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1. Machine Repair Model (from Tezt, Page 124, 5th Edition): Suppose that 72 = 4 and 8 = 2. Lifetimes of machines have distribution F(2) = 1 - (1+I/C: x 0, and repair times have distribution G(2) = 1 - T>0. Write up code and estimate E(C), where C denotes the time until a crash occurs. (Gen- erate m = 5000 iid copies of C to do so.) 2. Markovian Insurance Risk Model (from Text. Page 122, 5th Edition): Suppose that no = 3 and ao = 10. V/ = 1/2, A = 1/6, e = 1/20. c = 10. Claim size distribution is lognormal: Y ez. where Z ~ N(0,1). Write up code and estimate the probability of no ruin by time T = 100. (Generate m = 5000 iid copies of I to do so: I = 1 if no ruin by time T; I = 0 if Ruin by time T.)

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clear; clc; close all;
n=4; s=2;
f=@(x) 1-(1./((1+x).^6));%Machine lifetimes,   x>=0
fi=@(u) power(1./(1-u),1/6)-1;%Inverse of f
g=@(x) 1-exp(-2*sqrt(x));%Machine repair time, x>=0
gi=@(u) (-(1/2)*log(1-u)).^2;%Inverse of g
m=5000;%iid number for E(C), time to a crash
%x=0:0.1:10; ff=f(x); gg=g(x);
%plot(x,ff,x,gg);
T=[];
for i=1:m
    t=0; %Starting time is zero
    r=0; %Number of machines in repair initially is zero
    tstar=Inf;%Repair time for currently being repaired machine, Inf as no machine is under repair initially
    un=rand(1,n); tn=fi(un);
    tE=sort(tn);%event list
    while(1)
       if tE(1)<tstar
            t=tE(1); %r=r+1;...

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