Replacement cost

Calculate the replacement cost for priority actions in a project prioritization problem() (Moilanen et al. 2009). Actions associated with larger replacement cost values are more irreplaceable, and may need to be implemented sooner than actions with lower replacement cost values.

replacement_costs(x, solution, n = 1)

Arguments

x: project prioritization problem().
solution: base::data.frame() or tibble::tibble() table containing the solutions. Here, rows correspond to different solutions and columns correspond to different actions. Each column in the argument to solution should be named according to a different action in x. Cell values indicate if an action is funded in a given solution or not, and should be either zero or one. Arguments to solution can contain additional columns, and they will be ignored.
n: integer solution number to calculate replacement cost values. Since each row in the argument to solutions corresponds to a different solution, this argument should correspond to a row in the argument to solutions. Defaults to 1.

Value

A tibble::tibble() table containing the following columns:

"action": character name of each action.
"cost": numeric cost of each solution when each action is locked out.
"obj": numeric objective value of each solution when each action is locked out. This is calculated using the objective function defined for the argument to x.
"rep_cost": numeric replacement cost for each action. Greater values indicate greater irreplaceability. Missing (NA) values are assigned to actions which are not selected for funding in the specified solution, infinite (Inf) values are assigned to to actions which are required to meet feasibility constraints, and negative values mean that superior solutions than the specified solution exist.

Details

Replacement cost values are calculated for each priority action specified in the solution. Missing (NA) values are assigned to actions which are not selected for funding in the specified solution. For a given action, its replacement cost is calculated by (i) calculating the objective value for the optimal solution to the argument to x, (ii) calculating the objective value for the optimal solution to the argument to x with the given action locked out, (iii) calculating the difference between the two objective values, (iv) the problem has an objective which aims to minimize the objective value (only add_min_set_objective(), then the resulting value is multiplied by minus one so that larger values always indicate actions with greater irreplaceability. Please note this function can take a long time to complete for large problems since it involves re-solving the problem for every action selected for funding.

References

Moilanen A, Arponen A, Stokland JN & Cabeza M (2009) Assessing replacement cost of conservation areas: how does habitat loss influence priorities? Biological Conservation, 142, 575--585.

Examples

# \dontrun{
# load data
data(sim_projects, sim_features, sim_actions)

# build problem with maximum richness objective and $400 budget
p <- problem(sim_projects, sim_actions, sim_features,
             "name", "success", "name", "cost", "name") %>%
     add_max_richness_objective(budget = 400) %>%
     add_feature_weights("weight") %>%
     add_binary_decisions()

# solve problem
s <- solve(p)
#> Gurobi Optimizer version 9.5.2 build v9.5.2rc0 (linux64)
#> Thread count: 4 physical cores, 8 logical processors, using up to 1 threads
#> Optimize a model with 47 rows, 47 columns and 102 nonzeros
#> Model fingerprint: 0xa33f6587
#> Variable types: 0 continuous, 42 integer (42 binary)
#> Semi-Variable types: 5 continuous, 0 integer
#> Coefficient statistics:
#>   Matrix range     [9e-02, 1e+02]
#>   Objective range  [2e-01, 2e+00]
#>   Bounds range     [1e+00, 1e+00]
#>   RHS range        [1e+00, 4e+02]
#> Found heuristic solution: objective 0.6654645
#> Presolve removed 16 rows and 12 columns
#> Presolve time: 0.00s
#> Presolved: 31 rows, 35 columns, 64 nonzeros
#> Variable types: 0 continuous, 35 integer (35 binary)
#> Root relaxation presolved: 31 rows, 35 columns, 64 nonzeros
#> 
#> 
#> Root relaxation: objective 1.749045e+00, 11 iterations, 0.00 seconds (0.00 work units)
#> 
#>     Nodes    |    Current Node    |     Objective Bounds      |     Work
#>  Expl Unexpl |  Obj  Depth IntInf | Incumbent    BestBd   Gap | It/Node Time
#> 
#> *    0     0               0       1.7490448    1.74904  0.00%     -    0s
#> 
#> Explored 1 nodes (11 simplex iterations) in 0.00 seconds (0.00 work units)
#> Thread count was 1 (of 8 available processors)
#> 
#> Solution count 1: 1.74904 
#> 
#> Optimal solution found (tolerance 0.00e+00)
#> Best objective 1.749044775334e+00, best bound 1.749044775334e+00, gap 0.0000%

# print solution
print(s)
#> # A tibble: 1 × 21
#>   solution status    obj  cost F1_action F2_ac…¹ F3_ac…² F4_ac…³ F5_ac…⁴ basel…⁵
#>      <int> <chr>   <dbl> <dbl>     <dbl>   <dbl>   <dbl>   <dbl>   <dbl>   <dbl>
#> 1        1 OPTIMAL  1.75  395.         1       1       0       1       1       1
#> # … with 11 more variables: F1_project <dbl>, F2_project <dbl>,
#> #   F3_project <dbl>, F4_project <dbl>, F5_project <dbl>,
#> #   baseline_project <dbl>, F1 <dbl>, F2 <dbl>, F3 <dbl>, F4 <dbl>, F5 <dbl>,
#> #   and abbreviated variable names ¹F2_action, ²F3_action, ³F4_action,
#> #   ⁴F5_action, ⁵baseline_action

# calculate replacement cost values
r <- replacement_costs(p, s)

# print output
print(r)
#> # A tibble: 6 × 4
#>   name             cost    obj  rep_cost
#>   <chr>           <dbl>  <dbl>     <dbl>
#> 1 F1_action        300.   1.64    0.108 
#> 2 F2_action        397.   1.70    0.0458
#> 3 F3_action         NA   NA      NA     
#> 4 F4_action        399.   1.56    0.192 
#> 5 F5_action        398.   1.18    0.569 
#> 6 baseline_action  Inf  Inf    -Inf     

# plot histogram of replacement costs,
# with this objective, greater values indicate greater irreplaceability
hist(r$rep_cost, xlab = "Replacement cost", main = "")

# }

Arguments

Value

Details

References

See also

Examples