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 FICO Optimization Community FICO Xpress Optimization Home   Solve two related problems assigning pilots to crews under different constraints

Description

A crew consists of two pilots. Pilots are assigned to crews based on their language and aircraft skills. Two such matching problems are solved. The first is to determine whether there exists a perfect matching, i.e., one that assigns all pilots to some crew. The second problem is to find a perfect matching that maximizes a total crew score.

Further explanation of this example: This is a conversion of the Mosel example 'Composing Flight Crews'

Source Files
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flight_crew.R

#####################################
# This file is part of the          #
# Xpress-R interface examples       #
#                                   #
#   (c) 2021 Fair Isaac Corporation #
#####################################
#' ---
#' title: "Flight Crews"
#' author: Y. Gu
#' date: Jun. 2021
#' ---
#'
#'
#'
## ----setup, include=FALSE-----------------------------------------------------
knitr::opts_chunk$set(echo = TRUE) knitr::opts_chunk$set(results = "hold")
knitr::opts_chunk$set(warning = FALSE,message = FALSE) #' #' #' ## Brief Introduction To The Problem #' #' [This is a conversion of the Mosel example 'Composing Flight Crews'](https://www.fico.com/fico-xpress-optimization/docs/latest/examples/mosel/ApplBook/F_TransAir/f2crew.mos). #' Brief introduction to this problem is given below, and to see the full #' mathematical modeling of this problem you may refer to section 11.2, page 159 of the #' book 'Applications of optimization with Xpress'. #' #' There are 8 pilots and each of them has different levels (mark from 0 to 20) of #' language skills and experience with different aircraft. We would like to form valid #' flight crews that each consists of two pilots and both of the pilots have at least #' 10/20 for the same language and 10/20 on the same aircraft type. #' #' Two questions are raised, the first one asks whether all pilots can fly,and the second #' one wants to find the set of crews with maximum total score. The score of each valid #' crew is defined as the maximum of the sum of its pilot scores over all aircraft types #' for which both pilots are rated at least 10/20. #' #' To begin with, we find all the valid crews, and then we can think of this problem #' as an undirected compatibility graph with 8 nodes representing 8 pilots and #' arcs between each pair of valid crew pilots(nodes). We will visualize the solutions of #' both questions in network graphs to check the solutions more intuitively. #' #' For the first question, we actually want to find the maximum number of valid crews. #' After we find all the compatible pairs of pilots, we set binary variables 'fly' for #' each valid crew, and 'fly' indicates whether this crew is used or not. The objective #' we want to maximize is simply the sum of 'fly' variables. The only constraint here is #' that each pilot can only be contained in exactly one of the used crews. With this #' constraint, we are actually looking for maximum cardinality matching for this question, #' where matching is a set of arcs such that any two among them have no node in common. #' #' For the second question, firstly we need to calculate the score of each valid crew. #' Then, we can solve this question by just changing the objective coefficients of the #' first problem from 1 to the scores and keeping everything else the same. In this #' question, we are looking for a matching with maximum total weight. #' #' The full mathematical formulation and detailed explanations are included in the book #' 'Applications of optimization with Xpress'. #' #' #' For this example, we need packages 'xpress', 'dplyr' and 'igraph'. Besides, we use the #' function pretty_name to give the variables and constraints concise names. #' ## ----Load The Packages And The Function To Give Names------------------------- library(xpress) library(dplyr) library(igraph) pretty_name <- function(prefix, y) { "%s_%s" %>% sprintf(prefix, paste(lapply(names(y), function(name) paste(name, y[name], sep = "_")), collapse = "_")) } #' #' #' ## Question 1 #' #' Firstly, we create a new empty problem, set the objective sense as maximization and #' give the problem a suitable name. #' ## ----Create The Problem------------------------------------------------------- # create a new problem prob = createprob() # change this problem to a maximization problem chgobjsense(prob, objsense = xpress:::OBJ_MAXIMIZE) # set the problem name setprobname(prob, "FlightCrew") #' #' #' Add the values we need for this example. #' ## ----Data--------------------------------------------------------------------- # 1. pilots PILOT <- data.frame("pilot" = 1:8) cname <- apply(PILOT, 1, function(x) name = sprintf("P_%d", x["pilot"])) # 2. language scores of all the pilots LANG <- as.data.frame(matrix(c(20, 14, 0, 13, 0, 0, 8, 8, # English 12, 0, 0, 10, 15, 20, 8, 9, # French 0, 20, 12, 0, 8, 11, 14, 12, # Dutch 0, 0, 0, 0, 17, 0, 0, 16), # Norwegian byrow = TRUE, nrow = 4, ncol = 8)) rownames(LANG) <- c("English", "French", "Dutch", "Norwegian") colnames(LANG) <- cname # 3. scores of different plane types of all the pilots PTYPE <- as.data.frame(matrix(c(18, 12, 15, 0, 0, 0, 8, 0, # Reconnaissance 10, 0, 9, 14, 15, 8, 12, 13, # Transport 0, 17, 0, 11, 13, 10, 0, 0, # Bomber 0, 0, 14, 0, 0, 12, 16, 0, # Fighter-bomber 0, 0, 0, 0, 12, 18, 0, 18), # Supply plane byrow = TRUE, nrow = 5, ncol = 8)) rownames(PTYPE) <- c("Reconnaissance", "Transport", "Bomber", "Fighter-bomber", "Supply plane") colnames(PTYPE) <- cname #' #' #' We need to find all compatible crews. #' ## ----Compatible Crews--------------------------------------------------------- CREW <- data.frame() for (i in 1:(nrow(PILOT) - 1)) { for (j in (i + 1):nrow(PILOT)) { for (l in 1:nrow(LANG)) { if ((LANG[l, i] >= 10 && LANG[l, j] >= 10)) { for (p in 1:nrow(PTYPE)) { if (PTYPE[p, i] >= 10 && PTYPE[p, j] >= 10) { compatible <- c(i, j) CREW <- rbind(CREW, compatible) break } } } if (i == CREW[nrow(CREW), 1] & j == CREW[nrow(CREW), 2]) { break } } } } colnames(CREW) <- c("Member1", "Member2") #' #' #' Add binary variables 'fly' and store their indices in a newly created vector 'fly' in #' data frame 'CREW'. #' ## ----Add Columns-------------------------------------------------------------- # binary variable 'fly' to indicate whether this crew will fly or not CREW$fly <-
CREW %>% apply(1, function(x)
xprs_newcol(
prob,
lb = 0,
ub = 1,
coltype = "B",
name = pretty_name("fly", c(x["Member1"], x["Member2"])),
objcoef = 1
))

#'
#'
#' Add constraints that ensure every pilot is member of at most one single crew.
#'
for (i in PILOT$pilot) { crew <- CREW %>% filter(Member1 == i | Member2 == i) xprs_addrow( prob, colind = crew$fly,
rowcoef = rep(1, nrow(crew)),
rowtype = "L",
rhs = 1,
name = sprintf("Pilot_%d_in_1_crew", i)
)
}

#'
#'
#' Now we can solve the first question.
#'
## ----Solve The First Question-------------------------------------------------
setoutput(prob)
summary(xprs_optimize(prob))

#'
#'
#' Display solutions of the first question here.
#'
## ----Solution Of The First Question-------------------------------------------
CREW$sol_Q1 <- xprs_getsolution(prob) print(paste( "The maximum number of crews that can fly is:", getdblattrib(prob, xpress:::MIPOBJVAL) )) if (2 * sum(CREW$sol_Q1) == nrow(PILOT)) {
print("All pilots can fly.")
} else{
print("Not all pilots can fly.")
}

print("The crews are:")
CREW %>% filter(sol_Q1 == 1) %>% apply(1, function(x)
paste(x["Member1"], "-", x["Member2"]))

#'
#'

#'
#' Visualize the solutions in a graph, where the red edges represent the crews that can
#' fly.
#'
## ----Visualize The Solution Of Question 1-------------------------------------
# create the network object
elist <- data.frame(tails = CREW$Member1, heads = CREW$Member2)
graph <- graph_from_data_frame(d = elist, directed = F)

# set the solution edges(crews) in red color
g1.sol <- which(CREW$sol_Q1 == 1) ecolor1 <- rep("gray", nrow(CREW)) ecolor1[g1.sol] <- rep("red", length(g1.sol)) # plot plot( graph, vertex.color = "gold", vertex.size = 25, vertex.frame.color = "gray", vertex.label.color = "black", edge.color = ecolor1, layout = layout.circle(graph) ) #' #' ## Question 2 #' #' To solve question 2, we firstly need to calculate the scores of valid crews. #' ## ----Scores------------------------------------------------------------------- CREW$SCORE <- rep(0, nrow(CREW))

for (i in 1:nrow(CREW)) {
c1 <- cname[CREW$Member1[i]] c2 <- cname[CREW$Member2[i]]

crew_score <- PTYPE %>% select(all_of(c1), all_of(c2))
crew_score$sum <- rep(0, nrow(crew_score)) for (j in 1:nrow(crew_score)) { if (crew_score[j, c1] >= 10 && crew_score[j, c2] >= 10) { crew_score$sum[j] <- sum(crew_score[j, c1], crew_score[j, c2])
}
}

CREW$SCORE[i] <- max(crew_score$sum)
}

#'
#' Then based on the first problem, we change the objective coefficients to the scores.
#'
## ----Change The Objective Coefficients----------------------------------------
chgobj(prob, colind = CREW$fly, objcoef = CREW$SCORE)

#'
#'
#' Now we can solve the second question.
#'
## ----Solve The Second Question------------------------------------------------
setoutput(prob)
summary(xprs_optimize(prob))

#'
#'
#' Display the solutions Of the second question here.
#'
## ----Solution Of The Second Question------------------------------------------
CREW$sol_Q2 <- xprs_getsolution(prob) print(paste( "The maximum total score is:", getdblattrib(prob, xpress:::MIPOBJVAL) )) print(paste("The set of crews with maximum total score is :")) CREW %>% filter(sol_Q2 == 1) %>% select(Member1, Member2) %>% apply(1, function(x) paste(x["Member1"], "-", x["Member2"])) #' #' Visualize the solutions in a graph, where the red edges represent the crews that can #' fly and the width of each edge reflects the score of the crew it represents, an edge #' connecting crew with larger score will be wider and vice versa. #' ## ----Visualize The Solution Of Question 2------------------------------------- # set the solution edges(crews) in red color g2.sol <- which(CREW$sol_Q2 == 1)
ecolor2 <- rep("gray", nrow(CREW))
ecolor2[g2.sol] <- rep("red", length(g2.sol))

# plot
plot(
graph,
vertex.color = "gold",
vertex.size = 25,
vertex.frame.color = "gray",
vertex.label.color = "black",
edge.width = CREW$SCORE / 10, edge.label = CREW$SCORE,
edge.color = ecolor2,
layout = layout.circle(graph)
)

#'
#'   