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OMSA Midterm Exam 2 Cheat Sheet (DRAFT) by

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This is a draft cheat sheet. It is a work in progress and is not finished yet.

Joint Probab­ility Mass Function

E[XY]
Summe von x * y * f(xy)

Arena Templates

Arena Variables and Function

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.
TNOW
Current simulated time
NR(Res)
Res Servers currently in service
NQ(Queue)
Number of customers in Queue
Mod.Nu­mberOut
Customers who have left the mode
 

Expected Value, Variance, ...

Discrete E[X]
SUM[x * f(x)]
Continous E[X]
SUM[x * f(x) dx]
Variance of X
E[X2] - (E[X])2
Standard Deviation of X
SQRT[V­ar(X)]

Random Number Generators

Bad generators
Midsquare number generator, Random number tables, von Neumann's mid-square method, Fibonacci generator, Additive congru­ential generator, RANDU
Good generators
Linear Congru­ential Generators (modern cycle length > 2191; Mersenne Twister (219937)
Randu
65539Xi mod(231)
Desert island
16807Xi mod(231-1) mod(21­474­83647)

Inverse Transform Method Key Problems

If X ~ Normal­(0,1), what’s the distri­bution of Φ(X)?
Unif(0,1)
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)
-3ℓn(U²V²) where U, V ~ i.i.d. Unif(0,1)
= -6ℓn(U) - 6ℓn(V) ~ Exp(1/6) + Exp(1/6) ~ Erlang­₂(1/6)
 

Given CDF with two cases, generate X

Complete Distri­bution Reference Table

Box Muller Method

Z1
√(-2·l­n(U₁)) · cos(2π·U₂)
Z2
√(-2·l­n(U₁)) · sin(2π·U₂)
Radian­-Modus einsch­alten!

Chi-Square Distri­bution

If Z₁, Z₂, Z₃ are i.i.d. Nor(0,1), find c such that P(Z₁² + Z₂² + Z₃² < c) = 0.99

Calcul­ator: chiSqI­nv(­0.99, 3) → 11.34

Find distri­bution of U1 and U2

Find distri­bution of -4(U₁ + U₂) - 2
U₁ + U₂ ~ Tria(0, 1, 2)
Apply the transf­orm­ation -4(U₁ + U₂) = 4(Tria(0, 1, 2) )
The minimum becomes: -4(2) = -8; The mode becomes: -4(1) = -4; The maximum becomes: -4(0) = 0
4·Tria(0, 1, 2) = Tria(-8, -4, 0)
Subtract 2
Tria(-10, -6, -2)