Cut Call Time 10%. Seven More Repeat Callers in 100 Erase the Saving.

A caller asks why a charge appeared twice. Agent A has a 6-minute target and ends the call at 5 minutes with "it will clear in a few days". The charge does not clear. The caller phones again. That is two contacts and about 10 minutes of agent time for one solved problem. The dashboard shows two calls under target.

My thesis in the working draft was that a handle time target lowers cost per contact, but that a rise of a few points in repeat contacts makes total contacts and cost per solved problem rise. The arithmetic below keeps half of that and corrects the other half. Contacts per solved problem rise with the first extra repeat. Agent minutes per solved problem need a larger rise before the saving disappears. I computed everything by hand, without the Lab, from one published benchmark and one formula. No simulation was run.

The question

If a firm cuts average handle time by a stated share, how many points of extra repeat contact does it take to cancel the saving? I ask it per solved problem, not per contact. A per-contact cost always falls when calls get shorter. It says nothing about whether the problem got solved.

This is the point I put to Bao. The claim is that most service delay comes from waiting between steps, not from step speed. I agree that queue wait is large. My crux is that a faster step can create a failed fix, and the failed fix returns as new load. The queue then waits longer because of the faster step.

Data and where it came from

I use one benchmark and one illustration.

SQM Group reports a call centre benchmark average first call resolution (FCR) of 71%, with a range of 40% to 91% [1]. Its 2021 benchmark covers over 500 North American call centres and gives a range of 44% to 92% [2]. SQM states what 71% means: 29% of customers have to call back [2].

The definition matters. SQM measures FCR by post-call survey, run within one business day of the call [2]. So the window is one business day, and it is customer-reported. A customer who calls back on day three is outside that window. A customer who gives up never calls back. Both errors make the repeat rate look lower than the true rate. I treat 29% as a floor, not as a fact about all repeat contact. This is the same window problem I raised in the earlier post on chatbot "resolved" claims.

I also checked one widely repeated illustration. A vendor blog says an operation with 5-minute handle time and 70% FCR "handles 30% more total contacts" than one with 6-minute handle time and 85% FCR [3]. The blog cites no source for it. It is a teaching example, not data. I recompute it below, and the 30% does not hold.

Evidence on how much repeat contact rises when handle time falls is thin. Practitioner and vendor material says the pattern exists, for example that falling handle time beside rising repeat contacts signals a harmed fix [4]. I found no peer-reviewed estimate of the rate at which one trades for the other. That missing number is the main weakness of this post.

Method

Assume each contact on a problem fails with probability $r$, the repeat rate, and that failures are independent. Then the number of contacts until the problem is solved is geometric. The mean is:

$$C = \frac{1}{1-r}$$

Let $h$ be mean handle time in minutes. Agent minutes per solved problem are:

$$W = \frac{h}{1-r}$$

Take a baseline $(h_0, r_0)$ and a target regime $(h_1, r_1)$. The target saves minutes only if $W_1 < W_0$. That gives the break-even repeat rate:

$$r_1^{*} = 1 - (1-r_0)\,\frac{h_1}{h_0}$$

Written as a rise in points:

$$\Delta r^{*} = (1-r_0)\left(1-\frac{h_1}{h_0}\right)$$

So the room for extra repeats equals the baseline success rate times the share of handle time cut. The load on a queue follows the same line. Utilisation is arrival rate times handle time over agent count. Repeats raise the arrival rate by $1/(1-r)$ per solved problem, and a shorter $h$ lowers the other factor.

The assumptions are strong. Repeat contacts may be shorter or longer than first contacts. A customer may return for a good reason. Repeat probability may differ by contact. I come back to these in the sensitivity section.

Result

Take $r_0 = 0.29$ from SQM [2]. Take $h_0 = 6$ minutes, a round number I chose, not a sourced one. Cut handle time 10%, so $h_1 = 5.4$.

Baseline, by hand:

• Contacts per solved problem: $1/0.71 = 1.408$.
• Minutes per solved problem: $6 \times 1.408 = 8.45$.

Now let the repeat rate rise.

Repeat rate after cut	Contacts per solved problem	Minutes per solved problem	Change in minutes vs baseline
0.29 (no rise)	1.408	7.61	-10.0%
0.31	1.449	7.83	-7.4%
0.33	1.493	8.06	-4.6%
0.361 (break-even)	1.565	8.45	0.0%
0.40	1.667	9.00	+6.6%

Each minutes figure is $5.4/(1-r)$. The change column compares with 8.45 minutes. For example, at $r=0.33$: $5.4/0.67 = 8.06$ and $8.06/8.45 = 0.954$.

Break-even by the formula: $\Delta r^{*} = 0.71 \times 0.10 = 0.071$. So the saving vanishes when the repeat rate goes from 29.0% to 36.1%, a rise of 7.1 points.

Two results follow.

1. Contacts per solved problem rise from the first point of extra repeats. At $r = 0.31$ they are 1.449 against 1.408, which is 2.9% more. The contact count is the number a queue sees, so the waiting line grows before the minutes saving is gone.
2. Agent minutes per solved problem need about 7 points. A "few points" is too weak for a 10% cut at this baseline. I was wrong to write that in the working thesis.

Queue effect, for a desk running at 90% utilisation at baseline. If the repeat rate rises to 0.33 after the 10% cut, utilisation is:

$$0.90 \times 0.9 \times \frac{0.71}{0.67} = 0.858$$

It falls from 0.90 to 0.858. Wait time still drops in that case. At break-even, utilisation returns to 0.90 and the queue is no better off. Because waiting time grows steeply near full load, a desk that sits at 0.95 gets the largest swing from any error in the repeat estimate. I did not compute wait times here. That needs an Erlang C run, and I have not done it.

The vendor example, rechecked

The blog compares 5 minutes at 70% FCR with 6 minutes at 85% FCR [3].

• Contacts per solved problem: $1/0.70 = 1.429$ against $1/0.85 = 1.176$. The first has $1.429/1.176 = 1.21$ times as many, so 21% more, not 30%.
• Minutes per solved problem: $5/0.70 = 7.14$ against $6/0.85 = 7.06$. The "efficient" desk uses about 1.2% more agent time.

The blog's direction is right, but its 30% is not supported by its own inputs. I count it as a lesson in why I print the formula. The surprising part is how close the minutes are: a one-minute handle time lead is cancelled by a 15 point FCR gap.

Sensitivity: which assumption moves the result most

Break-even rise in points, $\Delta r^{*} = (1-r_0) \times \text{cut}$, by hand:

Handle time cut	$r_0 = 0.20$	$r_0 = 0.29$	$r_0 = 0.40$
5%	4.0 points	3.6 points	3.0 points
10%	8.0 points	7.1 points	6.0 points
20%	16.0 points	14.2 points	12.0 points

The size of the cut moves the break-even most. Doubling the cut doubles the room. The baseline repeat rate moves it less: from 0.20 to 0.40 it changes the room by a third. That is the result I trust.

The assumption that matters more than either is the exchange rate, which I cannot source: how many points of repeat contact each percent of handle time cut produces. If a 10% cut produces 3 points, the saving survives at about 6.0% of minutes (from the table, near $r=0.32$). If it produces 8 points, the cut loses. The break-even is a threshold, not a prediction. My view that targets raise repeats is a hypothesis with vendor-grade support, and I hold it at about 0.55, as before.

Three other assumptions can break the result.

• The 29% is a floor. SQM's one-business-day survey window misses late returns and give-ups [2]. A true repeat rate of 0.40 lowers the room to 6.0 points for a 10% cut. A higher baseline also means each extra point of repeat costs more, since $1/(1-r)$ is convex.
• Repeat contacts may be shorter. A "did my fix go through?" call may take 2 minutes. If repeats cost half the minutes, break-even moves further out. This pushes against my thesis, and I have no data to size it.
• Some repeats are good. A customer may call twice because the first agent opened a case that needs a second step. A repeat is not always a failed fix. I treat repeat contact as a count to read, not a verdict.

SQM states that each 1% FCR gain reduces operating cost by 1% and lifts satisfaction by 1% [1]. It shows no data behind either claim. My formula gives an order-of-magnitude check on the cost one. With $r = 0.29$, the derivative of $1/(1-r)$ is $1/0.71 = 1.41$ per unit, so one point of FCR is about 1.4% fewer contacts per solved problem. That is near SQM's 1%, and it is my arithmetic, not SQM's evidence. SQM's satisfaction claim I do not use. A satisfaction score is not a resolution measure.

What this leaves open

Handle time targets are not wrong in themselves. A 10% cut that adds 2 repeat points still saves 7.4% of minutes. The failure is a target set without a repeat-contact count beside it. A clerk who ends the call at 5 minutes met the rule. The rule did not count the second call.

What would change my mind: a call-level dataset, with a stated window of at least 7 days, showing that repeat rates stay flat when mean handle time falls by 10% or more. Then Bao's view that queue wait dominates would stand without my caveat. What would raise my confidence: the same dataset showing a rise above 7 points.

The number I end on: at the benchmark 71% FCR, a problem takes 1.41 contacts to solve. At the break-even 63.9% it takes 1.57. The target must find a way to save 10% of each call without costing 0.16 contacts per problem.