We know that |
We know that |
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z = x * se + y * (1-sp) |
z = x * se + y * (1-sp) |
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y = p * x |
x = y / p |
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which means that |
which means that |
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z = x * se + p * x * (1 - sp) |
z = (y / p) * se + y * (1 - sp) |
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we solve for x |
we solve for y |
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x = z / (-sp * p + se + p) |
y = p * z / (-sp * p + se + p) |
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For the previous example with z = 100 000 and p= 2.5 we get |
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x = |
215 054 |
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y = |
537 634 |
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I have again written a udf in VBA |
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Lyme2(se ; sp ; p ; z ; output) where output is either "x", "y" or "zz" |
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Lyme2(0.44;0.99;2.5;100000;"x") |
x = |
215 054 |
Lyme2(0.44;0.99;2.5;100000;"y") |
y = |
537 634 |
Lyme2(0.44;0.99;2.5;100000;"zz") |
zz = |
13.29 |
we know that |
we know that |
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zz = (z / (x + y)) * 100 |
zz = (z / (x + y)) * 100 |
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y = p * x |
x = y / p |
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which means that |
which means that |
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zz = (z / (x + p * x)) * 100 |
zz = (z / ((y / p) + y)) * 100 |
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we solve for x |
we solve for y |
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x = 100 * z / (p * zz + zz) |
y = 100 * p * z / (p * zz + zz) |
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For the previous example with z=100 000, zz =13.29 and p= 2.5 we get |
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x = |
215 054 |
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y = |
537 634 |
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I have again written a udf in VBA |
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Lyme3(se ; sp ; p ; z ; zz ; output) where output is either "x", "y" |
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Lyme3(0.44;0.99;2.5;100000;13.29;"x") |
x = |
215 054 |
Lyme3(0.44;0.99;2.5;100000;13.29;"y") |
y = |
537 634 |