- < < ) and variance > 0 has a normal distribution if it has the pdf f () = ep, The normal distribution is used so often that the notation X N (, ) has been adopted by many authors to indicate that the random variable X is normally distributed with mean and variance. Some of the special properties of the normal distribution are listed here.. lim - f () = 0 and lim f () = 0; the value of f () approaches zero as approaches negative infinity and, similarly, as approaches positive infinity.. f ( - ) = f ( + ); the pdf is symmetric about. 3. The maimum value of the pdf occurs at =. (Thus, the mean and mode are equal.)

f () The cdf for the normal distribution is given by F () = P (X ) = ep t dt It is not possible to evaluate this Equation. Numerical methods could be used, but it appears that it would be necessary to evaluate the integral for each pair (, ). However, a transformation of variables, z = (t - )/, allows the evaluation to be independent of and. If X N (, ), let Z = (X - ) / to obtain

F () = P (X ) = P ( ) / Z = e z / dz = ( ) / ( z ) dz Φ((X-µ)/σ) The pdf φ (z) = e z /, z is the pdf of a normal distribution with mean zero and variance. it is said that Z has a standard normal distribution. The cdf for the standard normal distribution is given by Φ (z) = The probabilities Φ (z) for Z z e t / dt 0 are given in Table Φ (z) = = 0 z z The pdf of the standard normal distribution Eample It is known that X N (50, 9). Determine F (56) = P (X 56). 56 50 3 F (56) = Φ () 0. 977 from Table [Cumulative Normal Distribution]. Eample The time required to load an oceangoing vessel, X, is distributed N (, 4). Find the probability that the vessel will be loaded in less than 0 hours.

The probability that the vessel will be loaded in less than 0 hours is given by F (0), where 0 F(0) = ( ) 0. 587 The probability that or more hours will be required to load the ship can also be determined by inspection, using the symmetry property of the normal pdf, determine P (X < )]. Now, P (X >) = F (). The standardized normal F () = Φ (0) = 0.50. Thus, P (X >) = 0.50 =0.50. The probability that between 0 and hours will be required to load a ship is given by P (0 X ) = F () F (0) = 0.5000 0.587 = 0.343 Eample: The time to pass through a queue to begin self-service at a cafeteria has been found to be N (5, 9). Find the probability that an arriving customer waits between 4 and 7 minutes. The probability that an arriving customer waits between 4 and 7 minutes is determined as follows: P (4 X 7) = F (7) F (4) = Φ 7 5 3 4 5 3 = Φ (0.667) Φ (-0.333) = Φ (0.667) (- Φ (0.333)) =0.7476-(-0.6304) =0.3780 The probability is 0.3780 that the customer will pass through the queue in a time between 4 and 7 minutes Eample Lead-time demand, X, for an item is approimated by a normal distribution with a mean of 5 and a variance of 9. It is desired to determine a value for lead time that will be eceeded only 5% of the time.

Here, the problem is to find 0 such that P (X > 0 ) = 0.05 P (X > 0 ) = 0 5 0 5 P Z 0. 05 3 3 i.e. Φ From Table [Cumulative Normal Distribution] it can be seen that Φ (.645) = 0.95. Thus, 0 can be determined by solving 0 3 5.645 Or 0 = 9.935 Therefore, in only 5% of the cases will demand during lead time eceed available inventory if an order to purchase is made when the shock level reaches 30. Eamples ) Demand for an item follows N(50, 7 )Determine the probability demand eceeds i) 45 units, ii)55 units, iii)65 units. ) The annual rainfall in Chennai is normally distributed with mean 9 cms and standard deviation 3 cms. Evaluate the probability i) of getting ecess rain( 40cms or above )in a given year. ii) of getting 80cms or below rain in a given year. 3) Let X be a normal variable with mean 0 and variance 4.Find the values of a and b such that P(a<X<b)=0.90 and µ-a = µ-b 4) Three shafts are made and assembled in to a linkage. The length of each shaft in cms is distributed as follows Shaft X --N(60, 0.09), Shaft X --N(40, 0.05), Shaft3 X 3 --N(50, 0.) i) Write down the distribution of the length of the linkage ii)determine the prob that length of the linkage will be longer than 50.cms. iii) The tolerance limits for assembly are(49.83, 50.), what proportion of the assembly are within the limits?

5)Given the following distribution evaluate P(6<X<8) i) Normal(0,4) ii)uniform (4,6) iii) eponential with parameter 0. iv) triangular (4,0,6) 8) Lognormal distribution. A random variable X has a lognormal distribution if its pdf is given by f () = 0, ep ( In ), > 0 where > 0. The mean and variance of a lognormal random variable are + E (X) = e / V (X) = e + (e ) The parameters and are not the mean and variance of the lognormal. These parameters come from the fact that when Y has a N (, ) distribution, then X = e Y has a lognormal distribution with parameters and. If the mean and variance of the lognormal are known to be respectively, then the parameters and are given by L and L, = n L L L L = n L L Eample: The rate of return on a volatile investment is modeled as having a lognormal distribution with mean 0% and standard deviation 5%. Determine the parameters for the lognormal distribution. From the information given we have L = 0 and L = 5. Since we have = L 0 n = n. 9654 0 5 L L

L 0 5 = n 0. 06 0 = n L L 9)Beta Distribution : (First Kind):. Beta integral of first type is given by ( m, n) 0 ( ) m n d ( m, n) m ( m n n) A random variable X has a beta distribution of first kind,if its pdf is given by ( m, n) m n f ( ) ( ) 0<< m>0, n>0 = 0 o.w. Mean=E(X)= m ( m n) V()= ( m mn n) ( m n ) Eample: In a certain country the proportion of highway sections requiring repairs in any given year is a r.v. having beta distribution of first kind with m = 3, n =. Find i. On the average what percentage of highway sections requires repairs in the given year? ii. The probability that at most half of the highway section will require repairs in the given year. Poisson Process: Consider random events such as arrivals at station, arrivals of calls, arrivals of jobs, number of breakdowns,.these events can be described by counting function N(t) defined for all t >0. N(t) denotes number of events occurred between [0,t] The counting process {N(t), t> 0} is said to be Poisson Process with mean rate λ if the following assumptions are fulfilled.

) Events should occur one at a time. ) {N(t), t> 0} has stationary increments. The number of events occurring in the interval depends only on the length of the interval. 3) {N(t), t> 0}has independent increments. Future arrivals occure completely at random, independent of the number of arrivals in the past time intervals. P[N(t) =n]=e -λt (λt) n /n! For t > 0 and n=0,,. P[N(t)-N(s) =n]=e -λ(t-s) (λ(t-s)) n /n! n=0,,. Consider first event occurs at t, second at t +t and so on Then P (t >t)= P[N(t)=0]= e -λt P (t <t) = - P(t >t)= - e -λt This is cdf of eponential distribution. Hence interarrival time is eponentially distributed with mean /λ. Empirical distribution: By collecting the data we can write the distribution which is not the standard distribution that is known as empirical distribution. Eg. The number of students coming in a group is observed for 00 groups.,,3,3,,,,5,5,,,6,7,.. And the data is complied as (Discrete variable) Number of students in a group Frequency Relative frequency= probability Cumulative probability 0 0.0 0.0 40 0.40 0.50 3 5 0.5 0.65 4 0 0.0 0.85 5 5 0.05 0.90 6 6 0.06 0.96

7 4 0.04.00 Inter arrival time is noted for 00 customers (continuous variable) Interval Frequency Relative frequency= probability Cumulative probability 0-5 0. 0. 5-0 0 0.0 0.4 0-5 9 0.9 0.70 5-0 9 0.9 0.89 0-5 0..00