Formula and accounting assumptions
Each hop uses a full-range constant-product calculation: output equals output reserve × fee-adjusted input divided by input reserve plus fee-adjusted input. First calculate B from the initial A input; then calculate C using that B quantity. The minimum output is the final modeled C amount × (1 − entered tolerance). Tolerance defines an illustrative acceptance guard; it is not an additional fee deducted from the expected quote.
Worked hypothetical example
Consider two fee-free pools, each with 100 units on both sides of its pair. Sending ten A through the first pool yields 100 ÷ 11, approximately 9.090909 B. Sending that quantity through the second yields 100 ÷ 12, approximately 8.333333 C. At a 10% output tolerance, the illustrative minimum is 7.5 C. Quoting the second hop using the original ten-unit amount would overstate the route's result and misrepresent the intermediate inventory.
Interpret the scenarios and limits
The two B reserves must refer to exactly the same intermediate asset and unit scale. No route discovery, competing paths, gas estimate, transfer-tax adjustment or pending transaction simulation is performed. Reserve snapshots are entered by the user and are assumed unchanged except for the modeled route. A tolerance cannot make a stale quote accurate or eliminate failed execution. Compare the sensitivity rows to identify whether increasing route size creates a disproportionate final-output shortfall.