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.

Applications of Group Replacement (real-life examples)

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



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:


The probabilities given in the problem are cumulative i.e. till week 1, till week 2 etc. Individual probabilities would be 0.10 in 1st week, 0.15 (0.25-0.10) in 2nd 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:

total of Individual and group cost

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).


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

Group Replacement Algorithm

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.


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