Why do special forces need to select soldiers with the same physical strength? Why do two adults of the same physique put the heavy object in the middle when lifting something? Why is it that when an adult and a child are lifting something, the weight should be placed closer to the adult? In fact, behind these practices lies the concept of energy efficiency optimization.

For the energy efficiency optimization of humans (or organisms), we can define it as that the physical arrangement method that continues to do work over a long period of time and ultimately maximizes the total amount of work, or minimizes the total physical ability. This is called human (or organisms) energy efficiency optimization Fig. 26.1.

Fig. 26.1
A photograph of a group of soldiers in uniform marching in formation. They carry equipment, wear helmets, and move in a coordinated manner across a flat, open area.

Physical distribution

Full size image

Let's take humans as an example to illustrate below, assuming that the situation of other living creatures is similar.

26.1 Human Energy Efficiency Curve

The human energy efficiency curve can be defined like this. The abscissa is the amount of work done per unit time W/t, labor intensity, which is similar to the output power P of a machine. We use P to represent the labor intensity of human work. The ordinate is the energy efficiencyη, which represents the ability to complete the total amount of labor within a period of longer time. The energy efficiency curve of the i-th person is shown in Fig. 26.2.

Fig. 26.2
A line graph plots the efficiency curve eta (P) versus power P. The curve rises steeply from the origin, reaches a peak efficiency eta e at power P e, and then gradually declines. Vertical dashed lines indicate the positions of P e and P lim.

Energy efficiency curve of person i

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Assume that the maximum energy efficiency of the first person and the i-th person are equal, but the amount of work done per unit time is different, which is equivalent to the different power of the machine, as shown in Fig. 26.3.

Fig. 26.3
A line graph plots the efficiency curve eta (P) with an additional red curve segment. The red curve peaks at P i plus and then drops before merging with the blue curve. Vertical dashed lines mark P i plus, P i minus 1, and P i plus 1.

Energy efficiency curves of the first person and the i-th person

Full size image

If the energy efficiency curves of person 1 and person i have the following relationship

$$ \eta_{i} \left( {P_{i} } \right) = \eta_{1} \left( {\frac{{P_{1} }}{{\beta_{i} }}} \right) $$
(26.1)

where βi is a constant, we call them similarly efficient people, and β1 = 1.

Assume that there are n people in a team to complete one task.

The total work to be done in a human team is P0 that is similar to the total energy output in a machine. Assuming P0 is a fixed value. The work completed by the i-th person is Pi, which is similar to the energy output. The physical energy input of the i-th person is Wi. The total physical energy consumed by all people in the team is Pt that is similar to the total energy input in a machine.

The expression of maximizing P0 is

$$ maxP_{0} = max\mathop \sum \limits_{i = 1}^{n} W_{i} \eta_{i} \left( {W_{i} } \right) $$
(26.2)
$$ P_{t} = \mathop \sum \limits_{i = 1}^{n} W_{i} $$

The expression of minimizing Pt is

$$ minP_{t} = min\mathop \sum \limits_{i = 1}^{n} \frac{{P_{i} }}{{\eta_{i} \left( {P_{i} } \right)}} $$
(26.3)
$$ P_{0} = \mathop \sum \limits_{i = 1}^{n} P_{i}. $$

Note that the arguments of the efficiency functions in the two expressions are different and they are not the identical function.

These are two different expressions of the same problem. We only need to analyze the subsequent minimization expression.

26.2 Energy Efficiency Optimal Control in a Human Team

  1. 1.

    If the n people are identical, Pt becomes

    $$ P_{t} = P_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{\theta_{i} }}{{\eta \left( {\theta_{i} P_{0} } \right)}} $$
    (26.4)

where

$$ \theta_{i} = \frac{{P_{i} }}{{P_{0} }} $$
(26.5)
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1. $$

Consider the minimization problem of Pt

$$ minP_{t} $$
(26.6)
$$ s.t.\theta_{i} > 0,i = 1,2,...n $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ P_{0} = cons\tan t. $$

This problem can also be written as

$$ minP_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{\theta_{i} }}{{\eta \left( {\theta_{i} P_{0} } \right)}} $$
(26.7)
$$ s.t.\theta_{i} > 0,i = 1,2,...n $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ P_{0} = cons\tan t. $$

We consider three cases:

  1. (1)

    n = 2

The team has two variables and has

$$ \theta_{1} + \theta_{2} = 1 $$
(26.8)
$$ \begin{gathered} \theta_{1} > 0 \hfill \\ \theta_{2} > 0 \hfill \\ \end{gathered}. $$

The objective function Pt can be expressed as

$$ P_{t} = P_{0} \left( {\frac{{\theta_{1} }}{{\eta \left( {\theta_{1} P_{0} } \right)}} + \frac{{\theta_{2} }}{{\eta \left( {\theta_{2} P_{0} } \right)}}} \right) = P_{0} \left( {\frac{{\theta_{1} }}{{\eta \left( {\theta_{1} P_{0} } \right)}} + \frac{{1 - \theta_{1} }}{{\eta \left( {\left( {1 - \theta_{1} } \right)P_{0} } \right)}}} \right). $$
(26.9)

The optimization condition is

$$ P_{t}{\prime} \left( {\theta_{1} } \right) = 0 $$
(26.10)

easy to draw

$$ \theta_{1} = \frac{1}{2} $$
(26.11)

for the optimization point. Then we have

$$ \theta_{2} = 1 - \theta_{1} = \theta_{1} = \frac{1}{2} = \frac{1}{n}. $$
(26.12)

That is, the optimal control method is to keep

$$ P_{1} = P_{2} = \frac{{P_{0} }}{2} = \frac{{P_{0} }}{n}. $$
(26.13)

The Pt is

$$ P_{t} = \frac{{P_{0} }}{{\eta \left( {\frac{{P_{0} }}{2}} \right)}} = \frac{{P_{0} }}{{\eta \left( {\frac{{P_{0} }}{n}} \right)}} $$
(26.14)

Since the shape of the overall efficiency curve of the team is the same as that of one person, so the second derivative of the Pt is also greater than zero

$$ P_{t}^{^{\prime\prime}} \left( {\theta_{1} } \right) > 0. $$
(26.15)

Pt is the only minimum value.

$$ {\text{min}}P_{t} = \frac{{P_{0} }}{{\eta \left( {\frac{{P_{0} }}{n}} \right)}} $$
(26.16)

The overall energy efficiency ηt of the human team is the only maximum value.

$$ max\eta_{t} = \eta \left( {\frac{{P_{0} }}{n}} \right) $$
(26.17)
  1. (2)

    n = 3

The team has three variables, based on known conditions, we have

$$ \theta_{1} + \theta_{2} + \theta_{3} = 1 $$
(26.18)
$$ \begin{array}{*{20}c} {\theta_{1} > 0} \\ {\theta_{2} > 0} \\ {\theta_{3} > 0} \\ \end{array}. $$

The Pt expression becomes

$$ P_{t} = P_{0} \left( {\frac{{\theta_{1} }}{{\eta \left( {\theta_{1} P_{0} } \right)}} + \frac{{\theta_{2} }}{{\eta \left( {\theta_{2} P_{0} } \right)}} + \frac{{\theta_{3} }}{{\eta (\left( {\theta_{3} P_{0} } \right)}}} \right). $$
(26.19)

Assuming that θ3 is fixed and an optimization point, only θ1 and θ2 are variables, we have

$$ \theta_{1} + \theta_{2} = 1 - \theta_{3} = constant. $$
(26.20)

Based on the conclusion of n = 2 above, we have

$$ \theta_{1} = \theta_{2} $$
(26.21)

is the optimal point.

Assuming that θ2 is fixed and is an optimization point, only θ1 and θ3 are variables, we have

$$ \theta_{1} + \theta_{3} = 1 - \theta_{2} = constant. $$
(26.22)

According to the conclusion of n = 2 above, we have

$$ \theta_{1} = \theta_{3} $$
(26.23)

is the optimal point.

Similarly, assuming that θ1 is fixed and is an optimization point, only θ2 and θ3 are variables, we have

$$ \theta_{2} + \theta_{3} = 1 - \theta_{1} = constant. $$
(26.24)

According to the conclusion of n = 2 above, we have

$$ \theta_{2} = \theta_{3} $$
(26.25)

is the optimal point.

So, we have the optimal point

$$ \theta_{1} = \theta_{2} = \theta_{3} = \frac{1}{3} = \frac{1}{n}. $$
(26.26)

That is, the optimal control method is to keep

$$ P_{1} = P_{2} = P_{3} = \frac{{P_{0} }}{3} = \frac{{P_{0} }}{n}. $$
(26.27)

The minimum Pt is

$$ minP_{t} = \frac{{P_{0} }}{{\eta \left( {\frac{{P_{0} }}{3}} \right)}} = \frac{{P_{0} }}{{\eta \left( {\frac{{P_{0} }}{n}} \right)}} $$
(26.28)

The maximum overall energy efficiency ηt is

$$ max \eta_{t} = \eta \left( {\frac{{P_{0} }}{3}} \right) = \eta \left( {\frac{{P_{0} }}{n}} \right). $$
(26.29)
  1. (3)

    n = k

The human team has k variables, the above conclusion can be extended to the case of n = k, the optimal point is

$$ \theta_{1} = \theta_{2} = ... = \theta_{k} = \frac{1}{k} $$
(26.30)

That is, the optimal method is to keep

$$ P_{1} = P_{2} = ... = P_{k} = \frac{{P_{0} }}{k}. $$
(26.31)

The minimum Pt is

$$ minP_{t} = P_{0} \frac{1}{{\eta \left( {\frac{{P_{0} }}{n}} \right)}}. $$
(26.32)

The maximum overall energy efficiency ηt is

$$ max \eta_{t} = \eta \left( {\frac{{P_{0} }}{k}} \right) = \eta \left( {\frac{{P_{0} }}{n}} \right) $$
(26.33)
  1. 2.

    If n people are the energy efficiency similarity people, the expression of Pt becomes

    $$ P_{t} = P_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{\theta_{i} }}{{\eta_{i} \left( {\theta_{i} P_{0} } \right)}} $$
    (26.34)

where

$$ \theta_{i} = \frac{{P_{i} }}{{P_{0} }} $$
(26.35)
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1. $$

Consider the minimization problem of total power consumption

$$ minP_{t} $$
(26.36)
$$ s.t.\theta_{i} > 0,i = 1,2,...n $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ P_{0} = cons\tan t. $$

This problem can also be written as

$$ minP_{0} \mathop \sum \limits_{i = 1}^{n} \frac{{\theta_{i} }}{{\eta_{i} \left( {\theta_{i} P_{0} } \right)}} $$
(26.37)
$$ s.t.\,\theta_{i} > 0,i = 1,2,...n $$
$$ \mathop \sum \limits_{i = 1}^{n} \theta_{i} = 1 $$
$$ P_{0} = cons\tan t. $$

We consider three cases:

  1. (1)

    n = 2

The human team has two variables and has

$$ \theta_{1} + \theta_{2} = 1 $$
(26.38)
$$ \begin{array}{*{20}c} {\theta_{1} > 0} \\ {\theta_{2} > 0} \\ \end{array}. $$

The objective function Pt can be expressed as

$$ P_{t} = P_{0} \left( {\frac{{\theta_{1} }}{{\eta_{1} \left( {\theta_{1} P_{0} } \right)}} + \frac{{\theta_{2} }}{{\eta_{2} \left( {\theta_{2} P_{0} } \right)}}} \right) = P_{0} \left( {\frac{{\theta_{1} }}{{\eta_{1} \left( {\theta_{1} P_{0} } \right)}} + \frac{{1 - \theta_{1} }}{{\eta_{2} \left( {\left( {1 - \theta_{1} } \right)P_{0} } \right)}}} \right). $$
(26.39)

The optimization condition is

$$ P_{t}{\prime} \left( {\theta_{1} } \right) = 0. $$
(26.40)

It is easy to see that

$$ \theta_{1} = \frac{1}{{1 + \beta_{2} }} = \frac{{\beta_{1} }}{{\beta_{1} + \beta_{2} }} $$
(26.41)

for the optimization point. Then we have

$$ \theta_{2} = 1 - \theta_{1} = \frac{{\beta_{2} }}{{1 + \beta_{2} }} = \frac{{\beta_{2} }}{{\beta_{1} + \beta_{2} }}. $$
(26.42)

That is, the optimal control method is to keep

$$ \begin{array}{*{20}c} {P_{1} = \frac{1}{{1 + \beta_{2} }}P_{0} } \\ {P_{2} = \frac{{\beta_{2} }}{{1 + \beta_{2} }}P_{0} } \\ \end{array}. $$
(26.43)

The Pt is

$$ P_{t} = \frac{{P_{0} }}{{\eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} }}} \right)}} $$
(26.44)

Since the shape of the overall efficiency curve of the human team is the same as that of one person, so the second derivative of the Pt is also greater than zero

$$ P_{t}^{^{\prime\prime}} \left( {\theta_{1} } \right) > 0. $$
(26.45)

Pt is the only minimum value with

$$ {\text{min}}P_{t} = \frac{{P_{0} }}{{\eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} }}} \right)}}. $$
(26.46)

The overall energy efficiency ηt of the team is the only maximum value.

$$ max\eta_{t} = \eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} }}} \right) $$
(26.47)
  1. (2)

    n = 3

The team has three variables, based on known conditions, we have

$$ \theta_{1} + \theta_{2} + \theta_{3} = 1 $$
(26.48)
$$ \begin{array}{*{20}c} {\theta_{1} > 0} \\ {\theta_{2} > 0} \\ {\theta_{3} > 0} \\ \end{array}. $$

The Pt expression becomes

$$ P_{t} = P_{0} \left( {\frac{{\theta_{1} }}{{\eta_{1} \left( {\theta_{1} P_{0} } \right)}} + \frac{{\theta_{2} }}{{\eta_{2} \left( {\theta_{2} P_{0} } \right)}} + \frac{{\theta_{3} }}{{\eta_{3} (\left( {\theta_{3} P_{0} } \right)}}} \right). $$
(26.49)

Assuming that θ3 is fixed and is an optimization point, only θ1 and θ2 are variables, we have

$$ \theta_{1} + \theta_{2} = 1 - \theta_{3} = constant. $$
(26.50)

Based on the conclusion of n = 2 above, there are

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{1}{{1 + \beta_{2} }}} \\ {\theta_{2} = \frac{{\beta_{2} }}{{1 + \beta_{2} }}} \\ \end{array}. $$
(26.51)

is the optimal point.

Assuming that θ2 is fixed and is an optimization point, only θ1 and θ3 are variables, we have

$$ \theta_{1} + \theta_{3} = 1 - \theta_{2} = constant. $$
(26.52)

According to the conclusion of n = 2 above, there are

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{1}{{1 + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{1 + \beta_{3} }}} \\ \end{array} $$
(26.53)

is the optimal point.

Similarly, assuming that θ1 is fixed and is an optimization point, only θ2 and θ3 are variables, we have

$$ \theta_{2} + \theta_{3} = 1 - \theta_{3} = constant. $$
(26.54)

According to the conclusion of n = 2 above, we have

$$ \begin{array}{*{20}c} {\theta_{2} = \frac{{\beta_{2} }}{{\beta_{2} + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{\beta_{2} + \beta_{3} }}} \\ \end{array} $$
(26.55)

to be the optimal point.

Thus, the optimal control method is to keep

$$ \begin{array}{*{20}c} {\theta_{1} = \frac{{\beta_{1} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ {\theta_{2} = \frac{{\beta_{2} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ {\theta_{3} = \frac{{\beta_{3} }}{{\beta_{1} + \beta_{2} + \beta_{3} }}} \\ \end{array} $$
(26.56)

The minimum Pt is

$$ minP_{t} = \frac{{P_{0} }}{{\eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} + \beta_{3} }}} \right)}} $$
(26.57)

The maximum overall energy efficiency ηt is

$$ max \eta_{t} = \eta_{1} \left( {\frac{{P_{0} }}{{1 + \beta_{2} + \beta_{3} }}} \right). $$
(26.58)
  1. (3)

    n = k

The human team has k variables, the above conclusion can be extended to the case of n = k, the optimal point and the optimal method is to keep

$$ \theta_{i} = \frac{{\beta_{i} }}{{\mathop \sum \nolimits_{l = 1}^{k} \beta_{l} }}. $$
(26.59)

The minimum Pt is

$$ minP_{t} = P_{0} \frac{1}{{\eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \nolimits_{l = 1}^{k} \beta_{l} }}} \right)}} $$
(26.60)

The maximum overall energy efficiency ηt is

$$ max\,\eta_{t} = \eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \nolimits_{l = 1}^{k} \beta_{l} }}} \right). $$
(26.61)

26.3 Optimal Number of Working People in a Human Team

  1. 1.

    A human team has m identical people, if the following equations are satisfied

    $$ \eta \left( {\frac{{P_{0} }}{n}} \right) \ge \eta \left( {\frac{{P_{0} }}{n - 1}} \right) $$
    (26.62)
    $$ \begin{array}{*{20}c} {\eta \left( {\frac{{P_{0} }}{n}} \right) \ge \eta \left( {\frac{{P_{0} }}{n + 1}} \right)} \\ {n \le m} \\ \end{array}. $$

Then the number n is the optimal number of working people.

  1. 2.

    A human team has m energy efficiency similarity people, if the following equations are satisfied

    $$ \eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \nolimits_{l = 1}^{n} \beta_{l} }}} \right) \ge \eta_{1} \left( {\frac{{P_{0} }}{{\mathop \sum \nolimits_{l = 1}^{n1} \beta_{l} }}} \right) $$
    (26.63)
    $$ \begin{array}{*{20}c} {n \le m} \\ {n1 \le m} \\ \end{array}. $$

n1 is any combination other than the optimal combination of n working people this time, and also include other combinations of n working people. The number n is the number of people with optimal working.

26.4 Optimal Scheduling for the Number of Working People

As long as people are sent out, they have to eat whether they work or not. In this regard, people are different from machines. Therefore, when a team is sent out to perform a task, as long as the number of people is determined, there is no problem of optimizing the number of people during the process. There is a problem of optimizing the number of people before departure.

  1. 1.

    A human team has m identical people, and the number n is the optimal number of working people. If P0 increases to P01, the following relation holds:

    $$ \eta \left( {\frac{{P_{01} }}{n}} \right) = \eta \left( {\frac{{P_{01} }}{n + 1}} \right) $$
    (26.64)
    $$ n \le m $$

Then P01 is the optimal scheduling point between n people working and n + 1 people working. When the total required power P0 is greater than P01, the optimal number of working people is adjusted from n to n + 1. If P0 increases until P0/n = P1m, there isn’t the point P01, then P0/n = P1m is the adjusting point from n to n + 1.

If P0 is reduced to P02, the following relation is established

$$ \eta \left( {\frac{{P_{02} }}{n}} \right) = \eta \left( {\frac{{P_{02} }}{n - 1}} \right). $$
(26.65)
$$ n \le m $$

Then P02 is the optimal adjusting point between the n working people and the n-1 working people. When the total required power P0 is less than P02, the optimal number of working people is adjusted from n to n-1. If P0 reduces until P0/(n-1) = P1m, there isn’t the point P02, then P0/(n-1) = P1m is the adjusting point from n to n-1.

The analysis process is shown in Fig. 26.4.

Fig. 26.4
A line graph plots the efficiency curve eta (P) for different segments. The curve rises, peaks at P e, and then falls. Additional segments show efficiency at P 0 slash (n plus 1), P 0 slash n, and P 0 slash (n minus 1). Vertical dashed lines and horizontal lines indicate these points and their corresponding efficiencies.

Energy efficiency comparison curve of the team with identical efficiency people

Full size image
  1. 2.

    A human team has m energy efficiency similarity people, and the number n is the number of the optimal working people. If P0 increases to P01, the following relation holds:

    $$ \begin{array}{*{20}c} {\eta_{1} \left( {\frac{{P_{01} }}{{\mathop \sum \nolimits_{l = 1}^{n} \beta_{l} }}} \right) = \eta_{1} \left( {\frac{{P_{01} }}{{\mathop \sum \nolimits_{l = 1}^{k1} \beta_{l} }}} \right)} \\ {n \le m} \\ \end{array}. $$
    (26.66)

Then P01 is the optimal adjusting point between the operation of n working people and the operation of k1 working people. When the total required power P0 is greater than P01, the optimal number of working people is adjusted from n to k1. If P0 increases until P0/\(\mathop \sum \limits_{l = 1}^{n} \beta_{l}\) = P1m, there isn’t the point P01, then P0/\(\mathop \sum \limits_{l = 1}^{n} \beta_{l}\) = P1m is the adjusting point from n to k1.

If P0 is reduced to P02, the following relation is established

$$ \begin{array}{*{20}c} {\eta_{1} \left( {\frac{{P_{02} }}{{\mathop \sum \nolimits_{l = 1}^{n} \beta_{l} }}} \right) = \eta_{1} \left( {\frac{{P_{02} }}{{\mathop \sum \nolimits_{l = 1}^{k2} \beta_{l} }}} \right)} \\ {n \le m} \\ \end{array}. $$
(26.67)

Then P02 is the optimal switching point between the operation of n transformers and the operation of k2 transformers. When the total required power P0 is less than P02, the optimal number of transformers in operation is switched from n to k2. If P0 reduces until P0/\(\mathop \sum \limits_{l = 1}^{k2} \beta_{l}\) = P1m, there isn’t the point P02, then P0/\(\mathop \sum \limits_{l = 1}^{k2} \beta_{l}\) = P1m is the adjusting point from n to k2.

The analysis process is shown in Fig. 26.5.

Fig. 26.5
A line graph plots the efficiency curve eta (P) with multiple segments. The curve shows efficiency at various points: P 0 slash summation i equals 1 n plus 1 A i, P 0 slash summation i equals 1 n A i, P e, P 0 slash summation i equals 1 n minus 1 A i, and P lim. Vertical dashed lines and horizontal lines mark these points and their associated efficiencies.

Energy efficiency comparison curve of the team with similarly efficiency people

Full size image

k1 and k2 are any combination other than the optimal combination of n working people this time, and also include other combinations of n working people. Point \( P_{0} /\mathop \sum \limits_{l = 1}^{k1} \beta_{l}\) is the point closest to \(P_{0} /\mathop \sum \limits_{l = 1}^{n} \beta_{l} \) to the left of point \(P_{0} /\mathop \sum \limits_{l = 1}^{n} \beta_{l}\). Point \(P_{0} /\mathop \sum \limits_{l = 1}^{k2} \beta_{l}\) is the point closest to \(P_{0} /\mathop \sum \limits_{l = 1}^{n} \beta_{l}\) to the right of point \(P_{0} /\mathop \sum \limits_{l = 1}^{n} \beta_{l}\). If all people are identical, we have k1 = n + 1 and k2 = n-1.