Time between process problems (Probability distributions)

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Homework Help Overview

The discussion revolves around the time between process problems in a manufacturing line, which is modeled using exponential and Erlang distributions. The original poster presents several questions regarding the distribution of waiting times, expected values, and probabilities related to these processes.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the distribution of waiting times for multiple problems, with some identifying the Erlang distribution as relevant. Questions arise about the parameters of this distribution and the expected waiting time calculations.
  • There is discussion about the probability of the fourth problem exceeding a certain time, with participants debating the interpretation of the problem statement and the appropriate use of Poisson distributions.
  • Some participants express uncertainty about how to approach the classification of problems into types and the associated costs, seeking clarification on the distribution for different problem types.

Discussion Status

The conversation is ongoing, with participants sharing insights and clarifications on various aspects of the problem. Some have provided guidance on specific calculations, while others are still grappling with certain components, particularly regarding the classification of problems and their costs.

Contextual Notes

Participants note the need for clarity on the parameters of the distributions involved and the implications of the memorylessness property of exponential distributions. There is also mention of potential ambiguities in the problem statements that could affect interpretations.

Spunkerspawn
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Homework Statement



The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days.

(a) Let T be the waiting time (in days) for four problems. What is the distribution of T?

(b) What is the expected waiting time for four problems?

(c) What is the probability that the time until the fourth problem exceeds 120 days?

(d) Suppose that the problems can be classified into three mutually exclusive classes: I, II and III. The probability that a problem is of type I, II, and III, are respectively, 0.75, 0.2, 0.05.

(i) Give the distribution of the waiting time for a process problem of type I.

(ii) Suppose that the problems I, II and III, cost (per problem): 1,000, 2,500 and 6,000 $, respectively. Give the mean total cost for process problems for a period of 90 days.

(e) There are no process problems for 30 days, what is the probability that there will be at least one process problem in the next 30 days?



Homework Equations





The Attempt at a Solution




For (a) (b) and (c) I have managed to answer the following:

(a) The waiting time distribution for T is Erlang since this is a sum of exponential distributions.

(b) E[X] = r.1/lambda = 4/lambda , where lambda = 1/30 ==> 4/0.033

(c) This would be a Poisson distribution P(X>120)

However for (d) and (e) I am kind of stuck. Any help would be greatly appreciated!

Thank you,

Ali.
 
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Spunkerspawn said:

Homework Statement



The time between process problems in a manufacturing line is exponentially distributed with a mean of 30 days.

(a) Let T be the waiting time (in days) for four problems. What is the distribution of T?

(b) What is the expected waiting time for four problems?

(c) What is the probability that the time until the fourth problem exceeds 120 days?

(d) Suppose that the problems can be classified into three mutually exclusive classes: I, II and III. The probability that a problem is of type I, II, and III, are respectively, 0.75, 0.2, 0.05.

(i) Give the distribution of the waiting time for a process problem of type I.

(ii) Suppose that the problems I, II and III, cost (per problem): 1,000, 2,500 and 6,000 $, respectively. Give the mean total cost for process problems for a period of 90 days.

(e) There are no process problems for 30 days, what is the probability that there will be at least one process problem in the next 30 days?



Homework Equations





The Attempt at a Solution




For (a) (b) and (c) I have managed to answer the following:

(a) The waiting time distribution for T is Erlang since this is a sum of exponential distributions.

(b) E[X] = r.1/lambda = 4/lambda , where lambda = 1/30 ==> 4/0.033

(c) This would be a Poisson distribution P(X>120)

However for (d) and (e) I am kind of stuck. Any help would be greatly appreciated!

Thank you,

Ali.

(a) An Erlang distribution has two parameters in it. What are those parameters in this case?
(b) The answer is not 4/0.033 = 121.212121... ; it is equal to 3/(1/30) = 120 exactly. There is no need to round off; just use the fact that 1/(1/30) = 30. If you are going to round off, keep more digits during the computation. For example, 3/0.03333 = 120.0120012..., which is much closer to the true answer.
(c) Your answer is meaningless unless you specify what Poisson distribution X has (in other words, the mean of X or some equivalent information). Anyway, you seem to be asking for the probability that more than 120 events have occurred; after all, the Poisson distribution is the distribution of the number of events!
(d) Read the textbook, or if it does not have the required material, look on-line. For example, the document http://www.netlab.tkk.fi/opetus/s38143/luennot/E_poisson.pdf will have everything you need to start (d) and get (d.i). Once you understand (d.i), (d.ii) will not be difficult.
(e) Are you sure you don't know how to do (e)? Surely you must know about some basic, very important properties of exponential distributions! I am going to assume that you really do know the answer but are just forgetting something or are afraid to commit yourself. However, if I am wrong, I suggest you read your textbook or look on line for properties of exponential distributions.

RGV
 
Actually I am ok with (e), correct me if I am wrong but we should be dealing with the "memorylessness" property of the exponential distribution right?

For (a) I said distribution Erlang since we are dealing with a sum of exponential distributions: the cdf of an erlang is: 1 - exp^-(mean.x) * sum(i=0 to r-1) * (mean.x)^i / i! (hopefully this formula is clear enough). Isn't this right?

For c) we want the fourth problem to exceed 120 days. So I think here we want P(N<=3) with parameter 120*(1/30)?

I will look at d and see how I can solve it.

Am I right for the others though?
 
For (a): your answer is OK; I just wanted you to specify the parameters, so you could either give the formula for the density or cdf, or you could say E4(30) instead of just "erlang". Actually, the density is a lot easier than the cumulative: the density of the sum of n independent expl(a) variables is
f(t) = \frac{a^n t^{n-1} e^{-at}}{(n-1)!}, \mbox{ for } t \geq 0.

You have (e) OK.

For (c) the only issue is whether or not you need {N <= 3} or {N = 3}; a lawyer could argue that the question is ambiguous, so {4th later than 120} could mean {1st later than 120} or {1st before 120, 2nd later} or {first two before 120 and third later} or {first 3 before 120 and 4th later}. Which one do you think is wanted, and which of {N <= 3} or {N = 3} should you use?

RGV
 
I completely understand what you mean for (c) as in just one problem might be before 120 days but the remaining 3 could happen after 120 days. But I believe, in this case, it is just asking for the exact fourth problem to happen after 120 days.
 
I'm having a bit of trouble with d.(ii)

For d.(i) the distribution for waiting time for Problem of Type 1 is exponential with a mean of 30/0.75.

How do I proceed with d.(ii)?
 
Spunkerspawn said:
I'm having a bit of trouble with d.(ii)

For d.(i) the distribution for waiting time for Problem of Type 1 is exponential with a mean of 30/0.75.

How do I proceed with d.(ii)?

You have a fixed span of time (90 days in this case). On average, how many problems of types I, II and III occur in that time span? What is the cost?

RGV
 
Would this be the product of the poisson distributions of each problem type?
 
Spunkerspawn said:
Would this be the product of the poisson distributions of each problem type?

Sorry: the Forum rules/guidelines prevent me from offering any more help on this question. You just have to do the best you can now.

RGV
 
  • #10
Figured it out. Thanks RGV.
 

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