Max drawdown

easyFinanceacceptance —~10 min

Implement max_drawdown(equity) — the largest peak-to-trough decline of an equity curve as a positive fraction of the peak: the maximum over time of (running_peak - value) / running_peak.

It is the number that ends careers before the Sharpe ratio gets a chance to. Note the running peak: a new high resets where the fall is measured from. Return the fraction rounded to 4 decimal places.

example 1

input

equity=[100.0, 95.0, 90.0, 97.0, 99.0]

output

0.1

example 2

input

equity=[100.0, 110.0, 120.0, 96.0, 105.0]

output

0.2

note

The peak is 120, not 100 — the fall to 96 is 20%.

constraints

2 ≤ len(equity) ≤ 10_000
every value > 0

teacher · 2 hints · the first is free

↗ read the theory

the chapter this problem is cut from

●solution.py· utf-8 · spaces:4

def max_drawdown(equity: list) -> float:
    """Largest peak-to-trough fall as a fraction of the peak, 4 dp."""
    # TODO: the peak moves — a new high resets where the fall is measured from
    peak = equity[0]
    worst = 0.0
    for v in equity:
        dd = (peak - v) / peak
        if dd > worst:
            worst = dd
    return round(worst, 4)

ln 1, col 1tab indent · esc leave editor

python 3.13 · pyodide

+ 6 hidden on submit

input · keyword arguments to max_drawdown()

equity=[100.0, 95.0, 90.0, 97.0, 99.0]

expected

0.1

the judge calls max_drawdown() with exactly these inputs · outputs are compared to 6 decimal places

● standing bypython 3.13 in your browser · pyodide 0.28.3 · judged on the server to 6 dp⌘↵ run · ⌘⇧↵ submit

paper only · educational · not advice