Factor of Utilization: Understanding Waiting Time in Queues

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Homework Help Overview

The discussion revolves around understanding the waiting time in queues, particularly in the context of a formula that relates waiting time to activity time and utilization. The original poster expresses uncertainty about the mathematical implications of utilization and its relationship to waiting time, seeking clarification on the underlying concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the ratio of utilization to idle time and its implications for waiting time. Questions arise regarding the interpretation of the formula and its application in real-world scenarios, particularly in service settings like oil change shops.

Discussion Status

Participants are actively engaging with the original poster's questions, providing examples and insights into the formula's application. Some participants suggest that the formula may not apply universally, particularly in multi-server scenarios, and highlight the need for further exploration of queue theory concepts.

Contextual Notes

There are indications of confusion regarding the assumptions underlying the formula, particularly in relation to different service scenarios and the nature of arrivals and service times. The original poster is also navigating the appropriateness of the problem's placement within the forum's categories.

IrinaK.
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Homework Statement


Hello!

I am not sure that I am posting the thread in the right place - maybe, it is more a pre-calculus issue.

Here is the formula (I am skipping the third factor in the formula because it is not relevant for the question):

Waiting time = Activity time * (Utilization / (1-Utilization)

Waiting time - how much time a person spends in a queue
Activity time - how much value-added time the resource adds (actual working hours)
Utilization - percentage of capacity that is used productively

Homework Equations


My question (simple math one):

1) Do I understand correctly that if I divide Utilization (for example, it equals 80%) by 1 - Utilization, that is by 20%, I find how many times does Utilization (productive time) covers unproductive time of 1 - Utilization? So, the bigger Utilization is, the bigger idle or waiting time of Utilization / 1-Utilization. How does math work? What is behind it?

2) If I understand correctly the simple math, then I don't understand what sense does Activity time * (Utilization / (1-Utilization) or, if in numbers, Activity time * 4 means.

Thank you!

The Attempt at a Solution

 
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For question 1, you are creating a ratio of in use : not in use. The more your resource is in use, the longer you might expect to wait for that resource.
So, what I get out of this is that if your resource is 100% in use, you will have infinite wait times, since there is no excess capacity to handle the queue.
If your resource is 80% in use, and takes 4 hours to complete an activity (at that level of utilization), then using the remaining 20% available will take you 80%/20%*4hours to complete the task.
Similarly you could rewrite the equation to equate the times (wait time ) * (idle %) = (activity time) * (utilization %) .
 
Perhaps I also misinterpreted this equation at first.
Consider the example of a shop that offers oil changes.
Each oil change takes 30 minutes for one tech to complete, and there are 4 techs working eight hours each.
Total capacity of the shop is 64 oil changes per day.
If they have booked 32 appointments for the day, then the utilization rate is 50%. If you were to drive in and ask for an oil change, you would expect to wait, on average, 30 minutes for an open spot for your car. wait time = 30 mins * 50%/50% = 30 mins.
If they had booked 0 appointments, you would be sure to get in right away. wait time = 30mins * 0%/100% = 0 mins.
If they had booked 48 appointments, you would expect to wait about 90 minutes. wait time = 30 mins * 75%/25% = 90 mins.
 
RUber said:
Perhaps I also misinterpreted this equation at first.
Consider the example of a shop that offers oil changes.
Each oil change takes 30 minutes for one tech to complete, and there are 4 techs working eight hours each.
Total capacity of the shop is 64 oil changes per day.
If they have booked 32 appointments for the day, then the utilization rate is 50%. If you were to drive in and ask for an oil change, you would expect to wait, on average, 30 minutes for an open spot for your car. wait time = 30 mins * 50%/50% = 30 mins.
If they had booked 0 appointments, you would be sure to get in right away. wait time = 30mins * 0%/100% = 0 mins.
If they had booked 48 appointments, you would expect to wait about 90 minutes. wait time = 30 mins * 75%/25% = 90 mins.
Dear RUber,
thank you very much for your detailed answers and your examples! These are very helpful! Thank you!
 
RUber said:
Perhaps I also misinterpreted this equation at first.
Consider the example of a shop that offers oil changes.
Each oil change takes 30 minutes for one tech to complete, and there are 4 techs working eight hours each.
Total capacity of the shop is 64 oil changes per day.
If they have booked 32 appointments for the day, then the utilization rate is 50%. If you were to drive in and ask for an oil change, you would expect to wait, on average, 30 minutes for an open spot for your car. wait time = 30 mins * 50%/50% = 30 mins.
If they had booked 0 appointments, you would be sure to get in right away. wait time = 30mins * 0%/100% = 0 mins.
If they had booked 48 appointments, you would expect to wait about 90 minutes. wait time = 30 mins * 75%/25% = 90 mins.

The formula the OP cites is false for a multiple-mechanic shop. It is true for a single-mechanic shop with (i) Poisson arrivals; (ii) exponential service times; (iii) first-come, first-served priorities; and (iv) unlimited waiting space. Basically, though, your explanation is the intuitively correct one: the average time I must wait for my car to be worked on is the average time per car, times the average number of cars in front. The real issue is: how to find the second factor.

To the OP:
Google "M/M/1 Queue". In particular, see http://www.win.tue.nl/~iadan/que/h4.pdf , especially Section 4.3
or http://ingforum.haninge.kth.se/armin/ALLA_KURSER/KOTEORI/EXER/repet3.pdf .

To see what happens in a finite-capacity garage (with limited waiting space), see, eg.
http://iitd.vlab.co.in/?sub=65&brch=182&sim=415&cnt=626

To see what happens in an infinite-capacity, multiple-mechanic garage, see, eg.
http://www.public.iastate.edu/~riczw/stat330s11/lecture/lec23.pdf

To see what happens in a 1-mechanic garage but with NON-exponential service times, see material on the M/G/1 queue, such as
http://www.richardclegg.org/previous/networks2/Lecture9_06.pdf or
http://web.mit.edu/modiano/www/6.263/lec8.pdf

Multiple-server models with non-exponential service times are essentially intractable, and must either be tackled by approximations or by Monte-Carlo simulation.
 
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