System selection and lifecycle cost

Example 1 Screen first, then score

The must requirement is mapping 4 km² within 3 flights. Area per flight comes from Module 2, and the 1–5 scores come from the team’s assessment (hypothetical).

import math

area_per_flight = {"multirotor": 0.73, "fixed-wing": 3.99, "VTOL": 2.42}   # km² per flight
AREA, MAX_FLIGHTS = 4.0, 3
W = {"coverage": 0.30, "night search": 0.25, "launch space": 0.15, "cost": 0.20, "ease of use": 0.10}
S = {
    "multirotor": {"coverage": 2, "night search": 5, "launch space": 5, "cost": 4, "ease of use": 5},
    "fixed-wing": {"coverage": 5, "night search": 2, "launch space": 1, "cost": 4, "ease of use": 2},
    "VTOL": {"coverage": 4, "night search": 4, "launch space": 5, "cost": 2, "ease of use": 3},
}

def score(w, names):
    return {k: round(sum(w[c] * S[k][c] for c in w), 2) for k in names}

print("without screening:", score(W, S))
passed = [k for k, a in area_per_flight.items() if math.ceil(AREA / a) <= MAX_FLIGHTS]
for k, a in area_per_flight.items():
    print(f"{k:<11} needs {math.ceil(AREA / a)} flights")
print("after screening:  ", score(W, passed))

without screening: {'multirotor': 3.9, 'fixed-wing': 3.15, 'VTOL': 3.65}
multirotor  needs 6 flights
fixed-wing  needs 2 flights
VTOL        needs 2 flights
after screening:   {'fixed-wing': 3.15, 'VTOL': 3.65}

Without screening, the multirotor wins, but it needs too many flights, which in a flood means data arriving hours late. After screening, VTOL beats fixed-wing because it can take off and land in small spaces and search better at night.

Example 2 Raising or lowering one weight at a time by 0.10

S = {
    "fixed-wing": {"coverage": 5, "night search": 2, "launch space": 1, "cost": 4, "ease of use": 2},
    "VTOL": {"coverage": 4, "night search": 4, "launch space": 5, "cost": 2, "ease of use": 3},
}
W = {"coverage": 0.30, "night search": 0.25, "launch space": 0.15, "cost": 0.20, "ease of use": 0.10}

def winner(w):
    tot = {k: sum(w[c] * v[c] for c in w) for k, v in S.items()}
    best = max(tot, key=tot.get)
    return best, round(tot["VTOL"] - tot["fixed-wing"], 2)

for c in W:
    for d in (-0.10, 0.10):
        w = dict(W)
        w[c] = max(0.0, w[c] + d)
        total = sum(w.values())
        w = {k: v / total for k, v in w.items()}          # renormalise to sum to 1
        best, gap = winner(w)
        print(f"{c:<13} {d:+.2f} -> {best:<10} (VTOL - fixed-wing {gap:+.2f})")

coverage      -0.10 -> VTOL       (VTOL - fixed-wing +0.67)
coverage      +0.10 -> VTOL       (VTOL - fixed-wing +0.36)
night search  -0.10 -> VTOL       (VTOL - fixed-wing +0.33)
night search  +0.10 -> VTOL       (VTOL - fixed-wing +0.64)
launch space  -0.10 -> VTOL       (VTOL - fixed-wing +0.11)
launch space  +0.10 -> VTOL       (VTOL - fixed-wing +0.82)
cost          -0.10 -> VTOL       (VTOL - fixed-wing +0.78)
cost          +0.10 -> VTOL       (VTOL - fixed-wing +0.27)
ease of use   -0.10 -> VTOL       (VTOL - fixed-wing +0.44)
ease of use   +0.10 -> VTOL       (VTOL - fixed-wing +0.55)

VTOL wins in every case tested, even when the weight on cost or coverage, where the fixed-wing is stronger, is raised. The result is fairly robust, but the margin is narrowest (+0.11) when the launch space weight is lowered, so the team should confirm that take-off sites in the flooded area really are small. If the result flips with a weight change of only 0.10, report to management that the decision depends on that weight.

Example 3 Own the system or contract a service?

A VTOL system with a thermal camera costs 1,200,000 THB with a 5-year life, fixed costs of 150,000 THB per year (insurance, training, maintenance, software) and 2,000 THB per mission. A contractor charges 25,000 THB per mission (hypothetical).

CAP, LIFE, FIXED, VAR, SVC = 1_200_000, 5, 150_000, 2_000, 25_000

def own(n):
    return CAP / LIFE + FIXED + VAR * n

break_even = (CAP / LIFE + FIXED) / (SVC - VAR)
print(f"break-even {break_even:.1f} missions per year")
for n in (6, 12, 17, 24):
    o, s = own(n), SVC * n
    print(f"{n:>2} missions/yr: own {o:>9,.0f}  contract {s:>9,.0f}  -> {'own' if o < s else 'contract'}")

break-even 17.0 missions per year
 6 missions/yr: own   402,000  contract   150,000  -> contract
12 missions/yr: own   414,000  contract   300,000  -> contract
17 missions/yr: own   424,000  contract   425,000  -> own
24 missions/yr: own   438,000  contract   600,000  -> own

If the province has about 12 missions per year, contracting is cheaper. But cost is not the only reason. Disaster work needs flights within a few hours; if the contractor cannot arrive in time, owning a system may be worthwhile even at a higher cost. Management should see both the numbers and the response-time terms in the contract.