# R SCRIPT FILE FOR CALCULATING THE STEADY-STATE GROWTH RATE, THE REAL AND
#    NOMINAL INTEREST RATES AND THE EQUILIBRIUM LEVEL OF LIQUIDITY IN RELATION
#    TO THE CAPITAL STOCK
#
# SET INITIAL VALUES OF THE GROWTH PARAMETERS
#
 hatm <- 0.14
 delta <- 0.05
 varpi <- .015
 alpha <- 20
 gamma <- .2
 tau <- .01
 oplkr <- .03
 Ups <- .2      # Ups = 1 - (1/3)*(varphi/lambda)^3, which, given that
#                  optlkr = varphi/lambda, implies values for varphi
#                  and lambda of 8.944 and 298.142 respectively which
#                  remain unchanged in all the analysis below
#
# For a situation where improved efficiency increases output by 4%
#  uncomment the following three lines lines
#
# newgamma <- .1822857
# oplkr <- oplkr*(hatm*(1 - newgamma) - delta)/(hatm * (1 - gamma) - delta)
# gamma <- newgamma
#
# For a situation where the implicit tax on consumption increases by
#  50 percent uncomment the following line
#
# tau <- .015
#
# FIRST CALCULATE THE EQUILIBRIUM VALUES OF g AND r WHEN THE 
#    LIQUIDITY/CAPITAL RATIO IS OPTIMUM IMPLYING THE FRIEDMAN RULE
#
# Calculate  mterm, tterm and dmterm from the formulas on page 3 in the document
#
 mterm <- hatm * (1 - gamma) - delta 
 dmterm <- 0                        # impose condition that dmterm = 0
 tterm <- 1                         # whence tterm = 1

 #
# To encorporate unemployment in the model, K is replaced by KU, the
#   fraction of the capital stock utilized.  Since we are dealing
#   entirely with full-employment here, we set  KU = 1, keeping
#   in mind that the units of capital are chosen so that K = 1.
#
 KU <- 1
#
# Full-employment output level equals  mterm x tterm
#
 Yf <- mterm * tterm
#
# SET INITIAL VALUE OF g = 0 AND CALCULATE acterm AND ssgterm
#    WHICH (equating (7) and (1) to eliminate r)  WILL 
#    SOLVE SIMULTANEOUSLY FOR g
#
 g <- 0.00
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
#
# THE CRITICAL VALUE TO BE MINIMISED NUMERICALLY TO FIND EQUILBIRUM g
#  BECOMES THE NEGATIVE NUMBER
#
 crit <- ssgterm - ((mterm / acterm) * tterm)
#
 crit
#
# FIND NUMERICALLY THE VALUE OF g FOR WHICH crit = 0
#
 repeat {
 g <- g + .000001
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm) * tterm)
 if (crit > .0) break
}
#
# The repeat function above keeps recalculating the critical number
#  adding progressively .000001 to g, stopping the calculations when 
#  the critical value passes 0.
#
# Write out the calculated critical value and the calculated level
#  of g
#
cat(crit);cat("\n")
cat(g);cat("\n")
#
# NOW CALCULATE r by substituting the calculated value of g into
#  equation (1) and then write that value of r
#
 r <- ssgterm
 cat(r);cat("\n")
#
# MODIFY THE CALCULATIONS TO ALLOW THE LEVEL OF LIQUIDITY TO BE
#  LESS THAN OPTIMAL -- WITH OUTPUT ALWAYS AT FULL-EMPLOYMENT
#  AND THE COST OF HOLDING MONEY SET EQUAL TO r SO THAT o = 0.
# 
# First, we need values for varphi and lambda.  With the full-
#  employment level of K normalized at unity, optimality of the
#  stock of liquidity implies that varphi = oplkr x lambda.  Given
#  these values, a zero stock of liquidity implies that 
#  Ups = 1 minus (oplkr x varphi-squared) [which should be corrected
#  to yield  Ups = 1 minus (oplkr x varphi-squared/3)]. Hence
#
 varphi <- sqrt((1 - Ups)*3/oplkr)
 lambda <- varphi/oplkr
 cat(varphi);cat("\n")
 cat(lambda);cat("\n")
#
# Set the initial value of L at oplkr (with K normalized at unity)
#
 L <- oplkr
cat(L);cat("\n")
# 
# Recalculate tterm, dmterm, ssgterm and Yf
#
 dmterm <- (varphi - (lambda * L))^2
 tterm <- 1 - (1/(3*lambda))*(varphi - (lambda * L))^3
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 Yf <- mterm * tterm
#
# Now calculate the value of o
#
o <- mterm*dmterm - r
 cat(o);cat("\n")
 cat(r);cat("\n")
#
# Set an initial value of g = 0 and equate equations C.1 and C.3   
#   substituting out r, and set crit equal to the left-side minus 
#   the right-side -- the same calculation as was done earlier
#
 g <- 0
 tterm <- 1 - (1/(3*lambda))*(varphi - (lambda * L))^3
 Yf <- mterm * tterm
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm)* tterm)         
#
 repeat {
 g <- g + .000001
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm) * tterm)
 if (crit > .0) break
}
# cat(g);cat("\n")
 r <- ssgterm
# cat(r);cat("\n")
o <- mterm*dmterm - r
 cat(Yf);cat("\n")
 cat(g);cat("\n")
 cat(o);cat("\n") 
 cat(r);cat("\n")
 cat(L);cat("\n")
 cat(L-oplkr);cat("\n")
 cat(o+r);cat("\n") 
#
# These calculations produce in more detail those generated the the previous
#  calculations
#
# INVESTIGATE WHAT HAPPENS TO o, r and g  WHEN L IS ADJUSTED DOWNWARD FROM
#   ITS OPTIMUM LEVE 
#
 L <- .03 - .00269         # This number was obtained through experimentation
 g <- 0
 dmterm <- (varphi - (lambda * L))^2
 tterm <- 1 - (1/(3*lambda))*(varphi - (lambda * L))^3
 Yf <- mterm * tterm
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm)* tterm)
 repeat {
 g <- g + .000001
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm) * tterm)
 if (crit > 0) break
} 
 r <- ssgterm
 o <- mterm*dmterm - r
 cat(g);cat("\n")
 cat(r);cat("\n")
 cat(o);cat("\n")
 cat(L);cat("\n")
 cat(L-oplkr);cat("\n")
#
# Inflation rate equals
#
 infrate <- o 
 cat(infrate);cat("\n")
#
#  which is virtually zero so the nominal interest rate equals
#
 i <- r + o
 cat(i);cat("\n")
#
#  which is virtually identical to the real interest rate
#
# To see how the value of L was actually obtained, choose o = 0 in
#   the now complete model below
#
#
# NOW GRADUALLY REDUCE L FROM ITS OPTIMUM LEVEL UNTIL o TAKES A PARTICULAR
#  VALUE AND THE COST OF HOLDING MONEY IS  r + o (which must equal mterm*dmterm). 
#
# TO SOLVE SIMULTANIOUSLY FOR THE EQUILBRIUM VALUES OF g, r and L
#   A DOUBLE LOOP IS REQUIRED --- optimize g at successively lower
#   (by tiny amount) values of L and check the value of o.  Stop
#   when o has risen to the point where it just passes zero.
#
# Set the critical value of o   --- which can be viewed as the equilibrium
#      expected inflation rate
#
crito <- 0      # uncomment to set the inflation rate equal to zero
#crito <- .02   # uncomment to alternatively set the inflation rate at 2 percent
#crito <- .50   # uncomment to alternatively set the inflation rate at 50 percent
#crito <- 1.00  # uncomment to alternatively set the inflation rate at 100 percent
 L <- oplkr
 o <- mterm*dmterm - r
 for (i in 1:2000){
 if (o < crito){L <- L - .00001}
 g <- 0
 dmterm <- (varphi - (lambda * L))^2
 tterm <- 1 - (1/(3*lambda))*(varphi - (lambda * L))^3
 Yf <- mterm * tterm
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm)* tterm) 
 repeat {
 g <- g + .00001
 Yf <- mterm * tterm
 acterm <- 1 + (2 * alpha) * (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm) * tterm)
 if (crit > 0) break
}
 o <- mterm*dmterm - r
}
 cat(Yf);cat("\n")
 cat(g);cat("\n")
 cat(o);cat("\n") 
 cat(r);cat("\n")
 cat(L);cat("\n")
 cat(L-oplkr);cat("\n")
 cat(o+r);cat("\n")
#
#
# CALCULATE THE ABOVE VALUES WHERE L = 0
#
 mterm <- hatm * (1 - gamma) - delta
 dmterm <- varphi^2
 tterm <- .2
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 Yf <- mterm * tterm
 KU <- 1
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
#
 g <- 0
 tterm <- .2
 Yf <- mterm * tterm
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm)* tterm)         
#
 repeat {
 g <- g + .000001
 acterm <- 1 + (2 * alpha)* (g + (KU - 1)* Yf)
 ssgterm <- ((1 + varpi)*(1 + g) / (1 - tau)) - 1
 crit <- ssgterm - ((mterm / acterm) * tterm)
 if (crit > .0) break
}
 r <- ssgterm
 o <- mterm*dmterm - r
#
 cat(Yf);cat("\n")
 cat(g);cat("\n")
 cat(o);cat("\n") 
 cat(r);cat("\n")
 cat(o+r);cat("\n")
#
# With zero government created liquidity in the economy
#   the growth rate goes to zero