Example
The probability of successful start of a certain engine is ¼ and
four trials are to be made. Evaluate the individual and
cumulative probabilities of success in this case.
n = 4,
p=¼,
q=1–¼=¾
(p+q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4
Number of
successes
0
1
2
3
4
failures
4
3
2
1
0
Individual probability
q4 = (3/4) 4
= 81/256
4pq3 = 4(1/4)(3/4) 3 = 108/256
6p2q2 = 6(1/4) 2 (3/4) 2 = 54/256
4p3q = 4(1/4) 3 (3/4) = 12/256
= 1/256
p4 = (1/4) 4
Σ=1
Cumulative
probability
81/256
189/256
243/256
255/256
256/256
Binomial Distribution Application
Example:
It is known that, in a certain manufacturing process, 1% of the products are
defective. If the a customer purchases 200 of these products selected at
random, what is the expected value and standard deviation of the number of
defects?
n = 50
q = 0.01
p = 1 – 0.01 = 0.99
E(defects) = n.q = 200 x 0.01 = 2
σ(defects) = √ npq = √200 x 0.01 x 0.99 = 1.41
1
Example:
The manufacturing company has a policy of replacing, free-of-charge,
all defective products that are purchased.
If the product manufacturing cost is $10 per unit and each product is
sold for $15, how much profit is made from a sale of 1000 products?
q = 0.01
For 1000 products, n = 1000
Expected # of defects, E(defects) = n.q = 10
Therefore, 1010 products must be manufactured to sell 1000 products.
Manufacturing cost = $10 x 1010 = $10,100
Income = $15 x 1000 = $15,000
Profit = $15,000 - $10,100 = $4900
Profit per unit = $4900/1000 = $4.90
Example:
If the company decides to increase the manufacturing cost to $10.05
per unit in order to decrease the probability of defects to 0.1%,
q = 0.001
For 1000 products, n = 1000
Expected # of defects, E(defects) = n.q = 1
Therefore, 1001 products must be manufactured to sell 1000 products.
Manufacturing cost = $10.05 x 1001 = $10,060.05
Income = $15 x 1000 = $15,000
Profit = $15,000 - $10,060.05 = $4939.95
Profit per unit = $4939.95/1000 = $4.94
2
Effect of Redundancy
Consider a system consisting of 4 identical components,
each having a failure probability of 0.1.
q = 0.1
(p = 0.9)
(p+q)4 = p4 + 4p3q + 6p2q2 + 4pq3 + q4
n=4
S y s te m st a t e
a ll c o m p o n e n t s w o r k in g
3 w o r k in g , 1 f a ile d
2 w o r k in g , 2 f a ile d
1 w o r k in g , 3 f a ile d
a ll c o m p o n e n t s fa ile d
I n d iv id u a l p r o b a b ilit y
p4
= (0 .9 ) 4
= 0 .6 5 6 1
4 p 3 q = 4 (0 .9 ) 3 (0 .1 ) = 0 .2 9 1 6
6 p 2 q 2 = 6 (0 .9 ) 2 (0 .1 ) 2 = 0 .0 4 8 6
4 p q 3 = 4 (0 .9 )( 0 .1 ) 3 = 0 .0 0 3 6
q4
= (0 .1 ) 4
= 0 .0 0 0 1
Σ = 1
Consider 4 criteria
System reliability, R
all components required for success
(no redundancy)
0.6561
3 components required for success
(partial redundancy)
0.6561+0.2916
= 0.9477
2 components required for success
(partial redundancy)
0.6561+0.2916+0.0486
= 0.9963
1 component required for success
(full redundancy)
0.6561+0.2916+0.0486+0.0036
= 0.9999
System with Derated States
Consider a generation plant with two 10 MW units, each having a
probability of failure (forced outage rate) of 10%.
q = 0.1, p = 0.9,
Binomial Distribution:
n=2
(p + q)2 = p2 + 2pq + q2
Capacity Outage Probability Table:
Units Out Cap Out (MW)
0
0
1
10
2
20
Cap In (MW) Probability
20
0.81
10
0.18
0
0.01
Cum. Prob
1
0.19
0.01
If the generation plant operates to supply a 15 MW load,
what is the probability of load loss (system failure)?
Probability of load loss = Loss of Load Probability (LOLP) = 0.19
Expected # of days of load loss = 0.19 x 365 = 69.35 days/yr
Loss of Load Expectation (LOLE)
3
System with Derated States
If the generation plant operates to supply a 15 MW load,
what is the Expected Load Loss (ELL)?
Capacity Outage Probability Table:
Units Out Cap Out (MW)
0
0
1
10
2
20
Cap In (MW) Probability
20
0.81
10
0.18
0
0.01
Cum. Prob
1
0.19
0.01
Units Out Cap Out (MW) Cap In (MW) Load Loss (MW) Probability Col.4 x Col.5
0
0
20
0
0.81
0
1
10
10
5
0.18
0.90
2
20
0
15
0.01
0.15
1.05
Expected Load Loss (ELL) = 1.05 MW
Example
A generating plant is to be designed to satisfy a constant 10 MW load.
Four alternatives are being considered:
a)
b)
c)
d)
1 x 10 MW unit
2 x 10 MW units
3 x 5 MW units
4 x 3.33 MW units
The probability of unit failure is assumed to be 0.02.
For each unit,
q = forced outage rate (FOR) = unavailability, U = 0.02
p = availability, A = 0.98
4
Capacity Outage Probability Tables
units
out
capacity
(MW)
out
in
Binom.
Distr.
Individual
prob
cum.
prob
0.98
0.02
1.00
0.01
0.9604
0.0392
0.0004
1.0000
0.0396
0.0004
LOLP = 0.0004
LOLE = 0.15 d/yr
0.941192
0.057624
0.001176
0.000008
1.000000
0.058808
0.001184
0.000008
LOLP = 0.001184
LOLE = 0.43 d/yr
0.92236816
0.07529536
0.00230496
0.00003136
0.00000016
1.00000000
0.07763184
0.00233648
0.00003152
0.00000016
(a) 1 x 10 MW
0
1
0
10
10
0
A
U
0
1
2
0
10
20
20
10
0
A
2AU
2
U
0
1
2
3
(d) 4 x 3.33 MW
0
1
2
3
4
0
5
10
15
15
10
5
0
A
2
3A U
2
3AU
3
U
0
3.33
6.66
10
13.33
13.33
10
6.66
3.33
0
A
3
4A U
2 2
6A U
3
4AU
4
U
(b) 2 x 10 MW
2
LOLP = 0.01
LOLE = 365 x 0.02
= 7.3 d/yr
(c) 3 x 5 MW
3
4
LOLP = 0.00233648
LOLE = 0.85 d/yr
Expected Load Loss
Capacity (MW)
Out
In
(a) 1 x 10 MW
0
10
10
0
Load Loss
Li (MW)
Prob
pi
Li x pi
(MW)
0
10
0.98
0.02
0
0.2
0.2
(b) 2 x 10 MW
0
10
20
20
10
0
0
0
10
0.9604
0.0392
0.0004
0
0
0.004
0.004
(c) 3 x 5 MW
0
5
10
15
15
10
5
0
0
0
5
10
0.941192
0.057624
0.001176
0.000008
0
0
0.00588
0.00008
0.00596
13.33
10
6.67
3.33
0
0
0
3.33
6.67
10
0.92236816
0.07529536
0.00230496
0.00003136
0.00000016
0
0
0.00768320
0.00020907
0.00000160
0.00789387
(d) 4 x 3.33 MW
0
3.33
6.67
10
13.33
ELL = 0.2 MW
= 200 kW
ELL = 0.004 MW
= 4.00 kW
ELL = 0.00596 MW
= 5.96 kW
ELL = 0.00789 MW
= 7.89 kW
5
Comparative Analysis
Effect of unit unavailability:
System
(a) 1 x 10 MW
(b) 2 x 10 MW
(c) 3 x 5 MW
(d) 4 x 3.33 MW
ELL (kW) at Different Unit FOR’s
2%
4%
6%)
200
400
600
4.00
16.00
36.00
5.96
23.68
52.92
7.89
31.12
69.09
System with Non-identical Components
All components must be identical to apply the Binomial Distribution.
If components of a system have non-identical capacities:
-
Units with identical capacities are grouped together
COPT is developed for each group
COPT for different groups are combined, one at a time
Final COPT for the system is used for reliability evaluation
A pumping station has 2 x 20 t/hr units, each having an unavailability
of 0.1, and 1 x 30 t/hr unit with an unavailability of 0.15.
Calculate the capacity outage probability table for this plant.
COPT for 2 x 20 t/hr units:
Units Cap Out Cap In
Out
(t/hr)
(t/hr)
0
0
40
1
20
20
2
40
0
Prob
0.81
0.18
0.01
COPT for 1 x 30 t/hr unit:
Units Cap Out Cap In
Out
(t/hr)
(t/hr)
0
0
30
1
30
0
Prob
0.85
0.15
6
System with Non-identical Components
Combining COPTs:
1 x 30
t/hr unit
30 / 0.85
0 / 0.15
2 x 20 t/hr units
40 / 0.81
20 / 0.18
0 / 0.01
70 / 0.6885 50 / 0.153 30 / 0.085
40 / 0.1215 20 / 0.027 0 / 0.0015
Each cell contains: Capacity In / probability
Overall System COPT:
Cap In
(t/hr)
70
50
40
30
20
0
Cap Out
(t/hr)
0
20
30
40
50
70
Prob
0.6885
0.1530
0.1215
0.0850
0.0270
0.0015
7