Formula Sheet - Quantitative Analysis

Formula Sheet for CPIT 603 (Quantitative Analysis) PROBABILITY

Probability of any event: 0  P (event)  1 P(A or B) = P(A) + P(B) – P(A and B) For Mutually exclusive events:

P(A or B) = P(A) + P(B) P(AB) = P(A | B) P(B) ) ( ) ( ) ( y Probabilit l Conditiona B B | A P AB P P  Independent Events: P(A and B) = P(A)P(B) P(A | B) = P(A)

Dependent Events:

P(A and B) = P(A) * P(B given A)

P(A and B and C) = P(A) * P(B given A) * P(C given A and B) Bayes’ Theorem ) ( ) ( ) ( ) ( ) ( ) ( ) ( A A | B A A | B A A | B B | A        P P P P P P P

A, B = any two events

A’ = complement of A Expected Value

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) X ( X ... ) X ( X ) X ( X X X X 2 2 1 1 1 n n n i i i P P P P E     

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

Xi = random variable’s possible values P(Xi) = probability of each of the random variable’s possible values

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

n

i 1

= summation sign indicating we are adding all n possible values

E(X) = expected value or mean of the random variable ) X ( )] X ( X [ 1 2 2

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    n i i i E P Variance 

Xi = random variable’s possible values E(X) = expected value of the random variable [Xi – E(X)] = difference between each value of the random variable and the expected value

P(Xi) = probability of each possible value of the random variable

2

Variance Deviation

Standard    

Discrete Uniform Distribution

For a series of n values, f(x) = 1/ n

For a range that starts from a and ends with b (a, a+1, a+2, …, b) and a  b

2 ) (ba   12 1 ) 1 ( 2 2 _{Variance} ba   Binomial Distribution r n r_q p r n r n _   )! ( ! ! n trials in success r of y Probabilit

Expected value (mean) = np Variance = np(1 – p) Geometric Distribution p p₎x 1 1 ( success first the until trials of number of y Probabilit _   Expected value (mean) = 1/p Variance = (1 – p)/p2 Poisson Distribution ! ) ( X X     e P x

P(X) = probability of exactly X arrivals or occurrences

 = average number of arrivals per unit of time

(the mean arrival rate), pronounced “lambda” e = 2.718, the base of natural logarithms X = number of occurrences (0, 1, 2, 3, …) Expected value = Variance = 

Normal Distribution (Continuous Distribution)

2 2 2 ) ( 2 1 ) (        x e f X

Completely specified by the mean, , and the standard

Exponential Distribution

x

e f_{( X)} 

X = random variable (service times)

 = average number of units the service facility can handle in a specific period of time

deviation,



    X Z

X = value of the random variable we want to measure  = mean of the distribution

= standard deviation of the distribution

Z = number of standard deviations from X to the mean, m

e = 2.718 (the base of natural logarithms)

2 1 = Variance time service Average = 1 = value Expected   t e t P_{(X )1}  formula by the given is t time to equal or than less is customer a serve to required (X) time d distribute lly exponentia an y that probabilit The

Negative Binomial Distribution

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

x r r p p r x f _ _      1 1 1 ) (x  = r/p _{= variance = r(1-p)/p}2

Continuous Uniform Distribution

For a series of n values, f(x) = 1/ (b –a)

For a range that starts from a and ends with b (a, a+1, a+2, …, b) and a  b

2 ) (ab   12 ) ( 2 2 _{Variance } ba  DECISION ANALYSIS Criterion of Realism

Weighted average = (best in row) + (1 – )(worst in row)

For Minimization:

Weighted average = (best in row) + (1 – )(worst in row)

Expected Monetary Value

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( )

= ative)

EMV(altern X_iP X_i

Xi = payoff for the alternative in state of nature i

P(Xi) = probability of achieving payoff Xi (i.e., probability of state of nature i)

∑ = summation symbol EMV (alternative i) = (payoff of first state of nature) x

(probability of first state of nature) + (payoff of second state of nature) x (probability of second state of nature) + … + (payoff of last state of nature) x (probability of last state of nature)

Expected Value with Perfect Information EVwPI = ∑(best payoff in state of nature i)

(probability of state of nature i)

EVwPI = (best payoff for first state of nature) x (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 Perfect Information EVPI = EVwPI – Best EMV

Expected Value of Sample Information

EVSI = (EV with SI + cost) – (EV without SI) EVPI100%

EVSI = n informatio sample of Efficiency

Utility of other outcome = (p)(utility of best outcome, which is 1) + (1 – p)(utility of the worst outcome, which is 0)

REGRESSION MODELS error random line regression the of slope 0) X when Y of (value intercept y) explanator or (predictor t variable independen X (response) variable dependent Y 1 0 1 0                X Y results sample on based , of estimate results sample on based , of estimate Y of value predicted ˆ ˆ 1 1 0 0 1 0        b b b b Y X Y

Error = (Actual value) – (Predicted value)

Y Y  ˆ  e X Y X X Y Y X X Y Y Y X X X 1 0 2 1 ) ( ) )( ( values of (mean) average values of (mean) average b b b n n          

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2 ) ( SST Total Squares of Sum  

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Y Y SSE + SSR SST

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   _SSE 2 ₍ ˆ₎2 Error Squares of Sum e Y Y

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  _{SSR (}ˆ ₎2 Regression Squares of Sum Y Y SST SSE – 1 SST SSR ion Determinat of t Coefficien _r2 _{ } Correlation Coefficient =_{r r}2 1 SSE MSE 2      k n s Error Squared Mean MSE  s Estimate of Error

Standard GenericLinearModelY 0 1X 

model in the s t variable independen of number k SSR MSR   k MSE MSR  F : Statistic F

degrees of freedom for the numerator = df1 = k degrees of freedom for the denominator = df2 = n – k – 1 0 : 0 : 1 1 1 0     H H Test Hypothesis 1 if Reject 2 1 , , _{1 2}      k n k F F_{calculated df df} df df      value -if Reject ) statistic test calculated ( value -p F P p 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 k k b b b b X X X Yˆ 0  1 1 2 2 ...

Yˆ = predicted value of Y

b0 = sample intercept (an estimate of 0) bi = sample coefficient of the i th variable (an estimate of i) ) 1 /( SST ) 1 /( SSE 1 Adjusted 2      n k n r FORECASTING n

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 forecast error (MAD) Deviation Absolute Mean n

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 2 ) error ( (MSE) Error Squared Mean

100% actual error (MAPE) Error Percent Absolute Mean n

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 n n n t t t t 1 1 1 ... periods n previous in demands of sum Forecast Average Mean           F Y Y Y n n t n t t t w w w w w w i            

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... ... ) Weights ( ) period in value Actual )( period in Weight ( 2 1 1 1 2 1 1 Y Y Y F : Average Moving Weighted forecast) s period’ Last – demand actual s period’ Last ( forecast s period’ Last forecast New ) ( 1         t t t t F Y F F : Smoothing l Exponentia 1 1 1 1 1 1 ) ( ) ( : Trend with Smoothing l Exponentia               t t t t t t t t t t t T F FIT FIT F T T FIT Y FIT F 

 X timeperiod(i.e.,X1,2,3, ,n) line the of slope b1 intercept b0 value predicted ˆ where ˆ 0 1         Y X Y b b 4 4 3 3 2 2 1 1 ˆ _{X X X X} Y ab b b b MAD error) (forecast MAD RSFE signal Tracking  

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INVENTORY CONTROL MODELS 2 = level inventory Average Q o o _QC D C order each in units of Number Demand Annual order) per cost (Ordering year per placed orders of Number cost ordering Annual     h C Q 2 year) per unit per cost (Carrying 2 quantity Order year) per unit per cost (Carrying Inventory Average cost holding Annual    

 Economic Order Quantity

Annual ordering cost = Annual holding cost

h o C 2 Q C Q D _ h o C DC Q 2 EOQ_ * _ Total cost (TC) = Order cost + Holding cost

h o C Q C Q D TC 2  

Cost of storing one unit of inventory for one year = Ch = IC, where C is the unit price or cost of an inventory item and I is Annual inventory holding charge as a percentage of unit price or cost

IC DC Q*_ 2 o ROP without Safety Stock:

Reorder Point (ROP) = Demand per day x Lead time for a new order in days

 d  L

Inventory position = Inventory on hand + Inventory on order

EOQ without instantaneous receipt assumption

Maximum inventory level  (Total produced during the production run) – (Total used during the production run)

 (Daily production rate)(Number of days production) – (Daily demand)(Number of days production)

       p d Q p Q d p Q p dt pt – – 1– Total produced Q  pt      p d Q – 1 2 inventory Average C_h p d Q      1– 2 cost holding Annual s C Q D  cost setup Annual Co Q D  cost ordering Annual

D = the annual demand in units

Q  number of pieces per order, or production run

Production Run Model: EOQ without instantaneous receipt assumption

Annual holding cost  Annual setup cost

s h _QC D C p d Q      _1– 2      p d C DC Q h s – 1 2 *

Quantity Discount Model IC

DC_o 2 EOQ

If EOQ < Minimum for discount, adjust the quantity to Q = Minimum for discount

Total cost  Material cost + Ordering cost + Holding cost

h o C Q C Q D DC 2 + + cost Total 

Holding cost per unit is based on cost, so Ch = IC Where I = holding cost as a percentage of the unit cost (C)

Safety Stock

ROP = Average demand during lead time + Safety Stock

Service level = 1 – Probability of a stockout Probability of a stockout = 1 – Service level

Safety Stock with Normal Distribution

ROP = (Average demand during lead time) + ZsdLT 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 variable but lead time is constant

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L

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Z L d  d  ROP days in time lead demand daily of deviation standard demand daily average    L d d 

Demand is constant but lead time is variable

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dL

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Z L d   ROP demand daily time lead of deviation standard time lead average    d L L  Both demand and lead time are variable

2 2 2

ROPdLZ Ld d L

Total Annual Holding Cost with Safety Stock

Total Annual Holding Cost = Holding cost of regular inventory + Holding cost of safety stock

h h C C Q (SS) 2 THC 

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 MP + ML ML  P

PROJECT MANAGEMENT

Expected Activity Time

6 + 4 + = a m b t 2 6 – = Variance       b a Earliest finish time = Earliest start time + Expected

activity time EF = ES + t

Earliest start = Largest of the earliest finish times of immediate predecessors

ES = Largest EF of immediate predecessors Latest start time = Latest finish time – Expected

activity time LS = LF – t

Latest finish time = Smallest of latest start times for following activities

LF = Smallest LS of following activities

Slack = LS – ES, or Slack = LF – EF Project Variance = sum of variances of activities on the critical path riance Project va deviation standard Project T  T  completion of date Expected date Due   Z Value of work completed = (Percentage of work

complete) x (Total activity budget)

Activity difference = Actual cost – Value of work completed Crash time time Normal Cost Normal cost Crash Period cost/Time Crash   

WAITING LINES AND QUEUING THEORY MODELS Single-Channel Model, Poisson Arrivals,

Exponential Service Times (M/M/1)

= mean number of arrivals per time period (arrival rate)

= mean number of customers or units served per time period (service rate)

The average number of customers or units in the system, L      L

The average time a customer spends in the system, W

   1 W

The average number of customers in the queue, Lq ) ( 2       q L

The average time a customer spends waiting in the queue, Wq ) (      q W

The utilization factor for the system,  (rho), the probability the service facility is being used

  

The percent idle time, P0, or the probability no one is in the system

Multichannel Model, Poisson Arrivals, Exponential Service Times (M/M/m)

m = number of channels open = average arrival rate

 = average service rate at each channel

The probability that there are zero customers in the system                            

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m m m m n P _m – m n n n for ! 1 ! 1 1 1 = 0 = 0

The average number of customers or units in the system        _   _{( – 1)!( )}2 0 ) / ( _P m m L m

The average time a unit spends in the waiting line or being served, in the system

       L P m m W m     1 ) ( )! 1 – ( ) / ( 0 2

The average number of customers or units in line waiting for service

   L Lq

The average number of customers or units in line waiting for service

  q q L W W  1 

   1 0 P

The probability that the number of customers in the system is greater than k, Pn>k

1       k k > n P  

waiting for service (Utilization rate) 

 

m 

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

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      N n n n – N N P 0 0 )! ( ! 1  

Average length of the queue

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1 P– 0

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N L_q            

Average number of customers (units) in the system

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1 P– 0

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L L _q 

Average waiting time in the queue  ) (N – L L W q q 

Average time in the system 

1

 Wq

W

Probability of n units in the system

N – n P n N N P n n _! for 0,1,..., ! 0        

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)

= (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

Constant Service Time Model (M/D/1)

Average length of the queue

) ( 2 2       q L

Average waiting time in the queue

) ( 2      q W

Average number of customers in the system 

  Lq

L

Average time in the system 

1

 Wq

W

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/

 (i) = vector of state probabilities for period i

= (1, 2, 3, … , n) where

n = number of states

1, 2, … , n = probability of being in state 1, state 2, …, state n

Pij = conditional probability of being in state j in the future given the current state of i

             mn m m n n P P P P P P P P P        2 1 2 22 21 1 12 11 P

For any period n we can compute the state probabilities for period n + 1

 (n + 1) =  (n)P Equilibrium condition  = P Fundamental Matrix F = (I – B)–1 Inverse of Matrix                             r a r c r b r d d c b a d c b a 1 1 -P P r = ad – bc

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

Partition of Matrix for absorbing states

       B A O I P I = identity matrix O = a matrix with all 0s

Computing lambda and the consistency index

1 CI    n n  Consistency Ratio RI CI CR

STATISTICAL QUALITY CONTROL x x z x z x       (LCL) limit control Lower (UCL) limit control Upper

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 =

n x  R A x R A x x x 2 2 LCL UCL    

R = average of the samples

A2 = Mean factor

x = mean of the sample means

R D R D R R 3 4 LCL UCL  

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 p p p p z p z p       LCL UCL

p_{= mean proportion or fraction defective in the sample}

examined records of number Total errors of number Total  p

z = number of standard deviations

p

 _{= standard deviation of the sampling distribution}

p

 _{is estimated by} ˆp

Estimated standard deviation of a binomial distribution n p p p ) 1 ( ˆ   

where n is the size of each sample c-charts

The mean is c and the standard deviation is equal to

c

To compute the control limits we use c3 c (3 is used for 99.7% and 2 is used for 95.5%)

c c c c c c 3 LCL 3 UCL    

OTHERS

Computing lambda and the consistency index

1 CI    n n  Consistency Ratio RI CI CR

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

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n n

 

n n

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n n a s b d c s 1     Return Equations

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n n

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n n

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n n a s b d c r      v s f    cost/unit Variable – Price/unit cost Fixed (units) point even

-Break Probability of breaking even



  break-even point Z

P(loss) = P(demand < break-even) P(profit) = P(demand > break-even)

costs Fixed demand) (Mean unit cost Variable – unit Price EMV                BEP X $0for BEP X X)for – point even -K(break Loss y Opportunit where

K = loss per unit when sales are below the break-even point

X = sales in units

Using the unit normal loss integral, EOL can be computed using

EOL = KN(D)

EOL = expected opportunity loss

K = loss per unit when sales are below the break-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

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C AB                         ce be ae cd bd ad e d c b a

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



ad be cf

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f e d c b a                                 dh cf dg ce bh af bg ae h g f e d c b a d c b a

Determinant Value = (a)(d) – (c)(b)

i h g f e d c b a

Determinant Value = aei + bfg + cdh – gec – hfa – idb t determinan r denominato of value Numerical t determinan numerator of value Numerical  X

                     a c b d a b c d cb ad d c b a matrix the of Adjoint cofactors of Matrix matrix original of t value Determinan matrix Original                       cb ad a cb ad c ad cb b cb ad d d c b a 1

Equation for a line Y = a + bX

where b is the slope of the line

Given any two points (X1, Y1) and (X2, Y2)

1 2 1 2 – – in Change in Change X X Y Y X Y X Y b     

For the Nonlinear function Y = X2 _{– 4X + 6}

Find the slope using two points and this equation 1 2 1 2 – – in Change in Change X X Y Y X Y X Y b      c b a c b a           ) ( ) ( X 2 2 2 1 X X X X Y X Y 2 1 2 Y ( X) 2 X( X) ( X) Y Y          b a c X X X X X X X X X X X Y                    c a b c a b X c a b 2 ) 2 ( ) ( ) ( 2 ) ( 2 ) ( ) ( ) ( ) ( 1 x h x g x h x g c n n n         Y Y X Y X Y X Y C Y ) ( ) ( ) ( ) ( 0 1 1 1 x h x g x h x g n cn n n n n                       Y Y X Y X Y X Y Y

Total cost  (Total ordering cost) + (Total holding cost) + (Total purchase cost)

DC C Q C Q D TC o h  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

Economic Order Quantity 2 – 2 o Ch Q DC Q TC _{ } d d h o C DC Q 2 3 2 2 Q DC Q TC _o  d d