A critical component in a production line is known to have a constant failure rate of 0.25
failures per month. If the cycle of procurement of this component takes at least 12 months,
what should be the minimum stock of this component in order to keep the probability of
stoppage of production down to 0.05.
Probability of stoppage = Probability of failure of all components including the stock.
X is the number of failures in the period of T= 12 months.
X is RANDOM variable distributed according to Poisson’s distribution with
Mean = λ T , where λ = 0.25 failures/month. λ T = 0.25*12 = 3 Failures/ 12 m0nth
Poisson’s distribution is given as follows:
M = Minimum Stock
P( X 

T  y
y) 
y!
F  P( X  M )  1  P( X  M )
F  0.05  1  e  T
K M

K 0
e  T
K M

K 0
T K
K!
 0.95
T K
K!
e  T
Evaluate M that satisfies the last equation
by TRIALS &ERROR
Minimum Stock = M = 6
Prof. Ahmed Farouk Abdul Moneim
e  T
K M

T K
K 0
K!
T  3
K M

K 0
T K
K!
 0.95
e T  20.0855
 0.95 * e T  19.08
λT=3
K
0
1
2
3
4
5
6
Sum
1
3
4.5
4.5
3.375
2.025
1.0125
19.413
Then M = 6
Prof. Ahmed Farouk Abdul Moneim
Solve the previous problem with the assumption that the failure density function is WEIBULL (η =
4.5 months and β = 1.5 )
MODIFIED POISSON’S Distribution
Poisson’s Distribution
P( X

T K
 K) 
K!
e  T
T 
put T   
 

then
K
 K
P( X  M )  e
T








K M

K 0
T 

 



K!
T 
 


P( X  K ) 
K!
e
T
  





 0.95
 K
K M

K 1
T 

 



K!
 0.95 * e
T








3
Prof. Ahmed Farouk Abdul Moneim
T

3
  1.5
e
T








 180.57
 K
K M

K 0
T 

 

 
K!
 0.95 * e
T








 171.54
(T/η)^β=5.1
96
K
0
1
2
3
4
5
6
7
8
9
Sum
1
5.196
13.50
23.38
30.37
31.56
27.33
20.29
13.18
7.61
173.42
Then M = 9
Prof. Ahmed Farouk Abdul Moneim