Question 7 of 7: Life-cycle cost analysis and flexible-pavement distress

Question 7: Life-cycle cost analysis and flexible-pavement distress (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.

Given. Two competing sections over a common 50-year analysis period at a 5 % discount rate:

Find. The net present worth of each option and the more cost-effective alternative; the correct comparison technique when the analysis periods differ; and the mechanisms and surface patterns of three flexible-pavement distress modes.

Approach. Discount every future cost and the salvage credit to year zero with the net-present-worth expression printed on appendix page 8, compare the two totals, and convert to an equivalent uniform annual cost for part (b).

(a) Which option is more cost effective? (8 marks)

1. Statement of the criterion. The appendix gives$$\text{NPW}=IC+\sum_{j=1}^{k}\left(M\&R_{j}\left[\dfrac{1}{1+i}\right]^{n_{j}}\right)-SV\left[\dfrac{1}{1+i}\right]^{AP}$$with $i=0.05$ and $AP=50$ years. The salvage value is a benefit and therefore carries a negative sign.
2. Discount the maintenance and rehabilitation stream. Each intervention is brought back to year zero individually. For Option A the largest single contribution is the year-9 mill-and-patch, $1\,406\,063/(1.05)^{9}=906\,362$; summing all ten entries gives$$\text{PW}(M\&R)_{A}=3\,068\,917$$and for Option B’s seven entries, dominated by the year-38 resurfacing ($5\,696\,000/(1.05)^{38}=892\,024$),$$\text{PW}(M\&R)_{B}=3\,215\,151$$
3. Discount the salvage credits. With $(1.05)^{50}=11.467$,$$\text{PW}(SV)_{A}=\dfrac{12\,000\,000}{11.467}=1\,046\,531\qquad\text{PW}(SV)_{B}=\dfrac{25\,000\,000}{11.467}=2\,180\,273$$The rigid option’s salvage value is more than twice as large in nominal terms, yet fifty years of discounting reduce the advantage to about 1.13 million dollars in present worth — a good illustration of how little distant terminal values matter.
4. Net present worth.$$\text{NPW}_{A}=71\,050\,000+3\,068\,917-1\,046\,531=\boxed{73\,072\,000}$$$$\text{NPW}_{B}=79\,504\,000+3\,215\,151-2\,180\,273=\boxed{80\,539\,000}$$in Canadian dollars.
5. Decision. Option A costs $80\,539\,000-73\,072\,000=7\,467\,000$ less in present worth, about 9.3 % of the total, so the flexible perpetual pavement (Option A) is the more cost-effective alternative. The conclusion is robust: the difference is dominated by the 8.45 million dollar gap in initial construction cost, which is not discounted at all, while the discounted maintenance streams differ by only 0.15 million and the salvage advantage of the rigid option recovers only 1.13 million of the gap. Because it rests on an undiscounted term, the ranking does not change under any plausible discount rate.

(b) Comparing options with different analysis periods (4 marks)

Net present worth is only meaningful when the alternatives are compared over the same period, because a shorter-lived option is credited with no costs in the years after it ends. When the analysis periods genuinely differ, three techniques are available and the first is normally preferred.

The standard method is to convert each net present worth into an equivalent uniform annual cost over its own period, using the capital-recovery factor given on appendix page 8:$$\text{EUAC}=\text{NPW}\left[\dfrac{i(1+i)^{AP}}{(1+i)^{AP}-1}\right]$$For the present case both periods are 50 years and the factor is $0.054777$, giving$$\text{EUAC}_{A}=4.00\ \text{million/yr}\qquad\text{EUAC}_{B}=4.41\ \text{million/yr}$$confirming the ranking of part (a). Because an annual cost is independent of how long the stream runs, two options with unequal lives can be compared directly on this basis, provided the pattern of costs is assumed to repeat.

The two alternatives to that approach are to adopt a common analysis period equal to the least common multiple of the two service lives, repeating each alternative’s cycle until the periods coincide; or to keep the shorter period and credit the longer-lived alternative with a remaining-service-life salvage value, prorated from its most recent rehabilitation. The last is what FHWA practice recommends for pavement LCCA, and it is why the salvage term appears in the expression at all.

(c) Fatigue cracking, rutting and low-temperature cracking (8 marks)

Fatigue (alligator) cracking is a load-associated distress driven by the repeated horizontal tensile strain at the bottom of the bound layer each time a wheel passes. No single pass is damaging; damage accumulates as micro-cracks that coalesce, propagate upward and finally break the surface. The pattern is diagnostic: it appears only in the wheel paths, beginning as isolated longitudinal hairlines that progressively interconnect into the many-sided “alligator hide” polygons, and ends in potholing once water enters. It is worst where the pavement is thinnest or the subgrade weakest, and it is the criterion that governs bound-layer thickness. A true perpetual pavement is designed so that the tensile strain never exceeds the endurance limit, which is why Option A above carries no structural rehabilitation in its 50-year schedule.

Rutting is also load-associated but is a permanent-deformation rather than a fracture mechanism: it is the accumulation of small unrecovered vertical and shear strains under each pass. It has two distinct sources with different signatures. Subgrade or subbase rutting is a structural failure — the vertical compressive strain at the top of the subgrade is too high — and produces a wide, shallow depression in the wheel path with little or no upheaval at the edges. Mixture rutting occurs within the asphalt layer itself when the mix lacks aggregate interlock or has too much binder, and produces a narrow, sharp-shouldered channel with shoving and humping at its edges. Rutting is worst in hot weather, at low speeds and at stopping locations, where the binder is least viscous and the load dwells longest; it is a safety problem as well as a serviceability one because ruts hold water and promote hydroplaning.

Low-temperature (transverse thermal) cracking is not load-associated at all, which is the key contrast. As the surface cools, the asphalt layer tries to contract but is restrained by friction against the layer beneath, so a tensile stress builds; when it exceeds the tensile strength of the binder at that temperature the layer cracks through in a single event. The pattern is therefore a series of roughly evenly spaced cracks running transversely across the full width of the pavement, including the shoulders and the areas between the wheel paths where no traffic runs. Spacing tightens over successive winters as thermal-fatigue cycling adds intermediate cracks. It is the dominant winter distress in Northern Ontario and on the prairies, and it is controlled by binder selection — the low temperature grade of the performance-graded binder — not by thickness. This is the practical distinction: fatigue cracking is cured by a thicker or stiffer structure, rutting by mixture design and subgrade protection, and thermal cracking only by a softer binder.