Weighted average of two frequencies

Question

This should be really simple but I'm getting stuck and I'm probably extremely dumb..

I know that a machine receives two kind of parts:

Type 1 -> with frequency 50 per week, each one is processed for 20 minutes Type 2 -> with frequency 100 per week, each one is processed for 8 minutes

It was told me to consider (in working hours) 1 week = 2400 minutes.

I have to calculate the average arrival rate of the machine and the average service time of the machine.

Average arrival rate of the machine = 50 pieces per week + 100 pieces per week = 150 pieces per week, this is simple since each piece counts as an arrival (they're all equal)

I'm confused with the Average service time per piece of the machine...

I tried:

the machine can serve 2400/20 type 1 pieces per week = 120 and 2400 / 8 type 2 pieces per week = 300

so average_service_rate = (120*50 + 300*100) / 150 = 240, but this is wrong according to my solution (which is 200)...

if I use service times it works: time per each type 1 piece: 20 min time per each type 2 piece: 8 min average_time_per_piece = (20*50 + 8*100)/150 = 12 min, i.e. 12/2400 week so 1/averaget_time_per_piece = average_service_rate = 2400/12 = 200. And this is the right value.

My question is: why my first approach with the average frequency didn't work? What's wrong with it?

Answer 1

I'm not sure what you are calculating in the first attempt. First off, you seem to be calculating the total number of pieces possible for the machine to serve in one week. In this statement:

the machine can serve 2400/20 type 1 pieces per week = 120 and 2400 / 8 type 2 pieces per week = 300

you have calculated that, if the machine only processes type 1 pieces, it will be able to serve 120 type 1 peices in one week, and if the machine only processes type 2 pieces it will be able to serve 300 type 2 pieces in one week. It does not say that the machine can process type 1 pieces and type 2 pieces at the same time, so although it would be able to process (for instance) 300 type 2 pieces if that is all that it did, it would not be able to process any type 1 (perhaps).

The more pertinent problem with this line of thought is that you are told that you only receive 50 type 1 pieces and 100 type 2 pieces each week, so you cannot serve more pieces than you receive.

If you want to find the average service time per piece for the machine, that looks like

$$\text{Average Service Time Per Piece} = \frac{\text{time spent processing parts}} {\text{number of parts}}$$

Which is why, if you are given 50 parts per week which take 20 minutes each to process, and 100 parts per week which take 8 minutes each to process, you have

$$\text{time spent processing parts}=50*20+100*8=1800$$

and $$\text{number of parts}=150$$

So $$\text{Average Service Time Per Piece}=\frac{1800}{150}=12\ \text{min/piece}$$

Just think about the units you would want for "average service time per piece" - it should be some measurement of time divided by the number of pieces. Units help a lot in applied problems like this - if you checked the units in your first approach you would see that you are dividing a number of pieces by a number of pieces, which does not yield the units that you want.

Answer 2

Your original wrong answer might be illustrated by

• the question of the average speed of a person who travels at 20 seconds per kilometre for 50 seconds and then at 8 seconds per kilometre for another 100 seconds,

compared with

• the question of the average speed person who travels at 20 seconds per kilometre for 50 kilometres and then at 8 seconds per kilometre for another 100 kilometres.

Average speed is total distance divided by total time so:

• For the first person the average speed is $\dfrac{\frac{50}{20} + \frac{100}{8}}{50 + 100} = \frac{1}{10}$ kilometres per second.

• For the second person the average speed is $\dfrac{50 + 100}{50 \times 20 + 100 \times 8} = \frac{1}{12}$ kilometres per second.

Multiply both answers by $2400$ seconds per week if it helps.

In your original question you were asked for the second method (the weights given are items not times) but initially answered using the first.