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% Document Info
\author{jack1982}
\pdfinfo{
  /Title (omsa-midterm-exam-2.pdf)
  /Creator (Cheatography)
  /Author (jack1982)
  /Subject (OMSA Midterm Exam 2 Cheat Sheet)
}

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% Header and Footer
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
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    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
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\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{OMSA Midterm Exam 2 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{jack1982} via \textcolor{DarkBackground}{\uline{cheatography.com/216403/cs/47268/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}jack1982 \\
  \uline{cheatography.com/jack1982} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 31st October, 2025.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
\raggedcolumns

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% from this page to resize cheat sheet text:
% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Arena Templates}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jack1982_1761861223_Bildschirmfoto 2025-10-30 um 22.53.31.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.04057 cm} x{2.93643 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Arena Variables and Function}}  \tn
% Row 0
\SetRowColor{LightBackground}
DISC(0.3, 1, 0.8, 2, 1.0, 3) & DISC generates random discrete values based on cumulative probabilities; pair each probability with its corresponding value. \tn 
% Row Count 6 (+ 6)
% Row 1
\SetRowColor{white}
TNOW & Current simulated time \tn 
% Row Count 7 (+ 1)
% Row 2
\SetRowColor{LightBackground}
NR(Res) & Res Servers currently in service \tn 
% Row Count 9 (+ 2)
% Row 3
\SetRowColor{white}
NQ(Queue) & Number of customers in Queue \tn 
% Row Count 11 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Mod.NumberOut & Customers who have left the mode \tn 
% Row Count 13 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.69218 cm} x{3.28482 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{messy cdf}}  \tn
% Row 0
\SetRowColor{LightBackground}
See F(X) & Replace with Uniform(0,1) \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
Multiply by 3 & U(0,3) \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
Subtract 1 & U(−1,2) \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
Mean & (-1+2)/2 = 1/2 \tn 
% Row Count 4 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{What is the mean of the random variable 3F(X)−1?}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.54287 cm} x{3.43413 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Inverse Transformation}}  \tn
% Row 0
\SetRowColor{LightBackground}
Given U, find Z & invNorm(U, 0, 1) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Given Z, find U & normCdf(-9999, Z, 0, 1) \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Exponential(λ)} & X = -ln(1-U)/λ \tn 
% Row Count 6 (+ 2)
% Row 3
\SetRowColor{white}
Uniform(a,b) & X = a + (b-a)·U \tn 
% Row Count 7 (+ 1)
% Row 4
\SetRowColor{LightBackground}
Weibull(a,b) & X = a * (-ln(U))\textasciicircum{}1/b\textasciicircum{} \tn 
% Row Count 8 (+ 1)
% Row 5
\SetRowColor{white}
Triangular & If U\textless{}0.5: √(2U); If U≥0.5: 2-√(2(1-U)) \tn 
% Row Count 10 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\seqsplit{Bernoulli(p)} & If U \textless{} 1-p → 0; Else → 1 \tn 
% Row Count 12 (+ 2)
% Row 7
\SetRowColor{white}
Poisson(λ) & Build CDF, match U \tn 
% Row Count 13 (+ 1)
% Row 8
\SetRowColor{LightBackground}
Discrete Unif(1,n) & {[}n·U{]} \tn 
% Row Count 15 (+ 2)
% Row 9
\SetRowColor{white}
Erlang(k,λ) & -(1/λ)ln(∏Ui) \tn 
% Row Count 16 (+ 1)
% Row 10
\SetRowColor{LightBackground}
Geometric & ln(1-U)/ln(1-p) \tn 
% Row Count 17 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{For discrete: Find smallest x where F(x) ≥ U \newline For continuous: Use inverse CDF formulas \newline Box-Muller generates TWO Normal(0,1) values from TWO Uniform(0,1) values}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.54747 cm} x{4.42953 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{XOR}}  \tn
% Row 0
\SetRowColor{LightBackground}
XOR & is only true if different \tn 
% Row Count 1 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.83689 cm} x{2.14011 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Expected Value, Variance, ...}}  \tn
% Row 0
\SetRowColor{LightBackground}
Discrete E{[}X{]} & SUM{[}x * f(x){]} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
Continous E{[}X{]} & SUM{[}x * f(x) dx{]} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
Variance of X & E{[}X\textasciicircum{}2{]} - (E{[}X{]})\textasciicircum{}2 \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
Standard Deviation of X & SQRT{[}Var(X){]} \tn 
% Row Count 5 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.84149 cm} x{3.13551 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Random Number Generators}}  \tn
% Row 0
\SetRowColor{LightBackground}
Bad generators & Midsquare number generator, Random number tables, von Neumann's mid-square method, Fibonacci generator, Additive congruential generator, RANDU \tn 
% Row Count 6 (+ 6)
% Row 1
\SetRowColor{white}
Good generators & Linear Congruential Generators (modern cycle length \textgreater{} 2\textasciicircum{}191\textasciicircum{}; Mersenne Twister (2\textasciicircum{}19937\textasciicircum{}) \tn 
% Row Count 10 (+ 4)
% Row 2
\SetRowColor{LightBackground}
Randu & 65539Xi mod(2\textasciicircum{}31\textasciicircum{}) \tn 
% Row Count 11 (+ 1)
% Row 3
\SetRowColor{white}
Desert island & 16807Xi mod(2\textasciicircum{}31\textasciicircum{}-1) mod(2147483647) \tn 
% Row Count 13 (+ 2)
% Row 4
\SetRowColor{LightBackground}
Desert island technique & Z = {[}SUM(Ui)-n/2{]} / {[}SQRT(n * 1/12{]} \tn 
% Row Count 15 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{2.4885 cm} x{2.4885 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Inverse Transform Method Key Problems}}  \tn
% Row 0
\SetRowColor{LightBackground}
If X \textasciitilde{} Normal(0,1), what's the distribution of Φ(X)? & Unif(0,1) \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
If U∼Uniform(0,1) and  Φ(x) is the CDF of the standard normal, what is the distribution of 2Φ −1 (U)+3? & Φ−1(U) turns a Uniform(0,1) into a Normal(0,1). Multiply by 2 → scales the standard deviation by 2 = Normal(0,4). Add 3 → shifts the mean to 3. = Normal(3,4) \tn 
% Row Count 12 (+ 9)
% Row 2
\SetRowColor{LightBackground}
-3ℓn(U$^{\textrm{2}}$V$^{\textrm{2}}$) where U, V \textasciitilde{} i.i.d. Unif(0,1) & = -6ℓn(U) - 6ℓn(V) \textbackslash{}\textasciitilde{} Exp(1/6) + Exp(1/6) \textasciitilde{} Erlang₂(1/6) \tn 
% Row Count 16 (+ 4)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.84609 cm} x{4.13091 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Joint Probability Mass Function}}  \tn
% Row 0
\SetRowColor{LightBackground}
E{[}XY{]} & Summe von x * y * f(xy) \tn 
% Row Count 1 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.54287 cm} x{3.43413 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Joint p.m.f.}}  \tn
% Row 0
\SetRowColor{LightBackground}
Are independent if & X and Y are independent if and only if pX,Y​(x,y) = pX​(x) ⋅ pY​(y) \tn 
% Row Count 3 (+ 3)
% Row 1
\SetRowColor{white}
For example & P(X=1, Y=0) = P(Y=0) * P(X=1) \tn 
% Row Count 5 (+ 2)
% Row 2
\SetRowColor{LightBackground}
E{[}XY{]} & Example: (1)(0)(0.2) + (1)(1)(0.0.0) + ... \tn 
% Row Count 7 (+ 2)
% Row 3
\SetRowColor{white}
Cov(X,Y) & E{[}XY{]} - E{[}Y{]}E{[}X{]} \tn 
% Row Count 8 (+ 1)
% Row 4
\SetRowColor{LightBackground}
Variance X + Y & Var(X) + Var(Y) + 2Cov(X,Y) \tn 
% Row Count 10 (+ 2)
% Row 5
\SetRowColor{white}
Variance X - Y & Var(X) + Var(Y) - 2Cov(X,Y) \tn 
% Row Count 12 (+ 2)
% Row 6
\SetRowColor{LightBackground}
Theorem & Cov(X,Y) = 0 if X, Y indepedent. Converse not true. \tn 
% Row Count 14 (+ 2)
% Row 7
\SetRowColor{white}
Correlation & p = Cov(X,Y) / SQRT(Var(X) * Var(Y)) \tn 
% Row Count 16 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.29402 cm} x{3.68298 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{pdf, cdf}}  \tn
% Row 0
\SetRowColor{LightBackground}
pdf -\textgreater{} cdf & integrate with x, 0 as limit \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
cdf F-1(U) & solve(F(x)=U, x) \tn 
% Row Count 2 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Given CDF with two cases, generate X}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jack1982_1761869107_Bildschirmfoto 2025-10-31 um 01.04.33.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Complete Distribution Reference Table}}  \tn
\SetRowColor{LightBackground}
\mymulticolumn{1}{p{5.377cm}}{\vspace{1px}\centerline{\includegraphics[width=5.1cm]{/web/www.cheatography.com/public/uploads/jack1982_1761863907_Bildschirmfoto 2025-10-30 um 23.36.38.png}}} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.64701 cm} x{4.32999 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Box Muller Method}}  \tn
% Row 0
\SetRowColor{LightBackground}
Z1 & √(-2·ln(U₁)) · cos(2π·U₂) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Z2 & √(-2·ln(U₁)) · sin(2π·U₂) \tn 
% Row Count 4 (+ 2)
% Row 2
\SetRowColor{LightBackground}
Z1/Z2 & cot(2π·U₂) \textasciitilde{}Cauchy \tn 
% Row Count 5 (+ 1)
% Row 3
\SetRowColor{white}
Z2/Z1 & tan(2π·U₂) \textasciitilde{}Cauchy \tn 
% Row Count 6 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{Radian-Modus einschalten!}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Chi-Square Distribution}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{If Z₁, Z₂, Z₃ are i.i.d. Nor(0,1), find c such that P(Z₁$^{\textrm{2}}$ + Z₂$^{\textrm{2}}$ + Z₃$^{\textrm{2}}$ \textless{} c) = 0.99 \newline % Row Count 2 (+ 2)
Calculator: chiSqInv(0.99, 3) → 11.34% Row Count 3 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Find distribution of U1 and U2}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Find distribution of -4(U₁ + U₂) - 2} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{U₁ + U₂ \textasciitilde{} Tria(0, 1, 2)} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Apply the transformation -4(U₁ + U₂) = 4(Tria(0, 1, 2) )} \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{The minimum becomes: -4(2) = -8; The mode becomes: -4(1) = -4; The maximum becomes: -4(0) = 0} \tn 
% Row Count 6 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{4·Tria(0, 1, 2) = Tria(-8, -4, 0)} \tn 
% Row Count 7 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Subtract 2} \tn 
% Row Count 8 (+ 1)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Tria(-10, -6, -2)} \tn 
% Row Count 9 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{p{0.74655 cm} x{4.23045 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Acceptance-Rejection}}  \tn
% Row 0
\SetRowColor{LightBackground}
Goal & Generate random samples from a hard-to-sample distribution (f(x)). \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
Idea & Sample from an easier distribution (h(x)), then accept or reject each sample based on how well it fits f(x). \tn 
% Row Count 6 (+ 4)
% Row 2
\SetRowColor{LightBackground}
\seqsplit{Example} & If a random variable X has the beta distribution, then its p.d.f. is of the form f(x) = Γ(α+β) Γ(α)Γ(β) xα−1(1−x)β−1, 0 \textless{}x\textless{}1, for parameters α and β \textgreater{}0, and where Γ(·) is the gamma function. How might you generate such a random variate? Pick the best answer. \tn 
% Row Count 15 (+ 9)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{x{1.59264 cm} x{3.38436 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Goodness of fit}}  \tn
% Row 0
\SetRowColor{LightBackground}
Find critical value & Inverse X2; Area = 1-alpha; df = n-1; invx2(0.9,3) \tn 
% Row Count 2 (+ 2)
% Row 1
\SetRowColor{white}
 & If critical value is bigger than accept H0 \tn 
% Row Count 4 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}