Chapter 6 — Calculus II
Lesson map
Lesson 06 published. Map, example, EX011.
flowchart TD
N0["Learning outcomes"]
N1["6.1 Exponential models"]
N2["6.2 Sums and integrals for exposure"]
N3["6.3 Optimization"]
N4["6.5 Practical activity"]
N5["Synthesis"]
N0 --> N1 --> N2 --> N3 --> N4 --> N5
N5 --> EX["Worked example"]
EX --> Q["Exercise EX011"]
Logical drawing
flowchart LR
N0["Learning outcomes"]
N1["6.1 Exponential models"]
N2["6.2 Sums and integrals for exposure"]
N3["6.3 Optimization"]
N4["6.5 Practical activity"]
N5["Synthesis"]
N0 --> N1 --> N2 --> N3 --> N4 --> N5 --> Q["EX011"]
[Learning outcomes]
|
v
[6.1 Exponential models]
|
v
[6.2 Sums and integrals for exposure]
|
v
[6.3 Optimization]
|
v
[6.5 Practical activity]
|
v
[Synthesis]
|
v
[Worked example] --> [Exercise EX011]
Learning outcomes
You will (1) use exponential and logarithmic models for growth and decay; (2) compute sums and integrals that appear in cost and risk models; (3) apply optimization to choose parameters under constraints; and (4) model the "harvest-now, decrypt-later" exposure window. It builds on Chapter 1 and supports performance (Chapter 36) and optimization (Chapter 41).
6.1 Exponential models
Many quantities in this field are modelled exponentially: the volume of recorded traffic, the decline in cost per qubit, the decay of a certificate's usefulness. A model f(t) = f₀·e^(rt) has doubling time ln 2 / r. Engineers should treat such models as scenarios: an assumed doubling time of qubit quality is an assumption, not a law.
Doubling time is the most intuitive way to state an exponential model: if recorded traffic doubles every 3.8 years, a reader can picture the growth without logarithms. Showing results under a fast and a slow scenario keeps the model honest. If the recommendation changes between scenarios, the memo must say which assumption drives the change.
6.2 Sums and integrals for exposure
Suppose BPA records traffic at a rate v(t) GB per year, and any traffic recorded before migration date T is exposed if a CRQC appears at time Z. The exposed volume is the integral of v from 0 to T. If v grows exponentially, ∫₀^T v₀e^(rt) dt = (v₀/r)(e^(rT) − 1): delaying migration grows exposure faster than linearly. The engineering lesson is that a one-year delay late in the programme costs more exposure than a one-year delay early.
6.3 Optimization
Parameter choice is constrained optimization. Example: choose an ML-KEM parameter set to maximize security subject to a handshake byte budget. ML-KEM-512, -768 and -1024 offer NIST categories 1, 3 and 5 with encapsulation keys of 800, 1,184 and 1,568 bytes (FIPS 203). If the budget for the client key share is 1,300 bytes, ML-KEM-768 is the largest that fits. Lagrange multipliers formalize such trade-offs for continuous variables; in discrete parameter sets you simply enumerate.
The byte budget comes from real limits: some networks fragment or drop handshake messages above certain sizes, and middleboxes may reject unusual records. Choosing ML-KEM-768 within a 1,300-byte budget gives category 3 security with room left. If the budget were 1,000 bytes, only ML-KEM-512 would fit, and the memo would have to discuss whether category 1 is acceptable.
6.4 Worked example
BPA's customs exchange sends 2 TB/year today, growing 20% per year (r ≈ 0.182). Exposed volume if migration completes after 3 years: (2/0.182)(e^0.547 − 1) ≈ 8.0 TB; after 5 years: ≈ 16.3 TB. Two extra years roughly double the exposed archive.
With r equal to 0.182, the exponential terms are about 1.73 after three years and 2.49 after five. Multiplying the excess over one by about eleven, the factor two divided by r, gives the two figures. The doubling shows why the programme plan front-loads the customs exchange: it has the fastest-growing volume and the longest confidentiality requirement.
6.5 Practical activity
Compute the exposed volume for migration after 1–8 years and plot (or tabulate) it. Identify the year where exposure exceeds 10 TB.
Use the formula from the worked example rather than adding up yearly volumes: summing 2, 2.4, 2.88 and so on treats each year's traffic as arriving on the first day and gives slightly lower totals, about 7.3 TB after three years instead of 8.0. Both methods agree on which year first crosses 10 TB, but only the integral matches the figures in the worked example and in the exercise set. Tabulating each year shows the gaps between consecutive years growing, which is the non-linear effect that makes late delays expensive. Students should state their growth assumption beside the table and rerun it at 10% and 30%.
6.6 Common errors and limits
- Treating growth assumptions as forecasts.
- Linear thinking about delay under exponential growth.
- Ignoring that exposure also depends on how long data stays sensitive (Chapter 20).
Growth rates are assumptions; a 20% rate may fall if a service is retired or rise if new partners join. Linear thinking underestimates the cost of delay because each later year adds more volume than the one before. Exposure also depends on lifetime: traffic that loses value after a year matters less than customs records that must stay confidential for twenty-five years.
Exercises
- EX011 — Exposure integral. Compute BPA's exposed volume for T = 4 years. Rubric: 3 = correct (≈ 11.8 TB) with working.
- EX012 — Budgeted choice. With a 900-byte client-share budget, which ML-KEM set fits, and what should the engineer do if category 3 is required? Rubric: 3 = ML-KEM-512 fits; proposes changing the budget or transport rather than downgrading silently.
Synthesis
Calculus turns migration timing into exposure: under growth, delay costs more than linearly; parameter choice is constrained optimization over published parameter sets.
Continue studying — chapter 06 of 50
← Chapter 05 (draft) · Course chapter index · Chapter 07 (draft) →