No BSD License
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E645_a
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[A]=phaseak(a,T,b,S)
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[D]=e332c
% Compute the renewal function with U(0,1) interarrival time
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[D]=e713
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[F,g]=c3_wibo1(alpha,beta,t)
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[F]=fpss_pro(P,m,n)
% For the Markov chain with transition matrix P,
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[H]=e589
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[PC]=mc_canon(P,g)
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[P]=e411_dat
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[P]=e413_dat(m,s,S)
% Find the transition matrix for the (s,S) inventory example
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[P]=e422_dat(m)
% Transition matrix for Example 4.2.2
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[P]=e426_dat
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[P]=e441_dat
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[Q,f]=mc_canop(P)
% For a periodic irreducible Markov chain with transition probability matrix
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[R]=e621
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[R]=gph1_sra(A)
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[T,t]=e331
% Simulating mean total life
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[X]=c3_mt_f(F,g,t)
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[X]=e2311
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[X]=e239b
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[X]=e446
% Example 4.4.6 - the coupon collection example
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[X]=e523
% The Emergency Service Example
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[X]=e631
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[X]=e632
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[X]=e633
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[X]=e645_c
% The main calling program for Example 6.4.5
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[X]=e649
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[X]=e711
% Simulate a standard Weiner process
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[X]=e712_c
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[X]=e721
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[X]=e722
% Find virtual delay distribution of PH/PH/1 queue
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[X]=e724a
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[Z]=e522
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[a,T]=e721a
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[b,S]=e721b
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[f]=invt_laq(input,t)
% Numerical Inversion of Laplace transform
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[g]=mc_equca(P,po)
% Find the equivalent classes a Markov Chain
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[gcd]=mc_GCD(a,b)
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[lm]=e2311b(t)
% Generate the arrival rate for Example 2.3.11
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[lm]=e236(t)
% Generate the arrival rate for Example 2.3.6
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[p]=c4_pos(m)
% It produces a Poisson mass function with mean m
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[p]=mc_limsr(Q)
% Find the startionary probability vector pi of an irreducible,
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[q,v]=mc_perdo(P)
% The algorithm is from "Periods of Connected Networks and Powers of
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[tx,qx]=invt_lap(t)
% Numerical Inversion of Laplace transform
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[u1,v]=phasemv(a,T)
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[w]=binomial(n,m)
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[xdot]=e2311d(t,x)
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[xdot]=e239a(t,x)
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[y]=e129pgf(z)
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[y]=e649_a(x)
global lm1 lm2 bt1 bt2 b
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[y]=e649_b(x)
global lm1 lm2 bt1 bt2 b
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[y]=e724b(x)
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[y]=e724c(x)
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[z]=c4_cpos(m,x)
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[z]=e135(x,y)
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[z]=e332b(x,y)
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[z]=e332g(x,w)
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[z]=e451_run
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[z]=e645_b(x,y)
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e435
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e441_run
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e443
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e541
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e571
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e573
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e739
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invt_pgf
% Inverting PGF to the time domain
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out=e332d(y,ymin,ymax)
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out=e332e(variable)
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out=e332f(fun,outvar,xmin,xma...
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View all files
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| [F,g]=c3_wibo1(alpha,beta,t) |
function [F,g]=c3_wibo1(alpha,beta,t)
%
% Find distribution for Weibull F(x) = 1 - exp(-(x/alpha)^beta)
% x in (0, t)
% 1/n is the step length
%
n=200; n1=n*t;
tn=1/n; i=1:1:n1; x=tn*(i+0.5); x=(1/alpha)*x;
g=1-exp(-(x.^beta));
i=1:1:n1; x=tn*i; x=(1/alpha)*x;
F=1-exp(-(x.^beta)); |
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