7.4 Cost Analysis, Uncertainty, Expected Value, and Decision Trees

7.4 Cost Analysis, Uncertainty, Expected Value, and Decision Trees

The NCEES Engineering Economics specification lists five sub-topics. Time value of money, economic analyses, and project selection are covered in the three preceding sections. This section covers the remaining two: cost analysis ("incremental, average, sunk, estimating") and uncertainty ("expected value and risk"). These items usually involve less arithmetic than a present-worth problem but punish sloppy reading, because the trap is almost always a number that should have been ignored.

Cost Classifications the Exam Actually Tests

The Sunk-Cost Trap

A company has spent $400{,}000 developing a prototype. Completing it requires $150{,}000 more and will yield revenue of $220{,}000. Should they finish?

Yes. The $400{,}000 is gone whichever way the decision goes; it appears in no future cash flow. The forward-looking comparison is $220{,}000 of revenue against $150{,}000 of remaining cost, a $70{,}000 gain. Total project accounting shows a $330{,}000 loss, and that loss is real — but abandoning the project makes it a $400{,}000 loss instead. Past money cannot be recovered by refusing to earn future money.

The mirror-image error is equally common: continuing a project because of what has already been spent ("we've come too far to stop"). Both errors come from letting a sunk cost enter the comparison. The test is mechanical — if a number is dated before the decision point, strike it out.

Marginal vs. Average Cost

Total cost of producing $Q$ units with fixed cost $F$ and unit variable cost $v$:

$TC = F + vQ, \qquad AC = \frac{F}{Q} + v, \qquad MC = \frac{d(TC)}{dQ} = v$

Average cost falls continuously as $Q$ rises because fixed cost spreads thinner, while marginal cost stays at $v$. Accept-or-reject decisions on incremental volume use marginal cost, not average cost. An order priced above $v$ but below $AC$ still improves profit.

Cost Estimating

Capacity Scaling (the Six-Tenths Rule)

Equipment cost does not scale linearly with capacity, because cost tracks surface area while capacity tracks volume:

$\frac{C_2}{C_1} = \left(\frac{Q_2}{Q_1}\right)^{n}$

where $n \approx 0.6$ for much process equipment. Doubling capacity costs about $2^{0.6} = 1.52$ times as much, not twice as much — the origin of economies of scale.

Cost Indexes for Escalation

$C_{\text{now}} = C_{\text{then}} \times \frac{I_{\text{now}}}{I_{\text{then}}}$

Worked Example: Combined Scaling and Escalation

A 5{,}000 L reactor cost $180{,}000 when the cost index was 480. Estimate the cost today of a 12{,}000 L reactor with the index at 624, using $n = 0.6$.

Scale for capacity:

$C = 180{,}000\left(\frac{12{,}000}{5{,}000}\right)^{0.6} = 180{,}000\,(2.4)^{0.6}$

$(2.4)^{0.6} = e^{0.6\ln 2.4} = e^{0.6(0.8755)} = e^{0.5253} = 1.691$

$C = 180{,}000(1.691) = \$304{,}400$

Then escalate:

$C_{\text{now}} = 304{,}400 \times \frac{624}{480} = 304{,}400(1.30) = \boxed{\$395{,}700}$

Trap: capacity rose 140% but cost rose only 69% before escalation. Candidates who scale linearly get $432{,}000 before escalation and $561{,}600 after — a 42% overestimate.

Uncertainty: Expected Value and Risk

Expected monetary value (EMV) is the probability-weighted average outcome:

$E[X] = \sum_{i} p_i x_i, \qquad \sum_i p_i = 1$

A risk-neutral decision maker chooses the alternative with the best EMV. Variance measures the dispersion around it:

$\sigma^2 = \sum_i p_i (x_i - E[X])^2$

Two alternatives with identical EMV but different variance are not equivalent to a real decision maker; the exam may ask which carries more risk, which is answered by variance or standard deviation, not by EMV.

Worked Example: Expected Cost of a Design Choice

A pipeline coating decision:

$E[\text{Standard}] = 220{,}000 + 0.18(1{,}400{,}000) = 220{,}000 + 252{,}000 = \$472{,}000$ $E[\text{Enhanced}] = 430{,}000 + 0.03(1{,}400{,}000) = 430{,}000 + 42{,}000 = \$472{,}000$

The expected costs are identical. A risk-neutral analyst is indifferent — but the enhanced coating caps the worst case at $1{,}830{,}000 with 3% probability rather than 18%, and reduces the standard deviation of the outcome substantially. Where the failure consequence includes public safety or environmental release, the lower-variance option is the defensible engineering choice even at equal EMV. This is where the Engineering Economics area meets the ethics area.

Decision Trees

A decision tree lays out sequential choices and chance events:

• Decision node (square): the analyst chooses; take the best value
• Chance node (circle): nature chooses; take the expected value
• Solve by rolling back right to left (backward induction)

Worked Example: Rolling Back a Two-Stage Tree

A firm can build a large plant or a small plant. Demand is high with probability 0.6 and low with probability 0.4.

Expected values at the two chance nodes:

$E[\text{Large}] = 0.6(900{,}000) + 0.4(-200{,}000) = 540{,}000 - 80{,}000 = \$460{,}000$ $E[\text{Small}] = 0.6(400{,}000) + 0.4(250{,}000) = 240{,}000 + 100{,}000 = \$340{,}000$

At the decision node, take the larger: build the large plant, EMV $460{,}000.

Expected Value of Perfect Information

Now suppose a market study could reveal demand with certainty before the plant decision. With perfect information you would build large if high (+$900{,}000) and small if low (+$250{,}000):

$EV\text{ with perfect information} = 0.6(900{,}000) + 0.4(250{,}000) = 540{,}000 + 100{,}000 = \$640{,}000$

$EVPI = 640{,}000 - 460{,}000 = \boxed{\$180{,}000}$

Interpretation: $180{,}000 is the maximum worth of any study, however good. A study quoted at $210{,}000 must be declined even if it is perfectly reliable, and a real study that is less than perfect is worth strictly less than $180{,}000.

Trap: EVPI uses the probabilities before the information arrives, weighting each branch by how likely you are to be told that branch. Candidates who compute $900{,}000 - 460{,}000$ get $440{,}000, treating the best possible outcome as if it were certain.