How to calculate manpower for preventive maintenance with varying checklist frequency

In this article we will learn about ‘How to calculate manpower for preventive maintenance with varying checklist frequency’.

Here I will explain the situation.

We have a lot of different types of machines in which preventive maintenance to be done. Each type of machines have different type of activities or checklists and its frequency also varies drastically. Now we want to calculate optimum manpower required for PM of these machines in this situation.

Earlier we have posted an article on ‘How to calculate manpower for Preventive Maintenance‘. Its very simplest situation and I recommend you to start with the simplest one.

No lets move to our subject. We shall explain this with an example.

Calculate manpower for preventive maintenance : Example

Let’s consider a time frame of one week.

Please refer following table…

Now, by reading the data in the above table I hope you understand the situation more clearly.

We shall calculate optimum manpower required.

Its very simple…

We will calculate manpower for each day

PM manpower requirement for Sunday

= (210+420)/420

= 1.5

= 2 People

Note : 420 is the working time per employee per shift.

PM manpower requirement for Monday

= (40)/420

= 0.10

= 1 People

PM manpower requirement for Tuesday

= (390+120)/420

= 1.21

= 2 People

PM manpower requirement for Wednesday

= (287)/420

= 0.68

= 1 People

PM manpower requirement for Thursday

= (700)/420

= 1.67

= 2 People

PM manpower requirement for Friday

= (467+350)/420

= 1.94

= 1 People

PM manpower requirement for Saturday

= (72+250)/420

= 0.77

= 1 People

We are done with the calculations.

But when we look at the manpower requirement for Monday, its is 0.10 and we need to allocate one manpower for this. And when we allocate 1 person, he will be idle for (420-40) 380 minutes per shift.

This points towards the requirement of proper scheduling of preventive maintenance activities, so that the work load is even for all days.

When we schedule activities perfectly the ideal manpower will be the average as mentioned below…

= 1.5 + 0.1 + 1.21 + 0.68 + 1.67 + 1.94 + 0.77

= 7.787/7

= 1.124

This is the optimum manpower, when scheduling is done perfectly.

But we need to allocate 2 people.

Again, still there will idleness since we are allocating 2 people instead of 1.124.

This is pointing towards the requirement of multiskilled manpower, handling different activities or sections.

I hope I communicated something.

Thank you.

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