Formula Sheet - Quantitative Analysis

Formula Sheet for CPIT 603 (Quantitative Analysis) PROBABILITY P(A or B) = P(A) + P(B) – P(A and B) Probability of any event: 0  P (event)  1 For Mutually exclusive events: P(A or B) = P(A) + P(B) P(AB) = P(A | B) P(B) Independent Events: P ( AB ) P(A and B) = P(A)P(B) Conditional Probability P ( A | B )  P(A | B) = P(A) P (B ) 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 Expected Value P( A | B ) 

P (B | A)  P( A) P(B | A)  P ( A)  P(B | A )  P ( A )

A, B A’

= any two events = complement of A

n

E  X    X i P X i  i 1

 X1P ( X1 )  X 2 P (X 2 )  ...  X n P (X n ) Xi = random variable’s possible values P(Xi) = probability of each of the random variable’s possible values n



= summation sign indicating we are adding

i 1

all n possible values E(X) = expected value or mean of the random variable n

 2  Variance   [X i  E (X)]2 P(X i ) i 1

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 Binomial Distribution n! Probability of r success in n trials  p r q nr r! (n  r )!

Expected value (mean) = np Variance = np(1 – p) Geometric Distribution Probability of number of trials until the first success

 (1  p ) x 1 p

Expected value (mean) = 1/p Variance = (1 – p)/p2 Normal Distribution (Continuous Distribution)

f (X) 

1

Variance   2

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 (b  a ) (b  a  1) 2  1   2  Variance  2 12 Poisson Distribution x e   P( X ) 

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 =  Exponential Distribution f ( X )  e  x

( x   ) 2 2 2

e  2 Completely specified by the mean,

Standard Deviation  



, and the standard

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)

Expected value =

1

= Average service time  1 Variance = 2 

The probability that an exponentially distributed time (X) required to serve a customer is less than or equal to time t is given by the formula P ( X  t )  1  e  t

Negative Binomial Distribution  x  1 xr  1  p  f (x )   pr  r  1

 = r/p = variance = r(1-p)/p2

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 ( a  b) (b  a ) 2   2  Variance  2 12

DECISION ANALYSIS Criterion of Realism Expected Monetary Value EMV(alternative) =  X i P ( X i ) Weighted average = (best in row) + (1 – )(worst in 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 Perfect Information EVPI = EVwPI – Best EMV 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)

Efficiency of sample information =

EVSI 100% EVPI

2

Y   0  1 X  

REGRESSION MODELS

Y  dependent variable (response) X  independent variable (predictor or explanatory)

 0  intercept (value of Y when X  0)  1  slope of the regression line   random error

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

Error = (Actual value) – (Predicted value)

X 

e  Y  Yˆ

Y  b1 

X n Y

 average (mean) of X values

 average (mean) of Y values n  ( X  X )(Y  Y )

(X  X )

2

b0  Y  b1 X Sum of Squares Total  SST   (Y  Y )

Sum of Squares Error  SSE   e 2   (Y  Yˆ ) 2

2

SST  SSR + SSE

Sum of Squares Regression  SSR   (Yˆ  Y ) 2

Correlation Coefficient = r   r 2 Standard Error of Estimate  s 

MSE

SSR MSR  k k  number of independent variables in the model

HypothesisTest H 0 :  1  0

degrees of freedom for the numerator = df1 = k degrees of freedom for the denominator = df2 = n – k–1 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

H 1 : 1  0

Reject if Fcalculated  F , df1 , df 2 df 1  k

SSR SSE 1– SST SST SSE Mean Squared Error  s 2  MSE  n  k 1 Generic Linear Model Y   0   1 X   MSR F Statistic : F  MSE Coefficient of Determination  r 2 

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 Yˆ = predicted value of Y

Adjusted r 2  1 

SSE /( n  k  1) SST /( n  1)

b0 = sample intercept (an estimate of 0) bi = sample coefficient of the i th variable (an estimate of i)

 Mean Absolute Deviation (MAD) 

FORECASTING forecast error n

 (error) Mean Squared Error (MSE) 

2

n

3

error

Mean Absolute Percent Error (MAPE) 

Mean Average Forecast 

 actual n

100%

Y  Yt 1  ...  Yt  n 1 sum of demands in previous n periods  Ft 1  t n n

Weighted Moving Average : Ft 1  Exponentia l Smoothing : Ft 1  Ft   (Yt  Ft )

 (Weight in period i)(Actual value in period)  w Y  ( Weights ) 1

t

 w2Yt 1  ...  wnYt  n 1 w1  w2  ...  wn

New forecast  Last period’ s forecast   (Last period’ s actual demand – Last period’ s forecast)

Exponential Smoothing with Trend : Ft 1  FITt   (Yt  FITt ) Tt 1  Tt   (Ft 1  FITt )

ˆ b b X Y 0 1 ˆ  predicted value where Y b0  intercept b1  slope of the line X  time period (i.e., X  1, 2, 3,  , n)

Yˆ  a  b1 X1  b2 X 2  b3 X 3  b4 X 4

FITt 1  Ft 1  Tt 1

Tracking signal 

RSFE  MAD

 (forecast error) MAD

INVENTORY CONTROL MODELS Q Average inventory level = 2

Annual ordering cost  Number of orders placed per year

Annual holding cost  Average Inventory

Economic Order Quantity Annual ordering cost = Annual holding cost

(Carrying cost per unit per year)

(Ordering cost per order) Annual Demand D  Co  Co Number of units in each order Q

D Q Order quantity  (Carrying cost per unit per year) Q Co  2 Ch 2 Q 2DCo  Ch EOQ  Q *  2 Ch 

Total cost (TC) = Order cost + Holding cost D Q TC  Co  Ch Q 2

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

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 2DCo Q*  IC 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)  (pt) – (dt) 4

 pt – dt  p

 Q Q d –d  Q 1 –  p p p 

Total produced Q  pt Q d  1 –   2 p D Annual setup cost  Cs Q Average inventory 

Production Run Model: EOQ without instantaneous receipt assumption Annual holding cost  Annual setup cost Q d D  1 –  Ch  Cs 2 p Q

Q* 

2DCs  d Ch  1 –  p 

Q d  1 –  Ch  2 p D Annual ordering cost  Co Q Annual holding cost 

D = the annual demand in units Q  number of pieces per order, or production run Quantity Discount Model 2DCo EOQ  IC If EOQ < Minimum for discount, adjust the quantity to Q = Minimum for discount Total cost  Material cost + Ordering cost + Holding cost Total cost  DC +

D Q Co + C h Q 2

Demand is variable but lead time is constant ROP  d L  Z  d L

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 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 constant but lead time is variable ROP  dL  Z  d L 

d  average daily demand  d  standard deviation of daily demand

L  average lead time  L  standard deviation of lead time

Safety Stock ROP = Average demand during lead time + Safety Stock Service level = 1 – Probability of a stockout Probability of a stockout = 1 – Service level





d  daily demand

L  lead time in days

Both demand and lead time are variable ROP  d L  Z L   d  2 d

2

2 L

Total Annual Holding Cost with Safety Stock Total Annual Holding Cost = Holding cost of regular inventory + Holding cost of safety stock THC 

Q Ch  (SS)Ch 2

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 P

ML ML + MP

5

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 variance

Value of work completed = (Percentage of work complete) x (Total activity budget) Crash cost/Time Period 

Crash cost  Normal Cost Normal time  Crash time

 b –a  6  

2

Variance = 

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 Z

Due date  Expected date of completion T

Activity difference = Actual cost – Value of work completed

WAITING LINES AND QUEUING THEORY MODELS Single-Channel Model, Poisson Arrivals, Multichannel Model, Poisson Arrivals, Exponential Exponential Service Times (M/M/1) Service Times (M/M/m) = mean number of arrivals per time period (arrival m = number of channels open = average arrival rate rate) = mean number of customers or units served per  = average service rate at each channel time period (service rate) The probability that there are zero customers in the The average number of customers or units in the system 1 system, L P  for m   L

  

The average time a customer spends in the system, W W 

1  

The average number of customers in the queue, Lq Lq 

2  (   )

The average time a customer spends waiting in the queue, Wq Wq 

  (   )

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

0

n = m –1



n =0







n 1   1         n!     m!   

m

m m  

The average number of customers or units in the system  ( /  ) m  L P0  2 (m – 1)!(m   )  The average time a unit spends in the waiting line or being served, in the system  ( /  ) m 1 L W  P0   2 (m – 1)!(m   )   The average number of customers or units in line waiting for service Lq  L 

 

The average number of customers or units in line waiting for service Wq  W 

1 Lq   

The average number of customers or units in line 6

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     

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

1 N

N!

    

 (N – n )!  n 0

n

Average length of the queue   Lq  N    1 – P0     Average number of customers (units) in the system L  Lq  1 – P0 

Average waiting time in the queue

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

Lq

Wq 

 m

k 1

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 P0 



(N – L) Average time in the system W  Wq 

1 

Probability of n units in the system Pn 

  N!    N – n !   

n

P0 for n  0,1,..., N

Constant Service Time Model (M/D/1) Average length of the queue Lq 

Little’s Flow Equations L = W (or W = L/) L q =  Wq (or Wq = Lq/)

2 2 (    )

Average waiting time in the queue

Average time in system = average time in queue + average time receiving service W = Wq + 1/

 Wq  2 (    )

Average number of customers in the system L  Lq 

 

Average time in the system W  Wq 

1 

MARKOV ANALYSIS 7

 (i) i

vector of state probabilities for period Pij = conditional probability of being in state j in the future given the current state of i  P11 P12  P1n  = (1, 2, 3, … , n)  P where P22  P2 n  21  P n = number of states      1, 2, … , n = probability of being in state 1,    Pm1 Pm 2  Pmn  state 2, …, state n For any period n we can compute the state Equilibrium condition probabilities for period n + 1  = P  (n + 1) =  (n)P Fundamental Matrix M represent the amount of money that is in each of the F = (I – B)–1 nonabsorbing states Inverse of Matrix M = (M1, M2, M3, … , Mn) n = number of nonabsorbing states  a b P  M = amount in the first state or category 1  c d M2 = amount in the second state or category d  b   M = amount in the nth state or category n 1  r   a b -1 r P    c a   c d   r   r r = ad – bc Partition of Matrix for absorbing states Computing lambda and the consistency index  n  I O =

P  A

B 

I = identity matrix O = a matrix with all 0s

CI 

n 1

Consistency Ratio CR 

CI RI

8

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

UCL x  x  A2 R

Lower control limit (LCL)  x  z x

LCL x  x  A2 R

x = mean of the sample means

= average of the samples A2 = Mean factor x = mean of the sample means R

z = number of normal standard deviations (2 for 95.5% confidence, 3 for 99.7%)  x = standard deviation of the sampling distribution x of the sample means = n

p-charts

UCL R  D4 R

UCL p  p  z p

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

LCL p  p  z p p = mean proportion or fraction defective in the sample

p

Total number of errors Total number of records examined

z = number of standard deviations  p = standard deviation of the sampling distribution  p is estimated by ˆ p Estimated standard deviation of a binomial distribution ˆ p 

p (1  p ) n

where n is the size of each sample c-charts Range of the sample = Xmax - Xmin The mean is c and the standard deviation is equal to c

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

c

(3 is

UCL c  c  3 c LCL c  c  3 c

9

OTHERS Computing lambda and the consistency index The input to one stage is also the output from  n another stage CI  sn–1 = Output from stage n n 1 The transformation function Consistency Ratio CI tn = Transformation function at stage n CR  General formula to move from one stage to RI 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 Fixed cost Probability of breaking even Break - even point (units)  break - even point   Price/unit – Variable cost/unit Z f  

sv

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

 Price Variable cost  –   (Mean demand) unit  unit   Fixed costs

EMV  

 K(break - even point – X)for X  BEPUsing the unit normal loss integral, EOL can be computed using $0for X  BEP 

Opportunity Loss  

where K = loss per unit when sales are below the break-even point X = sales in units

EOL = KN(D) EOL = expected opportunity loss K = loss per unit when sales are below the breakeven point  = standard deviation of the distribution N(D) = value for the unit normal loss integral for a given value of D D

 a  ad ae      AB   b    d e    bd be   C  c  cd ce       d    a b c    e    ad  be  cf   f    a b   e f   ae  bg af  bh           c d   g h   ce  dg cf  dh 

 – break even point 

a

b

c

d

Determinant Value = (a)(d) – (c)(b) a d g

b e h

c f i

Determinant Value = aei + bfg + cdh – gec – hfa – idb X

Numerical value of numerator determinant Numerical value of denominator determinant

10

 a  c

Original matrix  

b 

d 

 a b    c d

Determinant value of original matrix  ad  cb  d Matrix of cofactors    b  d  c

Adjoint of the matrix  

 c

 

d ad  cb c ad  cb

b   ad  cb  a   ad  cb 

 a 

Y2  a( X  X ) 2  b( X  X )  c

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)

Q = order quantity D = annual demand Co = ordering cost per order Ch = holding cost per unit per year C = purchase (material) cost per unit



a   b

Y1  aX 2  bX  c

D Q C o + C h  DC Q 2







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 Y – Y1 b   2 Change in X X X 2 – X1

TC 



1

For the Nonlinear function Y = X2 – 4X + 6 Find the slope using two points and this equation Change in Y Y Y – Y1 b   2 Change in X X X 2 – X1 Y  0

Y C Y X

n

Y  cX n 1 Y  Xn Y  g ( x)  h( x ) Y  g ( x)  h( x )

Y   nX n 1 Y   cnX n 1 n X n 1 Y   g ( x )  h( x ) Y   g ( x )  h( x ) Y 

Economic Order Quantity dTC – DCo Ch   dQ Q2 2 2DCo Q Ch d 2TC DCo  dQ 2 Q3

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