# Percent operating time of a machine with breakdowns and repairs

This problem normally requires a complex logical structure since we are trying to model many processes simultaneously.

Let's consider the following problem:

A power plant needs 2 water pumps operating at max capacity to cool its turbines with river water. Since the pumps may break down, the power plant has installed two additional pumps. These four pumps operate at 50% capacity and if one or two pumps break down, the power plant can still operate.

The calculating complexities arise because if we have failures of any pumps, the remaining pumps would have to work harder, thus increasing the remaining pumps' probability of failure.  In other words, if we have all 4 pumps running together, and we only need 2 for the station to operate, then the pumps are working at half capacity but as soon as one pump fails the remaining pumps are working at 2/3 capacity, and so they have a higher failure rate or, equivalently, a lower mean time between failure.

The four pumps are of varying age and therefore of varying reliability. The following table summarises the data:

Probability of failure (fail/day)

Pump

Pumps working

4

3

2

A

0.002

0.007

0.025

B

0.004

0.013

0.079

C

0.007

0.034

0.142

D

0.002

0.007

0.025

Each repair of the pump takes Lognormal(20,15) days.

The questions are: a) How long will it take before a shutdown occurs? b)How many shutdowns will the station have in a year? c)What is the probability of one or more shutdowns per year?

The solution to this problem is illustrated in the model: Power Station Pumps

The links to the Power Station Pumps software specific models are provided here: Power_station_pumps

There are several inter-linked tables in this model. First we need to convert the table above (probability of failure) into a table of mean time between failures using the following formula:

MTBF = -1/LN(1-P), where P is the probability of a failure.

So, we get the new table:

MTBF (days)

Pump

Pumps failed

0

1

2

A

499.4998332

142.3565575

39.49789021

B

249.499666

76.42198649

12.15137069

C

142.3565575

28.90888214

6.529495909

D

499.4998332

142.3565575

39.49789021

This table shows that the MTBF decreases dramatically with each new pump failure.

The leftmost column in the spreadsheet model (column B) shows the time t of the next event occurring: the event could either be the failure of a pump, or the completion of repairing of the pump.

Columns C to F show the status of the pumps when an event occurs, i.e. for every point in time t in column B, this table will show which pumps are broken.

The next table (columns G to N) use the MBTF table above to return the mean times between failures for each pump depending on how many pumps are broken at point in time t.

The table "Time to repair completion new t" (columns O to V) checks if the pump is broken and if it is, then returns the Lognormal(20, 15) distribution, otherwise returning the value of 1000000. This large value is just a dummy showing that the pump is not broken. We use it instead of zero because we need to find the next event to occur, i.e. the minimum time until the next event. As the value of 1000000 will never be a minimum, the algorithm will always return the required value.

The next table (columns W to AD) also checks for both the broken pump and for the shutdown of the station and puts the value of 1000000 if either of these factors are positive, otherwise returning the time until the next failure for the pump.

The last three columns of this spreadsheet calculate the shutdowns of the power station. Column AF returns the value of 1 if at any point in time t there are 3 failures. Column AG returns 1 if the corresponding cell in column AF is one and the time is column B is less than 366 (less than 1 year). Column AH returns the point in time of the first power plant shutdown.

The outcomes of the model are located in cells E10 to E12. Cell E12 shows the one positive value from column AH, providing the answer for question (a). Cell E10 returns a summation of the column AG, thus answering question (b). Cell E11 generates a discrete 1:0 distribution, a mean of which is the answer to question (c). Power_station_pumps

There are several inter-linked tables in this model. First we need to convert the table above (probability of failure) into a table of mean time between failures using the following formula:

MTBF = -1/LN(1-P), where P is the probability of a failure.

So, we get the new table:

MTBF (days)

Pump

Pumps failed

0

1

2

A

499.4998332

142.3565575

39.49789021

B

249.499666

76.42198649

12.15137069

C

142.3565575

28.90888214

6.529495909

D

499.4998332

142.3565575

39.49789021

This table shows that the MTBF decreases dramatically with each new pump failure.

The leftmost column in the spreadsheet model (column B) shows the time t of the next event occurring: the event could either be the failure of a pump, or the completion of repairing of the pump.

Columns C to F show the status of the pumps when an event occurs, i.e. for every point in time t in column B, this table will show which pumps are broken.

The next table (columns G to N) use the MBTF table above to return the mean times between failures for each pump depending on how many pumps are broken at point in time t.

The table "Time to repair completion new t" (columns O to V) checks if the pump is broken and if it is, then returns the Lognormal(20, 15) distribution, otherwise returning the value of 1000000. This large value is just a dummy showing that the pump is not broken. We use it instead of zero because we need to find the next event to occur, i.e. the minimum time until the next event. As the value of 1000000 will never be a minimum, the algorithm will always return the required value.

The next table (columns W to AD) also checks for both the broken pump and for the shutdown of the station and puts the value of 1000000 if either of these factors are positive, otherwise returning the time until the next failure for the pump.

The last three columns of this spreadsheet calculate the shutdowns of the power station. Column AF returns the value of 1 if at any point in time t there are 3 failures. Column AG returns 1 if the corresponding cell in column AF is one and the time is column B is less than 366 (less than 1 year). Column AH returns the point in time of the first power plant shutdown.

The outcomes of the model are located in cells E10 to E12. Cell E12 shows the one positive value from column AH, providing the answer for question (a). Cell E10 returns a summation of the column AG, thus answering question (b). Cell E11 generates a discrete 1:0 distribution, a mean of which is the answer to question (c).

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