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Exponential Backoff with Jitter

Python

Spread retries out so a recovering service isn't stampeded.

Z
Zainab Bello Sample Oct 3, 2026
Python · 14 lines 
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import random
import time


def backoff(attempt: int, base: float = 0.5, cap: float = 30.0) -> float:
    return random.uniform(0, min(cap, base * 2 ** attempt))


for attempt in range(5):
    try:
        call_service()
        break
    except TimeoutError:
        time.sleep(backoff(attempt))
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