Erlang B: nobody is waiting, and some were turned away

Assumptions

Poisson arrivals and exponential service; identical positions and no waiting room at all; the returning stream behaves like the arriving one; steady state, so the window converts rates into counts and does not run a clock.

Model version

1.0

The formula, in words

• A service with no waiting room blocks rather than queues. Offered load is the arrival rate divided by what one position finishes in an hour. The Erlang B recursion turns that load and the position count into the probability an arrival finds every position busy: start at one, then for each position in turn replace the running value with the load times it, divided by the position number plus the load times it. That form is stable where a factorial form overflows.
• Retries make the arrival stream depend on itself. A blocked attempt that returns is a new attempt, so the total attempt rate is the unique arrival rate divided by one minus the retry probability times the blocking probability. Blocking depends on the attempt rate and the attempt rate depends on blocking, so the page searches for the one attempt rate at which the demand served plus the demand abandoned adds back up to the demand that arrived, narrowing a bracket around it by halves a fixed number of times.
• A blocked attempt and a lost case are different counts. Every blocked attempt either returns, with the retry probability, or stops. The share of offered demand never served is the blocking probability times one minus the retry probability, all divided by one minus the retry probability times the blocking probability. Raising the retry probability raises the first count and lowers the second, and both are true at once.
• The population outside is a rate times a wait. The number of cases waiting to try again, averaged over time, is the rate of returning attempts multiplied by the retry delay. It is the only quantity the retry delay touches, and it is invisible to every other number on the page.

A worked example

The service, as it loads. $\lambda=6.0$ cases an hour, $\mu=1.5$ cases per position hour, $c=5$ positions, no waiting room. $H=8$ hours, $T=2.0$ hours, where $H$ is the horizon and $T$ the retry delay. Offered load $a = \frac{6.0}{1.5} = 4.0$ erlangs against 5 positions.

Blocking with no retries at all, by the recursion, five steps:

19.9 per cent of arrivals find every position busy, before any retry behaviour is assumed. That figure is the anchor: it is what the page reports at a retry assumption of zero, and every retry assumption pulls the lost share below it.

The root at a 30 per cent retry assumption. Bisection on $g(x)=x\left(1-0.3\cdot B\left(\frac{x}{1.5}\right)\right)-6.0$ returns $x=6.4337$ attempts an hour with $B=22.4691\%$. That is a root rather than an approximation and a reader can check it in one line without repeating the search:

$6.4337 \times \left(1 - 0.30 \times 0.224691\right) = 6.4337 \times 0.932593 = 6.0000$, the offered rate recovered.

The eight hour window at 30 per cent, every step shown:

unique     $6.0000 \times 8 = 48.0$ cases wanted service
attempts   $6.4337 \times 8 = 51.5$
blocked    $6.4337 \times 0.224691 \times 8 = 1.44559 \times 8 = 11.6$
retries    $0.30 \times 11.5647 = 3.5$, check: $51.469 - 48.000 = 3.469$
admitted   $51.469 - 11.565 = 39.9$
lost       $11.5647 \times 0.70 = 8.1$, check: $48.000 - 39.905 = 8.095$
share      $\frac{8.0953}{48.0} = 16.9\%$
occupancy  $\frac{\frac{6.4337 \times 0.775309}{1.5}}{5} = \frac{3.32543}{5} = 66.5\%$
queue      0
outside    $0.30 \times 1.44559 \times 2.0 = 0.9$

The same window at 70 per cent. The root moves to $x=7.4914$ attempts an hour with $B=28.4402\%$, and it checks the same way: $7.4914 \times \left(1 - 0.70 \times 0.284402\right) = 7.4914 \times 0.800919 = 6.0000$.

unique     48.0 unchanged, and it must be
attempts   $7.4914 \times 8 = 59.9$
blocked    $7.4914 \times 0.284402 \times 8 = 2.13057 \times 8 = 17.0$
retries    $0.70 \times 17.0446 = 11.9$, check: $59.931 - 48.000 = 11.931$
admitted   $59.931 - 17.045 = 42.9$
lost       $17.0446 \times 0.30 = 5.1$, check: $48.000 - 42.887 = 5.113$
share      $\frac{5.1134}{48.0} = 10.7\%$
occupancy  $\frac{\frac{7.4914 \times 0.715598}{1.5}}{5} = \frac{3.57387}{5} = 71.5\%$
queue      0
outside    $0.70 \times 2.13057 \times 2.0 = 3.0$

The comparison, which is the page.

The two readings are 6.2 percentage points apart, and the lower assumption returns 1.6 times the loss the higher one does. Nothing between them is measured. Blocked attempts and lost cases move in opposite directions across that range, so a report quoting the first gets worse as retries rise and a report quoting the second gets better, on one service, on one day, with the same 48 cases.

One more position does not close the gap. Chip 1 sets $c=6$. Blocking with no retries falls from 19.9 to 11.7 per cent. The lost share falls to 9.4% at 30 per cent and 5.1% at 70 per cent, so both answers improve and the two are still 4.3 percentage points apart. Adding capacity moves the answer. It does not remove the assumption.

Only the delay changes. Chip 2 sets $T=8.0$ and touches nothing else. x, B, admitted, blocked, lost, the share and the occupancy are identical to the last printed digit in both columns, because the retry delay appears in exactly one line of the model. The population outside goes from 0.9 to 3.5 and from 3.0 to 11.9, four times in both columns. A quantity that quadruples while every reported figure holds still is the quantity a report cannot see.

The widest the guess can be. Chip 3 sets the two rates to 0 and 100 per cent. At zero, nobody comes back: attempts equal the 48 unique cases, 9.6 never get in, the share is 19.9%, occupancy is 64.1 per cent and nobody is outside. At one hundred, nobody gives up: the fixed point settles at an offered load of 6.84 erlangs with blocking at 41.5 per cent, 82.1 attempts deliver all 48.0 cases, nothing is lost, occupancy reaches 80.0%, and 8.5 cases are standing outside at any moment against a visible queue of zero. The full range of the assumption is 19.9 percentage points wide, and every point in it is consistent with a queue of zero.

Check it yourself

Leave everything alone and change only the retry delay, from 2 hours to 8. Admitted, blocked, never served, the share and the occupancy do not move by a single printed digit in either column. The number waiting outside goes from 0.9 to 3.5 at the low assumption and from 3.0 to 11.9 at the high one, four times in both.

What this does not model

The retry probability itself, which is a guess the page asks you to make twice and never estimates for you; retry behaviour that depends on how congested the service is, or on how many times a case has already tried, both of which are real and both of which make the returning stream something other than the simple fraction used here; a retry delay that varies from case to case rather than sitting at one figure; the edge of the window, meaning that a case turned away near the end of it cannot return inside it, which the steady state arithmetic does not see and which matters most when the window is short against the retry delay; anything a wait costs, since nothing waits inside this service and the page prices nothing that happens outside it; blocking that propagates upstream, which a real constrained service does and a single stage model cannot; arrivals that vary through the day, which Poisson at a constant rate does not represent; balking and abandonment as distinct behaviours from blocking, which remove demand in different ways and are not separated here; and any real service, queue, organisation or place, none of which these figures describe.