Recourse

nimopt.models.recourse is commitment under uncertainty. The demand and the fuel price are a scenario, and the on-off decision is taken before either is given. on is indexed by hour and unit alone, and p and shed are indexed by the scenario as well. One commitment applies to every scenario, and the binary column is therefore taken under uncertainty.

minimize    Σ_{t,g}   no_load[g] · on[t,g]
            + Σ_{s,g,t} weight[s] · cost[s,g] · p[s,g,t]
            + Σ_{s,t}   weight[s] · voll[s]  · shed[s,t]
subject to  p[s,g,t] ≤ p_max[g] · on[t,g]                  for each s, g, t
            p[s,g,t] ≥ p_min[g] · on[t,g]                  for each s, g, t
            Σ_g p[s,g,t] + shed[s,t] == demand[s,t]        for each s, t
            on ∈ {0,1},  p, shed ≥ 0

The first stage costs the same in every scenario. The second and third sums are weighted by the probability of a scenario, and the objective is a commitment charge plus the expected cost of the recourse.

from nimopt.models import recourse

print(recourse.definition().explain())

Run this example

Output

recourse  min  not built
  sets        S · G · T
  parameters  p_max (G) · p_min (G) · no_load (G) · cost (S,G) · demand (S,T) · weight (S) · voll (S)
  variables   on (T×G) [0.0, 1.0] integer · p (S×G×T) [0.0, inf] · shed (S×T) [0.0, inf]
  constraint  capacity (S,G,T)  p[S, G, T] - p_max[G] * on[T, G] <= 0
  constraint  minimum (S,G,T)  p[S, G, T] - p_min[G] * on[T, G] >= 0
  constraint  balance (S,T)  Sum(G, p[S, G, T]) + shed[S, T] == demand[S, T]
  objective   min  Sum(T, G, no_load[G] * on[T, G]) + Sum(S, G, T, (weight[S] * cost[S, G]) * p[S, G, T]) + Sum(S, T, (weight[S] * voll[S]) * shed[S, T])

No row couples one hour to the next, and the commitment is chosen hour by hour. reference enumerates every on-off subset of the fleet and scores each by its expected recourse across the scenarios. That enumeration is exact and cheap: three units make eight subsets.

from nimopt.models import recourse

inputs = recourse.data()
solution = recourse.definition().build(inputs).solve()
print(solution.objective, recourse.reference(inputs))

Run this example

Output

16744.8125 16744.8125

A committed unit runs at least its minimum, and the model has no sink for unwanted energy. A unit whose minimum exceeds the demand of the mildest scenario is therefore not committed: its output would violate the balance row of that scenario. The first hour demands 38.25 in the mildest scenario. The minimum of base is 40.0 and its fuel is the cheapest of the three, and it is not committed in that hour. The third hour is the opposite case: the fleet supplies 260.0 against a coldest demand of 268.75, every unit is committed, and the remainder is shed.

import numpy as np

from nimopt.models import recourse

inputs = recourse.data()
solution = recourse.definition().build(inputs).solve()
print(inputs["demand"].round(2))
print(np.asarray(solution.primal("on").values()).reshape(4, 3))
print(np.asarray(solution.primal("shed").values()).reshape(3, 4))

Run this example

Output

[[ 38.25 119.   182.75  80.75]
 [ 45.   140.   215.    95.  ]
 [ 56.25 175.   268.75 118.75]]
[[0. 1. 0.]
 [1. 1. 0.]
 [1. 1. 1.]
 [1. 0. 0.]]
[[0.   0.   0.   0.  ]
 [0.   0.   0.   0.  ]
 [0.   0.   8.75 0.  ]]

The mean demand in the first hour is 43.875. base serves that demand at the lowest cost. A model given that one number commits base, and that commitment violates the balance row of the mildest scenario, whose probability is 0.5. The scenario dimension on demand excludes the commitment, and the missing scenario dimension on on applies that exclusion to every scenario.