Formula Sheet for CPIT 701 (Advanced Probability and Statistics)
PROBABILITY
Probability of any event: 0  P (event)  1
Permutation: subsets of r elements from n different
elements
n!
Prn  n  (n  1)  (n  2)    (n  r  1) 
(n  r )!
Permutation of similar objects: n1 is of one type, n2 is of
second type, … among n = n1+n2+…+nr elements
n!
n1 ! n 2 ! n3 ! n r !
Combinations: subsets of size r from a set of n elements
 n
n!
nC r  Crn    
 r  r!(n  r )!
Independent Events:
P(AB) = P(A or B) = P(A) + P(B) – P(A and B)
P(ABC) = P(A) + P(B) + P(C) – P(AB) – P(AC) - P(AB) = P(A and B) = P(A)P(B)
P(A|B) = P(A)
P(BC) + P(ABC)
P(B) = P(BA) + P(BA’) = P(B|A)P(A)+P(B|A’)P(A’) P(B|A)=P(B)
Dependent Events:
For Mutually exclusive events AB=:
P(A and B) = P(A) * P(B given A)
P(AB) = P(A or B) = P(A) + P(B)
P(AB)=P(A|B)P(B)=P(BA)=P(B|A)P(A)
P(A and B and C) = P(A) * P(B | A) * P(C given
A and B)
Bayes’ Theorem
P(AB) = P(AB) = P(A | B) P(B) = P(B | A) P(A)
P( AB)
P(B | A)  P( A)
Conditiona l Probabilit y P( A | B) 
P( A | B) 
P(B)
P(B | A)  P( A)  P(B | A )  P( A )
A, B
= any two events
A’
= complement of A
Markov’s Inequality
Chebyshev’s Inequality
If X is a non-negative random variable with mean , then If X is a random variable with a finite mean 
for any constant a > 0
and variance 2, then for any constant a > 0

2
P(X  a) 
P(
X


a)

a
a2
𝑁
Mean of Central Tendency
𝑁
1
𝑀𝑒𝑎𝑛 = ∑ 𝑥𝑖
𝑁
1
𝑀𝑒𝑎𝑛 𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = ∑|𝑥𝑖 − 𝑥̅ |
𝑁
𝑖=1
𝑖=1
Sample Variance (mean squared deviation from the mean)
𝑁
𝑠2 =
1
∑(𝑥𝑖 − 𝑥̅ )2
𝑁−1
𝑖=1
Set Basics
Intersection
𝐴 ∩ 𝐵 = {𝑥|𝑥 𝜖 𝐴 𝑎𝑛𝑑 𝑥 ∈ 𝐵}
Union
𝐴 ∪ 𝐵 = {𝑥|𝑥 𝜖 𝐴 𝑜𝑟 𝑥 ∈ 𝐵}
𝐷𝑖𝑠𝑗𝑜𝑖𝑛𝑡 𝐴 ∩ 𝐵 = ∅
𝑆𝑢𝑏𝑠𝑒𝑡 𝐴 ⊃ 𝐵 𝑖𝑓 𝐴 ∩ 𝐵 = 𝐴
𝑃𝑎𝑟𝑡𝑖𝑡𝑖𝑜𝑛 𝐴 ∪ 𝐵 ∪ 𝐶 ∪ 𝐷 = Ω
𝐶𝑜𝑚𝑝𝑙𝑖𝑚𝑒𝑛𝑡 𝐴𝑐 = 𝐴̅ = {𝑥 ∈ Ω | 𝑥 ∉ 𝐴}
DeMorgan’s Laws
Useful Identities
Ω𝑐 = ∅
𝐴 ∪ A𝑐 = Ω
(A𝒄 )𝒄 = 𝐴
𝐴 ∩ A𝑐 = ∅
Probabilistic Law
𝑛
𝑐
𝑛
𝑃(𝐸) =
(⋃ 𝐸𝑖 ) = ⋂ E𝑐𝑖
𝑖=1
𝑛
p𝑖
𝑎𝑙𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑖𝑛 𝐸
𝑛
𝑖=1
𝑛
𝑐
∑
𝑃 (⋃ 𝐸𝑖 ) = P(𝐸1 ) + P(𝐸2 ) + ⋯ + P(𝐸𝑛 )
𝑐
(⋂ 𝐸𝑖 ) = ⋃ E𝑖
𝑖=1
𝑖=1
𝑖=1
Probability Identities
Discrete Uniform Probability
𝑃(∅) = 0
𝑃(𝐸 𝑐 ) = 1 − P(E)
𝑃(𝐸) ≤ 1
𝑃(𝐸) =
|𝐸|
# 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑠 𝑖𝑛 𝐸
=
|Ω|
𝑁
If 𝐸 ⊂ 𝐹 𝑡ℎ𝑒𝑛 𝑃(𝐸) ≤ 𝑃(𝐹)
𝑃(𝐸 ∪ 𝐹) = P(E) + 𝑃(𝐹) − 𝑃(𝐸 ∩ 𝐹)
Stage Counting Method
Product Rule = 𝑛1 . 𝑛2 … 𝑛𝑟
For n sets
Total number of possible subsets = 2𝑛
Counting without replacement
(ordering without repitition)Permutation = n!
K-Permutations
Combinatorics
Draw k from a set of n with replacement
Order matters = 𝑛𝑘
Conditional Probability
𝑛𝑘
Order doesn’t matter = 𝑛!
Draw k from a set of n without replacement
𝑛!
Order matters = (𝑛−𝑘)!
𝑛
𝑛!
= 𝑛𝑃𝑘
(𝑛 − 𝑘)!
Combinations: Order does not matter
Selecting k out of n =
𝑛𝑃𝑘
𝑛!
𝑛
( )=
=
𝑘
𝑘!
𝑘! (𝑛 − 𝑘)!
𝑃(𝐴 ∩ 𝐵)
𝑃(𝐵)
𝐵 = (𝐴1 ∩ 𝐵) ∪ (𝐴2 ∩ 𝐵) ∪ … (𝐴𝑛 ∩ 𝐵)
𝑃(𝐵) = 𝑃(𝐴1 ∩ 𝐵) + 𝑃(𝐴2 ∩ 𝐵) + ⋯ + 𝑃(𝐴𝑛 ∩ 𝐵)
𝑃(𝐴|𝐵) =
= ∑ 𝑃(𝐴𝑖 ∩ 𝐵)
𝑃(𝐴𝑖 ∩ 𝐵) = 𝑃(𝐵|𝐴𝑖 )𝑃(𝐴𝑖 )
𝑛!
Order doesn’t matter = (𝑘 ) = 𝑘!(𝑛−𝑘)!
𝑃(𝐵) = ∑ 𝑃(𝐵|𝐴𝑖 )𝑃(𝐴𝑖 )
𝑃(𝐵) = 𝑃(𝐵|𝐴)𝑃(𝐴) + 𝑃(𝐵|𝐴𝑐 )𝑃(𝐴𝑐 )
𝑃(𝐵|𝐴)𝑃(𝐴)
𝑃(𝐴|𝐵) =
𝑃(𝐵)
𝑃(𝐵|𝐴)𝑃(𝐴)
𝑃(𝐴|𝐵) =
𝑃(𝐵|𝐴)𝑃(𝐴) + 𝑃(𝐵|𝐴𝑐 )𝑃(𝐴𝑐 )
Independent events
𝑃(𝐴|𝐵) = 𝑃(𝐴)
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)
Discrete Random Variable
Probability Mass Function (PMF)
Discrete Random Variable
Expectation Value
𝑃𝑋 (𝑥) = 𝑃(𝑋 = 𝑥)
𝐸[𝑋] = ∑ 𝑥 𝑝𝑋 (𝑥)
∑ 𝑝𝑋 (𝑘) = 1
𝑎𝑙𝑙 𝑥
Variance
𝑎𝑙𝑙 𝑘
𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) =
∑
𝑝𝑋 (𝑥)
𝑎𝑙𝑙 𝑎≤𝑋≤𝑏
Function of Random Variables
∑
𝑥|𝑔(𝑥)=𝑦
𝑎𝑙𝑙 𝑥
Bernoulli Random Variable
𝑝𝑋 (𝑥) 𝑎𝑛𝑑 𝑦 = 𝑔(𝑋)
𝑝𝑌 (𝑦) =
𝑣𝑎𝑟[𝑋] = ∑ (𝑥 − 𝐸[𝑋])2 𝑝𝑋 (𝑥)
𝑝𝑋 (𝑥)
𝑝 𝑖𝑓 𝑘 = 1
𝑃𝑋 (𝑘) = {
1 − 𝑝 𝑖𝑓 𝑘 = 0
𝐸[𝑋] = 𝑝
2
𝑣𝑎𝑟[𝑋] = 𝑝(1 − 𝑝)
Cumulative Distribution Function
𝐹𝑋 (𝑥) = 𝑃(𝑋 ≤ 𝑥) =
∑ 𝑝𝑋 (𝑘)
𝑎𝑙𝑙 𝑘≤𝑥
Binomial Distribution
Geometric Distribution
𝑃𝑋 (𝑥) = (1 − 𝑝)𝑘−1 𝑝
𝐸[𝑋] = 1/𝑝
𝑣𝑎𝑟[𝑋] = (1 − 𝑝)/𝑝2
Poisson Distribution
Uniform Distribution
𝑛
𝑃𝑋 (𝑥) = ( ) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘
𝑘
𝐸[𝑋] = 𝑛𝑝
𝑣𝑎𝑟[𝑋] = 𝑛𝑝(1 − 𝑝)
𝑊ℎ𝑒𝑛 𝑛 ≫ 𝑘, 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝐵𝑖𝑛𝑜𝑚𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑛𝑝 = 𝜆 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝜆𝑘
𝑛
𝑃𝑋 (𝑥) = lim ( ) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘 = 𝑒 −𝜆
𝑛→∞ 𝑘
𝑘!
𝐸[𝑋] = 𝜆
𝑣𝑎𝑟[𝑋] = 𝜆
Independence
For all cases
𝑝𝑋 (𝑥) = 1/𝑁
𝑁
𝐸[𝑋] = ∑ 𝑥 𝑝𝑋 (𝑥) = ∑
𝑎𝑙𝑙 𝑥
𝑣𝑎𝑟[𝑋] = ∑ (𝑥 −
𝑥𝑖
𝑁
𝑖=1
2
𝐸[𝑋]) 𝑝𝑋 (𝑥)
𝑎𝑙𝑙 𝑥
Continuous Random Variable
𝑏
𝐸[𝑋 + 𝑌] = 𝐸[𝑋] + 𝐸[𝑌]
𝐸[𝑎𝑋 + 𝑏] = 𝑎𝐸[𝑋] + 𝑏
𝑣𝑎𝑟[𝑎𝑋 + 𝑏] = 𝑎2 𝑣𝑎𝑟[𝑋]
If X and Y are independent random variables
𝐸[𝑋𝑌] = 𝐸[𝑋]𝐸[𝑌]
𝑣𝑎𝑟[𝑋 + 𝑌] = 𝑣𝑎𝑟[𝑋] + 𝑣𝑎𝑟[𝑌]
𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) = ∫ 𝑓𝑋 (𝑥)𝑑𝑥
𝑎
Probability Density Function (PDF)
𝑃(𝑥 ≤ 𝑋 ≤ 𝑥 + 𝛿)
𝛿⟶0
𝛿
Non-negativity 𝑓𝑋 (𝑥) ≥ 0
Normalization
𝑓𝑋 (𝑥) = lim
∞
∫ 𝑓𝑋 (𝑥)𝑑𝑥 = 1
−∞
Probability at a point 𝑃(𝑋 = 𝑎) = 0
Continuous Random Variable
Expectation Value
Cumulative Distribution Function (CDF)
∞
𝐹𝑋 (𝑏) = 𝑃(𝑋 ≤ 𝑏) = ∫ 𝑓𝑋 (𝑥)𝑑𝑥
∞
𝐸[𝑋] = ∫ 𝑥𝑓𝑋 (𝑥)𝑑𝑥
−∞
Variance
∞
𝑣𝑎𝑟[𝑋] = ∫ (𝑥 − 𝐸[𝑋])2 𝑓𝑋 (𝑥)𝑑𝑥
−∞
Normal Distribution
𝑓𝑋 (𝑥) =
−∞
CDF is monotonically non-decreasing
𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) = 𝐹𝑋 (𝑏) − 𝐹𝑋 (𝑎)
𝑑𝐹𝑋 (𝑥)
𝑓𝑋 (𝑥) =
𝑑𝑥
Standard Normal Distribution
1
(𝑥−𝜇)2
−
𝑒 2𝜎2
= 𝑁(𝜇, 𝜎 2 )
√2𝜋𝜎
1
𝑥−𝜇
𝐹𝑋 (𝑥) = [1 + 𝑒𝑟𝑓 (
)]
2
√2𝜎
𝐸[𝑋] = 𝜇
𝑣𝑎𝑟[𝑋] = 𝜎 2
Sample Mean
1
𝑋̅ = ∑ 𝑋𝑖
𝑛
𝑖=1,𝑛
Expectation of the sample mean
1
𝐸(𝑋̅) = ∑ 𝑋𝑖 = 𝐸(𝑋𝑖 ) = 𝜇
𝑛
𝑖=1,𝑛
Variance of the sample mean
𝑥−𝜇
𝜎
𝐼𝑓 𝑋 = 𝑁(𝜇, 𝜎 2 ) 𝑡ℎ𝑒𝑛 𝑍 = 𝑁(0,1)
𝑧=
Linear function of a normally distributed RV
produces a normally distributed RV
𝐿𝑒𝑡 𝑌 = 𝛼𝑋 + 𝛽
𝐼𝑓 𝑋 = 𝑁(𝜇, 𝜎 2 ) 𝑡ℎ𝑒𝑛 𝑌 = 𝑁(𝛼𝜇 + 𝛽, (𝛼𝜎)2 )
RV with a distribution that approaches N(0,1)
as 𝑛 ⟶ ∞
𝑍=
𝑋̅ − 𝜇
𝜎/√𝑛
Student’s t Distribution (t is RV with DF = n-1)
𝑡=
𝑋̅ − 𝜇
𝑠/√𝑛
3
𝑣𝑎𝑟(𝑋̅) =
1
𝜎2
𝑣𝑎𝑟 ( ∑ 𝑋𝑖 ) =
𝑛2
𝑛
𝑖=1,𝑛
Standard deviation of the sample mean
𝜎𝑋̅ =
𝜎
√𝑛
Sampling distribution of the Variance (for sample
variance S2), RV with a chi-square distribution and DF =
n -1
(𝑛 − 1)𝑆 2
𝜎2
2
𝐸[𝜒 ] = 𝑛 − 1
𝑣𝑎𝑟[𝜒 2 ] = 2(𝑛 − 1)
Sampling distribution of the Variance for iid
random variables with 𝑿𝒊 ~𝑁(𝜇, 𝜎2 )
𝐸[𝑆 2 ] = 𝜎 2
2𝜎 4
𝑣𝑎𝑟[𝑆 2 ] =
𝑛−1
𝜒2 =
Signal to noise,
𝐸[𝑆 2 ]
√𝑣𝑎𝑟[𝑆 2 ]
𝑛−1
=√
2
Confidence interval for the variance
(𝑛 − 1)𝑠 2
2
𝜒𝛼/2
< 𝜎2 <
(𝑛 − 1)𝑠 2
2
𝜒1−𝛼/2
Comparing two sample variances, RV that
follows F-distribution with n1-1 and n2-1
degrees of freedom
𝐹=
Confidence Interval of the mean
𝑥̅ − 𝑧𝛼⁄2
𝑠
√𝑛
𝑠
< 𝜇 < 𝑥̅ + 𝑧𝛼⁄2
√𝑛
𝑆12 /𝜎12
𝑆22 /𝜎22
̂𝑞̂
𝑝
Standard error 𝑆𝐸(𝑝̂ ) = √ 𝑛
𝑋
𝑛
𝑤ℎ𝑒𝑟𝑒 𝑝 𝑖𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑠𝑢𝑐𝑒𝑐𝑠𝑠 𝑎𝑛𝑑
𝑞 𝑖𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑓𝑎𝑖𝑙𝑢𝑟𝑒
Margin of error = 𝑧𝛼⁄2 𝑆𝐸(𝑝̂ )
95% CI (𝛼 = 0.05)
𝑝̂ =
𝑝̂ 𝑞̂
𝑝̂ 𝑞̂
(𝑝̂ − 1.96√ , 𝑝̂ + 1.96√ )
𝑛
𝑛
Confidence Interval of the mean
𝑥̅ − 𝑡𝛼⁄2
𝑠
√𝑛
< 𝜇 < 𝑥̅ + 𝑡𝛼⁄2
𝑠
√𝑛
For large sample, 95% confidence interval
(𝑥̅ − 1.96
𝑠
√𝑛
, 𝑥̅ + 1.96
𝑠
√𝑛
)
Pooled Samples
Correlation coefficient
̅̅̅1 − ̅̅̅
𝐸[𝑋
𝑋2 ] = 𝜇1 − 𝜇2
𝜎12 𝜎22
̅̅̅1 − ̅̅̅
𝑣𝑎𝑟[𝑋
𝑋2 ] =
+
𝑛1
𝑛2
̅̅̅
̅̅̅
𝑋1 − 𝑋2 − (𝜇1 − 𝜇2 )
𝑍=
~𝑁(0,1)
2
2
𝑆
𝑆
√ 1+ 2
𝑛1 𝑛2
Two Matched Samples
(𝑛1 − 1)𝑠12 + (𝑛2 − 1)𝑠22
𝑠𝑝2 =
(𝑛1 − 1) + (𝑛2 − 1)
For t-tests, DF = 𝑛1 + 𝑛2 − 2
̅̅̅1 − ̅̅̅
𝑣𝑎𝑟[𝑋
𝑋2 ] = √
Two independent Sample means
𝑆12
𝑛1
+
𝐷𝑖 = 𝑋𝑖 − 𝑌𝑖
For large samples
𝑍=
𝑆22
1
1
= 𝑆𝑝 √ +
𝑛2
𝑛1 𝑛2
̅ − 𝜇𝐷
𝐷
𝑆𝐷 /√𝑛
~𝑁(0,1)
Covariance
4
Pearson’s Sample Correlation Coefficient
𝑛
𝑟=
(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
1
∑ 𝑍𝑥𝑖 𝑍𝑦𝑖 =
(𝑛 − 1)𝑆𝑥 𝑆𝑦
𝑛−1
𝑖=1
𝑐𝑜𝑣(𝑋, 𝑌) = 𝐸[(𝑋 − 𝐸[𝑋])(𝑌 − 𝐸[𝑌])]
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
𝑐𝑜𝑣(𝑋, 𝑌) =
(𝑛 − 1)
Correlation coefficient
𝑟=
Covariance Properties
𝑐𝑜𝑣(𝑋, 𝑌)
𝑆𝑥 𝑆𝑦
Linear Model
𝑐𝑜𝑣(𝑋, 𝑌) = 𝐸[𝑋𝑌] − 𝐸[𝑋]𝐸[𝑌]
With a and b as scalars
𝑌 = 𝑎𝑋 + 𝑏
Conditional expectation
𝑐𝑜𝑣(𝑋, 𝑎) = 0
𝑐𝑜𝑣(𝑋, 𝑋) = 𝜎𝑋2
𝑐𝑜𝑣(𝑋, 𝑌) = 𝑐𝑜𝑣(𝑌, 𝑋)
𝑐𝑜𝑣(𝑎𝑋, 𝑏𝑌) = 𝑎𝑏𝑐𝑜𝑠(𝑋, 𝑌)
𝑐𝑜𝑣(𝑋 + 𝑎, 𝑌 + 𝑏) = 𝑐𝑜𝑣(𝑋, 𝑌)
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
Law of iterated expectations
𝐸[𝐸[𝑌|𝑋]] = 𝐸[𝑌]
Using expectation on both sides for
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
𝐸[𝑌] = 𝑎𝐸[𝑋] + 𝑏
𝑏 = 𝐸[𝑌] − 𝑎𝐸[𝑋]
If X and Y are independent, then
𝑐𝑜𝑣(𝑋, 𝑌) = 0
Covariance
𝑐𝑜𝑣(𝑋, 𝑌) = 𝑎 𝑣𝑎𝑟(𝑋)
𝑐𝑜𝑣(𝑋, 𝑌)
𝑎=
𝑣𝑎𝑟(𝑋)
Maximum Likelihood Estimator (MLE)
Suppose iid Xi ~𝑁(𝜇, 𝜎2 )
Least Square Error (LSE)
𝑦𝑖 = 𝑎𝑥𝑖 + 𝑏 + 𝜖𝑖
Sum of the squares of the residuals (SSE)
MLE for the mean 𝜇
1
𝑃(𝑋𝑖 = 𝑥𝑖 |𝜇) =
𝑒 −(𝑥−𝜇)
2 ⁄2𝜎 2
√2𝜋𝜎
𝑛
1 𝑛 − ∑(𝑥−𝜇)2⁄2𝜎2
∏ 𝑃(𝑋𝑖 = 𝑥𝑖 |𝜇) = (
) 𝑒
√2𝜋𝜎
𝑖=1
𝑛
𝜇̂ =
1
∑ 𝑥𝑖
𝑛
𝑛
𝑖=1
𝑛
𝑆𝑆𝐸 =
𝑛
∑ 𝜖𝑖2
𝑖=1
= ∑(𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏)2
𝑖=1
Properties of the LSE Solution
𝑛
∑ 𝜖𝑖 = 0
𝑖=1
Unbiased estimator for 𝑣𝑎𝑟(𝜖) is SSE/(n-2)
1
𝜎̂2 = ∑(𝑥𝑖 − 𝑥̅ )2
𝑛
𝑖=1
𝑛
1
̂ 𝑦) = ∑(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
𝑐𝑜𝑣(𝑥,
𝑛
𝑖=1
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
𝑎=
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2
𝑏 = 𝑦̅ − 𝑎𝑥̅
5
Linear Model
𝑌 = 𝑎𝑋 + 𝑏
Conditional expectation
REGRESSION MODELS
Maximum Likelihood Estimator (MLE)
Suppose iid Xi ~𝑁(𝜇, 𝜎2 )
MLE for the mean 𝜇
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
Law of iterated expectations
𝐸[𝐸[𝑌|𝑋]] = 𝐸[𝑌]
Using expectation on both sides for
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
𝐸[𝑌] = 𝑎𝐸[𝑋] + 𝑏
𝑏 = 𝐸[𝑌] − 𝑎𝐸[𝑋]
Covariance
1
𝑃(𝑋𝑖 = 𝑥𝑖 |𝜇) =
2 ⁄2𝜎2
√2𝜋𝜎
1 𝑛 − ∑(𝑥−𝜇)2 ⁄2𝜎2
∏ 𝑃(𝑋𝑖 = 𝑥𝑖 |𝜇) = (
) 𝑒
√2𝜋𝜎
𝑖=1
𝑛
𝑛
1
𝜇̂ = ∑ 𝑥𝑖
𝑛
𝑛
𝑐𝑜𝑣(𝑋, 𝑌) = 𝑎 𝑣𝑎𝑟(𝑋)
𝑐𝑜𝑣(𝑋, 𝑌)
𝑎=
𝑣𝑎𝑟(𝑋)
𝑒 −(𝑥−𝜇)
𝑖=1
1
𝜎̂2 = ∑(𝑥𝑖 − 𝑥̅ )2
𝑛
𝑖=1
𝑛
1
̂ 𝑦) = ∑(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
𝑐𝑜𝑣(𝑥,
𝑛
𝑖=1
𝐸[𝑌|𝑋] = 𝑎𝑋 + 𝑏
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )(𝑦𝑖 − 𝑦̅)
𝑎=
∑𝑛𝑖=1(𝑥𝑖 − 𝑥̅ )2
𝑏 = 𝑦̅ − 𝑎𝑥̅
Least Square Error (LSE)
Linear Regression
𝑦𝑖 = 𝑎𝑥𝑖 + 𝑏 + 𝜖𝑖
Sum of the squares of the residuals (SSE)
𝑛
𝑆𝑆𝐸 =
𝑛
∑ 𝜖𝑖2
𝑖=1
= ∑(𝑦𝑖 − 𝑎𝑥𝑖 − 𝑏)2
𝑀𝑜𝑑𝑒𝑙: 𝑦𝑖 = 𝑎𝑥𝑖 + 𝑏 + 𝜖𝑖
𝑃𝑟𝑒𝑑𝑖𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒: 𝑦̂𝑖 = 𝑎𝑥𝑖 + 𝑏
𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙: 𝜖𝑖 = 𝑦𝑖 − 𝑦̂𝑖
𝑖=1
Properties of the LSE Solution
𝑛
∑ 𝜖𝑖 = 0
𝑖=1
Unbiased estimator for 𝑣𝑎𝑟(𝜖) is SSE/(n-2)
Goodness of Fit
𝐺𝑜𝑜𝑑𝑛𝑒𝑠𝑠 𝑜𝑓 𝑓𝑖𝑡 𝑚𝑒𝑡𝑟𝑖𝑐: 𝑅 2 = 𝑟 2
𝑐𝑜𝑣(𝑋, 𝑌)
𝑟=
√𝑣𝑎𝑟(𝑋)𝑣𝑎𝑟(𝑌)
𝑐𝑜𝑣 2 (𝑋, 𝑌)
𝑅2 =
𝑣𝑎𝑟(𝑋)𝑣𝑎𝑟(𝑌)
𝑣𝑎𝑟(𝜖)
𝑅2 = 1 −
𝑣𝑎𝑟(𝑌)
Y   0  1 X  
Y  dependent variable (response)
X  independen t variable (predictor or explanator y)
 0  intercept (value of Y when X  0)
1  slope of the regression line
  random error
Uncertainty of Fit parameters
𝑦̂𝑖 = 𝑎𝑥𝑖 + 𝑏
Using t-distribution and DF = n - 2
𝑣𝑎𝑟(𝜖)
𝑛 𝑣𝑎𝑟(𝑋)
𝑣𝑎𝑟(𝜖)
𝑥̅ 2
𝑣𝑎𝑟(𝑏) =
(1 +
)
𝑛
𝑣𝑎𝑟(𝑋)
(𝑥𝑖 − 𝑥̅ )2
𝑣𝑎𝑟(𝜖)
𝑣𝑎𝑟(𝑦̂𝑖 ) =
+
(1
)
𝑛
𝑣𝑎𝑟(𝑋)
𝑣𝑎𝑟(𝑎) =
Yˆ  b0  b1 X
Yˆ  predicted value of Y
b0  estimate of  0 , based on sample results
b1  estimate of  1 , based on sample results
6
Error = (Actual value) – (Predicted value)
e  Y  Yˆ
X
Y
b1 
 X  average (mean) of X values
n
Y
 average (mean) of Y values
n
 ( X  X )(Y  Y )
(X  X)
2
b0  Y  b1 X
Sum of Squares Total  SST   (Y  Y ) 2
Sum of Squares Error  SSE   e 2   (Y  Yˆ ) 2
Sum of Squares Regression  SSR   (Yˆ  Y ) 2
SSR
SSE
1–
SST
SST
SSE
Mean Squared Error  s 2  MSE 
n  k 1
SST  SSR + SSE
Correlation Coefficient = r   r 2
Standard Error of Estimate  s  MSE
SSR
MSR 
k
k  number of independen t variable s in the model
Hypothesis Test H 0 :  1  0
H 1 : 1  0
Reject if Fcalculated  F ,df1 ,df2
df 1  k df 2  n  k  1
p - value  P( F  calculated test statistic )
Reject if p - value  
Yˆ  b0  b1 X 1  b2 X 2  ...  bk X k
Ŷ = predicted value of Y
b0 = sample intercept (an estimate of 0)
bi = sample coefficient of the i th variable (an
estimate of i)
Coefficien t of Determinat ion  r 2 
Generic Linear Model Y   0  1 X  
MSR
MSE
degrees of freedom for the numerator = df1 = k
degrees of freedom for the denominator = df2 = n –
k–1
F Statistic : F 
Y = 0 + 1X1 + 2X2 + … + kXk + 
Y=
dependent variable (response variable)
Xi = ith independent variable (predictor or
explanatory variable)
0 = intercept (value of Y when all Xi = 0)
i = coefficient of the ith independent variable
k=
number of independent variables
=
random error
SSE /( n  k  1)
Adjusted r 2  1 
SST /( n  1)
(∑𝑛𝑖=1 𝑦𝑖 )(∑𝑛𝑖=1 𝑥𝑖 )
𝑆𝑥𝑦
𝑛
Least Square Estimates β̂0 = 𝑦̅ − β̂1 𝑥̅ 𝑎𝑛𝑑 β̂1 =
=
2
𝑆𝑥𝑥
(∑𝑛 𝑥 )
∑𝑛𝑖=1 𝑥𝑖2 − 𝑖=1 𝑖
𝑛
1
𝑛
𝑛
∑
∑
Where 𝑥̅ = (𝑛) 𝑖=1 𝑥𝑖 𝑎𝑛𝑑 𝑦̅ = (1/𝑛) 𝑖=1 𝑦𝑖
∑𝑛𝑖=1 𝑦𝑖 𝑥𝑖 −
𝑛
𝐸𝑟𝑟𝑜𝑟 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠, 𝑆𝑆𝐸 =
2
𝑆𝑆𝐸
𝑈𝑛𝑏𝑖𝑎𝑠𝑒𝑑 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑜𝑟, 𝜎̂ = 𝑛−2
∑ 𝑒𝑖2
𝑖=1
𝑛
= ∑(𝑦𝑖 − 𝑦̂𝑖 )2
𝑖=1
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − β̂1 𝑆𝑥𝑦
SS 𝑇 = ∑𝑛𝑖=1(𝑦𝑖 − 𝑦̅)2
𝜎̂ 2
Estimated Standard error of the slope, se(β̂1 ) = √
𝑆𝑥𝑥
7
1 𝑥̅ 2
Estimated Standard error of the intercept, se(β̂0 ) = √𝜎̂ 2 [ +
]
𝑛 𝑆𝑥𝑥
𝐻𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑇𝑒𝑠𝑡𝑖𝑛𝑔 (𝑡 − 𝑡𝑒𝑠𝑡)
Null hypothesis: 𝐻0 : 𝛽1 = 𝛽1,0
𝐻1 : 𝛽1 ≠ 𝛽1,0
Test Statistic: : 𝑇0 :
̂1 −𝛽1,0
𝛽
Reject null hypothesis if |𝑡0 | > t α/2,n−2
̂ 2 /𝑆𝑥𝑥
√𝜎
Null hypothesis: 𝐻0 : 𝛽0 = 𝛽0,0
𝐻1 : 𝛽0 ≠ 𝛽0,0
Test Statistic: : 𝑇0 :
̂0 −𝛽0,0
𝛽
̅2
1 𝑥
]
𝑛 𝑆𝑥𝑥
Reject null hypothesis if |𝑡0 | > t α/2,n−2
√𝜎
̂ 2[ +
𝑛
𝑛
𝑛
2
Analysis of Variance Identity, ∑(𝑦𝑖 − 𝑦̅) = ∑(𝑦̂𝑖 − 𝑦̅) + ∑(𝑦𝑖 − 𝑦̂𝑖 )2
𝑖=1
2
𝑖=1
𝑖=1
𝑆𝑆𝑇 = 𝑆𝑆𝑅 + 𝑆𝑆𝐸
𝑆𝑆𝑅 /1
𝑀𝑆𝑅
𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝑠𝑖𝑔𝑛𝑖𝑓𝑎𝑛𝑐𝑒 𝑜𝑓 𝑅𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛, 𝐹0 =
=
𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝐹1,𝑛−2 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑆𝑆𝐸 /(𝑛 − 2) 𝑀𝑆𝐸
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠, 𝑖𝑓 𝑓0 > 𝑓𝛼,1,𝑛−2
2
̂
̂
𝜎
𝜎
100(1−∝)% CI of slope β0 is β̂1 − 𝑡𝛼,𝑛−2 √ ≤ 𝛽1 ≤ β̂1 + 𝑡𝛼,𝑛−2 √
2
𝑆𝑥𝑥
2
2
𝑆𝑥𝑥
𝑥̅ 2
2
1
1
𝑥̅
100(1−∝)% CI of slope β1 is β̂0 − 𝑡𝛼,𝑛−2 √𝜎̂ 2 [𝑛 + 𝑆 ] ≤ 𝛽0 ≤ β̂0 + 𝑡𝛼,𝑛−2 √𝜎̂ 2 [𝑛 + 𝑆 ]
𝑥𝑥
2
𝑥𝑥
2
100(1−∝)% CI about the mean response x = x0 , say μ𝑌|𝑥0 , is
1 (𝑥0 − 𝑥̅ )2
1 (𝑥0 − 𝑥̅ )2
μ̂𝑌|𝑥0 − 𝑡𝛼,𝑛−2 √𝜎̂ 2 [ +
] ≤ μ𝑌|𝑥0 ≤ μ̂𝑌|𝑥0 + 𝑡𝛼,𝑛−2 √𝜎̂ 2 [ +
]
𝑛
𝑆𝑥𝑥
𝑛
𝑆𝑥𝑥
2
2
𝑤ℎ𝑒𝑟𝑒 μ̂𝑌|𝑥 = β̂0 + β̂1 𝑥0 𝑖𝑠 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑓𝑖𝑡𝑡𝑒𝑑 𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑚𝑜𝑑𝑒𝑙
0
100(1−∝)% PI on a future observation Y0 at the value x0 is
1 (𝑥0 − 𝑥̅ )2
1 (𝑥0 − 𝑥̅ )2
ŷ0 − 𝑡𝛼,𝑛−2 √𝜎̂ 2 [ +
] ≤ Y0 ≤ ŷ0 + 𝑡𝛼,𝑛−2 √𝜎̂ 2 [ +
]
𝑛
𝑆𝑥𝑥
𝑛
𝑆𝑥𝑥
2
2
𝑤ℎ𝑒𝑟𝑒 ŷ0 = β̂0 + β̂1 𝑥0 𝑖𝑠 𝑐𝑜𝑚𝑝𝑢𝑡𝑒𝑑 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑟𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛 𝑚𝑜𝑑𝑒𝑙
𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙𝑠 𝑒𝑖 = 𝑦𝑖 − ŷ𝑖
d𝑖 = 𝑒𝑖 /√𝜎̂ 2
𝑆𝑆
𝑆𝑆
𝑅
Coefficient of determination, R2 =
=1− 𝐸
𝐶𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝜌 =
𝑆𝑆𝑇
𝑆𝑆𝑇
𝜎𝑋𝑌
𝜎𝑋 𝜎𝑌
𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝑧𝑒𝑟𝑜 𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛: 𝐻0 : 𝜌 = 0
𝑇0 =
𝑅√𝑛 − 2
ℎ𝑎𝑠 𝑡 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛 − 2 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
√1 − 𝑅 2
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠, 𝑖𝑓 |𝑡0 | > 𝑡𝛼,𝑛−2
2
𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝑐𝑜𝑟𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛: 𝐻0 : 𝜌 = 𝜌0
𝑍0 = (arctanh 𝑅 − arctanh 𝜌0 )(𝑛 − 3)1/2
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠, 𝑖𝑓 |𝑧0 | > 𝑧𝛼
2
100(1−∝)% CI for correlation coefficient is
𝑧𝛼
𝑧𝛼/2
2
tanh (arctanh 𝑟 −
) ≤ 𝜌 ≤ tanh (arctanh 𝑟 +
) 𝑤ℎ𝑒𝑟𝑒 tanh 𝑢 = (𝑒𝑢 −𝑒−𝑢 )/(𝑒𝑢 +𝑒−𝑢 )
√𝑛 − 3
√𝑛 − 3
1
𝐹𝑖𝑡𝑡𝑒𝑑 𝑙𝑜𝑔𝑖𝑠𝑡𝑖𝑐 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛, 𝑦̂ =
1 + 𝑒𝑥𝑝[−(𝛽0 + 𝛽1 𝑥)]
Multiple Regression
8
𝑦1
𝑦2
𝑦=[⋮]
𝑦𝑛
𝑦 = 𝑋𝛽 + 𝜖
𝑦𝑖 = 𝛽0 + 𝛽1 𝑥𝑖1 + 𝛽2 𝑥𝑖2 + ⋯ + 𝛽𝑘 𝑥𝑖𝑘 + 𝜖𝑖 𝑖 = 1,2, … , 𝑛
𝜖1
𝛽1
1 𝑥11 𝑥12 … 𝑥1𝑘
…
𝑥
𝑥
𝑥
𝜖
𝛽
21
22
2𝑘
2
𝑋 = [1
𝛽 = [ 2]
𝜖=[⋮]
⋮
⋮
⋮ ]
⋮
1 𝑥𝑛1
𝑥𝑛2
⋮
… 𝑥𝑛𝑘
𝑛
𝜖𝑛
𝛽𝑛
𝑛
𝑛
𝑛
𝑛 𝛽̂0 + 𝛽̂1 ∑ 𝑥𝑖1 + 𝛽̂2 ∑ 𝑥𝑖2 + ⋯ + 𝛽̂𝑘 ∑ 𝑥𝑖𝑘 = ∑ 𝑦𝑖
𝑖=1
𝑛
𝑛
𝛽̂0 ∑ 𝑥𝑖1 +
𝑖=1
𝑛
2
𝛽̂1 ∑ 𝑥𝑖1
𝑖=1
𝑛
𝑖=1
𝑛
𝑖=1
𝑛
𝑖=1
𝑛
+ 𝛽̂2 ∑ 𝑥𝑖1 𝑥𝑖2 + ⋯ + 𝛽̂𝑘 ∑ 𝑥𝑖1 𝑥𝑖𝑘 = ∑ 𝑥𝑖1 𝑦𝑖
𝑖=1
𝑖=1
𝑛
𝑖=1
𝑛
𝑛
2
𝛽̂0 ∑ 𝑥𝑖𝑘 + 𝛽̂1 ∑ 𝑥𝑖𝑘 𝑥𝑖1 + 𝛽̂2 ∑ 𝑥𝑖𝑘 𝑥𝑖2 + ⋯ + 𝛽̂𝑘 ∑ 𝑥𝑖𝑘
= ∑ 𝑥𝑖𝑘 𝑦𝑖
𝑖=1
𝑖=1
𝑛
𝑛
𝑛
𝑖=1
𝑁𝑜𝑟𝑚𝑎𝑙 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠: 𝑋 ′ 𝑋𝛽̂ = 𝑋′𝑌
𝑛
∑ 𝑥𝑖1
∑ 𝑥𝑖2
𝑖=1
𝑛
𝑖=1
𝑛
2
∑ 𝑥𝑖1
∑ 𝑥𝑖1 𝑥𝑖2
𝑖=1
𝑖=1
𝑖=1
𝑛
𝑛
∑ 𝑥𝑖𝑘
[ 𝑖=1
⋮
𝑛
𝑖=1
𝑛
∑ 𝑥𝑖1
⋮
𝑖=1
⋯
∑ 𝑥𝑖𝑘
𝛽̂0
𝑖=1
𝑛
⋯ ∑ 𝑥𝑖1 𝑥𝑖𝑘
∑ 𝑥𝑖𝑘 𝑥𝑖1
∑ 𝑥𝑖𝑘 𝑥𝑖2
𝑖=1
𝑖=1
⋮
𝑛
𝑖=1
𝑛
⋮
𝑖=1
[𝛽̂𝑘 ]
𝑛
⋮
2
∑ 𝑥𝑖𝑘
…
∑ 𝑦𝑖
∑ 𝑥𝑖1 𝑦𝑖
𝛽̂1 =
𝑖=1
⋮
𝑛
⋮
∑ 𝑥𝑖𝑘 𝑦𝑖
]
[ 𝑖=1
]
𝐿𝑒𝑎𝑠𝑡 𝑠𝑞𝑢𝑎𝑟𝑒 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝛽: 𝛽̂ = (𝑋 ′ 𝑋)−1 𝑋′𝑌
∑𝑛𝑖=1 𝑒𝑖2
𝑆𝑆𝐸
2
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑜𝑟 𝑜𝑓 𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒, 𝜎̂ =
=
𝑛−𝑝
𝑛−𝑝
𝑖=1
𝑛
𝐸𝑟𝑟𝑜𝑟 𝑠𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠, 𝑆𝑆𝐸 =
𝑛
2
∑ 𝑒𝑖 = ∑(𝑦𝑖
𝑖=1
𝑖=1
′
−1
− 𝑦̂𝑖 )2 = 𝑒′ 𝑒
𝐶𝑜𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒, 𝐶 = (𝑋 𝑋)
̂ 2 𝐶𝑖𝑗
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝑜𝑓 𝛽̂𝑗 = 𝑠𝑒(𝛽̂𝑗 ) = √𝜎
Hypothesis of ANOVA Test
Null hypothesis: 𝐻0 : 𝛽1 = 𝛽2 = ⋯ = 𝛽𝑘 = 0
𝐻1 : 𝛽𝑗 ≠ 0 𝑓𝑜𝑟 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑗
𝑆𝑆𝑅 /𝑘
𝑀𝑆𝑅
𝑇𝑒𝑠𝑡 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑓𝑜𝑟 𝐴𝑁𝑂𝑉𝐴, 𝐹0 =
=
𝑆𝑆𝐸 /(𝑛 − 𝑝) 𝑀𝑆𝐸
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠, 𝑖𝑓 𝑓0 > 𝑓𝛼,𝑘,𝑛−𝑝
(∑𝑛𝑖=1 𝑦𝑖 )2
(∑𝑛𝑖=1 𝑦𝑖 )2
′
′
̂
𝑆𝑆𝐸 = 𝑦 𝑦 −
− [ 𝛽𝑋 𝑦 −
]
𝑛
𝑛
(∑𝑛𝑖=1 𝑦𝑖 )2
𝑆𝑆𝑅 = 𝛽̂ 𝑋 ′ 𝑦 −
𝑛
𝑆𝑆𝑅
𝑆𝑆𝐸
2
Coefficient of determination, R =
=1−
𝑆𝑆𝑇
𝑆𝑆𝑇
𝑆𝑆
/(𝑛
−
𝑝)
𝐸
Adjusted R2 , R2𝑎𝑑𝑗 =
𝑆𝑆𝑇 /(𝑛 − 1)
Null hypothesis: 𝐻0 : 𝛽𝑗 = 𝛽𝑗,0
𝐻1 : 𝛽𝑗 ≠ 𝛽𝑗,0
9
Test Statistic: : 𝑇0 :
̂𝑗 −𝛽𝑗,0
𝛽
̂ 2 𝐶𝑗𝑗
√𝜎
=
̂𝑗 −𝛽𝑗,0
𝛽
̂𝑗 )
𝑠𝑒(𝛽
Reject null hypothesis if |𝑡0 | > t α/2,n−p
𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝑠𝑖𝑔𝑛𝑖𝑓𝑎𝑛𝑐𝑒 𝑜𝑓 𝑅𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛
𝐻0 : 𝛽1 = 0
𝐻1 : 𝛽1 ≠ 0
𝑆𝑆𝑅 (𝛽1 |𝛽2 )/𝑟
𝐹0 =
𝑀𝑆𝐸
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠, 𝑖𝑓 𝑓0 > 𝑓𝛼,𝑟,𝑛−𝑝 . 𝑇ℎ𝑖𝑠 𝑐𝑜𝑛𝑐𝑙𝑢𝑑𝑒𝑠 𝑡ℎ𝑎𝑡 𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑜𝑓 𝑝𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠 𝑖𝑛 𝛽1 𝑖𝑠 𝑛𝑜𝑡 𝑧𝑒𝑟𝑜.
100(1−∝)% CI on the regression coefficient β𝑗 j = 0,1,2, … , k is
β̂𝑗 − 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 𝐶𝑖𝑗 ≤ 𝛽𝑗 ≤ β̂𝑗 + 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 𝐶𝑖𝑗
2
2
100(1−∝)% CI about the mean response at point x01 , x02 , … , x0𝑘 , is
μ̂𝑌|𝑥0 − 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 𝑥0′ (𝑋 ′ 𝑋)−1 𝑥0 ≤ μ𝑌|𝑥0 ≤ μ̂𝑌|𝑥0 + 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 𝑥0′ (𝑋 ′ 𝑋)−1 𝑥0
2
2
100(1−∝)% PI on a future observation Y0 at the value x01 , x02 , … , x0𝑘 is
ŷ0 − 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 (1 + 𝑥0′ (𝑋 ′ 𝑋)−1 𝑥0 ) ≤ Y0 ≤ ŷ0 + 𝑡𝛼,𝑛−𝑝 √𝜎̂ 2 (1 + 𝑥0′ (𝑋 ′ 𝑋)−1 𝑥0 )
2
2
𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙𝑠 𝑒𝑖 = 𝑦𝑖 − ŷ𝑖
𝑒𝑖
d𝑖 =
= 𝑒𝑖 /√𝜎̂ 2
√𝑀𝑆𝐸
𝑒𝑖
𝑆𝑡𝑢𝑑𝑒𝑛𝑡𝑖𝑧𝑒𝑑 𝑅𝑒𝑠𝑖𝑑𝑢𝑎𝑙𝑠 𝑟𝑖 =
𝑊ℎ𝑒𝑟𝑒 𝑖 = 1,2, … , 𝑛
√𝜎̂ 2 (1 − ℎ𝑖𝑖 )
𝐻𝑎𝑡 𝑚𝑎𝑡𝑟𝑖𝑥: 𝐻 = 𝑋(𝑋 ′ 𝑋)−1 𝑋′
ℎ𝑖𝑖 = 𝑖𝑡ℎ 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝐻 𝑚𝑎𝑡𝑟𝑖𝑥 = 𝑥𝑖′ (𝑋 ′ 𝑋)−1 𝑥𝑖
′
𝑆𝑆𝐸 (𝑝)
𝐶𝑜𝑜𝑘 𝑠𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝐶𝑝 𝑆𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐, 𝐶𝑝 =
− 𝑛 + 2𝑝
′ ′
̂
̂
̂
̂
𝜎̂ 2
(𝛽(𝑖) − 𝛽 ) 𝑋 𝑋(𝛽(𝑖) − 𝛽 )
𝐷𝑖 =
𝑤ℎ𝑒𝑟𝑒 𝑖 = 1,2, … , 𝑛
𝑃𝑟𝑒𝑑𝑖𝑐𝑡𝑖𝑜𝑛 𝐸𝑟𝑟𝑜𝑟 𝑆𝑢𝑚 𝑜𝑓 𝑆𝑞𝑢𝑎𝑟𝑒𝑠
𝑝𝜎̂ 2
𝑛
𝑛
2
𝑒𝑖
2
𝑟𝑖2 ℎ𝑖𝑖
𝑃𝑅𝐸𝑆𝑆 = ∑(𝑦𝑖 − 𝑦̂(𝑡) ) = ∑ (
)
𝐷𝑖 =
𝑤ℎ𝑒𝑟𝑒 𝑖 = 1,2, … , 𝑛
1 − ℎ𝑖𝑖
𝑝 (1 − ℎ )
𝑖𝑖
𝑖=1
𝑆𝑡𝑒𝑝𝑤𝑖𝑠𝑒 𝑅𝑒𝑔𝑟𝑒𝑠𝑠𝑖𝑜𝑛
𝑆𝑆𝑅 (𝛽𝑗 |𝛽1 , 𝛽0 )
𝐹𝑗 =
𝑀𝑆𝐸 (𝑥𝑗 , 𝑥1 )
Single-Factor Experiments
𝑖=1
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 𝐼𝑛𝑓𝑙𝑎𝑡𝑖𝑜𝑛 𝐹𝑎𝑐𝑡𝑜𝑟 (𝑉𝐼𝐹)
1
𝑉𝐼𝐹(𝛽𝑗 ) =
𝑊ℎ𝑒𝑟𝑒 𝑗 = 1,2, … , 𝑘
(1 − 𝑅𝑗2 )
𝑎
𝑛
𝑎
𝑎
2
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦, ∑ ∑(𝑦𝑖𝑗 − 𝑦̅.. ) = 𝑎 ∑(𝑦𝑖. − 𝑦̅..
𝑖=1 𝑗=1
)2
𝑖=1
𝑆𝑆𝑇 = 𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 + 𝑆𝑆𝐸
𝑛
2
+ ∑ ∑(𝑦𝑖𝑗 − 𝑦̅𝑖. )
𝑖=1 𝑗=1
𝑎
𝐸(𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 ) = (𝑎 − 1)𝜎 + 𝑛 ∑ 𝜏𝑖2
2
𝑖=1
𝐸(𝑆𝑆𝐸 ) = 𝑎(𝑛 − 1)𝜎 2
𝑀𝑆𝐸 = 𝑆𝑆𝐸 /[𝑎(𝑛 − 1)]
𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 /(𝑎 − 1) 𝑀𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
𝐹0 =
=
𝑆𝑆𝐸 /[𝑎(𝑛 − 1)]
𝑀𝑆𝐸
𝑎 𝑛
2
𝑦..
2
𝑆𝑆𝑇 = ∑ ∑ 𝑦𝑖𝑗
−
𝑁
𝑖=1 𝑗=1
𝑎
𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 = ∑
𝑖=1
𝑦𝑖.2 𝑦..2
−
𝑛
𝑁
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
10
𝑇=
𝑌̅𝑖. − 𝜇𝑖
ℎ𝑎𝑠 𝑡 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑎(𝑛 − 1) 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
√𝑀𝑆𝐸 /𝑛
100(1−∝)% CI about the mean of the i𝑡ℎ treatment μ𝑖 , is
𝑀𝑆𝐸
𝑀𝑆𝐸
𝑦̅𝑖. − 𝑡𝛼,𝑎(𝑛−1) √
≤ μ𝑖 ≤ 𝑦̅𝑖. + 𝑡𝛼,𝑎(𝑛−1) √
𝑛
𝑛
2
2
100(1−∝)% CI on the difference in two treatment means μ𝑖 − μ𝑗 , is
2𝑀𝑆𝐸
2𝑀𝑆𝐸
𝑦̅𝑖. − 𝑦̅𝑗. − 𝑡𝛼,𝑎(𝑛−1) √
≤ μ𝑖 − μ𝑗 ≤ 𝑦̅𝑖. − 𝑦̅𝑗. + 𝑡𝛼,𝑎(𝑛−1) √
𝑛
𝑛
2
2
𝐹𝑖𝑠ℎ𝑒𝑟 ′ 𝑠 𝐿𝑒𝑎𝑠𝑡 𝑆𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑡 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 (𝐿𝑆𝐷)𝑚𝑒𝑡ℎ𝑜𝑑
𝑁𝑢𝑙𝑙 𝐻𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝐻0 : μ𝑖 = μ𝑗 𝑤ℎ𝑒𝑟𝑒 𝑖 ≠ 𝑗
𝑦̅𝑖. − 𝑦̅𝑗.
𝑡0 =
√2𝑀𝑆𝐸
𝑛
𝑅𝑒𝑗𝑒𝑐𝑡 𝑛𝑢𝑙𝑙 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑖𝑠 𝑖𝑓 |𝑦̅𝑖. − 𝑦̅𝑗. | > 𝐿𝑆𝐷
2𝑀𝑆𝐸
𝐿𝑆𝐷 = 𝑡𝛼,𝑎(𝑛−1) √
𝑛
2
𝐼𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒𝑠 𝑎𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡, 𝐿𝑆𝐷 = 𝑡𝛼,𝑁−𝑎 √𝑀𝑆𝐸 (
2
1
1
+ )
𝑛𝑖 𝑛𝑗
𝑃𝑜𝑤𝑒𝑟 𝑜𝑓 𝐴𝑁𝑂𝑉𝐴 𝑡𝑒𝑠𝑡, 1 − 𝛽 = 𝑃{𝑅𝑒𝑗𝑒𝑐𝑡 𝐻0 |𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒} = 𝑃{𝐹0 > 𝑓𝛼,𝑎−1,𝑎(𝑛−1) |𝐻0 𝑖𝑠 𝑓𝑎𝑙𝑠𝑒}
𝑛 ∑𝑎𝑖=1 𝜏𝑖2
𝜙2 =
𝑎𝜎 2
𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
𝐸(𝑀𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 ) = 𝐸 (
) = 𝜎 2 + 𝑛𝜎𝜏2
𝑎−1
𝑆𝑆𝐸
𝐸(𝑀𝑆𝐸 ) = 𝐸 [
] = 𝜎2
𝑎(𝑛 − 1)
𝜎̂ 2 = 𝑀𝑆𝐸
𝑀𝑆
𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 − 𝑀𝑆𝐸
𝜎̂𝜏2 =
𝑛
Randomized Block Experiment
𝑎
𝑏
2
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦, ∑ ∑(𝑦𝑖𝑗 − 𝑦̅.. )
𝑖=1 𝑗=1
𝑎
𝑏
= 𝑏 ∑(𝑦̅𝑖. − 𝑦̅..
𝑖=1
)2
𝑎
𝑏
2
2
+ 𝑎 ∑(𝑦̅.𝑗 − 𝑦̅.. ) + ∑ ∑(𝑦𝑖𝑗 − 𝑦̅.𝑗 − 𝑦̅𝑖. + 𝑦̅.. )
𝑗=1
𝑖=1 𝑗=1
𝑆𝑆𝑇 = 𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 + 𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠 + 𝑆𝑆𝐸
𝑏 ∑𝑎𝑖=1 𝜏𝑖2
𝐸(𝑀𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 ) = 𝜎 2 +
𝑎−1
𝑏
∑
𝑎
𝛽𝑗2
𝑗=1
2
𝐸(𝑀𝑆𝐵𝑙𝑜𝑐𝑘𝑠 ) = 𝜎 +
𝑏−1
𝐸(𝑀𝑆𝐸 ) = 𝜎 2
𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
𝑀𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 =
𝑎−1
𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠
𝑀𝑆𝐵𝑙𝑜𝑐𝑘𝑠 =
𝑏−1
11
𝑀𝑆𝐸 =
𝑆𝑆𝐸
(𝑎 − 1)(𝑏 − 1)
𝑎
𝑏
2
𝑆𝑆𝑇 = ∑ ∑ 𝑦𝑖𝑗
−
𝑖=1 𝑗=1
𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠
𝑎
1
𝑦..2
= ∑ 𝑦𝑖.2 −
𝑏
𝑎𝑏
𝑖=1
𝑏
𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠 =
𝑦..2
𝑎𝑏
1
𝑦..2
∑ 𝑦.𝑗2 −
𝑎
𝑎𝑏
𝑗=1
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝑇𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡𝑠 − 𝑆𝑆𝐵𝑙𝑜𝑐𝑘𝑠
𝑒𝑖𝑗 = 𝑦𝑖𝑗 − 𝑦̂𝑖𝑗
𝑦̂𝑖𝑗 = 𝑦̅𝑖. + 𝑦̅.𝑗 − 𝑦̅..
Two Factors experiment
𝑏
𝑛
𝑦𝑖.. = ∑ ∑ 𝑦𝑖𝑗𝑘
𝑗=1 𝑘=1
𝑦𝑖..
𝑦̅𝑖.. =
𝑤ℎ𝑒𝑟𝑒 𝑖 = 1,2, … , 𝑎
𝑏𝑛
𝑎 𝑛
𝑦.𝑗. = ∑ ∑ 𝑦𝑖𝑗𝑘
𝑖=1 𝑘=1
𝑦.𝑗.
𝑦̅.𝑗. =
𝑤ℎ𝑒𝑟𝑒 𝑗 = 1,2, … , 𝑏
𝑎𝑛
𝑛
𝑦𝑖𝑗. = ∑ 𝑦𝑖𝑗𝑘
𝑘=1
𝑦̅𝑖𝑗. =
𝑦𝑖𝑗.
𝑤ℎ𝑒𝑟𝑒 𝑖 = 1,2, … , 𝑎
𝑛
𝑎
𝑏
𝑛
𝑦... = ∑ ∑ ∑ 𝑦𝑖𝑗𝑘
𝑖=1 𝑗=1 𝑘=1
𝑦...
𝑦̅... =
𝑤ℎ𝑒𝑟𝑒 𝑗 = 1,2, … , 𝑏
𝑎𝑏𝑛
𝑎
𝑏
𝑛
2
𝑆𝑆𝑇 = ∑ ∑ ∑(𝑦𝑖𝑗𝑘 − 𝑦̅... )
𝑖=1 𝑗=1 𝑘=1
𝑎
𝑏
𝑛
2
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦, ∑ ∑ ∑(𝑦𝑖𝑗𝑘 − 𝑦̅... )
𝑖=1 𝑗=1 𝑘=1
𝑎
𝑏
= 𝑏𝑛 ∑(𝑦̅𝑖.. − 𝑦̅...
𝑎
)2
𝑎
𝑏
2
2
+ 𝑎𝑛 ∑(𝑦̅.𝑗. − 𝑦̅... ) + 𝑛 ∑ ∑(𝑦̅𝑖𝑗. − 𝑦̅𝑖.. − 𝑦̅.𝑗. + 𝑦̅... )
𝑖=1
𝑏 𝑛
𝑗=1
𝑖=1 𝑗=1
2
+ ∑ ∑ ∑(𝑦𝑖𝑗𝑘 − 𝑦̅𝑖𝑗. )
𝑖=1 𝑗=1 𝑘=1
𝑆𝑆𝑇 = 𝑆𝑆𝐴 + 𝑆𝑆𝐵 + 𝑆𝑆𝐴𝐵 + 𝑆𝑆𝐸
𝑆𝑆𝐴
𝑏𝑛 ∑𝑎𝑖=1 𝜏𝑖2
𝐸(𝑀𝑆𝐴 ) = 𝐸 (
) = 𝜎2 +
𝑎−1
𝑎−1
∑𝑏𝑗=1 𝛽𝑗2
𝑎𝑛
𝑆𝑆𝐵
𝐸(𝑀𝑆𝐵 ) = 𝐸 (
) = 𝜎2 +
𝑏−1
𝑏−1
12
𝑛 ∑𝑎𝑖=1 ∑𝑏𝑗=1(𝜏𝛽)2𝑖𝑗
𝑆𝑆𝐴𝐵
𝐸(𝑀𝑆𝐴𝐵 ) = 𝐸 (
) = 𝜎2 +
(𝑎 − 1)(𝑏 − 1)
(𝑎 − 1)(𝑏 − 1)
𝑆𝑆𝐸
𝐸(𝑀𝑆𝐸 ) = 𝐸 (
) = 𝜎2
𝑎𝑏(𝑛 − 1)
𝑀𝑆𝐴
𝑀𝑆𝐸
𝑀𝑆𝐵
𝐹 𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 𝐵, 𝐹0 =
𝑀𝑆𝐸
𝑀𝑆𝐴𝐵
𝐹 𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝐴𝐵 𝐼𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛, 𝐹0 =
𝑀𝑆𝐸
𝐹 𝑇𝑒𝑠𝑡 𝑓𝑜𝑟 𝐹𝑎𝑐𝑡𝑜𝑟 𝐴, 𝐹0 =
𝑎
𝑏
𝑛
2
𝑆𝑆𝑇 = ∑ ∑ ∑ 𝑦𝑖𝑗𝑘
−
𝑖=1 𝑗=1 𝑘=1
𝑎
𝑦𝑖..2
𝑦...2
𝑆𝑆𝐴 = ∑
−
𝑏𝑛 𝑎𝑏𝑛
𝑖=1
𝑏
2
𝑦.𝑗.
𝑦...2
𝑆𝑆𝐵 = ∑
−
𝑎𝑛 𝑎𝑏𝑛
𝑗=1
𝑎 𝑏
2
𝑦𝑖𝑗.
𝑦...2
𝑆𝑆𝐴𝐵 = ∑ ∑
𝑖=1 𝑗=1
𝑛
−
𝑎𝑏𝑛
𝑦...2
𝑎𝑏𝑛
− 𝑆𝑆𝐴 − 𝑆𝑆𝐵
𝑆𝑆𝐸 = 𝑆𝑆𝑇 − 𝑆𝑆𝐴𝐵 − 𝑆𝑆𝐴 − 𝑆𝑆𝐵
𝑒𝑖𝑗𝑘 = 𝑦𝑖𝑗𝑘 − 𝑦̂𝑖𝑗.
Three-Factor Fixed Effects Model
Source
Sum of Degrees of
of
Squares Freedom
Variation
A
SSA
a–1
Mean
Square
MSA
B
SSB
b–1
MSB
C
SSC
c–1
MSC
AB
SSAB
(a – 1)(b – 1)
MSAB
AC
SSAC
(a – 1)(c – 1)
MSAC
BC
SSBC
(b – 1)(c – 1)
MSBC
ABC
SSABC
(a – 1)(b – 1)(c – 1)
MSABC
abc(n – 1)
abcn – 1
MSE
Error
SSE
Total
SST
k
2 Factorial Designs
Expected Mean Squares
F0
𝑏𝑐𝑛 ∑ 𝜏𝑖2
𝜎 +
𝑎−1
∑ 𝛽𝑗2
𝑎𝑐𝑛
2
𝜎 +
𝑏−1
∑ 𝛾𝑘2
𝑎𝑏𝑛
2
𝜎 +
𝑐−1
∑
∑(𝜏𝛽)2𝑖𝑗
𝑐𝑛
2
𝜎 +
(𝑎 − 1)(𝑏 − 1)
𝑏𝑛 ∑ ∑(𝜏𝛾)2𝑖𝑘
𝜎2 +
(𝑎 − 1)(𝑐 − 1)
𝑎𝑛 ∑ ∑(𝛽𝛾)2𝑗𝑘
2
𝜎 +
(𝑏 − 1)(𝑐 − 1)
∑ ∑ ∑(𝜏𝛽𝛾)2𝑖𝑗𝑘
𝑛
𝜎2 +
(𝑎 − 1)(𝑏 − 1)(𝑐 − 1)
𝜎2
2
𝑀𝑆𝐴
𝑀𝑆𝐸
𝑀𝑆𝐵
𝑀𝑆𝐸
𝑀𝑆𝐶
𝑀𝑆𝐸
𝑀𝑆𝐴𝐵
𝑀𝑆𝐸
𝑀𝑆𝐴𝐶
𝑀𝑆𝐸
𝑀𝑆𝐵𝐶
𝑀𝑆𝐸
𝑀𝑆𝐴𝐵𝐶
𝑀𝑆𝐸
(l) Represents the treatment combination with both factors at the low level.
𝑀𝑎𝑖𝑛 𝐸𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝐹𝑎𝑐𝑡𝑜𝑟 𝐴: 𝐴 = 𝑦̅𝐴+ − 𝑦̅𝐴− =
𝑎 + 𝑎𝑏 𝑏 + (𝑙)
1
[𝑎 + 𝑎𝑏 − 𝑏 − (𝑙)]
−
=
2𝑛
2𝑛
2𝑛
13
𝑏 + 𝑎𝑏 𝑎 + (𝑙)
1
[𝑏 + 𝑎𝑏 − 𝑎 − (𝑙)]
−
=
2𝑛
2𝑛
2𝑛
𝑎𝑏 + (𝑙) 𝑎 + 𝑏
1
[𝑎𝑏 + (𝑙) − 𝑎 − 𝑏]
𝐼𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝐸𝑓𝑓𝑒𝑐𝑡 𝐴𝐵: 𝐴𝐵 =
−
=
2𝑛
2𝑛
2𝑛
𝐶𝑜𝑛𝑡𝑟𝑎𝑠𝑡 𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑎𝑙𝑤𝑎𝑦𝑠 + 1 𝑜𝑟 − 1. 𝐶𝑜𝑛𝑡𝑟𝑎𝑠𝑡𝐴 = 𝑎 + 𝑎𝑏 − 𝑏 − (𝑙)
𝐶𝑜𝑛𝑡𝑟𝑎𝑠𝑡
𝐸𝑓𝑓𝑒𝑐𝑡 =
𝑛2𝑘−1
(𝐶𝑜𝑛𝑡𝑟𝑎𝑠𝑡)2
𝑆𝑢𝑚 𝑜𝑓 𝑠𝑞𝑢𝑎𝑟𝑒𝑠 𝑓𝑜𝑟 𝑎𝑛 𝑒𝑓𝑓𝑒𝑐𝑡, 𝑆𝑆 =
𝑛2𝑘
𝑌 = 𝛽0 + 𝛽1 𝑥1 + 𝜖
𝑒𝑓𝑓𝑒𝑐𝑡 𝑦̅+ − 𝑦̅−
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑎𝑛𝑑 𝑒𝑓𝑓𝑒𝑐𝑡, 𝛽̂ =
=
2
2
𝑀𝑎𝑖𝑛 𝐸𝑓𝑓𝑒𝑐𝑡 𝑜𝑓 𝐹𝑎𝑐𝑡𝑜𝑟 𝐵: 𝐵 = 𝑦̅𝐵+ − 𝑦̅𝐵− =
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝑜𝑓 𝑎 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝛽̂ =
𝑡 − 𝑠𝑡𝑎𝑡𝑖𝑠𝑡𝑖𝑐 𝑓𝑜𝑟 𝑎 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡, 𝑡 =
𝜎̂
1
1
1
√ 𝑘−1 + 𝑘−1 = 𝜎̂√ 𝑘
2 𝑛2
𝑛2
𝑛2
𝛽̂
𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑒𝑟𝑟𝑜𝑟 𝛽̂
=
(𝑦̅+ − 𝑦̅− )/2
1
𝑛2𝑘
𝜎̂√
2k Factorial Designs for k  3 factors
1
[𝑎 + 𝑎𝑏 + 𝑎𝑐 + 𝑎𝑏𝑐 − (1) − 𝑏 − 𝑐 − 𝑏𝑐]
4𝑛
1
[𝑏 + 𝑎𝑏 + 𝑏𝑐 + 𝑎𝑏𝑐 − (1) − 𝑎 − 𝑐 − 𝑎𝑐]
𝐵 = 𝑦̅𝐵+ − 𝑦̅𝐵− =
4𝑛
1
[𝑐 + 𝑎𝑐 + 𝑏𝑐 + 𝑎𝑏𝑐 − (1) − 𝑎 − 𝑏 − 𝑎𝑏]
𝐶 = 𝑦̅𝐶+ − 𝑦̅𝐶− =
4𝑛
𝐴 = 𝑦̅𝐴+ − 𝑦̅𝐴− =
1
[𝑎𝑏𝑐 − 𝑏𝑐 + 𝑎𝑏 − 𝑏 − 𝑎𝑐 + 𝑐 − 𝑎 + (1)]
4𝑛
1
[(𝑎) − 𝑎 + 𝑏 − 𝑎𝑏 − 𝑐 + 𝑎𝑐 − 𝑏𝑐 + 𝑎𝑏𝑐]
𝐴𝐶 =
4𝑛
1
[(1) + 𝑎 − 𝑏 − 𝑎𝑏 − 𝑐 − 𝑎𝑐 + 𝑏𝑐 + 𝑎𝑏𝑐]
𝐵𝐶 =
4𝑛
1
[𝑎𝑏𝑐 − 𝑏𝑐 − 𝑎𝑐 + 𝑐 − 𝑎𝑏 + 𝑏 + 𝑎 − (1)]
𝐴𝐵𝐶 =
4𝑛
𝑌 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + 𝛽12 𝑥1 𝑥2 + 𝜖
𝐴𝐵 =
2
2
𝑆𝑆𝐶𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒 =
𝑛𝐹 𝑛𝐶 (𝑦̅𝐹 − 𝑦̅𝐶 )
=
𝑛𝐹 + 𝑛𝐶
𝑦̅𝐹 − 𝑦̅𝐶
1
1
√ +
( 𝑛𝐹 𝑛𝐶 )
𝐹𝑖𝑟𝑠𝑡 𝑜𝑟𝑑𝑒𝑟 𝑚𝑜𝑑𝑒𝑙: 𝑌 = 𝛽0 + 𝛽1 𝑥1 + 𝛽2 𝑥2 + 𝛽𝑘 𝑥𝑘 + 𝜖
𝑘
𝑘
𝑆𝑒𝑐𝑜𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 𝑚𝑜𝑑𝑒𝑙: 𝑌 = 𝛽0 + ∑ 𝛽𝑖 𝑥𝑖 + ∑ 𝛽𝑖𝑖 𝑥𝑖2 + ∑ ∑ 𝛽𝑖𝑗 𝑥𝑖 𝑥𝑗 + 𝜖
𝑖=1
𝑖=1
𝑖<𝑗
𝑘
𝑆𝑡𝑒𝑒𝑝𝑒𝑠𝑡 𝐴𝑠𝑐𝑒𝑛𝑡 𝑦̂ = 𝛽̂0 + ∑ 𝛽̂𝑖 𝑥𝑖
𝑘
𝑖=1
𝑘
𝐹𝑖𝑡𝑡𝑒𝑑 𝑆𝑒𝑐𝑜𝑛𝑑 𝑜𝑟𝑑𝑒𝑟 𝑚𝑜𝑑𝑒𝑙 𝑦̂ = 𝛽̂0 + ∑ 𝛽̂𝑖 𝑥𝑖 + ∑ 𝛽̂𝑖𝑖 𝑥𝑖2 + ∑ ∑ 𝛽̂𝑖𝑗 𝑥𝑖 𝑥𝑗
𝑖=1
𝑖=1
𝑖<𝑗
14
WAITING LINES AND QUEUING THEORY MODELS
Single-Channel Model, Poisson Arrivals, Exponential Multichannel Model, Poisson Arrivals, Exponential
Service Times (M/M/1)
Service Times (M/M/m or M/M/s)
m = number of channels open
= mean number of arrivals per time period (arrival
rate)
= average arrival rate
= mean number of customers or units served per time  = average service rate at each channel
period (service rate)
The probability that there are zero customers in the
The average number of customers or units in the system, system
L
1
P0 
for m  
n
m

n = m –1






1

1

m

L
     
 
 
 n = 0 n!     m!    m  
The average time a customer spends in the system, W
The average number of customers or units in the system
1
W
 ( /  )m

 
L
P 
2 0
(m – 1)!(m   )

The average number of customers in the queue, Lq
The average time a unit spends in the waiting line or
2
being served, in the system
Lq 
 (   )
 ( /  )m
1 L
W
P  
The average time a customer spends waiting in the
2 0
(m – 1)!(m   )
 
queue, Wq
The average number of customers or units in line

Wq 
waiting for service
 (   )

Lq  L 
The utilization factor for the system,  (rho), the

probability the service facility is being used
The average number of customers or units in line


waiting for service

1 L
Wq  W   q
The percent idle time, P0, or the probability no one is in
 
the system
The
average
number of customers or units in line

P0  1 
waiting for service (Utilization rate)

The probability that the number of customers in the
system is greater than k, Pn>k
Pn>k

  



m
k 1
𝜆 𝑛
𝐶 = ( ) = 𝜌𝑛
𝜇
𝑃0 = 1 − 𝜌
𝑃𝑛 = (1 − 𝜌)𝜌𝑛
Finite Population Model
(M/M/1 with Finite Source)
= mean arrival rate
 = mean service rate
N = size of the population
Probability that the system is empty
Total service cost = (Number of channels) x (Cost per
channel)
Total service cost = mCs
m = number of channels
Cs = service cost (labor cost) of each channel
Total waiting cost = (Total time spent waiting by all
arrivals) x (Cost of waiting)
15
P0 
1
N
N!

 (N – n)!   
n
n 0
Average length of the queue
 
Lq  N  
1 – P0 
  
Average number of customers (units) in the system
L  Lq  1 – P0 
= (Number of arrivals) x (Average wait per arrival)Cw
= (W)Cw
Total waiting cost (based on time in queue) = (Wq)Cw
Total cost = Total service cost + Total waiting cost
Total cost = mCs + WCw
Total cost (based on time in queue) = mCs + WqCw
Average waiting time in the queue
Lq
Wq 
(N – L)
Average time in the system
1
W  Wq 

Probability of n units in the system
n
N!   
  P for n  0,1,..., N
Pn 
N – n !    0
Constant Service Time Model (M/D/1)
Average length of the queue
Lq 
2
2 (    )
Average waiting time in the queue
Wq 

2 (    )
Little’s Flow Equations
L = W
(or W = L/)
Lq = Wq
(or Wq = Lq/)
Average time in system = average time in queue +
average time receiving service
W = Wq + 1/
Average number of customers in the system
L  Lq 


Average time in the system
1
W  Wq 

N(t) = Number of customers in queuing system at time t (t >= 0)
Pn(t) = Probability of exactly n customers in queuing system at time t, given number at time 0.
s = number of servers (parallel service channels) in queuing system
n= mean arrival rate (expected number of arrivals per unit time) of new customers when n customers are in
system
= expected interarrival time
 n= mean service rate for overall system (expected number of customers completing service per unit time) when n
customers are in system. Represents combined rate at which all busy servers (those serving customers) achieve
service completions
= expected service time
Utilization factor =  = /(s)
Steady-state condition
𝑃𝑛 = 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑜𝑓 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑛 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠𝑖𝑛 𝑞𝑢𝑒𝑢𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚
16
∞
𝐿 = 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠 𝑖𝑛 𝑞𝑢𝑒𝑢𝑖𝑛𝑔 𝑠𝑦𝑠𝑡𝑒𝑚 = ∑ 𝑛 𝑃𝑛
𝑛=0
∞
𝐿𝑞 = 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑞𝑢𝑒𝑢𝑒 𝑙𝑒𝑛𝑔𝑡ℎ (𝑒𝑥𝑐𝑙𝑢𝑑𝑒𝑠 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟𝑠 𝑏𝑒𝑖𝑛𝑔 𝑠𝑒𝑟𝑣𝑒𝑑) = ∑(𝑛 − 𝑠) 𝑃𝑛
𝑛=𝑠
𝑊 = 𝐸(𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 (𝑖𝑛𝑐𝑙𝑢𝑑𝑒𝑠 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑡𝑖𝑚𝑒)𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟)
𝑊𝑞 = 𝐸(𝑤𝑎𝑖𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒 𝑖𝑛 𝑞𝑢𝑒𝑢𝑒 (𝑒𝑥𝑐𝑙𝑢𝑑𝑒𝑠 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑡𝑖𝑚𝑒)𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑐𝑢𝑠𝑡𝑜𝑚𝑒𝑟)
𝐿𝑖𝑡𝑡𝑙𝑒 ′ 𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎: 𝐿 = 𝜆𝑊
𝐿𝑞 = 𝜆𝑊𝑞
1
𝑊 = 𝑊𝑞 +
𝜇
Impact of Exponential distribution on Queuing Model
𝛼𝑒 −𝛼𝑡 𝑓𝑜𝑟 𝑡 ≥ 0
𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑠𝑡𝑟𝑖𝑏𝑡𝑢𝑖𝑜𝑛′ 𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑓𝑇 (𝑡) = {
0
𝑓𝑜𝑟 𝑡 < 0
−𝛼𝑡
−𝛼𝑡
𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠: 𝑃{𝑇 ≤ 𝑡} = 1 − 𝑒 ; 𝑃{𝑇 > 𝑡} = 𝑒
𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 1: 𝑓𝑇 (𝑡) 𝑖𝑠 𝑎 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡 (𝑡 ≥ 0); 𝑠𝑜 𝑃{0 ≤ 𝑇 ≤ ∆𝑡} > 𝑃{𝑡 ≤ 𝑇 ≤ 𝑡 + ∆𝑡}
𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 2: 𝐿𝑎𝑐𝑘 𝑜𝑓 𝑚𝑒𝑚𝑜𝑟𝑦; 𝑠𝑜 𝑃{𝑇 > 𝑡 + ∆𝑡 | 𝑇 > ∆𝑡} = 𝑃{𝑇 > 𝑡}
𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 3: 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 𝑜𝑓 𝑠𝑒𝑣𝑒𝑟𝑎𝑙 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑟𝑎𝑛𝑑𝑜𝑚 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠
ℎ𝑎𝑠 𝑎𝑛 𝑒𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛.
𝑈 = 𝑚𝑖𝑛{𝑇1 , 𝑇2 , … , 𝑇𝑛 }
𝑛
𝑃{𝑈 > 𝑇} = 𝑒𝑥𝑝 (− ∑ 𝛼𝑖 𝑡)
𝑛
𝑖=1
𝛼 = ∑ 𝛼𝑖
𝑖=1
𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 4: 𝑅𝑒𝑙𝑎𝑡𝑖𝑜𝑛𝑠ℎ𝑖𝑝 𝑡𝑜 𝑃𝑜𝑖𝑠𝑠𝑜𝑛 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑛 −𝛼𝑡
(𝛼𝑡) 𝑒
𝑃{𝑋(𝑡) = 𝑛} =
, 𝑓𝑜𝑟 𝑛 = 0, 1, 2, … ; 𝐴𝑙𝑠𝑜 𝑚𝑒𝑎𝑛 𝐸{𝑋(𝑡)} = 𝛼𝑡; 𝑤ℎ𝑒𝑟𝑒 𝛼 𝑖𝑠 𝑚𝑒𝑎𝑛 𝑟𝑎𝑡𝑒
𝑛!
𝑋(𝑡) = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑜𝑐𝑐𝑢𝑟𝑟𝑒𝑛𝑐𝑒𝑠 𝑏𝑦 𝑡𝑖𝑚𝑒 𝑡 (𝑡 ≥ 0)
𝑃𝑟𝑜𝑝𝑒𝑟𝑡𝑦 5: 𝐹𝑜𝑟 𝑎𝑙𝑙 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑣𝑎𝑙𝑢𝑒𝑠 𝑜𝑓 𝑡, 𝑃{𝑇 ≤ 𝑡 + ∆𝑡|𝑇 > 𝑡} ≈ 𝛼∆𝑡, 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 ∆𝑡
𝐵𝑖𝑟𝑡ℎ − 𝑎𝑛𝑑 − 𝐷𝑒𝑎𝑡ℎ 𝑃𝑟𝑜𝑐𝑒𝑠𝑠
𝜆𝑛−1 𝜆𝑛−2 ⋯ 𝜆0
𝐶𝑛 =
, 𝑓𝑜𝑟 𝑛 = 1, 2, …
𝜇𝑛 𝜇𝑛−1 ⋯ 𝜇1
𝐹𝑜𝑟 𝑛 = 0, 𝐶𝑛 = 1
𝑃𝑛 = 𝐶𝑛 𝑃0 , 𝑓𝑜𝑟 𝑛 = 0, 1, 2, …
∞
∑ 𝑃𝑛 = 1
𝑛=0
∞
−1
𝑃0 = (∑ 𝐶𝑛 )
𝑛=0
∞
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑟𝑎𝑡𝑒 𝑜𝑣𝑒𝑟 𝑡ℎ𝑒 𝑙𝑜𝑛𝑔 𝑟𝑢𝑛, 𝜆̅ = ∑ 𝜆𝑛 𝑃𝑛
T1, T2, … be independent service-time random variables
having an exponential distribution with parameter, and
let
Sn+1 = T1 + T2 + …+ Tn+1, for n = 0, 1, 2, …
𝑛=0
M/M/s, with s > 1
17
Sn+1 represents the conditional waiting time given n
customers already in the system. Sn+1 is known to have
an Erlang distribution.
∞
𝑃{𝐸(𝑊) > 𝑡} = ∑ 𝑃𝑛 𝑃{𝑆𝑛+1 > 𝑡} = 𝑒 −𝜇(1−𝜌)𝑡 , 𝑓𝑜𝑟 𝑡
𝑛=0
≥0
∞
𝜆 𝑛
𝜇
( )
𝐶𝑛 =
𝜆 𝑠
𝜇
( )
{
𝑠!
𝜆
(𝑠𝜇)
=
𝜆 𝑛
𝜇
𝑠!𝑠𝑛−𝑠
( )
𝑓𝑜𝑟 𝑛 = 𝑠, 𝑠 + 1, …
𝑠−1
(𝜆/𝜇)𝑛 (𝜆/𝜇)𝑠
1
𝑃0 = 1⁄[∑
+
]
𝑛!
𝑠! 1 − 𝜆/(𝑠𝜇)
𝑛=0
𝑃{𝐸(𝑊𝑞 ) > 𝑡} = ∑ 𝑃𝑛 𝑃{𝑆𝑛 > 𝑡} = 𝜌𝑒 −𝜇(1−𝜌)𝑡 , 𝑓𝑜𝑟 𝑡
𝑛=0
≥0
𝑃{𝐸(𝑊𝑞 ) > 𝑡|𝐸(𝑊𝑞 ) > 0} = 𝑒 −𝜇(1−𝜌)𝑡 , 𝑓𝑜𝑟 𝑡 ≥ 0
𝑓𝑜𝑟 𝑛 = 1, 2, … , 𝑠
𝑛!
𝑛−𝑠
𝑃𝑛 =
𝜆 𝑛
(𝜇 )
𝑃
𝑖𝑓 0 ≤ 𝑛 ≤ 𝑠
𝑛! 0
𝜆 𝑛
(𝜇 )
𝑖𝑓 𝑛 ≥ 𝑠
{𝑠! 𝑠 𝑛−𝑠 𝑃0
𝑃0 (𝜆/𝜇)𝑠 𝜌
𝐿𝑞 =
𝑠! (1 − 𝜌)2
𝐿𝑞
𝑊𝑞 =
𝜆
1
𝑊 = 𝑊𝑞 +
𝜇
𝜆
𝐿 = 𝐿𝑞 +
𝜇
𝑃{𝐸(𝑊) > 𝑡} = 𝑒 −𝜇𝑡 [1
𝜆
−𝜇𝑡(𝑠−1− )
𝜇
𝑃0 (𝜆/𝜇)𝑠 1 − 𝑒
+
(
)]
𝑠! (1 − 𝜌)
𝑠 − 1 − 𝜆/𝜇
𝑃{𝐸(𝑊𝑞 ) > 𝑡} = (1 − 𝑃{𝐸(𝑊𝑞 ) = 0})𝑒 −𝑠𝜇(1−𝜌)𝑡
𝑠−1
𝑃{𝐸(𝑊𝑞 ) = 0} = ∑ 𝑃𝑛
𝑛=0
Finite Queue Variation of M/M/s (M/M/s/K model)
Finite Queue: Number of customers in the system cannot
exceed a specified number, K.
Queue capacity is K-s
𝜆 𝑓𝑜𝑟 𝑛 = 0, 1, 2, … , 𝐾 − 1
𝜆𝑛 = {
0 𝑓𝑜𝑟 𝑛 ≥ 𝐾
M/M/1/K
(𝜆/𝜇)𝑛 = 𝜌𝑛 𝑓𝑜𝑟 𝑛 = 1, 2, … , 𝐾
𝐶𝑛 = {
0
𝑓𝑜𝑟 𝑛 > 𝐾
1−𝜌
𝑃0 =
1 − 𝜌𝐾+1
1−𝜌
𝑃𝑛 =
𝜌𝑛 , 𝑓𝑜𝑟 𝑛 = 0, 1, 2, … , 𝐾
𝐾+1
1−𝜌
𝜌
(𝐾 + 1)𝜌𝐾+1
𝐿=
−
1−𝜌
1 − 𝜌𝐾+1
𝐿𝑞 = 𝐿 − (1 − 𝑃0 )
Finite Calling Population variation of M/M/s
M/M/1 with finite population
𝑁!
𝜆 𝑛
( )
𝑓𝑜𝑟 𝑛 ≤ 𝑁
𝐶𝑛 = {(𝑁 − 𝑛)! 𝜇
0
𝑓𝑜𝑟 𝑛 > 𝑁
𝑁
𝑁!
𝜆 𝑛
𝑃0 = 1⁄∑ [
( ) ]
(𝑁 − 𝑛)! 𝜇
𝑛=0
𝑁!
𝜆 𝑛
𝑃𝑛 =
( ) 𝑃0 , 𝑖𝑓 𝑛 = 1, 2, … , 𝑁
(𝑁 − 𝑛)! 𝜇
𝑁
𝐿𝑞 = ∑(𝑛 − 1)𝑃0 = 𝑁 −
𝑛=1
𝜆+𝜇
(1 − 𝑃0 )
𝜇
𝜇
𝐿 = 𝑁 − (1 − 𝑃0 )
𝜆
𝐿
𝑊=
𝜆̅
18
𝐿𝑞
𝜆̅
𝐿
𝑊=
𝜆̅
𝜆̅ = ∑ 𝜆𝑛 𝑃𝑛 = 𝜆(𝑁 − 𝐿)
𝜆̅ = ∑ 𝜆𝑛 𝑃𝑛 = 𝜆(1 − 𝑃𝐾 )
M/M/s with finite population and s > 1
𝑊𝑞 =
∞
𝑊𝑞 =
∞
𝑛=0
𝐶𝑛 =
𝑛=0
M/M/s/K
𝑁!
𝜆 𝑛
(
)
(𝑁−𝑛)!𝑛! 𝜇
(𝜆/𝜇)𝑛
𝑓𝑜𝑟 𝑛 = 1, 2, … , 𝑠
𝑛!
𝑛−𝑠
(𝜆/𝜇)𝑛
= (𝜆/𝜇)𝑠 𝜆
( )
=
𝑠!
𝑠𝜇
𝑠! 𝑠 𝑛−𝑠 𝑓𝑜𝑟 𝑛 = 𝑠, 𝑠 + 1, … , 𝐾
{
0 𝑓𝑜𝑟 𝑛 > 𝐾
𝑛
𝜆 𝑠 𝐾
𝑠 (𝜆 )
(𝜇 )
𝜆 𝑛−𝑠
𝜇
𝑃0 = 1⁄[∑
+
∑ ( ) ]
𝑛!
𝑠!
𝑠𝜇
𝑁!
𝜆 𝑛
(
)
(𝑁−𝑛)!𝑠!𝑠𝑛−𝑠 𝜇
𝐶𝑛
𝑛=0
𝜆
(𝜇 )
𝑛=𝑠+1
𝑛
𝑓𝑜𝑟 𝑛 = 0, 1, 2, … , 𝑠
𝑓𝑜𝑟 𝑛 = 𝑠, 𝑠 + 1, … , 𝐾
0 𝑓𝑜𝑟 𝑛 > 𝐾
{
𝑁!
𝜆 𝑛
𝑁!
𝜆 𝑛
𝑁
𝑃0 = 1⁄[∑𝑠−1
𝑛=0 (𝑁−𝑛)!𝑛! (𝜇 ) + ∑𝑛=𝑠 (𝑁−𝑛)!𝑠!𝑠𝑛−𝑠 (𝜇 ) ]
𝑁!
𝜆 𝑛
( ) 𝑃0
(𝑁 − 𝑛)! 𝑛! 𝜇
𝑃𝑛 =
𝑁!
𝜆 𝑛
( ) 𝑃0
(𝑁 − 𝑛)! 𝑠! 𝑠 𝑛−𝑠 𝜇
0 𝑖𝑓 𝑛 > 𝑁
{
𝑖𝑓 0 ≤ 𝑛 ≤ 𝑠
𝑖𝑓 𝑠 ≤ 𝑛 ≤ 𝑁
𝑁
𝑃
𝑓𝑜𝑟 𝑛 = 1, 2, … , 𝑠
𝑛! 0
𝑃𝑛 =
𝜆 𝑛
(𝜇 )
𝑓𝑜𝑟 𝑛 = 𝑠, 𝑠 + 1, … , 𝐾
𝑃
𝑠! 𝑠 𝑛−𝑠 0
{0 𝑓𝑜𝑟 𝑛 > 𝐾
𝑃0 (𝜆/𝜇)𝑠 𝜌
𝐿𝑞 =
[1 − 𝜌𝐾−𝑠 − (𝐾 − 𝑠)𝜌𝐾−𝑠 (1−)]
𝑠! (1 − 𝜌)2
𝑠−1
𝐿𝑞 = ∑(𝑛 − 𝑠)𝑃𝑛
𝑛=𝑠
𝑠−1
𝑛=0
𝑠−1
𝐿𝑞
𝑊𝑞 =
𝜆̅
𝐿
𝑊=
𝜆̅
M/G/1 Model
𝑃0 = 1 − 𝜌
∞
𝐿
𝑊=
𝜆̅
𝐿𝑞
𝑊𝑞 =
𝜆̅
𝑛=0
𝜆̅ = ∑ 𝜆𝑛 𝑃𝑛 = 𝜆(𝑁 − 𝐿)
𝑛=0
𝑃𝑜𝑙𝑙𝑎𝑐𝑧𝑒𝑘 − 𝐾ℎ𝑖𝑛𝑡𝑐ℎ𝑖𝑛𝑒 𝐹𝑜𝑟𝑚𝑢𝑙𝑎, 𝐿𝑞 =
𝑠−1
𝐿 = ∑ 𝑛𝑃𝑛 + 𝐿𝑞 + 𝑠 (1 − ∑ 𝑃𝑛 )
𝐿 = ∑ 𝑛𝑃𝑛 + 𝐿𝑞 + 𝑠 (1 − ∑ 𝑃𝑛 )
𝑛=0
𝐿𝑞
𝜆̅
𝑛=0
M/D/s Model
2 2
2
𝜆 𝜎 +𝜌
2(1 − 𝜌)
𝐿 = 𝜌 + 𝐿𝑞
𝐿𝑞
𝑊𝑞 =
𝜆
1
𝑊 = 𝑊𝑞 +
𝜇
𝐹𝑜𝑟 𝑎𝑛𝑦 𝑓𝑖𝑥𝑒𝑑 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑡𝑖𝑚𝑒 1⁄𝜇 ,
𝑛𝑜𝑡𝑖𝑐𝑒 𝑡ℎ𝑎𝑡 𝐿𝑞 , 𝐿, 𝑊𝑞 𝑎𝑛𝑑 𝑊 𝑎𝑙𝑙 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒
𝑎𝑠 𝜎 2 𝑖𝑠 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑑.
M/Ek/s Model
𝐸𝑟𝑙𝑎𝑛𝑔 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛′ 𝑠 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛
𝐹𝑜𝑟 𝑀/𝐷/1 𝑚𝑜𝑑𝑒𝑙, 𝐿𝑞 =
𝜌2
2(1 − 𝜌)
Nonpreemptive Priorities Model
𝑆𝑡𝑒𝑎𝑑𝑦 − 𝑠𝑡𝑎𝑡𝑒 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑤𝑒𝑖𝑔ℎ𝑡𝑖𝑛𝑔 𝑡𝑖𝑚𝑒
𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 (𝑖𝑛𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑡𝑖𝑚𝑒)𝑓𝑜𝑟 𝑎
19
(𝜇𝑘)𝑘 𝑘−1 −𝑘𝜇𝑡
𝑡 𝑒
, 𝑓𝑜𝑟 𝑡 ≥ 0
(𝑘 − 1)!
𝜇 𝑎𝑛𝑑 𝑘 𝑎𝑟𝑒 𝑠𝑡𝑟𝑖𝑐𝑡𝑙𝑦 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒
𝐴𝑁𝐷 𝑘 𝑚𝑢𝑠𝑡 𝑏𝑒 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
𝐸𝑥𝑐𝑙𝑢𝑑𝑖𝑛𝑔 𝑖𝑛𝑡𝑒𝑔𝑒𝑟 𝑟𝑒𝑠𝑡𝑟𝑖𝑐𝑡𝑖𝑜𝑛,
𝐸𝑟𝑙𝑎𝑛𝑔 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑔𝑎𝑚𝑚𝑎
1
𝑀𝑒𝑎𝑛 =
𝜇
1 1
𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 =
√𝑘 𝜇
T1, T2, …, Tk are k independent random variables with
an identical exponential distribution whose mean is
1/(k).
T = T1 + T2 + …+ Tk, has an Erlang distribution with
parameters  and k. Exponential and degenerate
(constant) are special cases of Erlang distribution with
k=1 and k=respectively.
M/Ek/1 Model
2
𝜆 /(𝑘𝜇 2 ) + 𝜌2 1 + 𝑘
𝜆2
𝐿𝑞 =
=
2(1 − 𝜌)
2𝑘 𝜇(𝜇 − 𝜆)
𝐿 = 𝜆𝑊
1+𝑘
𝜆
𝑊𝑞 =
2𝑘 𝜇(𝜇 − 𝜆)
1
𝑊 = 𝑊𝑞 +
𝜇
𝑓(𝑡) =
𝑚𝑒𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑦 𝑐𝑙𝑎𝑠𝑠 𝑘, 𝑊𝑘
1
1
𝑊𝑘 =
+ , 𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
𝐴𝐵𝑘−1 𝐵𝑘 𝜇
𝑠−1
𝑠𝜇 − 𝜆
𝑟𝑗
𝑊ℎ𝑒𝑟𝑒 𝐴 = 𝑠!
∑
+ 𝑠𝜇
𝑟𝑠
𝑗!
𝑗=0
𝐵0 = 1
∑𝑘𝑖=1 𝜆𝑖
𝐵𝑘 = 1 −
𝑠𝜇
𝑠 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑒𝑟𝑣𝑒𝑟𝑠
𝜇 = 𝑚𝑒𝑎𝑛 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑟𝑎𝑡𝑒 𝑝𝑒𝑟 𝑏𝑢𝑠𝑦 𝑠𝑒𝑟𝑣𝑒𝑟
𝜆𝑖 = 𝑚𝑒𝑎𝑛 𝑎𝑟𝑟𝑖𝑣𝑎𝑙 𝑟𝑎𝑡𝑒 𝑓𝑜𝑟 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑦 𝑐𝑙𝑎𝑠𝑠 𝑖
𝑁
𝜆 = ∑ 𝜆𝑖
𝑖=1
𝑟=
𝜆
𝜇
𝑘
∑ 𝜆𝑖 < 𝑠𝜇
𝑖=1
𝐿𝑘 = 𝜆𝑘 𝑊𝑘 , 𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
𝐴 = 𝑠!
𝑭𝒐𝒓 𝒔𝒊𝒏𝒈𝒍𝒆 𝒔𝒆𝒓𝒗𝒆𝒓, 𝒔 = 𝟏
𝐴 = 𝜇 2 ⁄𝜆
𝑾𝒊𝒕𝒉 𝒅𝒊𝒇𝒇𝒆𝒓𝒆𝒏𝒕 𝒆𝒙𝒑𝒐𝒏𝒆𝒏𝒕𝒊𝒂𝒍 𝒕𝒊𝒎𝒆𝒔
𝜇𝑘 = 𝑚𝑒𝑎𝑛 𝑠𝑒𝑟𝑣𝑖𝑐𝑒 𝑟𝑎𝑡𝑒 𝑓𝑜𝑟 𝑝𝑟𝑖𝑜𝑟𝑖𝑡𝑦 𝑐𝑙𝑎𝑠𝑠 𝑘,
𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
𝑎𝑘
1
𝑊𝑘 =
+ , 𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
𝑏𝑘−1 𝑏𝑘 𝜇𝑘
𝑘
𝑊ℎ𝑒𝑟𝑒 𝑎𝑘 = ∑
𝑖=1
𝜆𝑖
𝜇𝑖2
𝑏0 = 1
𝑘
𝑏𝑘 = 1 − ∑
𝑘
∑
𝑖=1
Jackson Networks
With m service facilities where facility i (i=1, 2, .,., m)
1. Infinite Queue
2. Customers arriving from outside the system
according to a Poisson input process with
parameters ai
3. si servers with an exponential service-time
distribution with parameter 
𝑖=1
𝜆𝑖
𝜇𝑖
𝜆𝑖
<1
𝜇𝑖
Preemptive Priorities Model
1/𝜇
𝐹𝑜𝑟 𝑠 = 1,
𝑊𝑘 =
, 𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
𝐵𝑘−1 𝐵𝑘
𝐿𝑘 = 𝜆𝑘 𝑊𝑘 , 𝑓𝑜𝑟 𝑘 = 1, 2, … , 𝑁
20
A Customer leaving facility i is routed next to facility j
(j= 1, 2, …, m) with probability pij or departs the system
with probability
𝑚
𝑞𝑖 = 1 − ∑ 𝑝𝑖𝑗
𝑗=1
Jackson network behaves as if it were an independent
M/M/s queueing system with arrival rate
𝑚
𝜆𝑗 = 𝑎𝑗 + ∑ 𝜆𝑖 𝑝𝑖𝑗
𝑖=1
𝑤ℎ𝑒𝑟𝑒 𝑠𝑗 𝜇𝑗 > 𝜆𝑗
𝜆𝑖
𝜌𝑖 =
𝑠𝑖 𝜇𝑖
21
MARKOV ANALYSIS
 (i) =
vector of state probabilities for period Pij = conditional probability of being in state j in the
future given the current state of i
i
=
(1, 2, 3, … , n)
 P11 P12  P1n 
P
where
P22  P2 n 
21

P
n
=
number of states
 
   
1, 2, … , n =
probability of being in state 1,


state 2, …, state n
 Pm1 Pm 2  Pmn 
For any period n we can compute the state
probabilities for period n + 1
 (n + 1) =  (n)P
Fundamental Matrix
F = (I – B)–1
Inverse of Matrix
a b 
P

c d 
P -1
a b 


c d 
1
 d

 r
c

 r
 b
r 
a 

r 
r = ad – bc
Partition of Matrix for absorbing states
 I O
P

 A B
I = identity matrix
O = a matrix with all 0s
Equilibrium condition
 = P
M represent the amount of money that is in each of the
nonabsorbing states
M = (M1, M2, M3, … , Mn)
n
= number of nonabsorbing states
M1
= amount in the first state or category
M2
= amount in the second state or category
Mn
= amount in the nth state or category
Computing lambda and the consistency index
 n
CI 
n 1
Consistency Ratio
CI
CR 
RI
Stochastic process {Xt} (t = 0, 1, …) is a Markov chain
if it has the Markovian property.
Stochastic process {Xt} is said to have the
Markovian property if P{Xt+1 = j | X0 = k0, X1 = k1,
…, Xt-1 = kt-1, Xt = i} = P{Xt+1=j|Xt=i}, for t=0,1, …
and every sequence i, j, k0, k1, …, kt-1.
Pij = P{Xt+1 = j | Xt = i}
n-step transition probabilities:
(𝑛)
𝑃𝑖𝑗 = 𝑃{𝑋𝑡+𝑛 = 𝑗|𝑋𝑡 = 𝑖}
(𝑛)
𝑃𝑖𝑗
≥ 0, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 𝑎𝑛𝑑 𝑗; 𝑛 = 0, 1, 2, …
𝑀
(𝑛)
∑ 𝑃𝑖𝑗
= 1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖; 𝑛 = 0, 1, 2, …
𝑗=0
n-step transition matrix:
𝑆𝑡𝑎𝑡𝑒
0
1 … 𝑀
(𝑛)
(𝑛)
(𝑛)
𝑃00
𝑃01 … 𝑃0𝑀
0
(𝑛)
(𝑛)
(𝑛)
𝑃(𝑛) = 1
𝑃11
𝑃10
… 𝑃1𝑀
⋮
…
… … …
(𝑛)
(𝑛)
(𝑛)
𝑀
[𝑃𝑀0 𝑃𝑀1 … 𝑃𝑀𝑀 ]
𝐶ℎ𝑎𝑝𝑚𝑎𝑛 − 𝐾𝑜𝑙𝑚𝑜𝑔𝑜𝑟𝑜𝑣 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛𝑠
22
𝑀
(𝑛)
𝑃𝑖𝑗
(𝑚)
= ∑ 𝑃𝑖𝑘
(𝑛−𝑚)
𝑃𝑘𝑗
, 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 = 0, 1, … , 𝑀; 𝑗 = 0, 1, … , 𝑀; 𝑚 = 1, 2, … , 𝑛 − 1; 𝑛 = 𝑚 + 1, 𝑚 + 2, …
𝑘=0
𝑃 (𝑛) = 𝑃𝑃(𝑛−1) = 𝑃 (𝑛−1) 𝑃 = 𝑃𝑃𝑛−1 = 𝑃𝑛−1 𝑃 = 𝑃𝑛
(𝑛)
(𝑛)
(𝑛)
𝑈𝑛𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑎𝑙 𝑠𝑡𝑎𝑡𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑖𝑒𝑠: 𝑃{𝑋𝑛 = 𝑗} = 𝑃{𝑋0 = 0}𝑃0𝑗 + 𝑃{𝑋0 = 1}𝑃1𝑗 + ⋯ + 𝑃{𝑋0 = 𝑀}𝑃𝑀𝑗
(𝑛)
lim 𝑝
𝑛→∞ 𝑖𝑗
𝑛
= 𝜋𝑗 > 0
1
(𝑘)
lim ( ∑ 𝑝𝑖𝑗 ) = 𝜋𝑗
𝑛→∞ 𝑛
𝑘=1
𝑀
𝜋𝑗 = ∑ 𝜋𝑖 𝑝𝑖𝑗 , 𝑓𝑜𝑟 𝑗 = 0, 1, … , 𝑀
𝑖=0
𝑀
∑ 𝜋𝑗 = 1
𝑗=0
𝑛
𝑀
𝑡=1
𝑗=0
1
lim 𝐸 [ ∑ 𝐶(𝑋𝑡 )] = ∑ 𝜋𝑗 𝐶(𝑗)
𝑛→∞
𝑛
𝐹𝑖𝑟𝑠𝑡 𝑃𝑎𝑠𝑠𝑎𝑔𝑒 𝑇𝑖𝑚𝑒
(1)
(1)
𝑓𝑖𝑗 = 𝑝𝑖𝑗 = 𝑝𝑖𝑗
(2)
(1)
𝑓𝑖𝑗 = ∑ 𝑝𝑖𝑘 𝑓𝑘𝑗
𝑘≠𝑗
(𝑛)
𝑓𝑖𝑗
(𝑛−1)
= ∑ 𝑝𝑖𝑘 𝑓𝑘𝑗
∞
𝑘≠𝑗
(𝑛)
∑ 𝑓𝑖𝑗
≤1
𝑛=1
𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑓𝑖𝑟𝑠𝑡 𝑝𝑎𝑠𝑠𝑎𝑔𝑒 𝑡𝑖𝑚𝑒 𝑓𝑟𝑜𝑚 𝑠𝑡𝑎𝑡𝑒 𝑖 𝑡𝑜 𝑠𝑡𝑎𝑡𝑒 𝑗
∞
(𝑛)
∞ 𝑖𝑓 ∑ 𝑓𝑖𝑗
𝜇𝑖𝑗 =
∞
∑
{𝑛=1
∞
(𝑛)
𝑖𝑓 ∑ 𝑓𝑖𝑗
𝑛=1
(𝑛)
𝑛𝑓𝑖𝑗
<1
∞
(𝑛)
𝑖𝑓 ∑ 𝑓𝑖𝑗
=1
𝑛=1
= 1, 𝑡ℎ𝑒𝑛 𝜇𝑖𝑗 = 1 + ∑ 𝑝𝑖𝑘 𝜇𝑘𝑗
𝑛=1
𝑘≠𝑗
𝐴𝑏𝑠𝑜𝑟𝑏𝑖𝑛𝑔 𝑆𝑡𝑎𝑡𝑒𝑠
𝑀
𝑓𝑖𝑘 = ∑ 𝑝𝑖𝑗 𝑓𝑗𝑘 , 𝑓𝑜𝑟 𝑖 = 0, 1, … , 𝑀,
𝑗=0
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛𝑠
𝑓𝑘𝑘 = 1
𝑓𝑖𝑘 = 0,
𝑖𝑓 𝑠𝑡𝑎𝑡𝑒 𝑖 𝑖𝑠 𝑟𝑒𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑎𝑛𝑑 𝑖 ≠ 𝑘
23
Name
When to Use
Approximations / Conditions
Probability Mass function, Mean and Variance
E(X) is Expected Value = Mean; Xi = random variable’s possible values; P(Xi) = Probability of each of the random variable’s possible values
n
n
i 1
i 1
E X      X i PX i  Variance   2   [X i  E (X)] 2 P(X i ) Standard Deviation   Variance
Uniform
(Discrete)
Equal probability
Finite number of possible values
Bernoulli Trials:
 Each trial is independent
 Probability of success in a trial is constant
 Only two possible outcomes
Unknown: Number of successes
Known: Number of trials
 Number of trials that result in a success
(b  a)
2
 2  Variance 
(b  a  1) 2  1
12
If n is large (np > 5, n(1-p) > 5),
approximate binomial to normal.
P(X x) = P(X  x+0.5)
P(x X) = P(x-0.5  X)
If n is large & p is small,
approximate to Poisson as = np
Binomial expansion
a  bn   
Negative
Binomial
(Discrete)
Hypergeometri
c (Discrete)
Poisson
(Discrete)
Probability of x success in n trials

n!
p x q n x
x!(n  x)!
x = 0,1,…,n, 0  p  1, n = 1,2,…
n
n!
nC x  C xn    
 x  x!(n  x)!
Expected value (mean) E(X) =  = np
Variance = V(X) = = np(1 – p)
n  r nr
 p q
k 0  r 
Bernoulli trial; Memoryless
f ( x)  (1  p) x 1 p
x = 1,2,…,n, 0  p  1
 Number of trails until first success
Expected value (mean) = E(X) =  = 1/p
Variance = V(X) = = (1 – p)/p2
Unknown: Number of trials
 x  1
1  p x r p r x = r,r+1,r+2,…, 0  p  1
f (x )  
Known: Number of success
 r  1
 Number of trials required to obtain r successes
E(X) =  = r/p
= Variance = r(1-p)/p2
n
Geometric
(Discrete)
i 1
x
For a series of n values, f(x) = 1/ n, a  b
For a range that starts from a and ends with b (a, a+1, a+2, …, b) and a  b

Binomial /
Bernoulli
(Discrete)
n
Cumulative F(x) = P( X  x)   f ( x i ) ; xi
Trials are not independent
Without replacement
 Number of success in the sample
Poisson Process:
 Probability of more than one event in a
subinterval is zero.
 Probability of one event in a subinterval is
constant & proportional to length of
subinterval
 Event in each subinterval is independent
 Number of events in the interval
V(X) = V(X) of binomial * ((N-n)/(N1)) where ((N-n)/(N-1)) is called
finite population correction factor
n << N or (n/N) < 0.1,
hypergeometric is equal to
binomial.
Approximated to normal if np > 5,
n(1-p) > 5 and (n/N) < 0.1
Arrival rate does not change over
time; Arrival pattern does not
follow regular pattern; Arrival of
disjoint time intervals are
independent.
Approximated to normal if  > 5
Z  X   

 K  N  K 

f (x )   
 x  n  x 
N
 
n
x =max(0,n-N+k) to min(K,n), K  N, n  N
K objects classed as successes; N – K objects
classified as failures; Sample size of n objects
E(X) =  = np where p = K/N
 N n

 N 1 
 2  np(1  p)
f ( x)  P ( X  x ) 
 x e 
x = 0,1,2,…, 0 < 
x!
P(X) = probability of exactly X arrivals or occurrences
Expected value = Variance = 

Taylor series: e 

k
 k!
k 0
Name
When to Use
E X  

 xf ( x)dx

Approximations / Conditions
Probability Density function, Mean and
Variance
P(x1  X  x2) = P(x1  X  x2) = P(x1  X  x2) = P(x1  X  x2)

 2   ( x   ) 2 f ( x)dx
Uniform
Equal probability
(Continuous)
xa
 0

F ( x)  ( x  a) /(b  a) a  x  b
1
bx

Normal
Notation: N(,)
(Continuous) X is any random variable
Cumulative (z) = P(Z < z); Z is
standard normal
x
Cumulative F(x) = P(Xx) =

 f (u )du ; for -< x <

For a series of n values, f(x) = 1/ (b –a); where a  x  b
For a range that starts from a and ends with b (a, a+1, a+2, …, b) and a  b
( a  b)

2
(b  a)2
 Variance  V ( X ) 
12
2
If n is large (np > 5, n(1-p) > 5), normal is
approximated to binomial.
f (X) 

x  0.5  np 
P ( X  x)  P( X  x  0.5)  P Z 

np(1  p) 

𝑃(𝑥 ≤ 𝑋) = 𝑃(𝑥 − 0.5 ≤ 𝑋) ≅ 𝑃 (
𝑥−0.5−𝑛𝑝
√𝑛𝑝(1−𝑝)
𝑍)
Adding or subtracting 0.5 is called continuity
correction.
Normal is approximated to Poisson if  > 5
𝑍=
𝑋−𝜆
𝜎
≤
( x )2
1
 2
e
2 2
- < x <  - <  < , 0 < 
E(X) = 
V(X) =
Standard normal means mean = = 0 and variance =
= 1
Z
X 

X  x
P ( X  x)  P 

  P( Z  z )
 
 
Cumulative distribution of a standard normal
variable  ( z )  P( Z  z )
- <  < + = 68% -2 <  < +2 = 95%
-3 <  < +3 = 99.7%
for 0 x  f ( x )  0 for x < 0
Exponential Memoryless
f (x )  e x
(Continuous) P ( X  t1  t 2 | X  t1 )  P( X  t 2 )
P ( X  x) 1  F ( x)  e x
P( X  x)  F ( x)  1  e x
 distance between successive
1
events of Poisson process with mean P (a  X  b)  F (b)  F (a ) Expected value   = 1 = Average service time
Variance = 2


> 0
 length until first count in a
The probability that an exponentially distributed time (X) required to serve a customer is less than or equal to
Poisson process
 t
time t is given by the formula, P(X  t )  1  e
 is the mean of the Poisson
process, which is the number of
events within the interval
25
Erlang
r  shape
  scale
(Continuous) Time between events are
independent
 length until r counts in a Poisson
process or exponential distribution
For mean and variance: Exponential multiplied by r gives Erlang. Expected value   =
f (x) 
P(X>0.1) = 1 – F(0.1)
r x r 1 e  x
(r  1)!
r

Variance =
r
2
for x  0 and r  1, 2, ...
If r = 1, Erlang random variable is an exponential random variable
Gamma
For r is an integer (r=1,2,…), gamma
(Continuous) is Erlang
Erlang random variable is time until
the rth event in a Poisson process
and time between events are
independent
For  = ½, r = ½, 1, 3/2, 2, … gamma
is chi-square
Weibull
Includes memory property
(Continuous) - Scale;  - Shape
 Time until failure of many
different physical systems


Gamma Function (r )  x r 1 e  x dx, for r  0
0
f (x) 
r 1  x
x e
r
( r )
Mean    r

, for x  0,   0 and r  0 and (r )  r  1!, (1)  0! 1, (1/ 2)   1 / 2
Variance   2  r 2
E(X) and V(X) = E(X) and V(X) of exponential distribution multiplied by r
 1
=1, Weibull is identical to exponential
  x  
 x
f ( x )    exp    
=2, Weibull is identical to Raleigh
  
    
x > 0, 
Cumulative F (x )  1  e

  E ( X )   1 

1
 where (r )  r  1!
 

2
    1     2


2
Lognormal
Includes memory property
(Continuous) X = exp(W); W is normally
distributed with mean  and
variance 
ln(X) = W; X is lognormal
Easier to understand than Weibull
2
 
1 
 1  
 
 
2
Weibull can be approximated to lognormal with  and 
ln( x)   

 ln( x)   
F ( x)  P X  x   P exp(W )  x   P W  ln( x)  P  Z 
 

 for x  0
 

  
F ( X )  0, for x  0
f (x) 
 (ln x   ) 2 
exp 
 for 0  x  
2 2 
x 2

1
E ( X )  e  
Beta
Flexible but bounded over a finite
(Continuous) range

x
 
 
2
2


V ( X )  e 2  e   1
(   )  1
f (x) 
x (1  x)  1 for 0  x  1,   0,   0
( )(  )
2
2
26
E (X ) 
Power Law
Called as ‘heavy-tailed’ distribution.
(Continuous) f(x) decreases rapidly with x but not
as rapid as exponential distribution.


V (X ) 

        1
2
A random variable described by its minimum value xmin and a scale parameter > 1 is is said to obey the
power law distribution if its density function is given by
(  1)  x

f (x) 
x min  x min





Normalize the function for a given set of parameters to ensure that
 f ( x ) dx  1

Central Limit
Theorem
If X1, X2, …, Xn is a random sample of size n taken from a population (either finite or infinite) with mean  and finite variance 2, and if X is the
sample mean, the limiting form of the distribution of
Z
X 
/ n
as n  , is the standard normal distribution.
27
Name
Two or more
Discrete Random
Variables
f X (x )  P( X  x)  
y
Probability Density function, Mean and Variance
f XY ( x, y )
f Y (y )  P(Y  y)   f XY ( x, y)
f Y | x ( y )  f XY ( x, y ) / f X ( x)
x
E Y | x    y f Y | x ( y )
V (Y | x)   ( y   Y | x ) 2 f Y | x ( y )
y
y
f XY ( x, y) f X ( x) f Y ( y)

 f Y ( y); f XY ( x, y)  f X ( x) f Y ( y) for all x and y
f X ( x)
f X ( x)
Joint Probability Mass Fn: f X1 X 2 X p ( x1 , x2 , , x p )  P( X 1  x1 , X 2  x2 , , X p  x p ) for all points (x1,x2,…,xp) in the range of X1,X2,…,Xp
Independence f Y | x (y ) 
Joint Probability Mass For subset: f X1 X 2 X k ( x1 , x2 , , xk )  P( X 1  x1 , X 2  x2 , , X k  xk ) 
 P( X
1
 x1 , X 2  x2 , , X k  xk ) for
all points in the range of X1,X2,…,Xp for which X1=x1, X2=x2,…, Xk=xk
Marginal Probability Mass Function: f X i ( xi )  P( X i  xi )   f X 1 X 2 X p ( x1 , x 2 , , x p )

Multinomial
Probability
Distribution
2
f X 1 X 2  X p ( x1 , x 2 ,  , x p )
The random experiment that generates the probability distribution consists of a series of independent trials. However, the results from each trial
can be categorized into one of k classes.
P( X 1  x1 , X 2  x 2 ,  , X k  x k ) 
n!
p1x1 p 2x2  p kxk
x1 ! x 2 ! x k !
E X i   npi
Two or more
Continuous
Random Variables

Variance: V ( X i )   x i   X i
Mean: E ( X i )   xi f X1 X 2 X p ( x1 , x 2 , , x p )
for x1  x 2    x k  n and p1  p 2    p k  1
V ( X i )  npi (1  pi )
Marginal Probability Density Function: f X (x ) 
f
XY
f Y (y )   f XY ( x, y ) dy
( x, y ) dy
y
f Y | x (y ) 
f XY ( x, y)
for f X ( x)  0
f X ( x)
y
E Y | x    y f Y | x ( y ) dy
V (Y | x)   ( y   Y | x ) 2 f Y | x ( y ) dy
y
Independence: f Y | x ( y )  f Y ( y ); f X | y ( x )  f X ( x)

y
f XY ( x, y )  f X ( x) f Y ( y ) for all x and y
   f
Joint Probability Density Fn: P ( X 1  x1 , X 2  x 2 , , X p  x p )  B 
X 1 X 2 X p
( x1 , x2 , , x p )dx1 dx 2  dx p
B
Joint Probability Mass For subset: f X 1 X 2  X k ( x1 , x 2 ,  , x k )      f X 1 X 2  X p ( x1 , x 2 ,  , x p )dx1 dx 2  dx p for all points in the range of
X1,X2,…,Xp for which X1=x1, X2=x2,…, Xk=xk
Marginal Probability Density Function: f X i ( xi ) 
  f
X 1 X 2 X p
( x1 , x 2 , , x p ) dx1 dx 2  dxi 1 dxi 1  dx p where the integral is over all
R
points of X1,X2,…,Xp for which Xi=xi
28
 

 
 


Mean: E ( X i ) 
   x
i
f X 1 X 2  X p ( x1 , x 2 ,  , x p ) dx1 dx 2  dx p

V ( X i )      xi   X i
 

2
Variance:
f X 1 X 2  X p ( x1 , x 2 ,  , x p ) dx1 dx 2  dx p

Covariance is a measure of linear relationship between the random variables. If the relationship between the random variables is nonlinear, the covariance might
not be sensitive to the relationship.
Two random variables with nonzero correlation are said to be correlated. Similar to covariance, the correlation is a measure of the linear relationship between
random variables.
Correlation:  XY 
Covariance:  XY  E[( X   X )(Y  Y )]  E( XY )   X Y
cov( X , Y )

V ( X )V (Y )
 XY
 XY
where  1   XY  1
If X and Y are independent random variables,  XY   XY  0
Bivariate Normal
f XY ( x, y;  X ,  Y ,  X ,  Y ,  ) 
1
2 X  Y
2

2 ( x   X )( y   Y ) (    Y ) 2  
  1 (x   x )

exp 



2 
2
2
2
 XY
Y

1 

 2(1   )   X

for - < x <  and - < y < , with parameters x > 0, y > 0, - < X < , - < Y <  and -1 <  < 1.
Marginal Distribution: If X and Y have a bivariate normal distribution with joint probability density fXY(x, y;X,Y,X,Y, ), the marginal
probability distribution of X and Y are normal with means x and y and standard deviation x and y, respectively.
Conditional Distribution: If X and Y have a bivariate normal distribution with joint probability density fXY(x, y;X,Y,X,Y, ), the conditional
probability distribution of Y given X = x is normal with mean
Y |x  Y   X 
Linear Functions
of random
variables
Y Y

 X and variance  2 Y | X   Y2 1   2 
X X
Correlation: If X and Y have a bivariate normal distribution with joint probability density function fXY(x, y;X,Y,X,Y, ), the correlation
between X and Y is 
If X and Y have a bivariate normal distribution with = 0, X and Y are independent
Given random variables X1, X2, …, Xp and constants c1,c2, …, cp, Y = c1X1 + c2X2+ … + cpXp is a linear combination of X1, X2, …, Xp
Mean E(Y) = c1E(X1) + c2E(X2) + … + cpE(Xp)
Variance: V (Y )  c1 V ( X 1 )  c 2 V ( X 2 )    c pV ( X p )  2
2
2
2
 c c
i j
i
j
cov( X i , X j )
If X1, X2, …, Xp are independent, variance: V (Y )  c12V ( X 1 )  c22V ( X 2 )    c 2pV ( X p )
Mean and variance on average: E( X )  ;
V (X )   2 p
E(Y )  c1 1  c2  2    c p  p ;
with E( X i )   and V ( X i )   2
V (Y )  c12 12  c22 22    c 2p p2
29
General Functions
of random
variables
Discrete: y = h(x) and x = u(y): fY (y )  f X u( y)
Continuous: y = h(x) and x = u(y):
f Y ( y )  f X [u ( y)] | J | where J  u ' ( y) is called the Jacobian of the transformation
and the absolute value of J is used
30
Name
Confidence Interval
Sample Size
𝑥̅ 𝑖𝑠 𝑎𝑛 𝑒𝑠𝑡𝑖𝑚𝑎𝑡𝑜𝑟 𝑜𝑓 𝜇; 𝑆 2 𝑖𝑠 𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
One-Sided Confidence Bounds
Type I Error: Rejecting the null Hypothesis H0 when it is true; Type II Error: Failing to reject null hypothesis H0 when it is false.
Probability of Type I Error =  = P(type I error) = Significance level = -error = -level = size of the test.
Probability of Type II Error =  = (type II error)
Power = Probability of rejecting the null hypothesis H0 when the alternative hypothesis is true = 1 -  = Probability of correctly rejecting a false null hypothesis.
P-value = Smallest level of significance that would lead to the rejection of the null hypothesis H0.
100(1−∝)% CI on μ is
100(1−
100(1-)% upper confidence bound for 
∝)% upper confident that the error
x − z∝/2 /√𝑛 ≤ 𝜇 ≤ x + z∝/2 /√𝑛
𝜇 ≤ 𝑢 = x + zσ /√𝑛
|𝑥̅
100(1-)% lower confidence bound for 
− 𝜇| will not exceed a specific amount E
z∝⁄2 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 100 ∝⁄2
x − zσ ⁄√𝑛 = 𝑙 ≤ 𝜇
when
sample
size
is
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑
𝑍𝛼/2 𝜎 2
𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑛= (
)
𝐸
Large scale sample size n: Using central limit theorem, X has approximately a normal distribution with mean  and variance /n.
Z
Normal
Distribution
with mean
unknown
and variance
known
X 
S/ n
x − z∝/2
Null hypothesis: H0:= 0
Test Statistic: Z 0 
Alternative
hypothesis
H1: 0
≤ 𝜇 ≤ x + z∝/2
S
√𝑛
Probability of Type II Error for a two-sided test
𝛿 √𝑛
𝛿 √𝑛
𝛽 = 𝜙 (𝑧𝛼/2 −
) − 𝜙 (−𝑧𝛼/2 −
)
𝜎
𝜎
X  0
/ n
p-value
S
√𝑛
2
Rejection Criteria
Sample size for a two-sided test 𝑛 =
(𝑧𝛼/2 +𝑧𝛽 ) 𝜎 2
𝛿2
2
(𝑧𝛼 +𝑧𝛽 ) 𝜎 2
𝑤ℎ𝑒𝑟𝑒 𝛿 = 𝜇 − 𝜇0
Sample size for a one-sided test 𝑛 =
𝑤ℎ𝑒𝑟𝑒 𝛿 = 𝜇 − 𝜇0
Probability above |z0| & 𝑧0 > zα/2 𝑜𝑟 𝑧0 <
𝛿2
below -|z0|,
Large set: for n > 40, replace sample standard deviation s for .
−zα/2
|𝜇−𝜇 |
|𝛿|
P = 2[1 - (|z0|)]
Parameter for Operating Characteristic Curves: 𝑑 = 𝜎 0 = 𝜎
Probability above z0
𝑧0 > zα
H1: 0
P = 1 - (z0)
Probability
below z0
𝑧0 < −zα
H1: 0
P = (z0)
t distribution (Similar to normal in symmetry and unimodal. But t distribution is heavier tails than normal) with n-1 degrees of freedom. k = n-1
T
X 
S/ n
𝑓(𝑥) =
Γ[(𝑘+1)/2]
1
.
√𝜋𝑘Γ(𝑘/2) [(𝑥 2⁄𝑘 )+1](𝑘+1)⁄2
Mean = 0
Variance=k/(k-2) for k>2
31
Normal
Distribution
with mean
unknown
and variance
unknown
100(1−∝)% CI on μ is
x − t ∝/2,n−1 s/√𝑛 ≤ 𝜇
≤ x + t ∝/2,n−1 s/√𝑛
Finding s: Can be obtained only using trial
and error as s is unknown until the data is
collected.
t ∝⁄2,𝑛−1 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 100 ∝⁄2
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑤𝑖𝑡ℎ 𝑛 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
Null hypothesis: H0:= 0
Test Statistic: T0 
Alternative
hypothesis
H1: 0
H1: 0
H1: 0
Probability of Type II Error for a two-sided test
𝛽 = 𝑃(−t α/2,n−1 ≤ T0 ≤ t α/2,n−1 |𝛿 ≠ 0)
= 𝑃(−t α/2,n−1 ≤ 𝑇0′ ≤ t α/2,n−1 )
When the true value of mean  =  + , the distribution for T0 is called
the noncentral t distribution with n-1 degrees of freedom and
noncentrality parameter 𝛿√𝑛/𝜎. If = 0, the noncentral t distribution
reduces to the usual central t distribution. 𝑇0′ denotes the noncentral t
random variable.
X  0
S/ n
p-value
Rejection Criteria
𝑡0 > t α/2,n−1 𝑜𝑟 𝑡0 <
−t α/2,n−1
𝑡0 > t α,n−1
𝑡0 < −t α,n−1
Probability above
|t0| & below -|t0|
Probability above t0
Probability below t0
Parameter for Operating Characteristic Curves: 𝑑 =
|𝜇−𝜇0 |
𝜎
=
100(1-)% upper confidence bound for 
𝜇 ≤ 𝑢 = x + t ∝,n−1 s/√𝑛
100(1-)% lower confidence bound for 
x − t ∝,n−1 𝑠⁄√𝑛 = 𝑙 ≤ 𝜇
|𝛿|
𝜎
Chi − square (χ2 ) distribution with n-1 degrees of freedom. k = n-1
2 
Normal
Distribution;
CI on
variance and
Standard
deviation
(n  1) S 2
2
1
𝑓(𝑥) = 2𝑘/2 Γ(𝑘/2) 𝑥 (𝑘/2)−1 𝑒 −𝑥/2
Mean = k
Variance = 2k
100(1−∝)% CI on σ2 is
(𝑛 − 1)𝑠 2
(𝑛 − 1)𝑠 2
2
≤
𝜎
≤
2
2
𝜒𝛼/2,𝑛−1
𝜒1−𝛼/2,𝑛−1
100(1-)% upper confidence bound for 𝜎 2
(𝑛 − 1)𝑠 2
𝜎2 ≤ 2
𝜒1−𝛼,𝑛−1
100(1-)% lower confidence bound for 𝜎 2
2
2
𝜒𝛼/2,𝑛−1 𝑎𝑛𝑑 𝜒1−𝛼/2,𝑛−1 𝑎𝑟𝑒 𝑡ℎ𝑒
(𝑛 − 1)𝑠 2
≤ 𝜎2
2
upper and lower 100/2 percentage points of chi-square distribution with
𝜒𝛼,𝑛−1
n-1 degrees of freedom.
2
2
Null hypothesis: H0 :    0
2
Test Statistic:  0 
Alternative hypothesis
(n  1) S 2
 02
Rejection Criteria
32
2
𝜒02 > 𝜒𝛼/2,𝑛−1
𝑜𝑟 𝜒02 <
2
−𝜒𝛼/2,𝑛−1
2
𝜒02 > 𝜒𝛼,𝑛−1
H1:  2   02
H1:  2   02
H1:  2   02
2
𝜒02 < −𝜒𝛼,𝑛−1
Parameter for Operating Characteristic Curves: 𝑑 =
|𝜇−𝜇0 |
𝜎
=
|𝛿|
𝜎
Normal approximation for a binomial proportion: If n is large, the distribution of
Z
Large scale
CI for a
population
proportion
X  np
np(1  p)

pˆ  p
p(1  p)
n
is approximately standard normal. p̂ is the proportional population. Mean = p. Variance = p(1-p)/n
100(1−∝)% CI on 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 𝑝 is
100(1-)% upper confidence bound for 𝜎 2
𝑍𝛼/2 2
𝑛
=
(
)
𝑝(1
−
𝑝)
𝑝̂(1−𝑝)
𝐸
𝑝̂ (1 − 𝑝)
𝑝̂ − 𝑧𝛼 √ 𝑛 ≤ 𝑝 ≤ 𝑝̂ +
p can be computed as 𝑝̂ from a
2
𝑝 ≤ 𝑝̂ + 𝑧𝛼 √
𝑛
𝑝̂(1−𝑝)
preliminary sample or use the maximum
𝑧𝛼/2 √
𝑛
100(1-)% lower confidence bound for 𝜎 2
value of p, which is 0.5.
z∝⁄2 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 100 ∝⁄2
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑
𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
Null hypothesis: H0 : p  p0
Test Statistic: z0 
Alternative
hypothesis
H1:p p0
H1: p p0
H1: p p0
p-value
X  np0
np0 (1  p0 )
Rejection Criteria
𝑝̂ (1 − 𝑝)
𝑝̂ − 𝑧𝛼 √
≤𝑝
𝑛
Probability of Type II Error for a two-sided test
𝛽 = 𝜙(
𝑝0 −𝑝+𝑧𝛼/2 √𝑝0 (1−𝑝0 )/𝑛
√𝑝(1−𝑝)/𝑛
)−𝜙(
𝑝0 −𝑝−𝑧𝛼/2 √𝑝0 (1−𝑝0 )/𝑛
√𝑝(1−𝑝)/𝑛
)
Sample size for a two-sided test
𝑛=[
𝑧𝛼/2 √𝑝0 (1−𝑝0 )+𝑧𝛽 √𝑝(1−𝑝)
𝑝−𝑝0
2
]
Sample size for a one-sided test
Probability above
|z0| & below -|z0|,
P = 2[1 - (|z0|)]
Probability above z0
P = 1 - (z0)
Probability below z0
P = (z0)
𝑧0 > zα/2 𝑜𝑟 𝑧0 < −zα/2
𝑛=[
𝑧𝛼 √𝑝0 (1−𝑝0 )+𝑧𝛽 √𝑝(1−𝑝)
𝑝−𝑝0
2
]
𝑧0 > zα
𝑧0 < −zα
33
100(1−∝)% CI on μ1 − μ2 is
z𝛼/2
x1 − x2 −
≤ 𝜇1 − 𝜇2 ≤ x1 −
2
2
𝜎
√𝜎1 + 2
𝑛1 𝑛2
2
𝜎2
1
2
𝜎
x2 + z∝/2 /√𝑛1 + 𝑛2
z∝⁄2 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 100 ∝⁄2
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑
𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
Inference of
the
difference in
means of
two normal
distributions
, variance
known
100(1-)% upper confidence bound for 
100(1−∝)% upper confident that
the error in estimating 𝜇1 − 𝜇2 by
x1 − x2
will not exceed a specific amount E
when sample size is
𝑍𝛼/2 2 2
𝑛= (
) (𝜎1 + 𝜎22 )
𝐸
Z
𝜇1 − 𝜇2 ≤ x1 − x2 + zσ √
𝜎22
+
𝑛1 𝑛2
𝜎21
100(1-)% lower confidence bound for 
𝜎12 𝜎2
x1 − x2 − zσ √ + 2 ≤ 𝜇1 − 𝜇2
𝑛1 𝑛2
X 1  X 2  1  2 
 12
n1

 22
n2
Probability of Type II Error for a two-sided test 𝛽 = 𝜙 𝑧𝛼/2 −
Δ−Δ0
2 𝜎2
√𝜎1 + 2
𝑛1 𝑛2
(
Sample size for a two-sided test, with n1n2, 𝑛 =
− 𝜙 −𝑧𝛼/2 −
Δ−Δ0
2
2
𝜎
√ 𝜎1 + 2
)2
(
𝜎1 +𝜎22
2
𝜎1 /𝑛1 +𝜎22 /𝑛2
𝑛1 𝑛2
)
2
(𝑧𝛼/2 +𝑧𝛽 ) (𝜎12 +𝜎22 )
Sample size for a two-sided test, with n1n2, 𝑛 =
(Δ−Δ0 )2
2
Sample size for a one-sided test, with n1=n2, 𝑛 =
Parameter for Operating Characteristic Curves: 𝑑 =
(𝑧𝛼 +𝑧𝛽 ) (𝜎12 +𝜎22 )
(Δ−Δ0 )2
|𝜇1 −𝜇2 −Δ0|
|Δ−Δ0 |
√𝜎12 +𝜎22
=
√𝜎12 +𝜎22
Null hypothesis: H0:= 0
Test Statistic: Z0 
X 1  X 2  0
 12
n1
Alternative
hypothesis
H1: 0
H1: 0

 22
n2
p-value
Rejection Criteria
Probability above |z0|
& below -|z0|,
P = 2[1 - (|z0|)]
Probability above z0
P = 1 - (z0)
𝑧0 > zα/2 𝑜𝑟 𝑧0
< −zα/2
𝑧0 > zα
34
H1: 0
Probability below z0
𝑧0 < −zα
P = (z0)
100(1−∝)% CI on μ1 − μ2 with assumed equal variance is
100(1−∝)% CI on μ1 − μ2 with assumed unequal variance
= is
1
1
x1 − x2 − 𝑡𝛼/2,𝑛1 +𝑛2 −2 𝑆𝑝 √ +
≤ 𝜇1 − 𝜇2
𝑠12 𝑠22
𝑛1 𝑛2
x1 − x2 − 𝑡𝛼/2,𝜐 √ +
≤ 𝜇1 − 𝜇2
𝑛1 𝑛2
1
1
≤ x1 − x2 + 𝑡𝛼/2,𝑛1 +𝑛2 −2 𝑆𝑝 √ +
𝑠12 𝑠22
𝑛1 𝑛2
≤ x1 − x2 + 𝑡𝛼/2,𝜐 √ +
𝑛1 𝑛2
𝑡
𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 ∝⁄2
𝛼/2,𝑛1 +𝑛2 −2
𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
𝑤𝑖𝑡ℎ 𝑛1 + 𝑛2 − 2 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
Inference of
the
difference in
means of
two normal
distributions
, variance
unknown
Z
X 1  X 2  1  2 
1 1


n1 n2
Pooled estimator of 2, denoted by 𝑆𝑝2 , is: 𝑆𝑝2 =
(𝑛1 −1)𝑆12 +(𝑛2 −1)𝑆22
𝑛1 +𝑛2 −2
Null hypothesis: H0:= 0
H1: 0
Test Statistic: T0 
X 1  X 2  0
has t distribution with n1+n2-2 degrees of freedom; Called as pooled t-test
1 1
Sp

n1 n2
Alternative
hypothesis
H1: 0
H1: 0
H1: 0
p-value
Rejection Criteria
Probability above |t0|
𝑡0 > t α/2,𝑛1 +𝑛2 −2 𝑜𝑟 𝑡0 < −t α/2,𝑛1 +𝑛2 −2
& below -|t0|
Probability above t0
𝑡0 > t α,𝑛1 +𝑛2 −2
Probability below t0
𝑡0 < −t α,𝑛1 +𝑛2 −2
If variances are not assumed equal
*
If H0:= 0 is true, the statistic T0 
X 1  X 2  0
S12 S22

n1 n2
with t degrees of freedom given by
35
2
𝑆2 𝑆2
( 1 + 2)
𝑛1 𝑛2
𝜈= 2
𝐼𝑓 𝜈 𝑖𝑠 𝑛𝑜𝑡 𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑒𝑟, 𝑟𝑜𝑢𝑛𝑑 𝑑𝑜𝑤𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑛𝑒𝑎𝑟𝑒𝑠𝑡 𝑖𝑛𝑡𝑒𝑔𝑒𝑟
(𝑆1 /𝑛1 )2 (𝑆22 /𝑛2 )2
𝑛1 − 1 + 𝑛2 − 1
Goodness of
Fit Test
Statistic
(Oi  Ei ) 2
X0  
Where Oi is the observed frequency and Ei is the expected frequency in the ith class.
E
i 1
i
Approximated to Chi-square distribution with k-p-1 degrees of freedom. p represents the number of parameters. If test statistic is large, we
2
reject null hypothesis. P-value is 𝑃(𝜒𝑘−𝑝−1
> 𝜒02 ).
r
1 r
1 c
1 c
uˆi   Oij vˆ j   Oij Expected Frequency of each cell Eij  nuˆi vˆ j   Oij  Oij
n i 1
n j 1
n j 1 i 1
r
c (O  E ) 2
ij
2
For large n, the statistic  02   ij
P-value is 𝑃(𝜒(𝑟−1)(𝑐−1)
> 𝜒02 ).
E
i 1 j 1
ij
k
2
100(1−∝)% prediction interval on a single future observation from a normal distribution is
Prediction
Interval
x − t ∝/2,n−1 s√1 +
Tolerance
Interval
Sign Test
1
1
≤ 𝑋𝑛+1 ≤ x + t ∝/2,n−1 s√1 +
𝑛
𝑛
Prediction interval for Xn+1 will always be longer than the CI for because there is more variability associated with the prediction error than with the
error of estimation.
Tolerance interval for capturing at least γ% of the values in a normal distribution with cofidence level 100(1−∝)% is
x − 𝑘𝑠, x + 𝑘𝑠
Where k is a tolerance interval factor for the given confidence.
~ differences = r+ If P-value is less than some preselected level , we will reject H0.
Number of positive X i  
0
Normal approximation for sign test statistic: z 0 
~
~
Null hypothesis: H0 : 
0
~
~
H :
1
1
0




P-value: P R  r when p 
Wilcoxon
Signed-Rank
Test
~
~
One-sided hypothesis: H0 : 
0
~
~
H :
Null hypothesis: H0 :   0
1

2


0


P-value: P R  r when p 
1

2
R   0.5n
0.5 n
~
~
Two-sided hypothesis: H0 : 
0
~
~
H :
1
0
1


2

1
 

If r+ > n/2, P-value: 2 P R  r when p  
2



If r+ < n/2, P-value: 2 P R  r when p 
Sort based on Xi  0 differences; Give the ranks the signs of their corresponding differences.
Sum of Positive Ranks: W+; Absolute value of Sum of Negative Ranks: W-. W = min(W+,W-)
36
H1 :   0
Reject Null hypothesis, if observed value of statistic w  w*
For one-sided tests, H1 :   0
Reject Null hypothesis, if observed value of statistic w  w*
For one-sided tests, H1 :   0
Reject Null hypothesis, if observed value of statistic w  w*
Normal approximation for Wilcoxon signed-rank test statistic: z0 
W   n(n  1) / 4
n(n  1)( 2n  1) / 24
Arrange all n1 and n2 observations in ascending order of magnitude and assign ranks to them. If two or more observations are tied(identical), use the
mean of the ranks that would have assigned if the observations differed. W1 is sum of ranks in smaller sample. W2 is sum of ranks in other sample.
𝑊2 =
(𝑛1 +𝑛2 )(𝑛1 +𝑛2 +1)
2
− 𝑊1
Reject null hypothesis, if w1 or w2 is less than or equal to tabulated critical value w.
For one-sided hypothesis: H1 : 1  2 reject H0 if w1 wFor H1 : 1  2 reject H0 if w2 w
Normal approximation when n1 and n2 > 8, Z0 =
𝑊1 −𝜇𝑤1
𝜎𝑤1
100(1−∝)% CI on 𝜇𝐷 = μ1 − μ2 is
̅
𝑑 − 𝑡𝛼,𝑛−1 𝑆𝐷 /√𝑛 ≤ 𝜇𝐷 ≤ 𝑑̅ + 𝑡𝛼,𝑛−1 𝑆𝐷 /√𝑛
Paired t-test
2
2
𝑡𝛼,𝑛−1 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 ∝⁄2 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑡 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛 − 1 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
2
Null hypothesis: H0:D= 0
Test Statistic: T0 
Alternative
hypothesis
H1:D 0
Inference on
the
variances of
two normal
distributions
(F
Distribution)
D  0
̅ is the sample average of n differences D1, D2, … , Dn,
where 𝐷
SD / n
and SD is the sample standard deviation of these differences.
p-value
Rejection Criteria
Probability above |t0|
𝑡0 > t α/2,𝑛−1 𝑜𝑟 𝑡0 < −t α/2,n−1
& below -|t0|
Probability above t0
𝑡0 > t α,n−1
H1:D 0
Probability
below
t
𝑡
0
H1:D 0
0 < −t α,n−1
Let W and Y be independent chi-square random variables with u and v degrees of freedom respectively.
Ratio F =
f(x) =
𝑊/𝑢
𝑌/𝑣
has the probability density function
𝑢 + 𝑣 𝑢 𝑢/2 (𝑢)−1
Γ ( 2 ) (𝑣 ) 𝑥 2
,0 < 𝑥 < ∞
(𝑢+𝑣)/2
𝑢
𝑣
𝑢
Γ (2) Γ (2) [(𝑣 ) 𝑥 + 1]
ν
2𝜈 2 (𝑢+𝜈−2)
Mean μ = ν−2 𝑓𝑜𝑟 𝜈 > 2
Variance σ2 = 𝑢(𝜈−2)2 (𝜈−4) , 𝑓𝑜𝑟 𝜈 > 4
Lower tail percentage point, 𝑓1−𝛼,𝑢,𝜈 = 𝑓
1
𝛼,𝜈,𝑢
37
F Distribution 𝐹 =
𝑆12 ⁄𝜎12
𝑆22 ⁄𝜎22
n1-1 numerator degrees of freedom and n2-1 denominator degrees of freedom
Null hypothesis: 𝐻0 : 𝜎12 = 𝜎22
𝑆2
Test Statistic: 𝐹0 = 𝑆12
2
Alternative
hypothesis
H1: 𝜎12  𝜎22
H1: 𝜎12 
Rejection Criteria
𝑓0 > fα,𝑛
−1,𝑛2 −1
2 1
𝜎22
𝑜𝑟 𝑓0 < −f1−α,𝑛
−1,𝑛2 −1
2 1
𝑓0 > f𝛼,𝑛1 −1,𝑛2 −1
𝑓0 < f1−𝛼,𝑛1 −1,𝑛2 −1
H1: 𝜎12  𝜎22
P-value is the area (probability) under the F distribution with n1-1 and n2-1 degrees of freedom that lies beyond the computed value of the test
statistic f0.
σ12
100(1−∝)% CI on the ratio 2 is
𝜎2
𝑠12
σ12 𝑠12
𝑓 𝛼
≤
≤ 𝑓𝛼
𝑠22 1−2 ,𝑛2 −1,𝑛1 −1 𝜎22 𝑠22 2 ,𝑛2 −1,𝑛1 −1
𝑓𝛼,𝑛 −1,𝑛 −1 𝑎𝑛𝑑 𝑓1−𝛼,𝑛 −1,𝑛 −1 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑎𝑛𝑑 𝑙𝑜𝑤𝑒𝑟 ∝⁄2 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝐹 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ 𝑛2 − 1 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 𝑎𝑛𝑑 𝑛1
2 2
1
2 2
1
− 1 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 𝑜𝑓 𝑓𝑟𝑒𝑒𝑑𝑜𝑚
Null hypothesis: 𝐻0 : 𝑝1 = 𝑝2
𝐻1 : 𝑝1 ≠ 𝑝2
̂
𝑃1 − 𝑃̂2 − (𝑝1 − 𝑝2 )
𝑍=
𝑝 (1 − 𝑝1 ) 𝑝2 (1 − 𝑝2 )
√ 1
+
𝑛1
𝑛2
Inference on
the
population
proportions
Test Statistic: 𝑍0 =
Alternative
hypothesis
H1:p1 p2
H1: p1 p2
H1: p1 p2
𝑃̂1 −𝑃̂2
1
1
√𝑃̂(1−𝑃̂)(𝑛 +𝑛 )
1
2
p-value
Probability above |z0| & below -|z0|,
P = 2[1 - (|z0|)]
Probability above z0
P = 1 - (z0)
Probability below z0
P = (z0)
Probability of Type II Error for a two-sided test
Rejection Criteria for FixedLevel tests
𝑧0 > zα/2 𝑜𝑟 𝑧0 < −zα/2
𝑧0 > zα
𝑧0 < −zα
38
𝛽 = 𝜙[
̅̅̅̅(1⁄𝑛1 +1⁄𝑛2 )−(𝑝1 −𝑝2 )
𝑧𝛼/2 √𝑝𝑞
𝜎𝑃1 −𝑃2
𝑛 𝑝 +𝑛 𝑝
Where 𝑝̅ = 1𝑛1 +𝑛2 2
1
2
]−𝜙[
and 𝑞̅ =
̅̅̅̅(1⁄𝑛1 +1⁄𝑛2 )−(𝑝1 −𝑝2 )
−𝑧𝛼/2 √𝑝𝑞
𝜎𝑃1 −𝑃2
𝑛1 (1−𝑝1 )+𝑛2 (1−𝑝2 )
𝑛1 +𝑛2
]
Sample size for a two-sided test
2
𝑛=
[𝑧𝛼/2 √(𝑝1 +𝑝2 )(𝑞1 +𝑞2 )/2+𝑧𝛽 √𝑝1 𝑞1 +𝑝2 𝑞2 ]
(𝑝1 −𝑝2 )2
where q1 = 1 – p1 and q2 = 1 – p2
100(1−∝)% CI on the difference in the true proportions p1 − p2 is
𝑝̂1 (1 − 𝑝̂1 ) 𝑝̂ 2 (1 − 𝑝̂2 )
𝑝̂1 (1 − 𝑝̂1 ) 𝑝̂ 2 (1 − 𝑝̂2 )
𝑝̂1 − 𝑝̂2 − zα/2 √
+
≤ 𝑝1 − 𝑝2 ≤ 𝑝̂1 − 𝑝̂2 + zα/2 √
+
𝑛1
𝑛2
𝑛1
𝑛2
z∝⁄2 𝑖𝑠 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 ∝⁄2 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑛𝑜𝑟𝑚𝑎𝑙 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛
Inferences:
1. Population is normal: Sign test or t-test.
a. t-test has the smallest value of  for a significance level , thus t-test is superior to other tests.
2. Population is symmetric but not normal (but with finite mean):
3.
4.
5.
6.
a. t-test will have the smaller  (or a higher power) than sign test.
b. Wilcoxon Signed-rank test is comparable to t-test.
Distribution with heavier tails:
a. Wilcoxon Signed-rank test is better than t-test as t-test depends on the sample mean, which is unstable in heavy-tailed distributions.
Distribution is not close to normal:
a. Wilcoxon Signed-rank test is preferred.
Paired observations:
a. Both sign test and Wilcoxon Signed-rank test can be applied. In sign test, median of the differences is equal to zero in null hypothesis. In Wilcoxon
Signed-rank test, mean of the differences is equal to zero in null hypothesis.
A
39
DECISION ANALYSIS
Criterion of Realism
Expected Monetary Value
Weighted average = (best in row) + (1 – )(worst in
EMV(altern ative) =  Xi P( Xi )
row)
Xi = payoff for the alternative in state of
nature i
For Minimization:
P(Xi) = probability of achieving payoff Xi (i.e.,
Weighted average = (best in row) + (1 – )(worst in
probability of state of nature i)
row)
∑ = summation symbol
EMV (alternative i) = (payoff of first state of nature) x
Expected Value with Perfect Information
(probability of first state of nature) + (payoff of second
EVwPI = ∑(best payoff in state of nature i)
state of nature) x (probability of second state of nature) +
(probability of state of nature i)
… + (payoff of last state of nature) x (probability of last
EVwPI = (best payoff for first state of nature) x
state of nature)
(probability of first state of nature) + (best
payoff for second state of nature) x (probability
of second state of nature) + … + (best payoff for
last state of nature) x (probability of last state of
nature)
Expected Value of Sample Information
EVSI = (EV with SI + cost) – (EV without SI)
Utility of other outcome = (p)(utility of best outcome,
which is 1) + (1 – p)(utility of the worst outcome, which
is 0)
Expected Value of Perfect Information
EVPI = EVwPI – Best EMV
EVSI
Efficiency of sample informatio n =
100%
EVPI
FORECASTING
forecast error
(error ) 2


Mean Absolute Deviation (MAD) 
Mean Squared Error (MSE) 
n
n
error
 actual
Mean Absolute Percent Error (MAPE) 
100%
n
Y  Yt 1  ...  Yt  n 1
sum of demands in previous n periods
Mean Average Forecast 
 Ft 1  t
n
n
Weighted Moving Average : Ft 1 
 (Weight in period i)(Actual value in period )  w Y  w Y
w w
 (Weights )
 ...  wnYt n1

...  wn
2
2 t 1
1 t
1
Exponential Smoothing : Ft 1  Ft   (Yt  Ft )
New forecast  Last period’ s forecast   (Last period’ s actual demand – Last period’ s forecast)
Exponentia l Smoothing with Trend :
Yˆ  b0  b1 X
Ft 1  FITt   (Yt  FITt )
where Yˆ  predicted value
Tt 1  Tt   (Ft 1  FITt )
b0  intercept
FITt 1  Ft 1  Tt 1
b1  slope of the line
X  time period (i.e., X  1, 2, 3, , n)
Tracking signal 
RSFE  (forecast error)

MAD
MAD
Yˆ  a  b1 X1  b2 X 2  b3 X 3  b4 X 4
41
INVENTORY CONTROL MODELS
Annual ordering cost  Number of orders placed per year
Q
Average inventory level =
2
(Ordering cost per order)
Annual Demand
D

Co  Co
Number of units in each order
Q
Annual holding cost  Average Inventory
Economic Order Quantity
Annual ordering cost = Annual holding cost
(Carrying cost per unit per year)
D
Q
Order quantity
C  C

 (Carrying cost per unit per year) Q o 2 h
2
2DCo
Q
EOQ  Q* 
 Ch
Ch
2
Total cost (TC) = Order cost + Holding cost
Cost of storing one unit of inventory for one year = Ch =
IC, where C is the unit price or cost of an inventory item
D
Q
TC  Co  Ch
and I is Annual inventory holding charge as a percentage
Q
2
of unit price or cost
2DCo
Q* 
IC
ROP without Safety Stock:
EOQ without instantaneous receipt assumption
Reorder Point (ROP) = Demand per day x Lead
Maximum inventory level  (Total produced during the
time for a new order in days
production run) – (Total used during the production run)
dL
 (Daily production rate)(Number of days production)
Inventory position = Inventory on hand +
– (Daily demand)(Number of days production)
Inventory on order
 (pt) – (dt)
 d
Q
Q
 pt – dt  p – d  Q1 – 
p
p
 p
Total produced Q  pt
Q d
Q d
Average inventory  1 – 
Annual holding cost  1 – Ch
2  p
2  p
D
D
Annual setup cost  Cs
Annual ordering cost  Co
Q
Q
D = the annual demand in units
Q  number of pieces per order, or production run
Production Run Model: EOQ without
Quantity Discount Model
instantaneous receipt assumption
2DCo
EOQ 
Annual holding cost  Annual setup cost
IC
Q d
D
If EOQ < Minimum for discount, adjust the quantity to Q
1 – Ch  Cs
2  p 
Q
= Minimum for discount
Total cost  Material cost + Ordering cost + Holding cost
2DCs
Q* 
D
Q
Total cost  DC + Co + Ch
 d
Ch 1 – 
Q
2
 p
Holding cost per unit is based on cost, so C = IC
h
Safety Stock
Where I = holding cost as a percentage of the unit cost
(C)
Safety Stock with Normal Distribution
ROP = (Average demand during lead time) + ZsdLT
42
ROP = Average demand during lead time + Safety
Stock
Service level = 1 – Probability of a stockout
Probability of a stockout = 1 – Service level
Demand is variable but lead time is constant
ROP  d L  Z  d L


d  average daily demand
Z
= number of standard deviations for a given
service level
dLT = standard deviation of demand during the lead
time
Safety stock = ZdLT
Demand is constant but lead time is variable
ROP  dL  Z d L 
L  average lead time
 d  standard deviation of daily demand
L  lead time in days
 L  standard deviation of lead time
d  daily demand
Both demand and lead time are variable
Total Annual Holding Cost with Safety Stock
Total Annual Holding Cost = Holding cost of regular
inventory + Holding cost of safety stock
Q
THC  Ch  (SS)Ch
2
ROP  d L  Z L  d2  d 2 L2
The expected marginal profit = P(MP)
The expected marginal loss = (1 – P)(ML)
The optimal decision rule
Stock the additional unit if P(MP) ≥ (1 – P)ML
P(MP) ≥ ML – P(ML)
P(MP) + P(ML) ≥ ML
P(MP + ML) ≥ ML
ML
P
ML + MP
43
PROJECT MANAGEMENT
Expected Activity Time t =
a + 4m + b
6
Earliest finish time = Earliest start time + Expected
activity time
EF = ES + t
Latest start time = Latest finish time – Expected
activity time
LS = LF – t
Slack = LS – ES, or Slack = LF – EF
Project standard deviation  T  Project va riance
Value of work completed = (Percentage of work
complete) x (Total activity budget)
Crash cost  Normal Cost
Crash cost/Time Period 
Normal time  Crash time
2
b –a
Variance = 

 6 
Earliest start = Largest of the earliest finish times of
immediate predecessors
ES = Largest EF of immediate predecessors
Latest finish time = Smallest of latest start times for
following activities
LF = Smallest LS of following activities
Project Variance = sum of variances of activities on the
critical path
Due date  Expected date of completion
Z
T
Activity difference = Actual cost – Value of work
completed
44
Upper control limit (UCL)  x  z x
STATISTICAL QUALITY CONTROL
UCL x  x  A2 R
Lower control limit (LCL)  x  z x
x = mean of the sample means
z = number of normal standard deviations (2 for
95.5% confidence, 3 for 99.7%)
 x = standard deviation of the sampling distribution
of the sample means =
LCL x  x  A2 R
R = average of the samples
A2 = Mean factor
x = mean of the sample means
x
n
UCL R  D4 R
LCL R  D3 R
UCLR = upper control chart limit for the range
LCLR = lower control chart limit for the range
D4 and D3 = Upper range and lower range
p-charts
UCL p  p  z p
LCL p  p  z p
p = mean proportion or fraction defective in the sample
Total number of errors
p
Total number of records examined
z = number of standard deviations
 p = standard deviation of the sampling distribution
 p is estimated by ˆ p
c-charts
The mean is c and the standard deviation is equal to
c
Estimated standard deviation of a binomial distribution
p(1  p)
ˆ p 
n
where n is the size of each sample
Range of the sample = Xmax - Xmin
To compute the control limits we use c  3 c (3 is
used for 99.7% and 2 is used for 95.5%)
UCL c  c  3 c
LCL c  c  3 c
Control Chart Model
k is the distance of control limits from the center line,
expressed in Standard Deviation units. Common choice
is k = 3.
𝑈𝐶𝐿 = 𝜇𝑊 + 𝑘𝜎𝑊
𝐶𝐿 = 𝜇𝑊
𝐿𝐶𝐿 = 𝜇𝑊 − 𝑘𝜎𝑊
𝜇𝑊 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑊
𝜎𝑊 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑎𝑚𝑝𝑙𝑒 𝑊
𝜎
𝜎𝑊 =
√𝑛
𝑅 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡
𝑋̅ 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡
𝑈𝐶𝐿 = 𝜇 + 3𝜎/√𝑛
𝐿𝐶𝐿 = 𝜇 − 3𝜎/√𝑛
𝐶𝐿 = 𝜇
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝑚𝑒𝑎𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝜇, 𝑔𝑟𝑎𝑛𝑑 𝑚𝑒𝑎𝑛,
𝑚
1
̿
𝜇̂ = 𝑋 =
∑ 𝑋̅𝑖
𝑚
𝑖=1
𝑋̿ 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑒𝑟 𝑙𝑖𝑛𝑒 𝑜𝑛 ̅𝑋 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑐ℎ𝑎𝑟𝑡
̅𝑋 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑐ℎ𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝑅̅
𝑈𝐶𝐿 = 𝑥̿ + 𝐴2 𝑟̅
𝐶𝐿 = 𝑥̿
𝐿𝐶𝐿 = 𝑥̿ − 𝐴2 𝑟̅
𝑆 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡
45
𝑚
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝑚𝑒𝑎𝑛 𝜇𝑅 𝑖𝑠 𝑅̅ =
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝜎, 𝑖𝑠 𝜎̂ =
1
∑ 𝑅𝑖
𝑚
𝑖=1
𝑅̅
𝑑2
𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑑2 𝑖𝑠 𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑜𝑟 𝑣𝑎𝑟𝑖𝑜𝑢𝑠 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒𝑠
3
𝑈𝐶𝐿 = 𝑋̿ +
𝑅̅
𝑑2 √𝑛
3
𝐿𝐶𝐿 = 𝑋̿ −
𝑅̅
𝑑2 √𝑛
3
𝐴2 =
𝑑2 √𝑛
𝑈𝐶𝐿 = 𝐷4 𝑟̅
𝐶𝐿 = 𝑟̅
𝐿𝐶𝐿 = 𝐷3 𝑟̅
𝑊ℎ𝑒𝑟𝑒 𝑟̅ 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑟𝑎𝑛𝑔𝑒.
𝐷3 𝑎𝑛𝑑 𝐷4 𝑎𝑟𝑒 𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑜𝑟 𝑣𝑎𝑟𝑖𝑜𝑢𝑠 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒𝑠
𝑀𝑜𝑣𝑖𝑛𝑔 𝑅𝑎𝑛𝑔𝑒 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡
𝑚
1
∑|𝑋𝑖 − 𝑋𝑖−1 |
𝑚−1
𝑖=2
̅̅̅̅̅
̅̅̅̅̅
𝑀𝑅
𝑀𝑅
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝜎, 𝑖𝑠 𝜎̂ =
=
𝑑2
1.128
𝐶𝐿, 𝑈𝐶𝐿 𝑎𝑛𝑑 𝐿𝐶𝐿 𝑓𝑜𝑟 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑐ℎ𝑎𝑟𝑡 𝑓𝑜𝑟 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑠
̅̅̅̅
𝑚𝑟
̅̅̅̅
𝑚𝑟
𝑈𝐶𝐿 = 𝑥̅ + 3
= 𝑥̅ + 3
𝑑2
1.128
𝐶𝐿 = 𝑥̅
̅̅̅̅
𝑚𝑟
̅̅̅̅
𝑚𝑟
𝐿𝐶𝐿 = 𝑥̅ − 3
= 𝑥̅ − 3
𝑑2
1.128
𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝑐ℎ𝑎𝑟𝑡 𝑓𝑜𝑟 𝑚𝑜𝑣𝑖𝑛𝑔 𝑟𝑎𝑛𝑔𝑒𝑠
𝑈𝐶𝐿 = 𝐷4 𝑚𝑟
̅̅̅̅ = 3.267𝑚𝑟
̅̅̅̅
𝐶𝐿 = 𝑚𝑟
̅̅̅̅
𝐿𝐶𝐿 = 𝐷3 𝑚𝑟
̅̅̅̅ = 0 𝑎𝑠 𝐷3 𝑖𝑠 0 𝑓𝑜𝑟 𝑛 = 2.
̅̅̅̅̅ =
𝑀𝑅
𝑆̅
𝑐4
𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑐4 𝑖𝑠 𝑡𝑎𝑏𝑢𝑙𝑎𝑡𝑒𝑑 𝑓𝑜𝑟 𝑣𝑎𝑟𝑖𝑜𝑢𝑠 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒𝑠
𝑠̅
𝑈𝐶𝐿 = 𝑠̅ + 3 √1 − 𝑐42
𝑐4
𝐶𝐿 = 𝑠̅
𝑠̅
𝐿𝐶𝐿 = 𝑠̅ − 3 √1 − 𝑐42
𝑐4
𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝜎, 𝑖𝑠 𝜎̂ =
̅𝑋 𝑐𝑜𝑛𝑡𝑟𝑜𝑙 𝑐ℎ𝑎𝑟𝑡 𝑓𝑟𝑜𝑚 𝑆̅
𝑠̅
𝑈𝐶𝐿 = 𝑥̿ + 3
𝑐4 √𝑛
𝐶𝐿 = 𝑥̿
𝑠̅
𝐿𝐶𝐿 = 𝑥̿ − 3
𝑐4 √𝑛
𝑈𝑆𝐿 − 𝐿𝑆𝐿
𝑃𝑟𝑜𝑐𝑒𝑠𝑠 𝐶𝑎𝑝𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑅𝑎𝑡𝑖𝑜 (𝑃𝐶𝑅) =
6𝜎̂
𝑟̅
𝜎̂ =
𝑑2
𝑈𝑆𝐿 − 𝜇 𝜇 − 𝐿𝑆𝐿
𝑂𝑛𝑒 − 𝑠𝑖𝑑𝑒𝑑 𝑃𝐶𝑅, 𝑖𝑠 𝑃𝐶𝑅𝑘 = 𝑚𝑖𝑛 [
,
]
3𝜎
3𝜎
𝐿𝑆𝐿 − 𝜇
𝑃(𝑋 < 𝐿𝑆𝐿) = 𝑃(𝑍 <
)
𝜎
𝑈𝑆𝐿 − 𝜇
𝑃(𝑋 > 𝑈𝑆𝐿) = 𝑃(𝑍 >
)
𝜎
𝑃 𝐶ℎ𝑎𝑟𝑡 (𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡 𝑓𝑜𝑟 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛𝑠)
𝑈 𝐶ℎ𝑎𝑟𝑡 (𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡 𝑓𝑜𝑟 𝐷𝑒𝑓𝑒𝑐𝑡𝑠 𝑝𝑒𝑟 𝑈𝑛𝑖𝑡)
1
1
𝑃̅ =
∑ 𝑃𝑖 =
∑ 𝐷𝑖
𝑚
𝑚𝑛
1
̅=
𝑈
∑ 𝑈𝑖
𝑚
𝑝̅ (1 − 𝑝̅ )
𝑈𝐶𝐿 = 𝑝̅ + 3√
𝑛
𝐶𝐿 = 𝑝̅
𝑢̅
𝑈𝐶𝐿 = 𝑢̅ + 3√
𝑛
𝑝̅ (1 − 𝑝̅ )
𝐿𝐶𝐿 = 𝑝̅ − 3√
𝑛
𝑊ℎ𝑒𝑟𝑒 𝑝̅ 𝑖𝑠 𝑡ℎ𝑒 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑎𝑣𝑒𝑟𝑎𝑔𝑒.
𝑢̅
𝐿𝐶𝐿 = 𝑢̅ − 3√
𝑛
𝑚
𝑚
𝑖=1
𝑖=1
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑅𝑢𝑛 𝐿𝑒𝑛𝑔𝑡ℎ, 𝐴𝑅𝐿 =
𝑚
𝑖=1
𝐶𝐿 = 𝑢̅
1
𝑝
𝑊ℎ𝑒𝑟𝑒 𝑝 𝑖𝑠 𝑡ℎ𝑒 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 𝑡ℎ𝑎𝑡 𝑎 𝑛𝑜𝑟𝑚𝑎𝑙𝑙𝑦
𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑒𝑑 𝑝𝑜𝑖𝑛𝑡 𝑓𝑎𝑙𝑙𝑠 𝑜𝑢𝑡𝑠𝑖𝑑𝑒 𝑡ℎ𝑒 𝑙𝑖𝑚𝑖𝑡𝑠
𝑤ℎ𝑒𝑛 𝑡ℎ𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠 𝑖𝑠 𝑖𝑛 𝑐𝑜𝑛𝑡𝑟𝑜𝑙
𝑊ℎ𝑒𝑟𝑒 𝑢̅ 𝑖𝑠 𝑡ℎ𝑒 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑓𝑒𝑐𝑡𝑠 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝐶𝑈𝑆𝑈𝑀 𝐶ℎ𝑎𝑟𝑡
(𝐶𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑆𝑢𝑚 𝐶𝑜𝑛𝑡𝑟𝑜𝑙 𝐶ℎ𝑎𝑟𝑡)
𝑈𝑝𝑝𝑒𝑟 𝑜𝑛𝑒 − 𝑠𝑖𝑑𝑒𝑑 𝐶𝑈𝑆𝑈𝑀 𝑓𝑜𝑟 𝑝𝑒𝑟𝑖𝑜𝑑 𝑖
𝑆𝐻 (𝑖) = 𝑚𝑎𝑥[0, 𝑥̅𝑖 − (𝜇0 + 𝐾) + 𝑠𝐻 (𝑖 − 1)]
𝐿𝑜𝑤𝑒𝑟 𝑜𝑛𝑒 − 𝑠𝑖𝑑𝑒𝑑 𝐶𝑈𝑆𝑈𝑀 𝑓𝑜𝑟 𝑝𝑒𝑟𝑖𝑜𝑑 𝑖
46
𝐸𝑥𝑝𝑜𝑛𝑒𝑛𝑡𝑖𝑎𝑙𝑙𝑦 𝑊𝑒𝑖𝑔ℎ𝑡𝑒𝑑 𝑀𝑜𝑣𝑖𝑛𝑔 𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝐶𝑜𝑛𝑡𝑟𝑜𝑙
𝐶ℎ𝑎𝑟𝑡 (𝐸𝑊𝑀𝐴)
𝑧𝑡 = 𝜆𝑥̅𝑡 + (1 − 𝜆)𝑧𝑡−1 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑡𝑖𝑚𝑒 𝑡
𝑈𝐶𝐿 = 𝜇0 + 3
𝜎
𝜆
[1 − (1 − 𝜆)2𝑡 ]
√𝑛 2 − 𝜆
𝐶𝐿 = 𝜇0
𝜆
√
[1 − (1 − 𝜆)2𝑡 ]
√𝑛 2 − 𝜆
𝑅̅
𝑆̅
𝜇̂ 0 = 𝑋̿ 𝑎𝑛𝑑 𝜎̂ =
𝑜𝑟 𝜎̂ =
𝑑2
𝑐4
̅̅̅̅̅
𝑀𝑅
𝐹𝑜𝑟 𝑛 = 1, 𝜇̂ 0 = 𝑋̅ 𝑎𝑛𝑑 𝜎̂ =
1.128
𝐿𝐶𝐿 = 𝜇0 − 3
𝜎
√
𝑆𝐿 (𝑖) = 𝑚𝑎𝑥[0, (𝜇0 − 𝐾) − 𝑥̅𝑖 + 𝑠𝐿 (𝑖 − 1)]
𝑊ℎ𝑒𝑟𝑒 𝑠𝑡𝑎𝑟𝑡𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒𝑠 𝑠𝐻 (0) = 𝑠𝐿 (0) = 0
𝐻 𝑖𝑠 𝑐𝑎𝑙𝑙𝑒𝑑 𝑑𝑒𝑐𝑖𝑠𝑖𝑜𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙
Δ
𝑅𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑣𝑎𝑙𝑢𝑒, 𝐾 =
2
𝜇1 = 𝜇0 + Δ
𝐾 = 𝑘𝜎; 𝐻 = ℎ𝜎
𝑠𝐻 (𝑖)
𝜇0 + 𝐾 +
, 𝑖𝑓 𝑠𝐻 (𝑖) > 𝐻
𝑛𝐻
𝜇̂ =
𝑠𝐿 (𝑖)
𝜇0 − 𝐾 −
, 𝑖𝑓 𝑠𝐿 (𝑖) > 𝐻
{
𝑛𝐿
47
OTHERS
Computing lambda and the consistency index
 n
CI 
n 1
Consistency Ratio
CI
CR 
RI
Fixed cost
Price/unit – Variable cost/unit
f

sv
P(loss) = P(demand < break-even)
P(profit) = P(demand > break-even)
Break - even point (units) 
The input to one stage is also the output from
another stage
sn–1 = Output from stage n
The transformation function
tn = Transformation function at stage n
General formula to move from one stage to
another using the transformation function
sn–1 = tn (sn, dn)
The total return at any stage
fn = Total return at stage n
Transformation Functions
sn1  an  sn   bn  d n   cn
Return Equations
rn  an  sn   bn  d n   cn
Probability of breaking even
break - even point  
Z

 Price Variable cost 
EMV  
–
  (Mean demand)
unit
 unit

 Fixed costs
K(break - even point – X)for X  BEP Using the unit normal loss integral, EOL can be
Opportunit y Loss  
computed using
$0for X  BEP

EOL = KN(D)
where
EOL = expected opportunity loss
K = loss per unit when sales are below the break-even point K = loss per unit when sales are below the breakX = sales in units
even point
 = standard deviation of the distribution
N(D) = value for the unit normal loss integral for a
given value of D
 – break even point
D

a
 
AB   b   d
c
 
 ad ae 


e    bd be   C
 cd ce 


d 
 
a b c   e   ad  be  cf 
f
 
a b   e

  
c d  g
f   ae  bg af  bh 


h   ce  dg cf  dh 
a b
c d
Determinant Value = (a)(d) – (c)(b)
a b c
d e f
g h i
Determinant Value = aei + bfg + cdh – gec – hfa –
idb
Numerical value of numerator determinan t
X
Numerical value of denominato r determinan t
48
a b 

Original matrix  
c
d


Determinan t value of original matrix  ad  cb
 d  c

Matrix of cofactors  

b
a


 d  b

Adjoint of the matrix  
 c a 
Equation for a line
Y = a + bX
where b is the slope of the line
Given any two points (X1, Y1) and (X2, Y2)
Change in Y Y Y2 – Y1
b


Change in X X X 2 – X1
a b 


c d 
1
 d

  ad  cb
 c

 ad  cb
b 

ad  cb 
a 

ad  cb 
For the Nonlinear function
Y = X2 – 4X + 6
Find the slope using two points and this equation
Change in Y Y Y2 – Y1
b


Change in X X X 2 – X1
Y1  aX 2  bX  c
Y C
Y  0
Y2  a( X  X ) 2  b( X  X )  c
Y  Xn
Y   nX n 1
Y  Y2  Y1  b(X )  2aX (X )  c(X ) 2
Y b(X )  2aX (X )  c(X ) 2

X
X
X (b  2aX  cX )

 b  2aX  cX
X
Total cost  (Total ordering cost) + (Total holding cost)
+ (Total purchase cost)
Y  cX n
Y   cnX n 1
n
Y   n 1
X

Y  g ( x)  h( x)
Y   g ( x)  h( x)
D
Q
Co + C h  DC
Q
2
Q = order quantity
D = annual demand
Co = ordering cost per order
Ch = holding cost per unit per year
C = purchase (material) cost per unit
TC 
1
Xn
Y  g ( x )  h( x )
Y  g ( x )  h( x )
Economic Order Quantity
dTC – DCo Ch


dQ
Q2
2
2DCo
Q
Ch
Y
d 2TC DCo
 3
dQ 2
Q
49
Series
Geometric series is a series for which the ratio of each two consecutive terms is a constant function of the summation index.
The geometric sequence is a number of terms when we can get each term by multiplying the previous term into constant like {a, ar, ar 2, ar3, ….}
where a is the first of first term and r is the ratio between terms.
Now we can define the geometric series like :
𝑛
1 − 𝑟 𝑛+1
𝑎 + 𝑎𝑟 + 𝑎𝑟 + 𝑎𝑟 + ⋯ + 𝑎𝑟 = ∑ 𝑎𝑟 𝑘 = 𝑎 (
)
1−𝑟
2
3
𝑛
𝑘=0
∞
2
3
𝑎 + 𝑎𝑟 + 𝑎𝑟 + 𝑎𝑟 + ⋯ = ∑ 𝑎𝑟 𝑘 =
𝑘=0
𝑎
, 𝑓𝑜𝑟 |𝑟| < 1
1−𝑟
With a = 1
1 + 𝑟 + 𝑟2 + 𝑟3 + ⋯ =
𝑚
𝑟 +𝑟
𝑚+1
+𝑟
𝑚+2
1
1−𝑟
(𝑟 𝑛+1 − 𝑟 𝑚 )
+ ⋯+ 𝑟 =
, 𝑓𝑜𝑟 𝑚 < 𝑛, 𝑎 = 1, 𝑟 > 1
𝑟−1
𝑛
(𝑟 𝑚 − 𝑟 𝑛+1 )
, 𝑓𝑜𝑟 𝑚 < 𝑛, 𝑎 = 1, 𝑟 < 1
1−𝑟
(𝑟 𝑚 − 𝑟 𝑛+1 )
𝑚
𝑚+1
𝑚+2
𝑛
𝑎𝑟 + 𝑎𝑟
+ 𝑎𝑟
+ ⋯ + 𝑎𝑟 = 𝑎
, 𝑓𝑜𝑟 𝑚 < 𝑛, 𝑟 < 1
1−𝑟
𝑟 𝑚 + 𝑟 𝑚+1 + 𝑟 𝑚+2 + ⋯ + 𝑟 𝑛 =
With n → ∞ :
𝑛×(𝑛−1)…×(𝑛−𝑥+1)
𝑛𝑥
→1
∞
λ𝑦
∑
= 𝑒𝜆
𝑦!
𝑦=0
𝑛
𝑐
𝐹𝑟𝑜𝑚 𝐶𝑎𝑙𝑐𝑢𝑙𝑢𝑠, (1 + ( )) → 𝑒 𝑐
𝑛
1+2+3+⋯+𝑛 =
𝑛(𝑛 + 1)
2
The Number e
The number e, sometimes called the natural number, or Euler’s number, is an important mathematical constant approximately equal
to 2.71828. When used as the base for a logarithm, the corresponding logarithm is called the natural logarithm, and is written
as ln(x). Note that ln(e)=1 and that ln(1)=0.
There are a number of different definitions of the number e. Most of them involve calculus. One is that e is the limit of the sequence
1 𝑛
whose general term is (1 + ) . Another is that e is the unique number so that the area under the curve y=1/x from x=1 to x=e
𝑛
is 1 square unit.
1
1
1
Another definition of e involves the infinite series 1 + + + + ⋯ It can be shown that the sum of this series is e.
1!
2!
3!
Importance of e
The number e is very important in mathematics, alongside 0, 1, i, and π. All five of these numbers play important and recurring roles
across mathematics, and are the five constants appearing in the formulation of Euler’s identity, which (amazingly) states that 𝑒 𝑖𝜋 +
1 = 0. Like the constant π, e is irrational (it cannot be written as a ratio of integers), and it is transcendental (it is not a root of any
non-zero polynomial with rational coefficients).
50
𝑚 𝑗 𝑚−𝑗
Binomial theorem says: (𝑎 + 𝑏)𝑚 = ∑𝑚
𝑗=0 ( 𝑗 ) 𝑎 𝑏
∞
∑(1 − 𝜃)𝑘 𝜃 =
𝑘=1
∞
∑ 2−𝑘
𝑘=1
1−𝜃
1 − (1 − 𝜃)
2𝑘 = 1 ⇒ 𝑘 = log 2 1 ⇒ 𝑘 = 0
1 ∞
∞
∞
𝑘
1 − (2)
1
1
1
1
1
1
= ∑ 2−𝑘 = ∑ ( ) = [
]= ×
=1
2
2
2
2 1−1
2 1/2
𝑘=0
𝑘=0
2
1 ∞
∵( ) =0
2
𝑛
1 − 𝑟 𝑛+1
𝑎
𝐺𝑒𝑜𝑚𝑒𝑡𝑟𝑖𝑐 𝑆𝑒𝑟𝑖𝑒𝑠: ∑ 𝑎𝑟 𝑘 = 𝑎 (
)=
, 𝑓𝑜𝑟 |𝑟| < 1
1−𝑟
1−𝑟
𝑘=0
51