Download The Poisson Distribution

Design and Uncertainty
Discussion Points
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Mean
Variance
Standard Deviation
Probabilities
Uniform Distribution
Normal Distribution
Computational Form for Variance and
Standard Deviation
 Poisson Distribution
 Manufacturing Scheduling and Uncertainty
Mean ( µ )
Given a set of N numbers, X1 , X2 , ... , XN , the mean, or
average, is calculated by summing the numbers and dividing by
N.
N
∑ Xi
i=1
µ = --------N
For example, for the numbers 3, 4, 5
µ = (3+4+5)/3
= 12 / 3
= 4
Variance ( σ2 )
The variance is a measure of the spread of a distribution of
numbers. It is computed as the average squared deviation of
each number from the mean of the distribution.
N
∑ ( X i - µ)2
i=1
σ = --------------N
2
For the numbers 3, 4, 5
σ2 = ( (3 – 4)2 + (4 – 4)2 + (5 – 4)2 ) / 3
= (1 + 0 + 1) / 3
= 0.667
Standard Deviation ( σ )
The standard deviation of a distribution of numbers is the
square root of the variance.
For the numbers 3, 4, 5
σ2 = 0.667
σ
= 0.816
Another Example
For the numbers 2, 4, 6
µ
= (2+4+6)/3
= 12 / 3
= 4
σ2 = ( (2 – 4)2 + (4 – 4)2 + (6 – 4)2 ) / 3
= (4 + 0 + 4) / 3
= 2.667
σ
= 1.633
Coin Toss
Possible Outcomes
(Head)
(Tail)
Each outcome has a probability of 1/2.
Single Standard Die
Possible Outcomes
(1) (2) (3) (4) (5) (6)
Each outcome has a probability of 1/6.
These are both Uniform Distributions.
Pair of Standard Dice
Possible Outcomes
(1,1)
(1,2)
(1,3)
(1,4)
(1,5)
(1,6)
(2,1)
(2,2)
(2,3)
(2,4)
(2,5)
(2,6)
(3,1)
(3,2)
(3,3)
(3,4)
(3,5)
(3,6)
(4,1)
(4,2)
(4,3)
(4,4)
(4,5)
(4,6)
(5,1)
(5,2)
(5,3)
(5,4)
(5,5)
(5,6)
(6,1)
(6,2)
(6,3)
(6,4)
(6,5)
(6,6)
Frequency Distribution
2
3
4
5
6
7
8
9
10
11
12
-
1
2
3
4
5
6
5
4
3
2
1
This is a Normal Distribution.
Probability
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
The 68-95-99.7 Rule
All normal density curves satisfy the following property, which
is often referred to as the Empirical Rule.
68%
of the observations fall within 1 standard deviation of the
mean, that is between  -  and  + .
95%
of the observations fall within 2 standard deviations of the
mean, that is between  - 2 and  + 2.
99.7%
of the observations fall within 3 standard deviations of the
mean, that is between  - 3 and  + 3.
For a pair of Standard 6-Sided Dice:
=7
2 = 5.833
 = 2.415
 -  to  +  = 4.585 - 9.415  24 / 36 = 66.666%
 - 2 to  + 2 = 2.17 - 11.83  34 / 36 = 94.444%
 - 3 to  + 3 = -0.245 - 14.245  36 / 36 = 100.0%
Computation Form for Variance
Computational form for the variance in a population.
All summations are assumed to be from i equals 1 to N.
∑ (X - µ)2
σ2 = --------------N
∑ (X2 - 2 X µ + µ2)
= ---------------------------N
∑ X2 - 2 µ ∑ X + ∑ µ2
= -------- ----------- -----N
N
N
∑ X2 - (2 µ) µ + µ2
= -------N
∑ X2 - µ2
σ2 = -------N
The Poisson Distribution
The Poisson distribution is used to model the number of random
occurrences of some phenomenon in a specified unit of space or
time. For example,

The number of phone calls in a 30-minute period.

The number of defective parts produced during an eight
hour shift.

The number emergency assistance calls received in an
hour.
For a Poisson random variable, the probability that X is some
value x is given by the formula:
x e-
P (X = x) = -------x!
x = 0, 1, . . .
where:  : mean number of occurrences in a
specified interval (time period).
x : number of occurrences of interest.
e : Base of the natural logarithmic function
ln (  2.71828 )
Example
On an average day in January, a furnace repair person receives 4
calls for emergency furnace repair service. Find the probability
that this person will receive 5 calls today. (i.e.,  = 4, X = 5)
45 e-4
P (X = 5) = -------- = 0.1563
5!
Find the probability that this person will receive fewer than 4
calls today. (i.e.,  = 4, X < 4)
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
= 0.0183 + 0.0733 + 0.1465 + 0.1954
= 0.4335
Problem Statement
The plant manager of a toy manufacturing plant has been given an order
for 10,000 new toys with a shipping date 10 weeks from now. He
currently has 8 machines that produce 100 toys per week per machine,
assuming one shift of 40 hours per week. He needs to know if the
machines need to run overtime to meet the order and if so how much
overtime is necessary. The goal is to minimize overtime.
Calculate how many toys the 8 machines will produce in 10 weeks
working one 40 hour shift. Do the machines need to run additional shifts
(each shift for each machine is 40 hours per week)?
Write a formula for calculating the number of weeks with an extra shift
to meet this order in 10 weeks.
Now, in actuality, the machines have been known to produce on the
average 2 defective toys in a week (one shift) per machine. Also, the
machines periodically require maintenance. This occurs on the average
of 1 machine down for maintenance per week (one shift), with the
maintenance requiring 1 hour. With these uncertainties calculate the
necessary extra weeks of overtime to meet the order.
Given:
Ttotal = 10,000 toys
8 machines
10 weeks production time
100 toys/machine . shift-week
Determine R, the rate of production for all 8 machines for one shiftweek
R = (8 machines * 100 toys/ machine . shift-week)
= 800 toys/shift-week
Determine Tp, the production for 10 one shift-weeks
Tp = 800 toys/shift-week * 10 shift-weeks
Tp = 8000 toys
But we need 10000, so we are 2000 short.
How many shift-weeks to add (as overtime) to production?
2000 toys * 1 shift-week/800 toys = 2.5 shift-weeks of overtime
Write a formula for calculating the number of weeks with an extra shift
to meet this order in 10 weeks.
S (shift-weeks) = T (toys) / R (toys/shift-week)
Dealing with uncertainty
The number of defective toys produced are the number of toys lost each
shift-week.
Average 2 defective toys in one shift-week per machine.
Use Poisson Distribution as model.
λ =2
P(X = 0) = 0.1353
P(X = 1) = 0.2707
P(X = 2) = 0.2707
P(X = 3) = 0.1804
P(X = 4) = 0.0902
---------------------P(X = 5) = 0.0361
Let L = toys lost/shift-week, based on above probability, choose X = 4
L = 8 machines * 4 toys lost/machine . shift-week
L = 32 toys lost/shift-week
Tlost = L * 12.5 shift-weeks
Tlost = 32 toys lost/shift-week * 12.5 shift weeks = 400 toys lost
Substitute into formula:
Shift-weeks = 400 toys / 800 toys/shift-week = 0.5 shift-weeks
Dealing with uncertainty, Part 2
Average 1 machine down for maintenance in one shift-week with
maintenance requiring 1 hour to complete.
Use Poisson Distribution as model.
λ =1
P(X = 0) = 0.3679
P(X = 1) = 0.3679
P(X = 2) = 0.1839
P(X = 3) = 0.0613
----------------------P(X = 4) = 0.0153
P(X = 5) = 0.0031
Let H = hours lost to maintenance, choose X = 3 machines
H = 3 machines * 1 hr/machine . shift-week = 3 hrs/shift-week
3 hr/shift-week * 13.0 shift-weeks = 39.0 hours
39.0 hours * 2.5 toys/hour = 97.5 toys lost
Substitute into formula:
Shift-weeks = 97.5 toys / 800 toys/ shift-weeks = 0.1219
Putting it all together:
Total production time =
12.5 shift-weeks + 0.5 shift-weeks + 0.1219 shift-weeks
Total production time = 13.1219 shift-weeks
Any time over 10 shift-weeks is overtime. Therefore the plant manager
will have to schedule 4 shift-weeks of overtime during the 10 week run
of production on this new toy.