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16-Civ-B4 Engineering Hydrology · May 2018

Question 6 of 7: Statistical Frequency and Probability Analysis

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Examination, May 2018, 16-Civ-B4 Engineering Hydrology, three hours’ duration, closed book with one two-sided candidate-prepared aid sheet (8½″ × 11″) and one approved Casio or Sharp calculator whose model designation must be written on the first inside left-hand sheet of the work book. Seven problems are printed. Page-1 Note 4 states that any five (5) questions constitute a complete paper and that only the first five answers appearing in the work book will be marked; Note 5 weights each problem at twenty (20) points, so the examinable total is 5 × 20 = 100 points. All seven problems are solved here, because this set is a study resource rather than a timed sitting. Sub-part mark values below are the printed ones from the page-6 marking scheme, which on this sitting is internally consistent — every problem’s sub-parts sum to twenty, and the marks printed in the page margins agree with the scheme.

Reference texts. V. T. Chow, D. R. Maidment and L. W. Mays, Applied Hydrology (hydrologic cycle, unit hydrographs, conceptual models, routing, frequency analysis); L. W. Mays, Water Resources Engineering, 3rd ed. (urban and highway drainage, stormwater management, reservoir operation); W. Viessman and G. L. Lewis, Introduction to Hydrology, 5th ed. (precipitation and streamflow measurement, hydrologic modelling); C. W. Fetter, Applied Hydrogeology, 4th ed. (Darcy’s law and aquifer hydraulics); V. T. Chow, Open-Channel Hydraulics (1959) (flood-wave propagation and unsteady flow). Canadian practice references: Environment and Climate Change Canada Engineering Climate Datasets (short-duration rainfall and IDF curves) and the Water Survey of Canada HYDAT archive; the WMO Manual on Stream Gauging (WMO-No. 1044) and ISO 1100-2 (stage–discharge ratings); the Transportation Association of Canada Guide to Bridge Hydraulics and provincial highway drainage manuals; provincial stormwater management planning and design manuals (e.g. Ontario MOECC 2003, British Columbia Stormwater Planning Guidebook); and the Canadian Dam Association Dam Safety Guidelines (inflow design flood and reservoir routing).

Check — which numbers come from the paper and which are the solver’s. This sitting supplies numerical data in only three places: Problem 2(iii) (the 7 ha suburban development, its 30-minute time of concentration and the IDF relation), Problem 5(iii) (the printed IDF chart, from which an intensity must be read graphically), and Problem 6(iii) (the 50-year return period and the 10-year exposure). Those three answers are computed from the paper’s own data. Every other number below appears inside a short illustrative example whose inputs are stated in an explicit Given line as the solver’s own representative Canadian values; they exist to make a discussion answer concrete and checkable, and they are not exam data. A candidate who assumed different but reasonable values and carried them through consistently would receive the same marks.

Question 6: Statistical Frequency and Probability Analysis (20 marks)

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.

6(i) — How frequency analysis works, with an example method (7 marks)

Frequency analysis is the procedure by which a finite record of past events is converted into a statement about the probability of future ones. Its logic is that the annual maxima of a hydrologic variable — the largest flood of each year, the deepest one-hour rainfall of each year — may be treated as a random sample from a single underlying probability distribution, so that fitting that distribution to the sample allows the magnitude corresponding to any exceedance probability to be estimated, including probabilities rarer than anything in the record. The steps are these.

  1. Assemble and screen the series. The annual maximum series is extracted from a homogeneous gauged record, and screened for trend, for shifts caused by regulation or land-use change, for outliers, and for mixed populations (a basin whose large floods come sometimes from snowmelt and sometimes from rain-on-snow does not have one parent distribution).
  2. Rank the data and assign plotting positions. The values are ranked from largest to smallest and each is assigned an empirical exceedance probability, most commonly by the Weibull formula $P = m/(n+1)$ for rank $m$ in a record of $n$ years, or by the Gringorten or Cunnane formulae. Plotting the ranked data on the probability paper of a candidate distribution shows at once whether that distribution is a plausible parent.
  3. Fit a distribution and compute quantiles. A distribution is fitted by the method of moments, L-moments or maximum likelihood, and the design quantile is obtained from Chow’s general frequency-factor equation $$x_T=\bar{x}+K_T\,s$$ where $\bar x$ and $s$ are the sample mean and standard deviation and $K_T$ depends on the distribution and the return period. In Canadian flood practice the usual candidates are the Gumbel (EV1), the generalised extreme value, the three-parameter log-normal and the log-Pearson type III.
  4. Report with confidence limits and check the extrapolation. The quantile is quoted with confidence limits, which widen rapidly for return periods beyond about twice the record length; where the record is short, a regional or index-flood analysis pools data from hydrologically similar gauged basins to stabilise the estimate.

Example method — the Gumbel (extreme value type I) distribution. Its frequency factor is available in closed form,

$$K_T=-\frac{\sqrt{6}}{\pi}\left[0.5772+\ln\ln\left(\frac{T}{T-1}\right)\right]$$

Given. An annual maximum flood series with mean $\bar{x}=420\ \text{m}^3/\text{s}$ and standard deviation $s=130\ \text{m}^3/\text{s}$. Find. The 100-year flood, and the return period corresponding to an observed magnitude.

  1. Evaluate the frequency factor for the design return period. With $T = 100$ years, $$K_{100}=-\frac{\sqrt{6}}{\pi}\left[0.5772+\ln\ln\left(\frac{100}{99}\right)\right]=3.1367$$
  2. Convert the frequency factor to a discharge. Substituting in the frequency-factor equation, $$x_{100}=420+(3.1367)(130)=\boxed{828\ \text{m}^3/\text{s}}$$ This is the magnitude side of the question: a discharge of about 828 $\text{m}^3/\text{s}$ has a one per cent chance of being exceeded in any year.
  3. Invert the relation to obtain a return period. The same fitted distribution is used in the other direction for an observed event: the standardised variate $K = (x-\bar x)/s$ is computed, the Gumbel expression is solved for the non-exceedance probability $F$, and the return period follows as $T = 1/(1-F)$. This is how a flood that has just occurred is described as, for instance, “a 1-in-70-year event on this river”.

6(ii) — The objective of probability analysis of precipitation, and two assumptions (8 marks)

The main objective. Probability analysis applied to precipitation exists to convert a finite historical rainfall record into a statement of risk: the probability that a rainfall of a given depth or intensity, for a given duration, will be equalled or exceeded in any year. Its purpose is to allow hydraulic structures — storm sewers, culverts, bridges, spillways, stormwater ponds, urban major systems — to be sized against an explicit and defensible probability of failure chosen to match the consequences of that failure, rather than against the largest storm that happens to appear in the record. Concretely, its products are the design storm and the intensity–duration–frequency curve, and they replace the statement “the biggest storm we have seen” with the statement “the storm with a two per cent chance in any year”. That reframing is what makes level of service a design decision, allows the risk over the life of a structure to be computed and compared across alternatives, and permits the design to be justified economically and legally. A secondary but important objective is that it makes rainfall records from different places and different lengths comparable, since a 50-year intensity means the same thing everywhere.

Assumption 1 — the data are a random sample of independent and identically distributed observations from one population. Successive annual maxima must be statistically independent, with no serial correlation carried from one year into the next; each must be drawn from the same population, meaning the same storm-generating mechanism, so that mixing convective summer thunderstorms with frontal or hurricane-remnant rainfall in a single series violates the assumption and produces a distribution with a broken tail; and where a partial-duration (peaks over threshold) series is used instead of annual maxima, successive events must be separated enough to be independent. The gauge record must also be homogeneous — the same instrument, the same exposure, the same site.

Assumption 2 — the series is stationary. The statistical properties of the record — its mean, its variance, its distributional form — must not change through time, since it is this that licenses using the past to describe the future. Stationarity is violated by urbanisation or land-use change around the gauge, by relocation or replacement of the instrument, and increasingly by climate change, which is altering short-duration rainfall extremes in much of Canada. Canadian practice now expects the assumption to be tested (for example by the Mann–Kendall trend test) and, if it fails, to be addressed either by non-stationary frequency analysis or by applying a climate-change uplift to the design intensities, as ECCC’s IDF_CC tool does.

Two further assumptions would also be credited: that the record is long enough for the target return period, the usual guidance being not to extrapolate much beyond twice the record length; and that the chosen distribution is in fact the correct parent, a choice that has more effect on a 100-year quantile than the choice of fitting method does.

6(iii) — Probability of at least one 50-year flood in 10 years (5 marks)

Given. Return period $T = 50$ years, so the annual exceedance probability is $p = 1/T = 0.02$; exposure period $N = 10$ years; the occurrences in successive years are independent Bernoulli trials with constant $p$.

Find. The probability that the 50-year flood is equalled or exceeded at least once during the next 10 years — the encounter probability or risk of failure.

Approach. Compute the probability of the complementary event, that the flood does not occur in any of the ten years, and subtract from one.

  1. State the annual probabilities. The probability of occurrence in any one year and the probability of non-occurrence are $$\begin{aligned} p&=\frac{1}{T}=\frac{1}{50}=0.02\\ 1-p&=0.98 \end{aligned}$$
  2. Form the probability of no occurrence in N independent years. Because the years are independent, the probabilities multiply: $$P(\text{no occurrence in }N\text{ years})=(1-p)^{N}=(0.98)^{10}=0.817073$$
  3. Take the complement. The risk of at least one occurrence is therefore $$R=1-(1-p)^{N}=1-0.817073=\boxed{R=0.1829\ \ \text{or about }18.3\ \text{per cent}}$$
  4. Check against the Poisson approximation. For small $p$ the occurrences may be treated as a Poisson process with mean $Np = 10(0.02) = 0.2$ events, giving $R \approx 1-e^{-0.2}=0.1813$, which agrees with the binomial result to within two parts in a thousand and confirms the arithmetic.
QuantitySymbolValue
Annual exceedance probability$p=1/T$0.02
Probability of no occurrence in 10 years$(1-p)^{N}$0.8171
Probability of at least one occurrence in 10 years$R$ 0.1829, i.e. 18.3 per cent
Poisson check$1-e^{-Np}$0.1813
Same risk over a 50-year design life$R$ 0.6358, i.e. 63.6 per cent

The last line is the part of this result that matters in practice, and it is worth stating explicitly: over a design life equal to the return period the risk of encountering the design event is not small but about 64 per cent. A structure designed to the 50-year flood is therefore more likely than not to see that flood during fifty years of service, which is why the choice of return period must be made against the consequences of exceedance and not against an intuition that a 50-year event is unlikely.