What is the difference between simple interest and simple discounting?
$begingroup$
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv frac{C}{1+in}$ now.
Simple discounting:
$C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv frac{C}{1-dn}$ in $n$ years.
Where $d=frac{i}{1+i}$"
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 SmithM Smith
1,216823
$endgroup$
add a comment |
$begingroup$
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv frac{C}{1+in}$ now.
Simple discounting:
$C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv frac{C}{1-dn}$ in $n$ years.
Where $d=frac{i}{1+i}$"
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 SmithM Smith
1,216823
$endgroup$
add a comment |
$begingroup$
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv frac{C}{1+in}$ now.
Simple discounting:
$C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv frac{C}{1-dn}$ in $n$ years.
Where $d=frac{i}{1+i}$"
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 SmithM Smith
1,216823
$endgroup$
I have been given the following statements:
"Simple interest: $C$ now $equiv (1+in)C$ in $n$ years; $C$ in $n$ years $equiv frac{C}{1+in}$ now.
Simple discounting:
$C$ in $n$ years $equiv (1−dn)C$ now; $C$ now $equiv frac{C}{1-dn}$ in $n$ years.
Where $d=frac{i}{1+i}$"
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
finance
asked May 18 '15 at 9:48
M SmithM Smith
1,216823
asked May 18 '15 at 9:48
M SmithM Smith
1,216823
asked May 18 '15 at 9:48
M SmithM Smith
1,216823
asked May 18 '15 at 9:48
M SmithM Smith
1,216823
asked May 18 '15 at 9:48
M SmithM Smith
1,216823
1,216823
add a comment |
add a comment |
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)
$C_{nc}=C_{0c}cdot (1+in)$
$C_{0d}=C_{nd}cdot frac{1}{1-dcdot n}=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
with $boxed{d=frac{i}{1+icdot n}}$
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.
$C_{0c}cdot (1+in)=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
$C_{0c}$ and $C_{nd}$ can be cancelled, because they are equal, too.
$(1+in)= frac{1}{1-frac{icdot n}{1+icdot n}}$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-frac{icdot n}{1+icdot n} right)=1$
Now you can proof, if the equation is true.
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
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)
$C_{nc}=C_{0c}cdot (1+in)$
$C_{0d}=C_{nd}cdot frac{1}{1-dcdot n}=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
with $boxed{d=frac{i}{1+icdot n}}$
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.
$C_{0c}cdot (1+in)=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
$C_{0c}$ and $C_{nd}$ can be cancelled, because they are equal, too.
$(1+in)= frac{1}{1-frac{icdot n}{1+icdot n}}$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-frac{icdot n}{1+icdot n} right)=1$
Now you can proof, if the equation is true.
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
$endgroup$
add a comment |
$begingroup$
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)
$C_{nc}=C_{0c}cdot (1+in)$
$C_{0d}=C_{nd}cdot frac{1}{1-dcdot n}=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
with $boxed{d=frac{i}{1+icdot n}}$
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.
$C_{0c}cdot (1+in)=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
$C_{0c}$ and $C_{nd}$ can be cancelled, because they are equal, too.
$(1+in)= frac{1}{1-frac{icdot n}{1+icdot n}}$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-frac{icdot n}{1+icdot n} right)=1$
Now you can proof, if the equation is true.
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
$endgroup$
add a comment |
$begingroup$
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)
$C_{nc}=C_{0c}cdot (1+in)$
$C_{0d}=C_{nd}cdot frac{1}{1-dcdot n}=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
with $boxed{d=frac{i}{1+icdot n}}$
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.
$C_{0c}cdot (1+in)=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
$C_{0c}$ and $C_{nd}$ can be cancelled, because they are equal, too.
$(1+in)= frac{1}{1-frac{icdot n}{1+icdot n}}$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-frac{icdot n}{1+icdot n} right)=1$
Now you can proof, if the equation is true.
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
$endgroup$
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)
$C_{nc}=C_{0c}cdot (1+in)$
$C_{0d}=C_{nd}cdot frac{1}{1-dcdot n}=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
with $boxed{d=frac{i}{1+icdot n}}$
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.
$C_{0c}cdot (1+in)=C_{nd}cdot frac{1}{1-frac{icdot n}{1+icdot n}}$
$C_{0c}$ and $C_{nd}$ can be cancelled, because they are equal, too.
$(1+in)= frac{1}{1-frac{icdot n}{1+icdot n}}$
Multiplying the equation by the denominator of the RHS.
$(1+in)cdot left( 1-frac{icdot n}{1+icdot n} right)=1$
Now you can proof, if the equation is true.
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
edited May 18 '15 at 23:57
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
answered May 18 '15 at 23:25
callculuscallculus
18.7k31428
18.7k31428
add a comment |
add a comment |
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