How to minimise a function with both linear and exponent variable?
$begingroup$
I am trying to minimise $N$ in the following equation with respect to $b$.
$$N = p^b + (1-p^b)(b+1)$$
Notes:
$ 0 le p le 1 $, because $p$ is a probability.
Background:
This is my interpretation of a problem concerning how a hospital can minimise the number of HIV blood tests $N$ it has to run by pooling blood samples into bundles $b$. The HIV test has a probability $p$ of returning negative (no HIV); by extrapolation, the bundle of $b$ samples has a probability of $p^b$ of returning negative. If the bundle returns a positive, each sample in the bundle will be tested individually, yielding $b+1$ tests, the bundle plus each individual sample in the bundle.
$N$ denotes the expected number of tests, and is the probability of a bundle negative multiplied by one test, plus the probability of a bundle negative multiplied by $b+1$ tests.
I wish to find an expression for the value of $b$ that minimises $N$ for a given $p$.
I attempted to use a Langrarian (I am new to the method) but had trouble isolating the $b$ term.
calculus optimization
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
$endgroup$
add a comment |
$begingroup$
I am trying to minimise $N$ in the following equation with respect to $b$.
$$N = p^b + (1-p^b)(b+1)$$
Notes:
$ 0 le p le 1 $, because $p$ is a probability.
Background:
This is my interpretation of a problem concerning how a hospital can minimise the number of HIV blood tests $N$ it has to run by pooling blood samples into bundles $b$. The HIV test has a probability $p$ of returning negative (no HIV); by extrapolation, the bundle of $b$ samples has a probability of $p^b$ of returning negative. If the bundle returns a positive, each sample in the bundle will be tested individually, yielding $b+1$ tests, the bundle plus each individual sample in the bundle.
$N$ denotes the expected number of tests, and is the probability of a bundle negative multiplied by one test, plus the probability of a bundle negative multiplied by $b+1$ tests.
I wish to find an expression for the value of $b$ that minimises $N$ for a given $p$.
I attempted to use a Langrarian (I am new to the method) but had trouble isolating the $b$ term.
calculus optimization
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
$endgroup$
$begingroup$
Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43
add a comment |
$begingroup$
I am trying to minimise $N$ in the following equation with respect to $b$.
$$N = p^b + (1-p^b)(b+1)$$
Notes:
$ 0 le p le 1 $, because $p$ is a probability.
Background:
This is my interpretation of a problem concerning how a hospital can minimise the number of HIV blood tests $N$ it has to run by pooling blood samples into bundles $b$. The HIV test has a probability $p$ of returning negative (no HIV); by extrapolation, the bundle of $b$ samples has a probability of $p^b$ of returning negative. If the bundle returns a positive, each sample in the bundle will be tested individually, yielding $b+1$ tests, the bundle plus each individual sample in the bundle.
$N$ denotes the expected number of tests, and is the probability of a bundle negative multiplied by one test, plus the probability of a bundle negative multiplied by $b+1$ tests.
I wish to find an expression for the value of $b$ that minimises $N$ for a given $p$.
I attempted to use a Langrarian (I am new to the method) but had trouble isolating the $b$ term.
calculus optimization
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
$endgroup$
I am trying to minimise $N$ in the following equation with respect to $b$.
$$N = p^b + (1-p^b)(b+1)$$
Notes:
$ 0 le p le 1 $, because $p$ is a probability.
Background:
This is my interpretation of a problem concerning how a hospital can minimise the number of HIV blood tests $N$ it has to run by pooling blood samples into bundles $b$. The HIV test has a probability $p$ of returning negative (no HIV); by extrapolation, the bundle of $b$ samples has a probability of $p^b$ of returning negative. If the bundle returns a positive, each sample in the bundle will be tested individually, yielding $b+1$ tests, the bundle plus each individual sample in the bundle.
$N$ denotes the expected number of tests, and is the probability of a bundle negative multiplied by one test, plus the probability of a bundle negative multiplied by $b+1$ tests.
I wish to find an expression for the value of $b$ that minimises $N$ for a given $p$.
I attempted to use a Langrarian (I am new to the method) but had trouble isolating the $b$ term.
calculus optimization
calculus optimization
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
edited Nov 29 '18 at 9:41
GNUSupporter 8964民主女神 地下教會
12.8k72445
12.8k72445
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
asked Nov 29 '18 at 9:40
Alex CraggsAlex Craggs
61
61
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Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
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– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43
add a comment |
$begingroup$
Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43
$begingroup$
Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43
$begingroup$
Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43
add a comment |
1 Answer
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$begingroup$
$$N(b) = p^b+(1-p^b)(b+1)$$
if $p>0$,
$$N(0) = 1$$
From you description, $b$ is a physical quantity, $b ge 0$.
If $b>0$, then $N(b)$ is a convex combination between $1$ and $b+1$, which is certainly bigger than $1$. Hence the minimal value is attained at $b=0$.
If we want to impose conditions such as $b ge b_0 >0$, then $b_0$ is the quantity that you are looking for by the convex combination argument.
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
add a comment |
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$begingroup$
$$N(b) = p^b+(1-p^b)(b+1)$$
if $p>0$,
$$N(0) = 1$$
From you description, $b$ is a physical quantity, $b ge 0$.
If $b>0$, then $N(b)$ is a convex combination between $1$ and $b+1$, which is certainly bigger than $1$. Hence the minimal value is attained at $b=0$.
If we want to impose conditions such as $b ge b_0 >0$, then $b_0$ is the quantity that you are looking for by the convex combination argument.
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
add a comment |
$begingroup$
$$N(b) = p^b+(1-p^b)(b+1)$$
if $p>0$,
$$N(0) = 1$$
From you description, $b$ is a physical quantity, $b ge 0$.
If $b>0$, then $N(b)$ is a convex combination between $1$ and $b+1$, which is certainly bigger than $1$. Hence the minimal value is attained at $b=0$.
If we want to impose conditions such as $b ge b_0 >0$, then $b_0$ is the quantity that you are looking for by the convex combination argument.
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
add a comment |
$begingroup$
$$N(b) = p^b+(1-p^b)(b+1)$$
if $p>0$,
$$N(0) = 1$$
From you description, $b$ is a physical quantity, $b ge 0$.
If $b>0$, then $N(b)$ is a convex combination between $1$ and $b+1$, which is certainly bigger than $1$. Hence the minimal value is attained at $b=0$.
If we want to impose conditions such as $b ge b_0 >0$, then $b_0$ is the quantity that you are looking for by the convex combination argument.
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
$endgroup$
$$N(b) = p^b+(1-p^b)(b+1)$$
if $p>0$,
$$N(0) = 1$$
From you description, $b$ is a physical quantity, $b ge 0$.
If $b>0$, then $N(b)$ is a convex combination between $1$ and $b+1$, which is certainly bigger than $1$. Hence the minimal value is attained at $b=0$.
If we want to impose conditions such as $b ge b_0 >0$, then $b_0$ is the quantity that you are looking for by the convex combination argument.
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
edited Nov 29 '18 at 9:59
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
answered Nov 29 '18 at 9:48
Siong Thye GohSiong Thye Goh
100k1465117
100k1465117
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
add a comment |
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
Ah, how would I go about finding b0, where b0 > 0 using the convex combination method? Would it be possible to find an expression of b0 in terms of p?
$endgroup$
– Alex Craggs
Nov 30 '18 at 14:41
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
$begingroup$
The reason why I mention that is because you are solving a physical problem, and I wasn't sure if $0$ will help you at all. So for your physical problem, to minimize that quantity, $b$ should be a positive or nonnegative number that is as small as possible given your physical constraint.
$endgroup$
– Siong Thye Goh
Nov 30 '18 at 14:49
add a comment |
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Welcome to Math.SE. Please use MathJax for objects of mathematical discourse.
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– GNUSupporter 8964民主女神 地下教會
Nov 29 '18 at 9:43