Publications by Dima
simulateBankSystems
Name:Shahad Al Jahwari, ID:133011 Name:Fatma Al Nadabi, ID:134288 Name:Dima Al Maawali, ID:135069 Introduction: In this report, we simulate a sequential, two-stage queuing system in a bank setting, with a single reception desk and a single bank employee. Customers must visit the reception first, followed by the bank employee for further assista...
4415 sym R (4774 sym/24 pcs) 3 img
fist2 services
Name:Dima Al Maawali, ID:135069 Name:……………., ID:…….. Name:……………., ID:…….. Introduction: In this report, we simulate a sequential, two-stage queuing system in a bank setting, with a single reception desk and a single bank employee. Customers must visit the reception first, followed by the bank employee for further ass...
3233 sym R (4774 sym/24 pcs) 3 img
2 services
Lab Exercise IV: Single Server Queuing System Given queuing system for 10 customers with the following inter-arrival time (ITA) and service time (ST): ST 4 1 4 3 2 4 5 4 5 3 ITA - 8 6 1 8 3 8 7 2 3 IAT = c(0,8,6,1,8,3,8,7,2,3) ST = c(4,1,4,3,2,4,5,4,5,3) AT= cumsum(IAT) TSB = c() TCW= c() TSE=c() TCSS=c() idl=c() for (i in 1:length(IAT...
897 sym 1 tbl
SP24- Assignment Iment
Question 1: [ 5 points] Toss a coin until you obtain \(k\) consecutive heads. Write a function with call form ngtm(k,m,nreps) that uses the Monte Carlo simulation to find and return the approximate probability that it takes more than \(m\) tosses to achieve the goal. set.seed(123) ngtm = function(k,m,nreps){ morem= 0 for (i in 1:nreps) ...
1051 sym 1 img
e2
Lab Exercise II: Generating Continuous Random Variables The Inverse Transform Method Let \(U\) be a uniform \((0, 1)\) random variable. For any continuous distribution function \(F\) the random variable \(X\) defined by \(X = F^{−1}(U)\) has distribution \(F\). [\(F^{−1}(u)\) is defined to be that value \(x\) such that \(F(x) = u\).] Alg...
3000 sym 4 img
ITM and ARM
The Inverse Transform Method (ITM) Acceptance-Rejection Method(ARM) # plot density func curve(4*exp(-4*x),from = 0, to = NULL) true.mean= 1/4 true.sd= sqrt(1/4^2) inf.fn = function(u){ x= -(1/4)*log(1-u) x } inf.fn(u= 0.8) ## [1] 0.4023595 N= 1000 n= 500 mean.est = numeric(N) sd.est = numeric(N) for (i in 1:N) { u = runif(n,0,1)...
2411 sym Python (5966 sym/108 pcs) 4 img