ECLSS Spares Provisioning Cost Curve (2D)

Component & Mission

Mean time between failures 5,000 h

Mission duration 180 days

Spare unit cost $8,000

Stockout cost $80,000

Optimal Provisioning

Optimal spares N*

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Min total cost

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Expected failures λ

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P(stockout) at N*

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Cost split at N*: spares vs. expected stockout

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Chart Key

Total expected cost Spares cost (N × unit cost) Expected stockout cost Demand distribution (top panel)

How it works

Failures of this component follow a Poisson process over the mission's operating hours. With mean time between failures MTBF and mission duration in hours H = days × 24:

λ = H / MTBF
P(D = k) = e^-λ λ^k / k!

Stocking N spares costs N × unit_cost up front. If demand D exceeds N, each unit short is a stockout — no way to resupply mid-mission — costing stockout_cost per missing unit. The total expected cost as a function of N is:

E[shortfall(N)] = Σ(k>N) (k − N) · P(D = k)
cost(N) = N · unit_cost  +  stockout_cost · E[shortfall(N)]

This is a classic newsvendor-style trade-off: too few spares and a random failure strands the mission; too many and every extra unit costs as much to launch as fuel or science payload for no benefit. The curve below is swept across every integer N so the minimum — the optimal spares count N* — falls out of the numbers rather than a guess. Hover the cost chart to read any N's breakdown.

MTBF — lower MTBF (a flakier part) raises λ and pushes N* up.
Mission duration — longer missions accumulate more operating hours, raising λ the same way.
Spare unit cost — the launch-mass cost of one spare; a pricier unit makes the curve's left (spares) term steeper, pulling N* down.
Stockout cost — the cost of a mid-mission failure with nothing to replace it; a harsher penalty pulls N* up.