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% Document Info
\author{musikdr}
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  /Title (ise-362-exam-1.pdf)
  /Creator (Cheatography)
  /Author (musikdr)
  /Subject (ISE 362 EXAM 1 Cheat Sheet)
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    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{ISE 362 EXAM 1 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{musikdr} via \textcolor{DarkBackground}{\uline{cheatography.com/93594/cs/20746/}}}
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  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}musikdr \\
  \uline{cheatography.com/musikdr} \\
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  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 15th October, 2019.\\
   Page {\thepage} of \pageref{LastPage}.
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  Measure your website readability!\\
  www.readability-score.com
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\begin{multicols*}{2}

\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Reliability}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Rseries}} = R\textasciicircum{}n\textasciicircum{} \newline % Row Count 1 (+ 1)
{\bf{Rparallel}} = 1-(1-R)\textasciicircum{}n\textasciicircum{} \newline % Row Count 2 (+ 1)
{\bf{2 Parallel Components in Series}} = (1-(1-R)\textasciicircum{}2\textasciicircum{})\textasciicircum{}2\textasciicircum{} \newline % Row Count 4 (+ 2)
{\bf{2 Series Components in Parallel}} = 1-(1-R\textasciicircum{}2\textasciicircum{})\textasciicircum{}2\textasciicircum{} \newline % Row Count 6 (+ 2)
{\bf{ Legitimate Values of R}} 0\textless{}=R\textless{}=1 \newline % Row Count 7 (+ 1)
{\bf{Consider f(R) = R(2-R) where does f(R) achieve it's max and what' the max value?}}  \newline % Row Count 9 (+ 2)
g(R) = R(2-R0 = 2R-R\textasciicircum{}2\textasciicircum{} taking it's derivative and setting equal to 0 we get \newline % Row Count 11 (+ 2)
dg(R)/dR = 2-2R=0 Its max is achieved when R=1 \newline % Row Count 12 (+ 1)
Plugging this in for R in g(R) yields 2-1=1\textgreater{}0 \newline % Row Count 13 (+ 1)
{\bf{k out of N Redundancy}} - system that operates if at least k out of N components function properly.  \newline % Row Count 16 (+ 3)
R=Sum(I=k to n) of R\textasciicircum{}I\textasciicircum{}(1-R)\textasciicircum{}n-i\textasciicircum{} (ex Sum(I=2 to 3(for ito 3)(.0.9)\textasciicircum{}i\textasciicircum{}(0.1)\textasciicircum{}3-i)% Row Count 18 (+ 2)
} \tn 
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Economic Analysis}}  \tn
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\mymulticolumn{1}{x{8.4cm}}{{\bf{How much would you need to invest on September 27th, 2019 to have \$10,000 on September 27th, 2027 given an interest rate of 5\%?}} \newline % Row Count 3 (+ 3)
PV = FV(P/F,i,n)=10k(P/F,5,7) \newline % Row Count 4 (+ 1)
{\bf{How much would you need to deposit on January 1, 2020 in a fund that yields 5\% annually in order to draw out \$250.00 at the end of each year starting December 31, 2020 for 7 years, leaving nothing in the fund at the end?}} \newline % Row Count 9 (+ 5)
PV=250(P/A,5\%,7) \newline % Row Count 10 (+ 1)
{\bf{Dominate}} - when one of the outcome paths is clearly the better choice for every case% Row Count 12 (+ 2)
} \tn 
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Review Session Material}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{P34 = P{[}In State 4|Current State{]} - Take Pmatrix to the 4th power and look at position 34. \newline % Row Count 2 (+ 2)
EMV(i) = SUM(j=1 to N) Pj*rij \newline % Row Count 3 (+ 1)
How much does P have to change before another alternative is best. Add a z to one state and subtract z from another state, then set the equations = to each other and solve for z. If z is (-) then it's not better.  \newline % Row Count 8 (+ 5)
EV of Perfect information means you take the largest value at each state and multiply it by it's probability. \newline % Row Count 11 (+ 3)
Expected Loss of Sales cost (When D exceeds inventory) \newline % Row Count 13 (+ 2)
E{[}lost sales cost | Demand=d,Inv=I{]}  \newline % Row Count 14 (+ 1)
MAX(d-i,0) \newline % Row Count 15 (+ 1)
To remove a condition \newline % Row Count 16 (+ 1)
E{[}lost sales{]} = SUM(d=0 to nd)SUM(I=0 to ni){\emph{max(d-i,0)}}Pd(d)*Pi(i) \newline % Row Count 18 (+ 2)
Can also use PMF (PI)% Row Count 19 (+ 1)
} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Markov Case}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{{\bf{Markov Case}} Stochastic process where we only take into account the present to predict the future. That is the probability of going to state j(future) from state I(present) \newline % Row Count 4 (+ 4)
-If there is one (and only one) closed, aperiodic communicating class, the process is ergodic. \newline % Row Count 6 (+ 2)
Pi(n) = P{[}In state after transitions{]} \newline % Row Count 7 (+ 1)
Pi(1) = Pi1(0){\emph{p11 + Pi2(0)}}P21+...+ Pin(0)*Pn1% Row Count 8 (+ 1)
} \tn 
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\begin{tabularx}{8.4cm}{X}
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\mymulticolumn{1}{x{8.4cm}}{\bf\textcolor{white}{Floyd's Barbershop}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{8.4cm}}{Haircuts take exactly 15min, we have  \newline % Row Count 1 (+ 1)
r.v. Y = \# customers who arrive during a 15m interval \newline % Row Count 3 (+ 2)
Haircuts cost 10; each customer consumes \$1 of snacks every 15min \newline % Row Count 5 (+ 2)
Pi(n) = Pi(n-1)*P \newline % Row Count 6 (+ 1)
=Pi(0)*p\textasciicircum{}n\textasciicircum{} \newline % Row Count 7 (+ 1)
{\bf{Q1}} - We have 3 customers in the shop, what's the Probability there will be 4 customers in the shop 4 periods from now? A - P\textasciicircum{}4\textasciicircum{}34 \newline % Row Count 10 (+ 3)
What is the probability in the long run, a customer comes to the shop but leaves because there is no seat available? \newline % Row Count 13 (+ 3)
P{[}A{]}=SUM(I=0 to 4{]}P(A|Statei)*P(Statei) \newline % Row Count 14 (+ 1)
What is the Long run expected number of customers that come to the shop but leaves? E{[}A{]} = SUM(I=1 to n) E{[}A|B\}*P{[}Bi{]} \newline % Row Count 17 (+ 3)
E{[}a|state=0{]} = 0(.2) +0(.7) ......% Row Count 18 (+ 1)
} \tn 
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\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}