# Factor of Utilization

1. Oct 28, 2014

### IrinaK.

1. The problem statement, all variables and given/known data
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

2. Relevant 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!

3. The attempt at a solution

2. Oct 28, 2014

### RUber

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 %) .

3. Oct 28, 2014

### RUber

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.

4. Nov 20, 2014

### IrinaK.

Dear RUber,

5. Nov 20, 2014

### Ray Vickson

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:
or http://ingforum.haninge.kth.se/armin/ALLA_KURSER/KOTEORI/EXER/repet3.pdf [Broken] .

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 [Broken]

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.

Last edited by a moderator: May 7, 2017