Maximum Decision-Logic Delay Before Tank Spill-Over: 19-Soft-A4 Real-Time Systems, May 2013, Q2

Question 2: Maximum Decision-Logic Delay Before Tank Spill-Over (20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Cylindrical tank, diameter $D = 1.0\ \text{m}$, height (rim / spill level) $H = 5.0\ \text{m}$; inlet flow rate $Q = 400\ \text{L/min}$; shut-off command is issued at level $h_{\text{shutoff}} = 4.9\ \text{m}$; the closed loop is over-damped (no overshoot in the level response).

Quantity	Value
Diameter, $D$	1.0 m
Rim / spill level, $H$	5.0 m
Shut-off trigger level, $h_{\text{shutoff}}$	4.9 m
Inlet flow rate, $Q$	400 L/min

Find. The largest time delay the decision-logic function can take, between the shut-off level being reached and the valve actually closing, without the tank spilling over.

Figure 1 — cylindrical tank: the safety margin between the shut-off level (4.9 m) and the rim (5.0 m) is the only buffer available to the decision logic before spill-over.

Approach. Because the loop is over-damped, the level rises monotonically with no overshoot, so the worst case is simply that the inlet keeps flowing at its full rated rate for the whole delay: the decision logic's response time is bounded by how long it takes to fill the remaining 0.1 m margin between the shut-off level and the rim, at the full inlet rate.

Tank cross-sectional area. $$A = \frac{\pi D^2}{4} = \frac{\pi (1.0\ \text{m})^2}{4} = \boxed{0.7854\ \text{m}^2}.$$
Remaining safety-margin volume. Between the shut-off level and the rim, the free volume that must not be exceeded is $$\Delta V = A \times (H - h_{\text{shutoff}}) = 0.7854\ \text{m}^2 \times 0.1\ \text{m} = 0.07854\ \text{m}^3 = \boxed{78.54\ \text{L}}.$$
Inlet flow rate in consistent units. $$Q = \frac{400\ \text{L/min}}{60\ \text{s/min}} = 6.667\ \text{L/s}.$$
Largest tolerable decision-logic delay. Over-damped, monotonic rise means the worst-case fill of the margin happens at the full inlet rate, so $$t_{\max} = \frac{\Delta V}{Q} = \frac{78.54\ \text{L}}{6.667\ \text{L/s}} = \boxed{11.78\ \text{s}}.$$ Substituting the shipped result back: $6.667 \times 11.78 = 78.5\ \text{L} \approx \Delta V$. ✓

Quantity	Result
Tank cross-sectional area, $A$	0.7854 m²
Safety-margin volume, $\Delta V$	78.54 L (0.07854 m³)
Largest tolerable decision-logic delay, $t_{\max}$	≈ 11.78 s

Concept & Theory

Topic. Deriving a hard real-time deadline from a physical safety margin (level-control decision logic).

Key relations. Volume–level relation $V = A\,h$ for a constant cross-section tank; worst-case deadline $t_{\max} = \Delta V / Q_{\max}$, i.e. safety margin divided by the fastest rate at which the margin can be consumed.

Why this works. This is the same pattern used throughout hard real-time design: a physical safety margin (here, 0.1 m of freeboard) is converted into a worst-case rate-of-consumption problem, and the resulting time is the deadline the software must guarantee — not "usually," but every time, because the over-damped assumption removes any oscillatory overshoot that could otherwise shrink the margin further. Sensor sampling, computation and valve-actuation delay together must sum to less than 11.78 s for the design to be safe.

Common pitfall. Using the full tank height (5 m) instead of the 0.1 m margin actually available once the shut-off level has already been reached, or forgetting to convert L/min to L/s before dividing.

Source. Liu, Real-Time Systems, Ch. 1 (deriving timing requirements from the controlled physical process).

Question 2 of 6: Maximum Decision-Logic Delay Before Tank Spill-Over

Question 2: Maximum Decision-Logic Delay Before Tank Spill-Over (20%)