R help - Jan 2014 - Reversing the Equation to find value of variable

Dear R forum

I have following variables -

EAD = 10000
LGD = 0.45
PD = 0.47
M = 3

# Equation 1

R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) + 0.24*(1-(1-exp(-50*PD))/(1-exp(-50)))

b = (0.11852 - 0.05478 * log(PD))^2

K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 * qnorm(0.999))
- PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b)

RWA = K * 12.5 * EAD

> RWA
[1] 22845.07

# _________________________________________________________________

# MY Problem

In the above part, knowing values of LGD, EAD, M and PD, the value of RWA was
calculated. However, I need to go reverse way in the sense knowing the values of
LGD, EAD, M and RWA, I need to find value of PD.

So I have tried to use uniroot as (RWA - K * 12.5 * EAD and used the above
equations i place of K and R)

RWA = 22845.07
LGD = 0.45
EAD = 10000
M = 3

f = function(x) RWA -
 (LGD*pnorm((1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50)))))^(-0.5)*qnorm(x)+((0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))))/(1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))))))^0.5*qnorm(0.999))-x*LGD)
* (1-1.5*((0.11852-0.05478 * log(x))^2))^(-1)*(1+(M-2.5)*((0.11852-0.05478 *
log(x))^2))*12.5*EAD

uniroot(f, c(0,1), tol = 0.0000000001)

I get following error -
> uniroot(f, c(0,1), tol = 0.0000000001)
Error in uniroot(f, c(0, 1), tol = 1e-10) : f.lower = f(lower) is NA

Kindly guide as I am not sure if uniroot is the correct way of doing it or not.
Ideally, I should be getting the PD value of 0.47.

With regards

Katherine
	[[alternative HTML version deleted]]

Hi

Reading the error message carefully you can see that f() is not defined at 0:
> uniroot(f, c(0, 1))
Error in uniroot(f, c(0, 1)) : f.lower = f(lower) is NA
> f(0)
[1] NaN

If you plot f() in the interval (0,1) then you'll see there is two
solutions:

> uniroot(f, c(0.0001, 1))
Error in uniroot(f, c(1e-04, 1)) : 
  f() values at end points not of opposite sign
> uniroot(f, c(0.0001, 0.2))
$root
[1] 0.1533901

$f.root
[1] 0.3414232

$iter
[1] 6

$estim.prec
[1] 6.103516e-05

> uniroot(f, c(0.3, 1))
$root
[1] 0.4699984

$f.root
[1] -0.04112121

$iter
[1] 8

$estim.prec
[1] 6.103516e-05
>

Yours sincerely / Med venlig hilsen


Frede Aakmann T?gersen
Specialist, M.Sc., Ph.D.
Plant Performance & Modeling

Technology & Service Solutions
T +45 9730 5135
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If you have received this e-mail in error please contact the sender. 
> -----Original Message-----
> From: r-help-bounces at r-project.org [mailto:r-help-bounces at
r-project.org]
> On Behalf Of Katherine Gobin
> Sent: 6. januar 2014 12:42
> To: r-help at r-project.org
> Subject: [R] Reversing the Equation to find value of variable
> 
> Dear R forum
> 
> I have following variables -
> 
> EAD = 10000
> LGD = 0.45
> PD = 0.47
> M = 3
> 
> # Equation 1
> 
> R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) + 0.24*(1-(1-exp(-50*PD))/(1-exp(-
> 50)))
> 
> b = (0.11852 - 0.05478 * log(PD))^2
> 
> K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 *
> qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b)
> 
> RWA = K * 12.5 * EAD
> 
> 
> > RWA
> [1] 22845.07
> 
> #
> __________________________________________________________
> _______
> 
> # MY Problem
> 
> In the above part, knowing values of LGD, EAD, M and PD, the value of RWA
> was calculated. However, I need to go reverse way in the sense knowing the
> values of LGD, EAD, M and RWA, I need to find value of PD.
> 
> So I have tried to use uniroot as (RWA - K * 12.5 * EAD and used the above
> equations i place of K and R)
> 
> RWA =?22845.07
> LGD = 0.45
> EAD = 10000
> M = 3
> 
> f = function(x) RWA - ?(LGD*pnorm((1-(0.12*(1-exp(-50*x))/(1-exp(-
> 50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50)))))^(-0.5)*qnorm(x)+((0.12*(1-
> exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))))/(1-(0.12*(1-
> exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-
> 50))))))^0.5*qnorm(0.999))-x*LGD) * (1-1.5*((0.11852-0.05478 *
> log(x))^2))^(-1)*(1+(M-2.5)*((0.11852-0.05478 * log(x))^2))*12.5*EAD
> 
> uniroot(f, c(0,1), tol = 0.0000000001)
> 
> I get following error -
> 
> > uniroot(f, c(0,1), tol = 0.0000000001)
> Error in uniroot(f, c(0, 1), tol = 1e-10) : f.lower = f(lower) is NA
> 
> Kindly guide as I am not sure if uniroot is the correct way of doing it or
not.
> Ideally, I should be getting the PD value of 0.47.
> 
> With regards
> 
> Katherine
> 	[[alternative HTML version deleted]]

On 06-01-2014, at 12:41, Katherine Gobin <katherine_gobin at yahoo.com>
wrote:
> Dear R forum
> 
> I have following variables -
> 
> EAD = 10000
> LGD = 0.45
> PD = 0.47
> M = 3
> 
> # Equation 1
> 
> R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) +
0.24*(1-(1-exp(-50*PD))/(1-exp(-50)))
> 
> b = (0.11852 - 0.05478 * log(PD))^2
> 
> K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 *
qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b)
> 
> RWA = K * 12.5 * EAD
> 
> 
>> RWA
> [1] 22845.07
> 
> # _________________________________________________________________
> 
> # MY Problem
> 
> In the above part, knowing values of LGD, EAD, M and PD, the value of RWA
was calculated. However, I need to go reverse way in the sense knowing the
values of LGD, EAD, M and RWA, I need to find value of PD.
> 
> So I have tried to use uniroot as (RWA - K * 12.5 * EAD and used the above
equations i place of K and R)
> 
> RWA = 22845.07
> LGD = 0.45
> EAD = 10000
> M = 3
> 
> f = function(x) RWA - 
(LGD*pnorm((1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50)))))^(-0.5)*qnorm(x)+((0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))))/(1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))))))^0.5*qnorm(0.999))-x*LGD)
* (1-1.5*((0.11852-0.05478 * log(x))^2))^(-1)*(1+(M-2.5)*((0.11852-0.05478 *
log(x))^2))*12.5*EAD
> 
> uniroot(f, c(0,1), tol = 0.0000000001)
> 
> I get following error -
> 
>> uniroot(f, c(0,1), tol = 0.0000000001)
> Error in uniroot(f, c(0, 1), tol = 1e-10) : f.lower = f(lower) is NA
> 
> Kindly guide as I am not sure if uniroot is the correct way of doing it or
not. Ideally, I should be getting the PD value of 0.47.
To make life easier and more readable define a function f, where x is a value
for PD

f <- function(x)  {  
    PD <- x
    R <- 0.12*(1-exp(-50*PD))/(1-exp(-50)) +
0.24*(1-(1-exp(-50*PD))/(1-exp(-50)))
    b <- (0.11852 - 0.05478 * log(PD))^2
    K <- (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 *
qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b)
    RWA - K * 12.5 * EAD
}

where a value of x must be found such that f becomes 0.

Draw a curve for f(x)

curve(f,from=0,to=1)

So is 0 a valid lower bound for f? Try f(0)

So try a little differently

curve(f,from=0.1,to=1) 

Have a careful look at the curve.  The endpoints for uniroot must be give
function values of opposite sign.

So try this:

uniroot(f,c(0.1,.4), tol=1e-8)
uniroot(f,c(0.4,.6), tol=1e-8)

where the second uniroot will give what you seem to want. But beware: your
function has two solutions.

Berend