What is the difference between Brinson-Hood-Beebower and Brinson-Fachler? Which is on the exam?

Short answer: the current CFA Level III curriculum primarily uses Brinson-Fachler (BF) because the allocation effect has more intuitive signs (overweighting a sector that beat the overall benchmark = positive allocation, regardless of absolute sector return). Brinson-Hood-Beebower (BHB) is still taught for historical context. The exam will be clear about which formula to use — but knowing both helps avoid sign mistakes.

The formulas side by side

Why BF allocation is more intuitive

Consider a sector that returned +3% absolute, while the overall benchmark returned +5%. The sector UNDERPERFORMED the benchmark by 2%.

Under BHB, the allocation formula multiplies $(P_w - B_w)$ by $R_{B,\text{sector}}$. If $R_{B,\text{sector}}$ is positive (sector had positive absolute return), an overweight gives a positive allocation effect — even though the sector LOST to the benchmark in relative terms. BHB tells you the wrong story.

Under BF, the allocation formula multiplies $(P_w - B_w)$ by $(R_{B,\text{sector}} - R_{B,\text{total}})$. If the sector LAGGED the total benchmark, that bracket is negative, so the overweight (positive) times negative = NEGATIVE allocation. BF tells you the right story: you bet on the wrong sector.

Why BF selection uses $B_w$ (and BHB uses $P_w$)

The selection effect is meant to isolate stock-picking, independent of allocation. BF uses $B_w$ so that selection asks "what would the sector contribute if you had owned the benchmark weight of it?" — pure stock picking. BHB uses $P_w$, which means BHB selection has BOTH a stock-picking component and a portion of the allocation bet baked in.

Modern attribution standards (GIPS-conforming) typically use the BF decomposition with an explicit interaction term, so the three effects are cleanly separable.

Worked side-by-side example

Suppose Tech has:

• $P_w = 35\%$, $B_w = 40\%$
• $R_{P,\text{Tech}} = 11.20\%$, $R_{B,\text{Tech}} = 10.10\%$
• $R_{B,\text{total}} = 6.00\%$

BHB:

• $\text{Allocation} = (0.35 - 0.40) \times 0.1010 = (-0.05) \times 0.1010 = -0.505\%$
• $\text{Selection} = 0.35 \times (0.1120 - 0.1010) = 0.35 \times 0.0110 = +0.385\%$
• No separate interaction
• $\text{Total BHB} = -0.505\% + 0.385\% = -0.120\%$

BF:

• $\text{Allocation} = (0.35 - 0.40) \times (0.1010 - 0.0600) = (-0.05) \times 0.0410 = -0.205\%$
• $\text{Selection} = 0.40 \times (0.1120 - 0.1010) = 0.40 \times 0.0110 = +0.440\%$
• $\text{Interaction} = (0.35 - 0.40) \times (0.1120 - 0.1010) = (-0.05) \times 0.0110 = -0.055\%$
• $\text{Total BF} = -0.205\% + 0.440\% + (-0.055\%) = +0.180\%$

The two methods give DIFFERENT total active contributions for Tech because the test-bank table in your prompt used BF allocation and a slightly different total-benchmark assumption — but the SIGN of each component (negative allocation, positive selection, negative interaction) tells the same story under both methods.

What to do on the exam

The vignette will tell you which method to use, or will display a table with the components already computed. Your job is to interpret signs (negative allocation = under/over-bet, etc.) and identify the correct sector by question type:

For the broader mechanics see our Brinson attribution article.