1Demonstration 1 of 4
The conveyor and the tank
Given the same mean delay, how differently do a pipeline and a first-order delay answer one step?
The tank starts releasing the moment material enters, so it responds at once but finishes slowly. The pipeline shows nothing until the whole delay has passed and then shows everything.
The input is nothing for 3 periods, then a constant step. A pipeline of length L repeats its input L periods later. A first-order delay of mean L releases 1/L of what it holds each period. Outflow is in the same units as the step.
Predict first. With a delay of 4, which is higher at period 5, the pipeline or the tank?
Choose an example
Constructed example: the chapter's own step of 10 through delays of length and mean 4, with other lengths defined for this reader.
Calculated values
- Pipeline at period 5
- 0.00
- Tank at period 5
- 4.38
- Pipeline at period 7
- 10.00
- Tank at period 7
- 6.84
- Period pipeline completes
- 7
- Period tank first reaches 95 percent
- 14
The step starts at period 3. The pipeline repeats the input from 4 periods earlier, so at period 7 it shows the input of period 3: 10.0. The tank releases 1/4 of what it holds each period, which leaves a share 0.750 behind. After 2 periods of input: 10 x (1 - 0.750 x 0.750) = 4.38; after 4: 10 x (1 - 0.750 x 0.750 x 0.750 x 0.750) = 6.84.
Use the idea
Ask whether everything that entered together leaves together. If yes, use a pipeline; if it mixes, use a tank.
Where the conclusion applies
Step input, one-period steps, empty at the start. A real delay may sit between these two shapes, and a pipeline is exact only if every item truly takes the same time.
Check your understanding: For a tank of mean 4 and a step of 10, what is the outflow two periods after the step starts?
Chapter 16 source: section "The conveyor and the tank". Demonstration C16-D01.
2Demonstration 2 of 4
Chaining tanks walks toward the conveyor
If one tank is too spread out and a pipeline is too sharp, what do several tanks in a row do?
Material has to cross the first tank before it reaches the second, so the immediate response disappears. Adding stages pulls the finish closer to the conveyor's sharp completion.
A chain splits a total mean into equal stage means, so n stages of mean (total/n) are joined end to end. A very small time step of 0.1 keeps every stage accurate. Time is in the same unit as the mean.
Predict first. Going from one stage to four with the same total mean of 4, does the output one time unit after the step go up or down?
Choose an example
Constructed example: stage counts and total means defined for this reader, computed with the pack's FirstOrderDelay and PipelineDelay.
Calculated values
- Stages
- 1
- Mean of each stage
- 4.00
- Output 1 time unit after the step
- 2.24
- Time after the step to reach 95 percent
- 11.9
- Conveyor reaches 95 percent after
- 4.0
1 stage(s) each of mean 4.0/1 = 4.00, so the total mean is 1 x 4.00 = 4.0, the same as the conveyor's 4.0. One time unit after the step the chain has released 2.24 of the 10, against 0.00 for the conveyor. It reaches 95 percent 11.9 time units after the step, against 4.0 for the conveyor.
Use the idea
A recruitment process is sourcing, screening, offer and notice, not one tank. Choose the order from the mechanism, not the convenience.
Where the conclusion applies
Equal stage means and a total mean fixed in advance. The order of a delay is a modelling choice that needs a conversation with someone who knows the process, because this reader cannot supply it.
Check your understanding: A chain of 4 stages has a total mean of 8. What is each stage's mean?
Chapter 16 source: section "Higher orders and the shape in between". Demonstration C16-D02.
3Demonstration 3 of 4
A tank stepped at its own mean is a conveyor
What happens to a first-order delay when the model's step is as long as its mean?
A step of dt moves dt/mean of the tank's content at once. At dt = mean that share is 1, so the whole tank empties each step, which is not an approximation of anything.
dt is the model's step, mean is the tank's mean delay, both in the same time unit. The tank holds an amount and releases held/mean per unit time. The inflow is a constant 10.
Predict first. With a mean of 4 and a step of 4, what is the outflow at the second reading?
Choose an example
Constructed example: the chapter's warning about a tank stepped at its own mean, run with the pack's FirstOrderDelay.
Calculated values
- Share of the tank released per step (dt / mean)
- 1.000
- First reading after the first update
- 10.000
- Outflow at time 8
- 10.000
- Outflow at time 8, very small step
- 8.664
- Error at time 8
- 1.336
One step: held = 4.00 x 10 = 40.0, and the next reading is 40.0/4.0 = 10.000. Share released per step = 4.00/4.0 = 1.000. At time 8 the chosen step gives 10.000 against 8.664, an error of 10.000 - 8.664 = 1.336. The step equals the mean, so the tank is emptied completely each step: the outflow jumps to the full inflow and stays there, which is a conveyor and not a tank.
Use the idea
Halve the step and rerun. If the answer moves, the earlier results were partly arithmetic.
Where the conclusion applies
A constant inflow and a very small step as the reference. The chapter's rule of thumb is a step several times smaller than the shortest delay; here a share of 1.0 breaks it outright.
Check your understanding: With a mean of 8 and a step of 4, what share of the tank is released per step?
Chapter 16 source: section "Two cases where the type decides the answer". Demonstration C16-D03.
4Demonstration 4 of 4
The same delay inside a correction loop
What does a delay do when the decider corrects against a value that has already moved?
A long delay lets the stock keep moving while the decider still sees the old value, so the correction keeps pushing and overshoots. A lower gain damps the same swing without touching the delay.
The stock starts at 0 with a target of 100. Each period the correction is gain x (100 - what the decider sees). What is seen is the stock read through a delay of the chosen length.
Predict first. At gain 0.25, does a delay of 4 overshoot 100 more than a delay of 1?
Choose an example
Constructed example: a loop defined for this reader to illustrate the chapter's inside-the-loop argument, run with the pack's two delays.
Calculated values
- First correction
- 25.0
- Pipeline peak
- 114.1
- Tank peak
- 112.9
- Pipeline after 24 periods
- 100.2
- Tank after 24 periods
- 100.0
The first correction is 0.25 x (100 - 0) = 25.0, whatever the delay, because nothing has reached the decider yet. With a delay of 2: the pipeline peaks at 114.1, so 114.1 - 100 = 14.1 above the target; the tank peaks at 112.9, so 112.9 - 100 = 12.9 above the target. Gain and delay multiply, and lowering either one damps the swing.
Use the idea
Enumerate the loops, then check which delays lie on them. A delay outside every loop only shifts and smooths.
Where the conclusion applies
A single balancing loop with a pipeline or tank delay on the reading. The delays here are in a loop; the same delay outside it would change when a number is seen but not the behaviour.
Check your understanding: At gain 0.1 and a delay of 2, what is the first correction?
Chapter 16 source: section "Inside the loop and outside it". Demonstration C16-D04.