Transcript PDF-4-ExpoPoissx

Basics Probability Distributions- Uniform
Ardavan Asef-Vaziri
Jan.-2016
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Basic Probability Distributions
How can it be that mathematics, being
after all a product of human thought
independent of experience, is so
admirably adapted to the objects of
reality
Albert Einstein
Some parts of these slides were prepared based on
Essentials of Modern Busines Statistics, Anderson et al. 2012, Cengage.
Managing Business Process Flow, Anupindi et al. 2012, Pearson.
Project Management in Practice, Meredith et al. 2014, Wiley
Continuous Probability
Distributions
Exponential and Poisson
Distributions
Poisson Probability Distribution1
Poisson Probability Distribution 2
Exponential and Poisson Relationship
Exponential: 10 minutes time interval
Poisson: 1 per 10 minutes
0.12
0.40
0.35
0.10
0.30
0.08
0.25
0.06
0.20
0.15
0.04
0.10
0.02
0.05
0.00
0.00
0
5
10
15
20
25
30
35
0
1
2
3
4
5
Poisson: 3 per 30 minutes
0.25
0.20
0.15
0.10
0.05
0.00
0
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2
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Exponential Probability Distribution
The exponential random variables can be used to describe:
 Time between vehicle arrivals at a toll booth.
 Distance between major defects in a highway.
 Time required to complete a questionnaire.
 Time it takes to complete a task.
In waiting line applications, the exponential distribution is often
used for interarrival times and service times.
In real-life applications it is valid for interarrival times but not for
service times. However, since it has some neat mathematical
futures, it is used for service times too.
A property of the exponential distribution is that the mean and
standard deviation are equal.
The exponential distribution is skewed to the right. Its skewness
measure is 2.
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Exponential Probability Distribution
𝑓 𝑥 =
1 −𝑥/𝜇
𝑒
𝜇
for x ≥0
µ= expected or mean in terms of time
Rate per unit of time = 1/
Mean = StdDev
If µ= 5 min, compute f(2)
f(2) =EXPON.DIST(2,1/5,0) =0.134
𝑃 𝑥 ≤ 𝑋1 = 1 − 𝑒 −𝑥/𝜇
Exponential: 10 minutes time interval
0.12
0.10
𝑃 𝑥 ≤ 2 = 1 − 𝑒 −2/5
0.08
0.06
=EXPON.DIST(2,1/5,1)
0.04
0.02
=0.32968
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0
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Exponential Probability Distribution
=EXPON.DIST(2,1/5,1) =0.32968
𝑃 𝑥 ≥ 𝑋1 = 1 − 𝑥 ≤ 𝑋1 = 1 − 1 + 𝑒 −𝑥/𝜇
𝑃 𝑥 ≥ 𝑋1 = 𝑒 −𝑥/𝜇
=EXP(-2/5) =0.67032
𝑥
𝑃 𝑥 ≤ 𝑋1 = 1 − 𝑒
−𝜇
= 1 − 0.67032 = 0.32968
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Exponential Probability Distribution
The time between arrivals of cars at Al’s gas station follows an
exponential probability distribution with a mean time between
arrivals of 3 minutes. Al would like to know the probability that
the time between two successive arrivals will be 2 minutes or less.
f(x)
=EXPON.DIST(2,1/3,1)
P(x < 2) = 1- e-2/3 =EXP(-2/3)
.4
0.4865829
0.5134171
.3
P(x ≥ 2) =1- P(x < 2) = 1-1+ e-2/3 = e-2/3
.2
.1
x
0
1
2
3
4
5
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7
8
9 10
Time Between Successive Arrivals (mins.)
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Exponential Probability Distribution
Average trade time in Ameritrade is one second. Ameritrade has
promised its customers if trade time exceeds 5 second it is free (a
$10.99 cost saving. The same promises have been practiced by
Damion Pizza (A free regular pizza) and Wells Fargo ($5 if
waiting time exceeds 5 minutes). There are 150,000 average daily
trade. What is the cost to Ameritrade”
P(x≥ X0) = e-X0/= e-5/1 = 0.006738
Probability of not meeting the promise is 0.6738%
0.006738*150,000* = 1011 orders
@10.99 per order = 10.99*1011 = $11111 per day
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Exponential Probability Distribution
What was the cost if they had improved their service level by 50%
that is to make it free for transactions exceeding 2.5 secs.
e-2.5/1 = 0.082085
8.2085%*150,000*10.99 = $135317 per day
We cut the promised time by half, our cost increased 12 times.
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Exponential Probability Distribution
In a single phase single server service process and exponentially
distributed interarrival time and service times, the actual total
time that a customer spends in the process is also exponentially.
Suppose total time the customers spend in a pharmacy is
exponentially distributed with mean of 15 minutes. The pharmacy
has promised to fill all prescriptions in 30 minutes. What
percentage of the customers cannot be served within this time
limit?
P(x≥30) = EXP(-30/15) = 0.1353
13.53% of customers will wait more than 30 minutes.
= P(x≤30) = EXPON.DIST(30,1/15,1)
= P(x≤30) = 0.864665
P(x ≥ 30) = 1- P(x≤30) = 1- 0.864665 = 0.1353
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Exponential Probability Distribution
90% of customers are served in less than what time interval?
1-e-X0/ = 0.9
Find X0
0
0.064493
0.124827
0.181269
0.234072
0.283469
0.32968
0.372911
0.413354
0.451188
0.486583
0.519695
0.550671
0.57965
0.606759
0.632121
0
1
2
3
4
5
6
7
8
9
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Chart Title
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20
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0
0
0.1
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0.3
0.4
SOLVER
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Exponential Random Variable
P(x ≤ X0) = 1-e(-X0/µ)
P(x ≤ X0) = rand() = 1-e(-X0/µ)
1-rand() = e(-X0/µ)
1-rand() by itself is a rand()
rand() = e(-X0)/µ)
e(-X0/µ) = rand()
X0= -µrand()
x= -µrand()
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Exponential & Poisson
The Poisson distribution provides an appropriate
Description of the number of occurrences per interval
The exponential distribution provides an appropriate description
of the length of the interval between occurrences
One customer arrives per 15 minutes.
The average number of customers arriving in 30 mins is 2.
This is Poisson distribution.
=POISSON.DIST(3,2,1) =0.857123
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Poisson Probability Distribution
 The number of knotholes in 14 linear feet of pine board
 The number of vehicles arriving at a toll booth in one hour
 Bell Labs used the Poisson distribution to model the arrival of
phone calls.
 A Poisson distributed random variable is often useful in
estimating the number of occurrences over a specified interval
of time or space.
 It is a discrete random variable that may assume an infinite
sequence of values (x = 0, 1, 2, . . . ).
 The probability of an occurrence is the same for any two
intervals of equal length.
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Poisson Probability Distribution
 The occurrence or nonoccurrence in any interval is
independent of the occurrence or nonoccurrence in any other
interval.
 Since there is no stated upper limit for the number of
occurrences, the probability function f(x) is applicable for
values x = 0, 1, 2, … without limit.
 In practical applications, x will eventually become large enough
so that f(x) is approximately zero and the probability of any
larger values of x becomes negligible.
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Poisson Probability Distribution
f ( x) 
 x e
x!
x = the number of occurrences in an interval
f(x) = the probability of x occurrences in an interval
 = mean number of occurrences in an interval
e = 2.71828
x! = x(x – 1)(x – 2) . . . (2)(1)
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Poisson Probability Distribution
More than 50 million guests stay at bed and breackfacts each year.
The websit for B&B Inns of North America which averages
7
visitors per minute,
enables many B&B to attract quests
a) Compute the probability of
0
website visotor in a one muinute period
b) Compute the probability of
2
or more website visotor in a one muinute period
c) Compute the probability of
1
or more website visotor in a 30 second period
d) Compute the probability of
5
or more website visotor in a one muinute period
e) Compute the probability that the interarrival between two consequtive customers exceed
0.000912
0.007295
0.030197
0.172992
20 seconds
0.000912
0.992705
0.969803
0.827008
Patients arrive at the emergency room of Mercy Hospital at the average rate of
7
per hour on weekend evenings.
enables many B&B to attract quests
a) Compute the probability of
4
arrivals in 30 minutes on a weekend evening?
b) Compute the probability of
2
or less arrivals in 30 minutes on a weekend evening?
c) Compute the probability of
3
or more arrivals in 30 minutes on a weekend evening? 0.188812
d) Compute the probability of
12
or more arrivals in two hous on a weekend evening? 0.320847
e) Compute the probability that the interarrival between two consequtive customersis less than 0.679153
0.73996
10 minutes.
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Simulation of Break-Even Analysis
F
V
P
Q
Uniform Uniform Normal Exponential
80000
10
40
9000
120000
30
8
Fixed Cost =INT(A$3+(A$4-A$3)*RAND())
Variable Cost=INT(B$3+(B$4-B$3)*RAND())
Sales Price=INT($C$3+$C$4*NORM.S.INV(RAND()))
Sales =-INT($D$3*LN(RAND()))
Uniform
F
80000
120000
80433
88984
108015
111706
110011
99404
95878
97437
82867
83784
83399
98222
91725
101480
Uniform
V
10
30
10
18
22
20
19
28
28
18
21
17
23
28
18
23
Normal
P
40
8
41
43
49
54
33
36
42
37
59
40
47
27
39
41
Exponential
Q
9000
profit
Profit Sorted
Cumulati
17614
503
13510
7484
9725
3850
4391
16448
10671
5008
2041
1643
7566
6859
465601
-76409
256755
142750
26139
-68604
-34404
215075
322631
31400
-34415
-99865
67161
21982
-312776
-299437
-274902
-269786
-256009
-213930
-210306
-197208
-191990
-191812
-182333
-176777
-173914
-170164
0.0002
0.0004
0.0006
0.0008
0.001
0.0012
0.0014
0.0016
0.0018
0.002
0.0022
0.0024
0.0026
0.0028
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Simulation of Break-Even Analysis
Probability of Failure: 48.8%
Min Profit:-312776 Max Profit:2460856
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-500000
0
500000
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1000000
1500000
2000000
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Simulation



Simulation helps us to overcome our shortcomings in analysis
of complex systems using statistics, and also to see the
dynamics of the system.
Statistics vs. Simulation: To compute probability of
completion time or cost of a network of activities.
 Both must enumerate all the paths to compute the
probability
 Statistics assume path interdependence while simulation
does not
For Simplicity, Triangular distribution is used to estimate
Beta distribution.
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