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I have a simple algebra formula, proven to work. But I need help in understanding why it works.

The Scenario: I work at a call center, and am trying to calculate the time free in-between calls. I have the 3 variables, provided by Live data:

  • Staff Available (not on calls)
  • Staff Busy (on calls)
  • Average call length of 5 minutes

So once I end a call, I go to the back of the line of available staff before I get the next call.

This is the formula, tested to work:

: (Staff Available / Staff Busy) * 5 minutes call length = Time in-between calls

Example: 100 staff. 80 busy, 20 available. [20/80 * 5 = 1.25 minutes]

Example: 100 staff. 50 busy, 50 available. [50/50 * 5 = 5 minutes] (Which is expected, as we are double staffed.)

Example: 100 staff. 20 busy, 80 available. [80/20 * 5 = 20 minutes]


Question - Why does this equation work? I must be taking shortcuts. Why do we divide Available/Busy instead of Available/Total?

I'd greatly appreciate any explanation. Thank you very much. -Brennan

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  • $\begingroup$ (abstract-algebra) is not a good tag for this question. You may want to consider (statistics) or something related instead. $\endgroup$
    – Element118
    Nov 10, 2015 at 1:17

2 Answers 2

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There are different ways of looking at it. One way is this: $${\mathrm{Average \ Gap} } = {{\mathrm{Idle \ Time}} \over {\mathrm{Number \ of \ Calls}}}$$

Now take your 80% busy, 20% available case. In an hour, your idle time will be 20% of the time, or 12 minutes. The average number of calls will be the remaining time (48 minutes) divided by the average length of a call (5 minutes.)

This results in 1.25 minutes. If you clean it up by removing the artifice of using one hour, you will get your shortcut version.

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Let's suppose any particular staff's workload can be represented by the busy staff and the available staff. Then, the staff spends $\frac{\text{Busy}}{\text{Total}}$ of the time taking calls and $\frac{\text{Available}}{\text{Total}}$ free. Since the calls are on average $5$ minutes long, if he take $5$ minutes taking calls, it follows that the amount of time free he has on average is $t$, where:

$$\frac{5}{\frac{\text{Busy}}{\text{Total}}}=\frac{t}{\frac{\text{Available}}{\text{Total}}}$$

Simplifying, we get the formula you stated:

$$t=\frac{\text{Available}}{\text{Busy}}\times5$$

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  • $\begingroup$ Thank you element. I'm still trying to visualize why we set those two parts equal to eachother. But I think that will come in time $\endgroup$ Nov 10, 2015 at 21:45
  • $\begingroup$ Here's the best I can figure $\endgroup$ Nov 11, 2015 at 4:18
  • $\begingroup$ Say there's some hidden common variable (C) in both parts. C[busy %]=5 min. And C[avail %]=t min. Solve for C, substitute. 5/[busy%] *[avail%] = t min $\endgroup$ Nov 11, 2015 at 4:26
  • $\begingroup$ It's just that 5 minutes is proportional to the Busy time and t minutes is proportional to the Available time. $\endgroup$
    – Element118
    Nov 11, 2015 at 10:08

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