Water Resources HubFE section 3 of 16 · free theory
Computational Tools
Computational Tools is the FE's "how do you actually compute things" section — 4–6 questions on spreadsheets, algorithms, and the errors that creep into every calculation. It rewards careful reading more than heavy mathematics.
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Spreadsheet logic and functions
The exam treats a spreadsheet as a small programming language. Two things matter most: how cell references behave when copied, and how the logic functions evaluate:
=IF(test, value_if_true, value_if_false) =AND(…), =OR(…), =NOT(…)
| A1 | relative reference — both row and column shift when copied |
| $A1, A$1 | mixed reference — the $ locks the column (first) or the row (second) |
| $A$1 | absolute reference — always points at the same cell |
=SUM(range), =AVERAGE(range), =POWER(x, n), =MOD(x, y), =ABS(x), =ROUND(x, digits)
A $ locks whatever sits directly after it. In $A1 the column is frozen but the row still moves when you copy down — reading the $ position is the whole question.
Algorithms and flowcharts
Flowchart questions are bookkeeping, not programming. Read the shapes, then trace the values step by step as if you were the computer:
rounded rectangle → start/stop parallelogram → input/output
rectangle → process (do this) diamond → decision (branch on true/false)
For a loop question, make a small trace table: write down the counter, the condition value, and the running result at each pass. The answer is whatever the table says at the exit — most errors come from skipping the final pass or the final check.
Error types
Every measurement and every calculation carries error. The exam wants you to name the type and know what to do about it:
absolute error = |measured − true| relative error = absolute error / |true|
percent error = relative error × 100%
| blunder | a mistake — misreading, transposing digits; caught by checking, not by statistics |
| systematic error | a bias that pushes every reading the same way; averaging will not remove it — calibrate or correct |
| random error | noise that scatters both ways; averages out with enough repeats |
| truncation error | chopping a method short (finite series terms, finite step size) |
| round-off error | finite digits in the machine; grows as the step size shrinks |
Error propagation
When you combine uncertain quantities, the errors combine too. The standard engineering move is the root-sum-of-squares (RSS) of the individual errors:
sums/differences: δS = √(δa² + δb²) (absolute errors)
products/quotients: δQ/Q = √((δa/a)² + (δb/b)²) (relative errors)
powers: z = xn ⇒ δz/z = n · δx/x (the exponent multiplies the relative error)
RSS assumes the errors are independent and random. If the errors are worst-case bounds instead, the conservative answer is the plain linear sum — check which the question asks for.
Numerical-methods concepts
The exam tests the ideas behind numerical methods rather than long hand iterations: what converges, how fast, and what can go wrong:
iteration: repeat until |change| < tolerance or |f(x)| is small enough
bisection: always converges, slowly (one binary digit per step); needs a sign change across the bracket
Newton's method: converges fast near the root; stalls or diverges where f′(x) ≈ 0 or the guess is poor
central difference (f(x+h) − f(x−h))/(2h) beats forward difference (f(x+h) − f(x))/h — error ≈ h² vs ≈ h
Simpson's 1/3 rule is exact for polynomials up to degree 3
See these ideas running on real functions with the free numerical-methods calculator: bisection and Newton root finding, plus trapezoidal and Simpson integration.
Worked example Spreadsheet logic
Given: Cells A1 through A5 hold 2, 4, 6, 8, 10. Cell B1 contains =AVERAGE(A1:A5) and cell C1 contains =IF(B1>5,"HIGH","LOW"). What do B1 and C1 display?
Solution:
- AVERAGE sums the range and divides by the count: (2 + 4 + 6 + 8 + 10)/5 = 30/5 = 6. B1 displays 6.
- The IF test is B1 > 5, i.e. 6 > 5, which is TRUE, so C1 displays the value_if_true argument: "HIGH".
Answer: B1 = 6, C1 displays "HIGH".
Worked example Error propagation in an area
Given: A rectangle measures W = 12.0 ± 0.1 m by H = 8.0 ± 0.05 m. Give the area with its RSS uncertainty.
Solution:
- Area: A = 12.0 × 8.0 = 96.0 m².
- Product → relative errors add in quadrature: δA/A = √((0.1/12.0)² + (0.05/8.0)²) = √((0.008333)² + (0.006250)²) = √(0.00010851) = 0.0104167.
- Absolute uncertainty: δA = 96.0 × 0.0104167 = 1.0 m².
Answer: 96.0 ± 1.0 m².
Worked example Forward vs central difference
Given: f(x) = x³. Approximate f′(2) with step h = 0.5 using the forward and central differences. (Exact: f′(2) = 3(2)² = 12.)
Solution:
- Needed values: f(2) = 8, f(2.5) = 15.625, f(1.5) = 3.375.
- Forward: (f(2.5) − f(2))/0.5 = (15.625 − 8)/0.5 = 7.625/0.5 = 15.25. Error: 15.25 − 12 = 3.25.
- Central: (f(2.5) − f(1.5))/(2 × 0.5) = (15.625 − 3.375)/1.0 = 12.25. Error: 12.25 − 12 = 0.25.
Answer: Forward gives 15.25, central gives 12.25 — the central difference is far closer to the exact 12.