1Demonstration 1 of 4
Exchange frequency is a modeling decision
If only the exchange interval changes, how different is the staffing the coupled model produces?
A stale reading is an information delay inside a feedback loop. The rule keeps acting on an old wait, so staffing overshoots and swings on the interval's own period.
Staff is the number of clinicians in the aggregate model. The queue serves nine patients per period. Exchanging every k periods means the staffing rule sees a wait refreshed only every k periods. Volatility is the standard deviation of staff over the 60 periods; wait is in hours.
Predict first. Does exchanging every 8 periods give a larger staffing range than exchanging every period?
Choose an example
Constructed example: the chapter's own two runs with seed 5, plus intervals 2 and 4 defined for this reader.
Calculated values
- Exchange every (periods)
- 8
- Staffing volatility
- 4.75
- Staff range
- 6.6 to 27.7
- Mean wait (hours)
- 0.61
- Volatility against exchanging every period
- 4.72
Volatility ratio = 4.75/1.01 = 4.72. Staff range 27.7 - 6.6 = 21.1 against 13.2 - 7.9 = 5.3 when exchanging every period. Mean wait 0.61 against 0.52, a difference of 0.09 hours. The rule corrects against a wait measured up to 7 periods ago, an information delay inside a feedback loop.
Use the idea
Choose the exchange interval from the system's timescales and write it beside the equations.
Where the conclusion applies
One seed, 60 periods and the pack's rule. Neither run settles; an interval picked for speed can report a behaviour that belongs to the interval.
Check your understanding: If the interval is 8 periods, how old can the wait the rule acts on be, at most?
Chapter 18 source: section "Exchange frequency is a modeling decision". Demonstration C18-D01.
2Demonstration 2 of 4
Nine patients per period become nine arrival events
With the staffing side held fixed, how does the queue behave as capacity moves relative to arrivals?
Going down the seam, an aggregate rate becomes individual events. Each period that capacity falls short of arrivals adds the shortfall to the queue.
Arrivals are patients per period, each given a random time inside the period. Servers each serve one patient per period. Waiting is patients still queued at the end of a period; wait is in hours, from arrival to the end of the period in which a patient is served.
Predict first. With 9 arrivals and 8 servers, how many patients are waiting after 12 periods?
Choose an example
Constructed example: the pack's patient queue with arrival and server counts defined for this reader.
Calculated values
- Arrivals per period
- 9
- Servers
- 9
- Patients waiting after 12 periods
- 0
- Mean wait in the last period (hours)
- 0.50
- Mean wait in the first period (hours)
- 0.31
Backlog change per period is 9 - 9 = 0, so it stays at 0 patients. Mean wait of those served went from 0.31 to 0.50 hours. Capacity equals arrivals exactly, a tie: nothing queues, and the wait is only the spread of arrival times within a period.
Use the idea
Test the queue with a constant server count before closing the loop, so a defect is not hidden inside the coupling.
Where the conclusion applies
Seed 5, one hour per service, 12 periods, and arrivals spread evenly at random within a period. Spreading them differently would change the waits, which the chapter notes is a modelling choice.
Check your understanding: With 10 arrivals and 8 servers, how many patients wait after 12 periods?
Chapter 18 source: section "What each side may assume". Demonstration C18-D02.
3Demonstration 3 of 4
The staffing rule alone has an equilibrium
With the wait held constant, where does the staffing rule settle and how fast?
The target is anchored to the baseline, not the current staff, so a constant wait gives a constant target and the staff moves smoothly toward it.
Staff starts at the baseline of 10. The target is 10 times the wait over the target wait of 0.5 hours, capped at three times the baseline. Each period staff closes 1/(adjustment time) of the gap.
Predict first. If the wait is held at 0.75 hours, what staffing level does the rule head toward?
Choose an example
Constructed example: the pack's staffing rule with the three constant waits of the chapter's test and others defined for this reader.
Calculated values
- Staff the rule aims at
- 15.0
- Staff after 1 period
- 11.25
- Staff after 40 periods
- 15.00
- Gap still open after 40 periods
- 0.00
Target = 10 x (0.75/0.5) = 15.0. First period: 10 + (15.0 - 10)/4 = 11.25. After 40 periods staff is 15.00, leaving a gap of 15.0 - 15.00 = 0.00. A slower adjustment time closes the gap more slowly.
Use the idea
A constant input should give a stable output; if the aggregate side alone cannot settle, no coupling can fix it.
Where the conclusion applies
The wait is an input held fixed, which the coupled model never does. The cap and the floor of one clinician are the pack's rules, not facts about clinics.
Check your understanding: If the wait is held at 0.25 hours, what staffing does the rule head toward?
Chapter 18 source: section "Testing across the seam". Demonstration C18-D03.
4Demonstration 4 of 4
Does the frequency finding survive another seed?
If the random seed changes, does exchanging less often still multiply the staffing swing?
The chapter reports one seed. Another seed gives different arrival times and, with them, a different size of effect, so a single pair of runs does not settle the matter.
A seed fixes the random arrival times. Volatility is the standard deviation of staff. The ratio is the volatility with an exchange every 8 periods divided by the volatility with an exchange every period.
Predict first. With seed 7 over 60 periods, is the volatility ratio near the chapter's 4.7?
Choose an example
Constructed example: the chapter's two coupled runs repeated with other seeds and run lengths defined for this reader.
Calculated values
- Volatility, every period
- 1.01
- Volatility, every 8 periods
- 4.75
- Ratio
- 4.72
- Chapter's seed 5 over 60 periods
- 1.01 and 4.75
Ratio = 4.75/1.01 = 4.72 for seed 5 over 60 periods. The chapter's seed 5 over 60 periods gives volatilities 1.01 and 4.75, about 4.7 times. The slower exchange swings clearly wider, as the chapter's run does.
Use the idea
Where a finding depends on the exchange interval, report it across several seeds, as the chapter suggests for the frequency.
Where the conclusion applies
Four seeds and two run lengths; this is a robustness check on a constructed model, not an estimate of how real clinics behave.
Check your understanding: With volatilities of 0.80 and 1.39, what is the ratio?
Chapter 18 source: section "The seam is part of the model". Demonstration C18-D04.