我有以下内容：
library(financialMath)
library(reshape2)
library(scales)
################################################################################
mortgage <- function(P = 500000, I = 6, L = 30, amort = TRUE) {
J <- I / (12 * 100)
N <- 12 * L
M <- P * J / (1 - (1 + J)^(-N))
monthPay <<- M
# Calculate Amortization for each Month
if (amort == TRUE) {
Pt <- P # current principal or amount of the loan
currP <- NULL
while (Pt >= 0) {
H <- Pt * J # this is the current monthly interest
C <- M - H # this is your monthly payment minus your monthly interest, so it is the amount of principal you pay for that month
Q <- Pt - C # this is the new balance of your principal of your loan
Pt <- Q # sets P equal to Q and goes back to step 1. The loop continues until the value Q (and hence P) goes to zero
currP <- c(currP, Pt)
}
monthP <- c(P, currP[1:(length(currP) - 1)]) - currP
aDFmonth <<- data.frame(
Month = 1:length(currP),
Year = sort(rep(1:ceiling(N / 12), 12))[1:length(monthP)],
Balance = c(currP[1:(length(currP))]),
Payment = monthP + c((monthPay - monthP)[1:(length(monthP))]),
Principal = monthP,
Interest = c((monthPay - monthP)[1:(length(monthP))])
)
aDFmonth <<- subset(aDFmonth, Year <= L * 12)
aDFyear <- data.frame(
Amortization = tapply(aDFmonth$Balance, aDFmonth$Year, max),
Annual_Payment = tapply(aDFmonth$Payment, aDFmonth$Year, sum),
Annual_Principal = tapply(aDFmonth$Principal, aDFmonth$Year, sum),
Annual_Interest = tapply(aDFmonth$Interest, aDFmonth$Year, sum),
Year = as.factor(na.omit(unique(aDFmonth$Year)))
)
aDFyear <<- aDFyear
}
return(aDFmonth)
# if (plotData == TRUE) {
# aDFyear2 <- aDFyear %>%
# rename(
# Interest = Annual_Interest,
# Payment = Annual_Payment,
# Principal = Annual_Principal
# )
# aDFyear2$Year <- as.factor(aDFyear2$Year)
# aDFyear2 <- melt(aDFyear2[, c("Interest", "Principal", "Year")], id.vars = "Year")
#
# ggplot(aDFyear2, aes(x = Year, y = value, fill = variable)) +
# geom_bar(position = "fill", stat = "identity") +
# labs(y = "Payment") +
# scale_y_continuous(labels = percent) +
# theme_minimal() +
# theme(legend.title = element_blank(), legend.position = "top")
# }
}
m = mortgage(P = 200000, I = 2, L = 25, amort = TRUE)
m %>%
head()
Month Year Balance Payment Principal Interest
1 1 1 199485.6 847.7087 514.3753 333.3333
2 2 1 198970.4 847.7087 515.2326 332.4760
3 3 1 198454.3 847.7087 516.0914 331.6173
4 4 1 197937.3 847.7087 516.9515 330.7572
5 5 1 197419.5 847.7087 517.8131 329.8956
6 6 1 196900.9 847.7087 518.6761 329.0326
amort.table(Loan = 200000, n = 12*25, i = 0.02, pf = 12) %>%
.[[1]] %>%
head()
Year Payment Interest Paid Principal Paid Balance
1 0.08 845.95 330.32 515.63 199484.4
2 0.17 845.95 329.46 516.48 198967.9
3 0.25 845.95 328.61 517.34 198450.5
4 0.33 845.95 327.76 518.19 197932.4
5 0.42 845.95 326.90 519.05 197413.3
6 0.50 845.95 326.04 519.90 196893.4
amort.table()financialMathmortgage我正在尝试使用
amort.table()paymentprincipalSummary
Principal (loan amount): 200,000
Annual interest rate: 2%
Term: 25 years (300 months)
Monthly payment: 847.71
Total cost: 200,000 (principal) + 54,312.6 (interest) = 254,312.6
但是，当我运行以下命令时：
> amort.table(Loan = 200000, n = 12*25, i = 0.02, pf = 12) %>%
+ .[[2]]
Details
Loan 200000.00000000
Total Paid 253784.16000000
Total Interest 53784.16000000
Eff Rate 0.02000000
i^(12) 0.01981898
支付的利息总额减少了 1,000 美元（54,312.6 vs 53,784.16）
如何使
amort.table()mortgage我们注意到amort.table中的
i^(12)将
iamort.table(Loan = 200000, n = 12*25, i = (1+0.02/12)^12-1, pf = 12) %>%
.[[1]] %>%
head()
Year Payment Interest Paid Principal Paid Balance
1 0.08 847.71 333.33 514.38 199485.6
2 0.17 847.71 332.48 515.23 198970.4
3 0.25 847.71 331.62 516.09 198454.3
4 0.33 847.71 330.76 516.95 197937.4
5 0.42 847.71 329.90 517.81 197419.5
6 0.50 847.71 329.03 518.68 196900.9
amort.table(Loan = 200000, n = 12*25, i = (1+0.02/12)^12-1, pf = 12) %>%
.[[2]]
Details
Loan 2.000000e+05
Total Paid 2.543126e+05
Total Interest 5.431260e+04
Eff Rate 2.018436e-02
i^(12) 2.000000e-02