How to resolve a circular reference / livelock
$begingroup$
I seem to have myself an equation that is what MS Excel refers to as a circular reference; that I believe is also called a "Livelock".
I have composed the formula:
A = M + D - R - (0.05(A - 250000)+2500)
Where:
A = Offer Price
M = Mortgage
D = Deposit
R = Renovation costs
Is it even possible to resolve A?
Background
I'm trying to calculate an offer price to make on a property, based on certain costs, including Stamp Duty Land Tax (SDLT), but SDLT requires the offer price.
I'm afraid my GCSE maths was so long ago, I can't think of a way that I might achieve this, or what it might be called in mathematics (unless it's a simultaneous equation (hence tag)).
TIA.
algebra-precalculus
edited Jan 7 at 11:01
Harry Peter
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
$endgroup$
add a comment |
$begingroup$
I seem to have myself an equation that is what MS Excel refers to as a circular reference; that I believe is also called a "Livelock".
I have composed the formula:
A = M + D - R - (0.05(A - 250000)+2500)
Where:
A = Offer Price
M = Mortgage
D = Deposit
R = Renovation costs
Is it even possible to resolve A?
Background
I'm trying to calculate an offer price to make on a property, based on certain costs, including Stamp Duty Land Tax (SDLT), but SDLT requires the offer price.
I'm afraid my GCSE maths was so long ago, I can't think of a way that I might achieve this, or what it might be called in mathematics (unless it's a simultaneous equation (hence tag)).
TIA.
algebra-precalculus
edited Jan 7 at 11:01
Harry Peter
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
$endgroup$
1
$begingroup$
Expand the right-hand side, move all the terms withAto the left side and divide through by the resulting coefficient.
$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14
add a comment |
$begingroup$
I seem to have myself an equation that is what MS Excel refers to as a circular reference; that I believe is also called a "Livelock".
I have composed the formula:
A = M + D - R - (0.05(A - 250000)+2500)
Where:
A = Offer Price
M = Mortgage
D = Deposit
R = Renovation costs
Is it even possible to resolve A?
Background
I'm trying to calculate an offer price to make on a property, based on certain costs, including Stamp Duty Land Tax (SDLT), but SDLT requires the offer price.
I'm afraid my GCSE maths was so long ago, I can't think of a way that I might achieve this, or what it might be called in mathematics (unless it's a simultaneous equation (hence tag)).
TIA.
algebra-precalculus
edited Jan 7 at 11:01
Harry Peter
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
$endgroup$
I seem to have myself an equation that is what MS Excel refers to as a circular reference; that I believe is also called a "Livelock".
I have composed the formula:
A = M + D - R - (0.05(A - 250000)+2500)
Where:
A = Offer Price
M = Mortgage
D = Deposit
R = Renovation costs
Is it even possible to resolve A?
Background
I'm trying to calculate an offer price to make on a property, based on certain costs, including Stamp Duty Land Tax (SDLT), but SDLT requires the offer price.
I'm afraid my GCSE maths was so long ago, I can't think of a way that I might achieve this, or what it might be called in mathematics (unless it's a simultaneous equation (hence tag)).
TIA.
algebra-precalculus
algebra-precalculus
edited Jan 7 at 11:01
Harry Peter
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
edited Jan 7 at 11:01
Harry Peter
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
edited Jan 7 at 11:01
Harry Peter
5,48911439
edited Jan 7 at 11:01
Harry Peter
5,48911439
edited Jan 7 at 11:01
Harry Peter
5,48911439
5,48911439
asked Jan 6 at 20:12
woter324woter324
1031
asked Jan 6 at 20:12
woter324woter324
1031
asked Jan 6 at 20:12
woter324woter324
1031
1031
1
$begingroup$
Expand the right-hand side, move all the terms withAto the left side and divide through by the resulting coefficient.
$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14
add a comment |
1
$begingroup$
Expand the right-hand side, move all the terms withAto the left side and divide through by the resulting coefficient.
$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14
1
1
$begingroup$
Expand the right-hand side, move all the terms with
A to the left side and divide through by the resulting coefficient.$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Expand the right-hand side, move all the terms with
A to the left side and divide through by the resulting coefficient.$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is elementary algebra; we treat everything else as a constant and work to isolate $A$. First, we expand out that expression on the right using the distributive property:
$$ A = M+D-R - (0.05(A-250000)+2500)$$
$$ A = M+D-R - 0.05(A-250000)-2500$$
$$ A = M+D-R - 0.05A +0.05cdot 250000-2500$$
Next, we move all the multiples of $A$ over to one side, and apply the distributive property in reverse. I'll also consolidate the constant terms here:
$$A + 0.05A = M+D-R + 12500-2500$$
$$1.05cdot A = M+D-R + 10000$$
Finally, divide by the coefficient of $A$:
$$A = frac1{1.05}(M+D-R + 10000)$$
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
$endgroup$
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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$begingroup$
This is elementary algebra; we treat everything else as a constant and work to isolate $A$. First, we expand out that expression on the right using the distributive property:
$$ A = M+D-R - (0.05(A-250000)+2500)$$
$$ A = M+D-R - 0.05(A-250000)-2500$$
$$ A = M+D-R - 0.05A +0.05cdot 250000-2500$$
Next, we move all the multiples of $A$ over to one side, and apply the distributive property in reverse. I'll also consolidate the constant terms here:
$$A + 0.05A = M+D-R + 12500-2500$$
$$1.05cdot A = M+D-R + 10000$$
Finally, divide by the coefficient of $A$:
$$A = frac1{1.05}(M+D-R + 10000)$$
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
$endgroup$
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
add a comment |
$begingroup$
This is elementary algebra; we treat everything else as a constant and work to isolate $A$. First, we expand out that expression on the right using the distributive property:
$$ A = M+D-R - (0.05(A-250000)+2500)$$
$$ A = M+D-R - 0.05(A-250000)-2500$$
$$ A = M+D-R - 0.05A +0.05cdot 250000-2500$$
Next, we move all the multiples of $A$ over to one side, and apply the distributive property in reverse. I'll also consolidate the constant terms here:
$$A + 0.05A = M+D-R + 12500-2500$$
$$1.05cdot A = M+D-R + 10000$$
Finally, divide by the coefficient of $A$:
$$A = frac1{1.05}(M+D-R + 10000)$$
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
$endgroup$
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
add a comment |
$begingroup$
This is elementary algebra; we treat everything else as a constant and work to isolate $A$. First, we expand out that expression on the right using the distributive property:
$$ A = M+D-R - (0.05(A-250000)+2500)$$
$$ A = M+D-R - 0.05(A-250000)-2500$$
$$ A = M+D-R - 0.05A +0.05cdot 250000-2500$$
Next, we move all the multiples of $A$ over to one side, and apply the distributive property in reverse. I'll also consolidate the constant terms here:
$$A + 0.05A = M+D-R + 12500-2500$$
$$1.05cdot A = M+D-R + 10000$$
Finally, divide by the coefficient of $A$:
$$A = frac1{1.05}(M+D-R + 10000)$$
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
$endgroup$
This is elementary algebra; we treat everything else as a constant and work to isolate $A$. First, we expand out that expression on the right using the distributive property:
$$ A = M+D-R - (0.05(A-250000)+2500)$$
$$ A = M+D-R - 0.05(A-250000)-2500$$
$$ A = M+D-R - 0.05A +0.05cdot 250000-2500$$
Next, we move all the multiples of $A$ over to one side, and apply the distributive property in reverse. I'll also consolidate the constant terms here:
$$A + 0.05A = M+D-R + 12500-2500$$
$$1.05cdot A = M+D-R + 10000$$
Finally, divide by the coefficient of $A$:
$$A = frac1{1.05}(M+D-R + 10000)$$
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
answered Jan 6 at 20:38
jmerryjmerry
16.1k1633
16.1k1633
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
add a comment |
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
$begingroup$
Thank you @jmerry. I guess like most things in life, it's simple when you know how.
$endgroup$
– woter324
Jan 6 at 21:16
add a comment |
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$begingroup$
Expand the right-hand side, move all the terms with
Ato the left side and divide through by the resulting coefficient.$endgroup$
– amd
Jan 6 at 20:32
$begingroup$
Circular references of formulas are not locks of any kind. They generally just lead to simplification of formulas.
$endgroup$
– Matt Samuel
Jan 6 at 21:14