What is the difference between simple interest and simple discounting?
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I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.
Simple discounting:
$C$ in $n$ years $equiv (1âÂÂdn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.
Where $d=fraci1+i$"
These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?
Thanks
finance
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up vote
1
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I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.
Simple discounting:
$C$ in $n$ years $equiv (1âÂÂdn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.
Where $d=fraci1+i$"
These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?
Thanks
finance
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.
Simple discounting:
$C$ in $n$ years $equiv (1âÂÂdn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.
Where $d=fraci1+i$"
These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?
Thanks
finance
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv fracC1+in$ now.
Simple discounting:
$C$ in $n$ years $equiv (1âÂÂdn)C$ now; $C$ now $equiv fracC1-dn$ in $n$ years.
Where $d=fraci1+i$"
These statements imply that simple interest and simple discounting are not equivalent, since the statements do not equate to one another. Why is this, and what is the difference between simple interest and simple discounting?
Thanks
finance
asked May 18 '15 at 9:48
M Smith
1,151621
1,151621
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1 Answer
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I think you have the wrong formula for d.
$C_0c$: present value (simple compound)
$C_0d$: present value (simple discount)
$C_nd$: future value (simple discount)
$C_nc$: future value (simple compound)
$C_nc=C_0ccdot (1+in)$
$C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$
with $boxedd=fraci1+icdot n$
Here I have a different expression for d.
The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.
$C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$
$C_0c$ and $C_nd$ can be cancelled, because they are equal, too.
$(1+in)= frac11-fracicdot n1+icdot n$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$
Now you can proof, if the equation is true.
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I think you have the wrong formula for d.
$C_0c$: present value (simple compound)
$C_0d$: present value (simple discount)
$C_nd$: future value (simple discount)
$C_nc$: future value (simple compound)
$C_nc=C_0ccdot (1+in)$
$C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$
with $boxedd=fraci1+icdot n$
Here I have a different expression for d.
The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.
$C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$
$C_0c$ and $C_nd$ can be cancelled, because they are equal, too.
$(1+in)= frac11-fracicdot n1+icdot n$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$
Now you can proof, if the equation is true.
add a comment |Â
up vote
0
down vote
I think you have the wrong formula for d.
$C_0c$: present value (simple compound)
$C_0d$: present value (simple discount)
$C_nd$: future value (simple discount)
$C_nc$: future value (simple compound)
$C_nc=C_0ccdot (1+in)$
$C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$
with $boxedd=fraci1+icdot n$
Here I have a different expression for d.
The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.
$C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$
$C_0c$ and $C_nd$ can be cancelled, because they are equal, too.
$(1+in)= frac11-fracicdot n1+icdot n$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$
Now you can proof, if the equation is true.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think you have the wrong formula for d.
$C_0c$: present value (simple compound)
$C_0d$: present value (simple discount)
$C_nd$: future value (simple discount)
$C_nc$: future value (simple compound)
$C_nc=C_0ccdot (1+in)$
$C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$
with $boxedd=fraci1+icdot n$
Here I have a different expression for d.
The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.
$C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$
$C_0c$ and $C_nd$ can be cancelled, because they are equal, too.
$(1+in)= frac11-fracicdot n1+icdot n$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$
Now you can proof, if the equation is true.
I think you have the wrong formula for d.
$C_0c$: present value (simple compound)
$C_0d$: present value (simple discount)
$C_nd$: future value (simple discount)
$C_nc$: future value (simple compound)
$C_nc=C_0ccdot (1+in)$
$C_0d=C_ndcdot frac11-dcdot n=C_ndcdot frac11-fracicdot n1+icdot n$
with $boxedd=fraci1+icdot n$
Here I have a different expression for d.
The future value of the simple compound has to be equal to the present value of the simple discount. This is your asked connection.
$C_0ccdot (1+in)=C_ndcdot frac11-fracicdot n1+icdot n$
$C_0c$ and $C_nd$ can be cancelled, because they are equal, too.
$(1+in)= frac11-fracicdot n1+icdot n$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-fracicdot n1+icdot n right)=1$
Now you can proof, if the equation is true.
edited May 18 '15 at 23:57
answered May 18 '15 at 23:25
callculus
16.6k31427
16.6k31427
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