**Replacement of items that fail suddenly**

There are certain items which do not deteriorate but fail completely after certain amount of use. These kinds of failures are analysed by the method called as group replacement theory. Here, large numbers of items are failing at their average life expectancy. This kind of items may not have maintenance costs as such but they fail suddenly without any prior warning. Also, in case of sudden breakdowns immediate replacement may not be available. Few examples are fluorescent tubes, light bulbs, electronic chips, fuse etc.

Let’s consider the example of street lights. We often see street-lights being repaired by the corporation staff using extendable ladders. If a particular light is beyond repairs, then it is replaced. This kind of policy of replacement is called as ‘replacement of items as-and-when they fail’ or *‘Individual Replacement’*. On the other hand, if all the street lights in a particular cluster are replaced as and when they fail and also simultaneously in groups, then the policy is called as *‘Group Replacement’*. It should be noted that, **group replacement does involve periodic simultaneous replacements along with individual replacements in between**.

It is found that replacing these random failing items simultaneously at specific intervals is economical as compared to replacing them only when an item fails. A long period between group replacements results in increase in cost of individual replacements, while frequent group replacements are definitely costly. There lies the need to balance this and find an optimum replacement time for optimum cost of replacement.

**Illustration of Group Replacement**

**___________________________________________________________________________**

**Problem:**

A factory has 1000 bulbs installed. Cost of individual replacement is Rs. 3/- while that of group replacement Re. 1/-per bulb respectively. It is decided to replace all the bulbs simultaneously at fixed interval & also to replace the individual bulbs that fail in between. Determine optimal replacement policy. Failure probabilities are as given below:

**Solution:**

** **The probabilities given in the problem are cumulative i.e. till week 1, till week 2 etc. Individual probabilities would be 0.10 in 1^{st} week, 0.15 (0.25-0.10) in 2^{nd} week, and so on. (as shown in the below table)

**Policy-I: Individual Replacement**

**Step 1) Cost of Individual Replacements**

Individual Failures/week = Total Quantity / Mean Life = 1000 / 3.45 = 289.9

Individual Replacement Cost = (Individual Failures per week) x (Individual replacement cost)

= 289.9 x 3 = **Rs. 869.6**

**Policy-II: Group Replacement**

**Step 2) Individual failures per week**

In the first week: 10 % (0.10) of the bulbs will fail out of 1000 bulbs i.e. 100

In the second week: 15 % of the bulbs will fail out of 1000 bulbs i.e. 150. Also, 10% of 100 replaced in the first week i.e. 10. TOTAL bulbs failed until second week = 160 (150+10)

Rest of the calculation is as shown in the below table:

**Step 3) Calculating the total cost & time of replacement:**

Thus, replacing all the bulbs simultaneously at fixed interval & also to replace the individual bulbs that fail in between will be economical or optimal after 4 weeks (optimal interval between group replacements).

**Interpretation:**

- The cost of only individual replacements id Rs. 869.6 (As seen in the Policy-I)
- The cost of combine policy i.e. group and individual replacement id Rs. 863.6 (see last column of table 3)
- Hence the Policy-II is the optimum replacement policy

Hence, the bulbs shall be replaced every four weeks individually as well as in groups which combine would cost Rs. 863.6 per week (lesser than individual cost of Rs. 869.6 per week)

*Click the following link to Download the Excel Solver*

**Practice Problem:**

**___________________________________________________________________________**

The following mortality rates have been observed for a special type of light bulbs. There are 1000 such bulbs in the concerned unit of the industry.

It costs Rs 10 to replace an individual bulb that has burnt out. If the bulbs were replaced simultaneously, it would cost Rs. 2.50 per bulb. It is proposed to replace all the bulbs at fixed interval, whether are not they have burnt out, and to continue replacing the burnt out bulbs as they fail. At what intervals of time should the manager replace all the bulbs? Decide the optimum replacement policy.

**Solution:**

** **Step 1) Download the Excel Solver

Step 2) Over-write the new problem values in the orange cells.

Step 3) Compare:

- Individual Cost = Rs. 2985 (Cell D20)
- Individual & Group Cost = Rs. 2550 (Cell Q10)

Hence, the bulbs need to be replaced completely after every two months along with individual replacements as and when they fail.

**PowerPoint Presentation **

**Older Posts**

**___________________________________________________________________________**

Operations Research Techniques: Operations Research Theories

Replacement – Basic Example: Replacement Theory Algorithm

Present Worth Factor (pwf) : Replacement with Time-Value of Money

BPR in US, Europe and India: Business Process Re-engineering

Operations Research Software: TORA, SIMNET & LINDO

Product Inspection to Business Excellence: History of Quality Control

Inventory – Introduction: Inventory Control

thank you sir

but i didn’t understand table2 ie Individual failures per week

how 24.00 came in the 3rd week

please help me

@ Nishanthi,

Table 2: Week 4

200 : 20% (0.20) of 1000 bulbs failing in the present week i.e 4th week

25.00 : 25% (0.25) of 100 bulbs failing from 1st row, i.e. 3rd week for the 100 bulbs to fail at probability of 0.25

24.00 : 15% (0.15) of 160 bulbs failing from 2nd row i.e. 2nd week for the 160 bulbs to fail at probability of 0.15

28.10 : 10% (0.10) of 281 bulbs failing from 3rd row i.e. 1st week for the 281 bulbs to fail at probability of 0.10

Sir,

First of all; thank you. Can you post algorithm for Dynamic Programming? I tried on your lines but, something is wrong with my algorithm. I HAVE MAILED YOU THE SAME.

@ Ravi,

I will sure post something on Dynamic Programming.

sir

i have a doubt regarding the replacement theory

in the above solved replacement problem in the 2nd week how com 160 bulbs getting failed?

why is it 1000*.15+100*.10

i understood (100*.10)part

out of 1000 bulbs at the beginning of 2nd week only 900 is older so shouldnt it be 900*.15. because the failure probability for the newly replaced 100 bulbs which has completed only one month is 0.1. why for those 100 bulbs cumulated probability is taken for both 2nd nd 1st week, because when we take 1000*.15 the new 100 bulbs are also included.

kinldy give me a reply

Thank you

Merlin

@ Merlin

“Bulk Discount and Policy of Individual & Group Replacement”The individual cost of replacement is Rs.3 while group cost is Re.1. So, 100 bulbs require Rs.300 and so on. So instead of only opting for these individual replacements, even if all bulbs are replaced at any point of time it would cost Rs.1000 only.

Now, by the failure rates probabilities we would have 900 bulbs in 1st week, 750, 500, 300 and 0 bulbs at the end of 5 weeks if they are not replaced.

In the example, in the second para you can read “It should be noted that, group replacement does involve periodic simultaneous replacements along with individual replacements in between.”

So, every-time it is assumed that a policy of individual as well as group replacement is followed and cost is calculated.

So, even though 100 individual replacements are done in the 1st week, the simultaneous group replacement cost is also calculated for 1000 bulbs.

See the table in step 3 now. For the first week calculations, there is a cost of Rs.300 in the 1st week along with Rs.1000 i.e. the total cost is Rs.1300.

So what we are calculating here is, what would it cost every-time (every week) for individual + group replacement.

and so on…. hope i have answered you.

Thank you Mr.Rajesh

But my doubt was not regarding the cost of replcement but regarding the number of items getting failed at the end of each week.

thats ok.

Thank you once again

Merlin

Rajesh, please check your mailbox. Sonali mailed you the details about the problem she’s facing with replacing resistors. Thank you in advance.

Regards

Thank you Mr Rajesh, pls i want you to guide on how to develop a thesis on suitable replacement theory model for street lighting. I’ll appreciate it. Thank you. Pls send me a contact with wish i can get you.

@ Mohammed,

Have send you my contact details. Let me know about your whereabouts.

thank you very much. i just got what i needed.

It is a nice resource that you are providing and that too for free. Thanks Buddy !

can you please send me excel solver…. becoz my exam will be on 10th may so i need workout this algorithm..please if you can

@ Keshala, Elli & Laxmi,

Thanks!

I need Maximization of profit technique,plz help me if i can use it for trading purpose

Please send me the problem. You can mail it at: timane.rajesh @ gmail.com

Can you mail me the details?

thnx rajesh your page helped me secure 20 marks in my majors

That’s Great Parth!

well written.

I have a one query..

how to calculate for week 3 & 4 in step-2??

i am confused why you take 5 week in individual replacement and 7 weeks in group replacement and the probability of failure in 5 week is 1 so all bulbs fail in 5 weeks