Packages

Here are the pacakges needed for this study.

library(ggmap)
library(tbart)
library(dplyr)
library(raster)
library(rgeos)
library(tigris)
library(ggplot2)
library(GISTools)
library(acs)
library(leaflet)
library(rgdal)
library(tidycensus)

Data

My study area is Erie couty of NEW YORK state.

1.locations of the fixed clinics
The locations of fixed clinics are acquired from Erie County, NY Department of Health
http://www2.erie.gov/health/index.php?q=locations
To get the lontitude and latitude of these locations, geocode function in ggmap package is used based on Google Map. Because the searching results are not stable, they are saved to be output as a file(csv). Their locations are shown below.

point=read.csv("data/points.csv")
point
EireMap=qmap('erie county',zoom=13,color='bw')
EireMap+geom_point(aes(x=lon,y=lat),colour='red',size=4,data=point)

2.census data
The census data are from American Community Survey (ACS) for 2011-2015 five years. Package tidycensus is used to load the data. Since the attributes are a little bit more, they are saved as an shapefile.

## 
erie=readOGR(dsn="data",layer ="ErieData_tract")

Methods

This study use p-Median problem to sovele the question. P-median problem is the problem of locating P “facilities” relative to a set of “customers” such that the sum of the shortest demand weighted distance between “customers” and “facilities” is minimized.

P-Median problem has practical applications in a wide variety of planning problems: locating telephone switching centers [5], school districting [6], and bank location [7]. Chardaire Nooradelena and Noraida [8] used the p-median problem to determine the best location to place a limited number of emergency medical services taking into account the uncertainty in the demand for the emergency services. This study will test if p-Median model adapts to clinic problem.

1.Calculate weight distance
The objective equation of p-Median peoblem is:
wi: weight at location i
dij: distance between i and j
xij= 0,1

This study mainly considers the persons who are potential patients. I would like to choose several factors to determine the potential locations including households with persons under 18 years and older than 65 years, poverty, peole under disability status, and Women who had a birth in the past 12 months.The equation of weight is:
Wi = (Pi + Di + Yi+ Oi+ Bi)
i: index of a cell
Pi: people under poverty level
Di: people with disability status
Yi: households with person under 18 years
Oi: households with person older than 65 years
Bi: women who had a birth in the past 12 months

To get the weight of each tract, I need to transform each attribute into numeric values. Then, use the equation above to create a new column as weight. euc.dists function is used to calculate the distances between tracts. Finally, create a matrix by using weight multiplying distance. This matrix serves as weighted distance.

erie$B00001e1=as.numeric(as.character(erie$B00001e1))
erie$B11005e2=as.numeric(as.character(erie$B11005e2))
erie$B11007e2=as.numeric(as.character(erie$B11007e2))
erie$B13002e2=as.numeric(as.character(erie$B13002e2))
erie$B17001e2=as.numeric(as.character(erie$B17001e2))
erie$disability=as.numeric(as.character(erie$disability))
## get weight
erie2=mutate(erie@data,weight=B11005e2+B11007e2+B13002e2+B17001e2+disability)
weight=dplyr::select(erie2,GEOID,weight)
erie3=geo_join(erie,weight,"GEOID","GEOID")
eucdist<-euc.dists(erie3,erie3)
dist<-matrix(0,237,237)
for (i in 1:237)
{
  dist[i,]<-erie3$weight[i]*eucdist[i,]
}

2.Create potential sites
For this research, tracts are used as spatial resolution. First, It is necessary to identify the tracts which contain the fixed clinics. And then create new spatial objects to be used as potential sites by deleting the fixed clinics. It shows that there are four tracts that already have clinics.

## transfer points into spatial points
coordinates(point)=~lon+lat
projection(point)=erie3@proj4string  
crs=CRS("+proj=longlat +datum=NAD83 +no_defs +ellps=GRS80 +towgs84=0,0,0")
## find the tracts that contains points 
point_over=sp::over(point,erie3)
all_OBJECT=dplyr::select(erie3@data,OBJECTID)
all_OBJECT$OBJECTID=as.numeric(as.character(all_OBJECT$OBJECTID))
OBJECT=dplyr::select(point_over,OBJECTID)
OBJECT$OBJECTID=as.numeric(as.character(OBJECT$OBJECTID))
point_number=matrix(0,nrow = 5,ncol=2)
p1=which(all_OBJECT == OBJECT$OBJECTID[1], arr.ind=T)
p2=which(all_OBJECT == OBJECT$OBJECTID[2], arr.ind=T)
p3=which(all_OBJECT == OBJECT$OBJECTID[3], arr.ind=T)
p4=which(all_OBJECT == OBJECT$OBJECTID[4], arr.ind=T)
p5=which(all_OBJECT == OBJECT$OBJECTID[5], arr.ind=T)
point_number=rbind(p1,p2,p3,p4,p5)
point_number
erie_sub=erie3[-c(26,180,213,214),]

3.Determine the value of p
To use p-median model, it is important to determine the value of p which indicates the number of facilities. Here, this research selects p value varying from 1 to 20, and then, calculate the sum of all distances from tracts to selected locations to serves as score. This study makes a tradeoff curve to decide which value of p is most efficient.

##p-median model 
score<-c()
for (i in 1:20){ans<-allocations(swdf1 = erie3,swdf2=erie_sub,metric=dist,p=i,verbose=T)
ans_unique <- unique(ans$allocation)
score[i]<- sum(ans$allocdist)}

Here, this study chooses p equals to five to do the analysis. Assuming there is no clinics, 10 is also used as the value of p to make a comparison. In addition, use p equals to 9 and force the model to select the 4 tracts with clinics inside to ba a comparison.

ans5=allocations(swdf1=erie3,swdf2=erie_sub,metric=dist,p=5,verbose=T)
ans_10=allocations(erie3,metric=dist,p=10,verbose = T)
ans_9=allocations(erie3,metric = dist,p=9,force = c(26,180,213,214),verbose = T)