“Oct 13”, “Nov 13”, “Dec 13”, “Jan 14”, “Feb 14”, “Mar 14”, “Apr 14”, “May 14”, “Jun 14”, “Jul 14”, “Aug 14”, “Sept 14”
## Month Received Resolved Reopen
## 1 1 104 105 5
## 2 2 105 99 2
## 3 3 84 69 1
## 4 4 80 66 2
## 5 5 122 115 4
## 6 6 157 153 3
## 7 7 162 157 1
## 8 8 130 122 3
## 9 9 139 134 5
## 10 10 151 145 4
## 11 11 118 131 3
## 12 12 109 108 2
Plotting graph:
g1 <- ggplot(ind1, aes(x=Month, y=NumberOfCases))
g1 <- g1 + geom_line(aes(y=Received), colour="purple")
g1 <- g1 + geom_line(aes(y=Resolved), colour="maroon")
g1 <- g1 + geom_line(aes(y=Reopen), colour="red")
g1
In order to evaluate performance, it requires number of cases resolved on time, average first response time and average resolution time, which is currently not available.
load<- data.frame(k=c(1,2,3,4,5,6,7,8,9,10,11,12), x=c(104, 105, 84, 80, 122, 157, 162, 130, 139, 151, 118, 109), y=c(105, 99, 69, 66, 115, 153, 157, 122, 134, 145, 131, 108), i= c(0,6,15,14,7,4,5,8,5,6,0,1))
nc2<-c("Month","Received","Resolved","Backlog")
nr2<-c(1,2,3,4,5,6,7,8,9,10,11,12)
colnames(load)<-nc2
rownames(load)<-nr2
load
## Month Received Resolved Backlog
## 1 1 104 105 0
## 2 2 105 99 6
## 3 3 84 69 15
## 4 4 80 66 14
## 5 5 122 115 7
## 6 6 157 153 4
## 7 7 162 157 5
## 8 8 130 122 8
## 9 9 139 134 5
## 10 10 151 145 6
## 11 11 118 131 0
## 12 12 109 108 1
Plotting graph:
gL1 <- ggplot(load, aes(x=Month, y=NumberOfCases)) + geom_line(aes(y=Received, colour="Received"))+labs(title="Load Analysis") + geom_line(aes(y=Resolved, colour="Resolved")) + geom_line(aes(y=Backlog, colour="Backlog")) + scale_colour_manual("", breaks=c("Received","Resolved","Backlog"), values=c("purple","maroon","orange"))
gL1
library(lattice)
library(caret)
meanReceived <- mean(ind1$Received)
stdev <- sd (ind1$Received)
n <- 12
meanReceived + c(-1,1) * qt(.975, n-1) * stdev / sqrt(n)
## [1] 104.6 138.9
Therefore, there is a 95% chance that the number of Received cases for the next month will be between 104.6 and 138.9.
meanResolved <- mean(ind1$Resolved)
stdevRes <- sd (ind1$Resolved)
meanResolved + c(-1,1) * qt(.975, n-1) * stdevRes / sqrt(n)
## [1] 98.18 135.82
Therefore, there is a 95% chance that the number of Resolved cases for the next month will be between 98.18 and 135.82.
meanReopen <- mean(ind1$Reopen)
stdevReo <- sd (ind1$Reopen)
meanReopen + c(-1,1) * qt(.975, n-1) * stdevReo / sqrt(n)
## [1] 2.041 3.793
Therefore, there is a 95% chance that the number of Reopened cases for the next month will be between 2.041 and 3.793.