NAME: SOYEMI ADEMOLA OLUWOLE MATRIC MATRIC NUMBER: 13CF015171
QUESTION 1.
A manufacturer provides a warranty against failure ofa carbon steel product within the rst 30 days after sale. 10 Out of 1000 sold were found to have failed by corrosion during the warranty period. Total Total cost of o f replacement for each failed product is approximately approximately 1!"00!000! including the cost of environmental clean#up! loss of product! downtime! repair! and replacement. a.$ %alculate the ris& of failure by corrosion! in aira. SOLUTION. R P × C =
'( 'is& )( )robability )robability %( %onse*uence %alculating probability! +f 10 out of 1000 carbon#steel products fail! the probability of one product failing is P=
10 1000
0.01
=
The cost of replacement of one product is therefore the conse*uence of the ris& incurred C = N 1,200,000
Therefore ris& is! R 0.01 × N 1,200,000 N 12,000 per one product =
=
,or 10 failed products ' is simply! R=10 × N 12,000 = N 120,000
b.$+f a corrosion#resistant alloy would prevent failure by corrosion! is an incremental cost of 1"00 to manfacture the product using such an alloy -ustied/hat would be the maximum incremental cost that would be -ustied in using an alloy that would prevent failures by corrosion SOLUTION.
sing a basis of 1000 products! 10 products will fail based on the probabilistic calculations above. +f the incremental cost is to be incurred! an additional 1!"00!0001!"0021000( 1!"00!000$ goes into producing the product corrosion#resistant product$. +t is more economical to spend an extra 1!"00!000 on production of 1000 corrosion#resistant products than it is to spend 1!"00!000 on a failed product! therefore the incremental cost is -ustied.
QUESTION 2
inings of tan&s can fail because of salt contamination of the surface that remains after the surface is prepared for the application of the lining. 4etween 156 and 706 of coating failures have been attributed to residual salt contamination.
The cost of rewor&ing a failed lining of a specic tan& has been estimated at 8!000!000.9'eference:;.)eters!
5 5$ 30 "00?$@ a.$%alculate the ris& due to this type of failure assuming that "06 of failures are caused by residual salt contamination. SOLUTION R P ×C =
'( 'is& )( )robability %( %onse*uence %alculating probability! +f "06 of all failures are caused by salt contamination! then the probability of failure due to salt contamination is P=
20 100
0.2
=
The cost of rewor&ing a failed lining is therefore the conse*uence of the ris& incurred C N 9 , 0 00,000 =
Therefore ris& is! R 0.2 × N 9,0 00,000 N 1,800 , 00 0 per lining =
=
b.$+f the cost of testing and removal of contaminating salts is >50!000 is this additional cost -ustied based on the ris& calculation in a$. SOLUTION
,rom the above ris& calculation! it can be seen that the probability of a lining failing due to salt contamination as well as its conse*uence in monetary terms$ is high. This means that the ris& involved is also on high. +f the cost of testing and removal of contaminating salts is >50!000! to ensure that no lining is aected by salts the cost is >50!000 B where B is the number of tan&s$. +f the probabilty of a number of tan&s failing out of a given number of tan&s due to salt contamination is "06 ta&ing a basis of 100 tan&s! then "0 fail will fail due to salt contamination$! then the cost of removing the ris& of failure is N 450,000 X = N 450,000 × 100= N 45,000,000
%omparing this to the hypothetical situation whereby you would to replace all failed tan&s would be N 9,000,000 × 20 N 180,000,000 =
,rom the above calculations! a cost of >50!000 to remove or signicantly reduce the ris& of failure is -ustied since it woud ta&e a much smaller amount to prevent failure than to replace failed e*uipment. c.$%alculate the minimum percentage of failures caused by residual salt contamination at which the additional cost of >50!000 for testing and removal is -ustied. SOLUTION.
,rom the above hypothetical situation it would ta&e >5!000!000 to prevent failure due to salt contamination in 100 tan&s. ow! for us to -ustify the incremental cost of >50!000 per tan& ! we have to calculate the amount of tan&s that would have to fail for the cost of replacement to e*ual the cost of prevention of failure.
et C be the number of failed tan&s N 9,000,000 × Y = N 45,000,000
Y =
45,000,000 9,000,000
5
=
,rom the above calcutations! using a basis of 100 tan&s! only 5 tan&s would have failed for the cost of rewor&ing tan&s 8!000!000 per tan&$ to e*ual the cost of preventing failure in all 100 tan&s. Therefore! the minimum percentage of failure caused by salt contamination at which the incremental cost is -ustied is the number of failed tan&s divided by the total number of tan&s! i.e. number of failed tanks total number of tanks
.
=
5 100
=
5
