I've been trying to solve this exercise about Poisson but it seems my reasoning is not correct.
It goes like this:
Failures in a system are modeled according to a Poisson process with a rate of $0.015$ failures/hour. Every time a fault occurs, the system stops to be repair
Since repairs are expensive, and requires 10 hours for each failure, it is of great interest to estimate by simulation the proportion of the system's inoperable time (that is, the ratio of total repair time to total active operating time).1) Simulate $5$ times between failures from the following random values: $0.944$, $0.5253$, $0.6571$, $0.951$, $0.3383$.
Write the $5$ simulated values separated by a space.
Note: in a Poisson process the times between events are exponentially distributed.2) From the $5$ simulated values from point 1), calculate the proportion of time that the system was not operable.
3) We now want to simulate $10$ non-operable time ratio values. For this, we will repeat the previous steps $9$ more times, using the following random numbers:
0.944 0.5253 0.6571 0.951 0.3383 0.2735 0.7894 0.3067 0.7617 0.9497 0.5512 0.5871 0.4472 0.0338 0.8519 0.0793 0.6877 0.8679 0.2803 0.6606 0.9513 0.7587 0.8264 0.4401 0.065 0.1108 0.3068 0.2353 0.1592 0.3166 0.3171 0.0158 0.9257 0.9631 0.8393 0.1995 0.8858 0.1165 0.3954 0.3075 0.6685 0.8801 0.6398 0.7918 0.309 0.1345 0.5291 0.3983 0.633 0.9775Use each row of $5$ values to generate a simulated non-operable time ratio value. Answer the $10$ simulated total values separated by spaces.
Note: the first row is the one given in the first item, so the first value of the answer has already been calculated.4) Calculate the average value among the $10$ simulated values from point 3) for the proportion of non-operable time.
This is my reasoning:
0.015 failures per hour means that on average there is 1/0.015 = 66.66 hours between failures, right? The five random numbers give the times between successive failures (random number x 66.66)
So, in order to solve point 1) i would need to do this:
0.944 * 66.66 = 62.93
0.5253 * 66.66 = 35.02
0.6571 * 66.66 = 43.81
0.951 * 66.66 = 63.40
0.3383 * 66.66 = 22.55
The answer for point 1) would be:
62.93 . . . 35.02 . . . 43.81 . . . 63.40 . . . 22.55
Next, point 2). It says: "the proportion of time that the system was not operable"
Then, the operating time is the sum of 62.93 + 35.02 + 43.81 + 63.40 + 22.55 = 227.71.
Operating time = 227.71 hours
The downtime is 5 periods of 10 hours = 50 hours.
and this is a proportion of 50/227.1 = 21.95%
Hence non-operating time = 21.95% (not sure of this one)
Mental note: If I were operating this system, I would be more interested in the proportion of wasted time over the entire available time (not simply repair time over operating time), which means that the percentage should be 50/(227.71 + 50) = 18.00% So if 21.95% is not an acceptable answer, then it is possible that the required answer may be 18.00%.. Again, not sure about it.
Next, point 3)
What am i supposed to do here? Just do the same for the other 9 sets of random numbers? Find the average of the ten percentages? Don't get it. :/
Finally, point 4)
I do not understand this one, i'm a little confused.
Could you please tell me if my answers for point 1 and 2 are correct?
How would you solve points 3) and 4)?
Thanks
EDIT:
So, i would need to do this in point 1):
[-log(1-0.944)]/[(0.015)] --> 192.160
[-log(1-0.5253)]/[(0.015)] --> 49.6715
[-log(1-0.6571)]/[(0.015)] --> 71.3544
[-log(1-0.951)]/[(0.015)] --> 201.062
[-log(1-0.3383)]/[(0.015)] --> 27.529
Answer shoud be:
192.160|49.6715|71.3544|201.062|27.529
For point 2) i would need to do this:
[-log(1-0.944)]/[(0.015)]+[-log(1-0.5253)]/[(0.015)]+[-log(1-0.6571)]/[(0.015)]+[-log(1-0.951)]/[(0.015)]+[-log(1-0.3383)]/[(0.015)]
I mean:
0.944 0.5253 0.6571 0.951 0.3383 --> 541.7769
It would give me this number: 541.7769
Then, i should do this calculation:
50/(541.7769 + 50) = 0.08449
For point 3)
[-log(1-0.944)]/[(0.015)]+[-log(1-0.5253)]/[(0.015)]+[-log(1-0.6571)]/[(0.015)]+[-log(1-0.951)]/[(0.015)]+[-log(1-0.3383)]/[(0.015)]
[-log(1-0.2735)]/[(0.015)]+[-log(1-0.7894)]/[(0.015)]+[-log(1-0.3067)]/[(0.015)]+[-log(1-0.7617)]/[(0.015)]+[-log(1-0.9497)]/[(0.015)]
[-log(1-0.5512)]/[(0.015)]+[-log(1-0.5871)]/[(0.015)]+[-log(1-0.4472)]/[(0.015)]+[-log(1-0.0338)]/[(0.015)]+[-log(1-0.8519)]/[(0.015)]
[-log(1-0.0793)]/[(0.015)]+[-log(1-0.6877)]/[(0.015)]+[-log(1-0.8679)]/[(0.015)]+[-log(1-0.2803)]/[(0.015)]+[-log(1-0.6606)]/[(0.015)]
[-log(1-0.9513)]/[(0.015)]+[-log(1-0.7587)]/[(0.015)]+[-log(1-0.8264)]/[(0.015)]+[-log(1-0.4401)]/[(0.015)]+[-log(1-0.0650)]/[(0.015)]
[-log(1-0.1108)]/[(0.015)]+[-log(1-0.3068)]/[(0.015)]+[-log(1-0.2353)]/[(0.015)]+[-log(1-0.01592)]/[(0.015)]+[-log(1-0.3166)]/[(0.015)]
[-log(1-0.3171)]/[(0.015)]+[-log(1-0.0158)]/[(0.015)]+[-log(1-0.9257)]/[(0.015)]+[-log(1-0.9631)]/[(0.015)]+[-log(1-0.8393)]/[(0.015)]
[-log(1-0.1995)]/[(0.015)]+[-log(1-0.8858)]/[(0.015)]+[-log(1-0.1165)]/[(0.015)]+[-log(1-0.3954)]/[(0.015)]+[-log(1-0.3075)]/[(0.015)]
[-log(1-0.6685)]/[(0.015)]+[-log(1-0.8801)]/[(0.015)]+[-log(1-0.6398)]/[(0.015)]+[-log(1-0.7918)]/[(0.015)]+[-log(1-0.3090)]/[(0.015)]
[-log(1-0.1345)]/[(0.015)]+[-log(1-0.5291)]/[(0.015)]+[-log(1-0.3983)]/[(0.015)]+[-log(1-0.6330)]/[(0.015)]+[-log(1-0.9775)]/[(0.015)]
So, after performing the calculations (using wolfram) i got this results:
0.944 0.5253 0.6571 0.951 0.3383 --> 541.778
0.2735 0.7894 0.3067 0.7617 0.9497 --> 444.505
0.5512 0.5871 0.4472 0.0338 0.8519 --> 281.516
0.0793 0.6877 0.8679 0.2803 0.6606 --> 312.007
0.9513 0.7587 0.8264 0.4401 0.065 --> 456.133
0.1108 0.3068 0.2353 0.01592 0.3166 --> 76.5910
0.3171 0.0158 0.9257 0.9631 0.8393 --> 541.649
0.1995 0.8858 0.1165 0.3954 0.3075 --> 225.788
0.6685 0.8801 0.6398 0.7918 0.309 --> 412.346
0.1345 0.5291 0.3983 0.633 0.9775 --> 413.479
Anwers for point 3) should be:
541.778|444.505|281.516|312.007|456.133|76.5910|541.649|225.788|412.346|413.479
I'm not sure about point 4) :/
I don't think you have the right interpretation of part $1$. The cumulative distribution function of the exponential distribution is $$F(x)=1-e^{-\lambda x},\ x\geq0$$ and when you're given a random number $r$, I think you are supposed to find $x$ such that $F(x)=r$, that is $$x=\frac{-\log(1-r)}\lambda.$$ This is the usual way of simulating distributions other than the uniform distribution.
For part $2$, I'm sure your second thought is right. You should be looking at the total time.
For part $3$ yes, just do it $10$ more times. This gives you $10$ averages.
For part $4$ it says to take the average of the $10$ averages computed in part $3$. This is a bit odd, It seems like the right thing to do is to compute total downtime over total time, but that's not what the exercise says.they will all cover different lengths of time.