Driver Trees 2026: Where the Number Actually Comes From

The identity is the whole point

Write the top line as an equation before drawing anything. Revenue for a scheduled service is sailings times seats times load factor times fare, plus onboard spend per passenger times passengers. Every term in that sentence is a branch, and no branch exists that is not a term.

This discipline rules things out, which is most of its value. Brand, punctuality and crew quality all matter to a ferry business and none of them is a term. They enter through load factor, because that is the only channel available to them, and that is where they belong in the discussion. A branch that cannot be connected to a term will not survive the first follow-up question.

Oliver Wyman's published method has a step called using numbers, and a separate step that asks you to show why your approach beats the alternatives, checked 30 September 2026 at oliverwyman.com. An identity does both at once: it produces the number, and it explains itself, because the reader can see which factor moved.

Charles River Associates puts the obligation more bluntly. Its case interview guidance says to "be prepared to explain how you came to your answer and show your math", checked 30 September 2026 at crai.com. A driver tree is the cheapest way to comply, because the work is already written down.

A driver tree, an issue tree, and a decision tree

Three decompositions get called trees, and they close in three different ways. Which one a prompt deserves follows from that difference.

Only the first of the three is an identity. An issue tree can be complete and still produce no number, which is the correct behavior: its output is a shortlist of things to test. A decision tree computes nothing until you supply probabilities it cannot get from the data. A driver tree computes the moment the bottom row is filled, and it breaks the moment a branch stops being a term in the equation.

Pick by what the prompt asks for. A question about why something moved wants sub-questions, so issue trees. A question about which option to take wants a rollback, so decision trees. A question about how big something is, or where a number comes from, wants the identity on this page.

Two firms publish the warning that belongs with that choice. Kearney's case page tells candidates to think top down and, in the same paragraph, to "avoid forcing your solution to fit a standard framework. The only things it needs to fit are the problem, the business, and the objectives", checked 30 September 2026 at kearney.com. Deloitte's published five-step approach opens with "Understand the issue" and names no framework anywhere in it, checked 30 September 2026 at deloitte.com.

Read those two lines together and they describe the same discipline a driver tree needs. It is top down and it is forbidden as a template, so drawing a revenue identity before establishing that the question is about revenue is exactly the error both pages name.

A published driver tree, recalculated line by line

Oliver Wyman publishes a driver tree on its Poseidon Water Park case, in which a client wants the daily revenue of an established Dubai competitor. The firm names three revenue streams: entry tickets, food and beverages, merchandise. Then it fills in the volume branch.

Its published chain runs as follows. Four major slides. Each runs from 10am to 5pm, so seven hours. Each takes 100 riders an hour. So "Total riders per slide per day = 7 hours * 100 riders/hour = 700 riders", and "With four slides, total daily capacity = 4 slides * 700 riders per slide = 2,800 riders". Then the conversion that makes it a visitor count: "Estimated total visitors = total rider capacity / average rides per visitor = 2,800 / 2 = 1,400 visitors", checked 30 September 2026 at oliverwyman.com.

The revenue branches then run off that single volume. Tickets: 70 percent of 1,400 is 980 visitors at 300 dirhams, which is 294,000, and 30 percent is 420 at 250 dirhams, which is 105,000. Total ticket revenue 399,000. Ancillary: 1,400 times 50 dirhams at an 80 percent purchase rate is 56,000, and 1,400 times 20 dirhams at 50 percent is 14,000, for 70,000. The firm's total is 469,000 dirhams a day.

We recalculated all eight operations by hand on 30 September 2026 and every one of them lands. That is worth saying, because it is not always the case on published cases.

One thing the tree assumes and the recommendation does not revisit. The 1,400 visitors are the maximum throughput of four slides, and the firm's own assumption list says each slide "operates continuously without downtime". So the number is a capacity ceiling rather than an attendance figure. Treating capacity as demand is the most common way a driver tree overstates a market, and it is our reading here rather than the firm's.

Building a driver tree on a ferry route

A regional operator runs one coastal route. It wants to know where its 12 million dollars of revenue actually comes from before it buys a second vessel.

Fill in the bottom row first, then multiply upward.

1. Sailings. Twelve crossings a day across 340 operating days is 4,080 sailings a year.
2. Seat capacity. 4,080 sailings at 220 seats is 897,600 seats offered.
3. Load factor. At 62 percent, that is 556,512 passengers.
4. Fare. 556,512 passengers at 18 dollars is 10,017,216 dollars of ticket revenue.
5. Onboard spend. At 3.10 dollars a passenger, 556,512 passengers produce 1,725,187 dollars.

Total revenue is 11,742,403 dollars, or about 11.74 million. Note the unit of every line: sailings, seats, passengers, dollars. Each level converts into the next, and the conversion is where the unit changes.

That is exactly where units go wrong. An associate at L.E.K. describes his own market sizing on the firm's careers blog, in which he sized the wedding market as though each person married separately instead of two people per wedding, checked 29 September 2026 at lek.com. The multiplications were fine. The unit at the top was not, and everything below inherited it.

The sensitivity that reverses

Now the question the tree exists to answer: which driver do you pull. Run each one in isolation first, because that is what most people do, and then link it.

The load factor line: one point of 897,600 seats is 8,976 passengers, each worth 18 plus 3.10, which is 21.10 dollars, so 189,394 dollars. The sailings line: 340 extra crossings at 220 seats and 62 percent is 46,376 passengers at 21.10 dollars, or 978,534 dollars, and that one arrives with fuel, crew and berth cost attached, so it does not belong in a revenue comparison on its own. The fare line: 556,512 passengers at one more dollar each is 556,512 dollars, on the assumption that every one of them still sails. That assumption is doing all the work.

Link it and the picture changes. A dollar on an 18 dollar fare is a 5.6 percent price rise. At a fare elasticity of minus 0.8, volume falls 4.4 percent, so 556,512 passengers become 531,778. Ticket revenue is 531,778 times 19, which is 10,103,782. Onboard spend is 531,778 times 3.10, which is 1,648,512. Total 11,752,294 dollars, against 11,742,403 before. The gain is 9,891 dollars, so roughly 9,900.

A dollar of fare looked worth 556,500 dollars and is worth about 9,900. The isolated figure overstates it by a factor of 56. The load factor line has no such linkage, because filling an empty seat on a sailing that is already leaving costs nothing beyond the onboard cost of goods. So of the two moves that need no capital, the one that looked twice as powerful ends up worth five percent of the other.

Where the marks sit on a chain like that

Our own grader reports case arithmetic on four lines, and no consulting firm publishes a marking scheme of its own. The weights below are ours, read from the rubric code, each out of one hundred.

Two drills that expose a weak driver tree

First, rebuild the ferry tree in seat-miles instead of passengers. The route is 31 nautical miles, so capacity is 897,600 seats times 31, which is 27,825,600 seat-miles, and yield is 10,017,216 divided by 27,825,600, or 36 cents a seat-mile. Nothing about the business changed and every number did. If your conclusions change too, they were conclusions about the unit.

Second, take the Poseidon chain and replace the capacity assumption with an attendance assumption. If the park runs at 70 percent of slide throughput, visitors fall from 1,400 to 980. Tickets become 686 adults at 300 and 294 children at 250, so 205,800 plus 73,500, which is 279,300. Ancillary becomes 980 times 50 times 0.80, or 39,200, plus 980 times 20 times 0.50, or 9,800, so 49,000. Total 328,300 dirhams against 469,000, a fall of 30 percent that exactly tracks the volume change, because every branch is linear in visitors.

Time the chain itself in the case math drill, which returns Answer, Working, Units and Interpretation separately, and rough out the top row in the market size calculator. Both are free and need no account. Neither will choose your drivers, and neither knows whether your load factor is plausible, which is the part a client pays for. For the version of this tool that decomposes a question instead of a quantity, see issue trees.