Along this paper, some slight approaches on the the accuracy of the model for different branch sizes have been made. The detailed discussion about this will be developed in this section with a comparative statics exercise on the optimal policy m
^{∗} (or, alternatively,{\underline{m}}^{\ast}) in terms of the primitive parameters R and μ.
The branch size
Let us start this paragraph by making some remarks about the concept of branch size.
Intuitively, the notion of branch size is identified with the volume of its turnover. Actually, there are many criteria to quantify the size of a branch: the volume of its credits, the number of its clients, the number of its staff or the volume of its deposits, among others. In practice (i.e., among bank managers), the most accepted criterion to measure the size of a branch is simply to quantify its total needs of cash and then, to state that branch size as an increasing function of these needs. In short, bank managers assume that the bigger branch sizes correspond to the mayor needs of cash of the branch, related to mayor moves of money (entries and exits).
Even though we used R before in order to shorten the notation (of the quartic equation), we do make now a formal definition: R is defined as
R:=E[{X}_{1}]E[{Y}_{1}]+K,
and it represents the total needs of cash of the branch. For the reasons exposed before, we will assume from now on that the parameter R quantifies specifically the size of the branch. As a parameter, R takes values on [0,+∞), although for simplicity, in practice R is considered as belonging to one of the three categories: small, medium and big size,
Then, regarded simply as the branch size, the parameter R will allow us to calibrate for which sizes our model is more or less accurate. Recall that R is the second indicator we define since actually we introduced at the end of section “The mathematical model” the parameter μ as an index for measuring the cash flow fluctuations of the branch.
As we mentioned before, the main objective of this section is to do some comparative static exercises on the optimal policy m
^{∗} (or, alternatively,{\underline{m}}^{\ast}) in terms of the primitive parameters R and μ. Let us start with R, by finding the sign of\frac{\partial m}{\partial R} to state in which cases m is increasing or decreasing on R.
Let us observe first that m is function of y, m = m(y), via the objective function
\gamma \frac{A}{\mathit{\text{xy}}}+\frac{\nu}{3}(x+2y)+\mathit{\text{By}},
with the change x = R − y, that is,
m(y)=\frac{\gamma A}{(Ry)y}+\frac{\nu}{3}(R+y)+\mathit{\text{By}},
so we may calculate\frac{\partial m}{\partial y}=\frac{\mathit{\text{dm}}}{\mathit{\text{dy}}}. This partial derivative is always positive since this is the result of algebraic operations (sum and product) among positive quantities. Specifically,
\begin{array}{ll}\phantom{\rule{5pt}{0ex}}\frac{\partial m}{\partial y}& =\frac{\gamma A(R2y)}{{(Ry)}^{2}{y}^{2}}+(\frac{\nu}{3}+B)=\\ =\underset{+}{\underset{\u23df}{\frac{\gamma A(2yR)}{{(Ry)}^{2}{y}^{2}}}}+\underset{+}{\underset{\u23df}{(\frac{\nu}{3}+B)}}>0\end{array}
which is positive as all terms involved so are as we claimed. Actually, since the second addend is positive with no doubt, we focus on the first addend: while it is clear that the denominator of the first term is positive, note that also it is the numerator of the fraction since y(=C
_{
z
}) ≥ R for well functioning of the branches, so
y\ge R\Rightarrow y\ge R>\frac{R}{2}\Rightarrow 2yR\ge 0.
This shows that the optimal value m
^{∗} is increasing on the optimal policy y
^{∗}.
Let us note that this fact, m
^{∗} is increasing on y
^{∗}, is not related with the parameters of the model but it is simply a logical consequence of the model: mayor will be this upper bound (maximum of cash){C}_{z}^{\ast}, higher will be the optimal value for costs. This statement should be kept in mind for the bank companies when deciding (or not) to add the “bit more” over{C}_{z}^{\ast} for precautionary motives according with their internal rules of functioning, independently either of the size of the branches or the fluctuations of their cash balance.
On the other hand, the function m() may also be regarded as a function of R:
m(R)=\frac{\gamma A}{\mathit{\text{yR}}{y}^{2}}+\frac{\nu}{3}R+(\frac{\nu}{3}+B)y.
Henceforth we may calculate\frac{\partial m}{\partial R}=\frac{\mathit{\text{dm}}}{\mathit{\text{dR}}}:
\begin{array}{ll}\phantom{\rule{5pt}{0ex}}\frac{\partial m}{\partial R}& =\frac{\gamma \mathit{\text{Ay}}}{{(Ry)}^{2}{y}^{2}}+\frac{\nu}{3}=\\ =\frac{\gamma A}{{(Ry)}^{2}y}+\frac{\nu}{3}\ge 0\phantom{\rule{.3em}{0ex}}\text{or}\phantom{\rule{0.3em}{0ex}}\le 0\end{array}
which could be positive or negative, depending on the values of both\frac{\gamma A}{{{x}^{\ast}}^{2}{y}^{\ast}} and\frac{\nu}{3}.
We note that we have the following chain of logical implications:
\text{when}\phantom{\rule{2.77626pt}{0ex}}{{x}^{\ast}}^{2}{y}^{\ast}\to 0\Rightarrow \frac{\gamma A}{{{x}^{\ast}}^{2}{y}^{\ast}}\to \infty \Rightarrow \frac{\partial m}{\partial R}<0,
since
\frac{\partial m}{\partial R}=\frac{\nu}{3}\frac{\gamma A}{{{x}^{\ast}}^{2}{y}^{\ast}}<0,
because the subtrahend → ∞. A similar argument allows us to state that
\text{when}\phantom{\rule{2.77626pt}{0ex}}{{x}^{\ast}}^{2}{y}^{\ast}\to \infty \Rightarrow \frac{\gamma A}{{{x}^{\ast}}^{2}{y}^{\ast}}\to 0\Rightarrow \frac{\partial m}{\partial R}>0.
On the other hand, it is nor difficult to see that x
^{2}
y is increasing on y since
\begin{array}{l}\frac{\partial {x}^{2}y}{\partial y}={x}^{2}>0,\end{array}
that is, if y↗ ⇒x
^{2}
y↗. Thus, combining both results, we have that

1.
when y\to 0\Rightarrow {x}^{2}y\to 0\Rightarrow \frac{\partial {m}^{\ast}}{\partial R}<0.

2.
When y\to \infty \Rightarrow {x}^{2}y\to \infty \Rightarrow \frac{\partial {m}^{\ast}}{\partial R}>0.
In other words, since y = C
_{
z
}, the above result states that

1.
When C _{
z
} → 0, then m ^{∗} is decreasing on R.

2.
When C _{
z
}→∞ then m ^{∗} is increasing on R.
Along this paper we have rejected for simplicity some features of C
_{
z
} to center our attention on the main requisite on C
_{
z
} of minimizing costs to the bankťs company. At this point, we return to the fact that C
_{
z
} (maximum of cash) is fixed by the bankťs company for each branch, attending some parameters, particularly, the size of the branch.
Although other considerations (like geographic location) may influence, in the reality C
_{
z
} is fixed by the bank company in a directly proportional way to the branch size: higher is the branch size, bigger will be C
_{
z
}. That is, if
R\nearrow \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{C}_{z}\nearrow .
This result states that in the category of small branches (branches for which size R takes small values) the optimal value for branch costs m
^{∗} decreases in despite of R increases. This may be interpreted in the sense that the bank company can expected to have the same optimal value for costs m
^{∗}
for all branches of this category, even for branches with sizes in a neighborhood of small size. That is, for small/medium branches.
However, the opposite phenomena occurs for the category of big branches: dealing with them, the bank company may expect that the optimal costs will blow up as the size of the branches increase.
Now we return to the central optimization problem of this paper, which yields us to discuss the features of the Lagrange multiplier associated to this program. This allows us to state some complementary properties of the index R.
While we have solved the minimization program
\begin{array}{cl}\text{Minimize}:& \gamma \frac{A}{\mathit{\text{xy}}}+\frac{\nu}{3}(x+2y)+\mathit{\text{By}}\\ \text{s.a.}& x+y=E[{X}_{1}]E[{Y}_{1}]+K\end{array}
by substituting x = E[X
_{1}] − E[Y
_{1}] + K − y into the objective function, one also would have resolved this by applying the Lagrange method: let us see then which features the associated Lagrange multiplier exhibits and which economic implications could be derived from this.
When the objective is to minimize the cost function subject to the “output constraint”, it is well known that the Lagrange multiplier turns out to be the marginal cost of “production” that is, the increase in total costs when one more unit of output is permitted. Note that the “output” in our constrain corresponds to the branch needs of cash (or alternatively, the branch size), R.
By incorporating R, the minimization problem written in terms of x and y as variables (as before) turns out to be
\begin{array}{cl}\text{Minimize}:& \gamma \frac{A}{\mathit{\text{xy}}}+\frac{\nu}{3}(x+2y)+\mathit{\text{By}}\\ \text{s.a.}& \begin{array}{ll}x+y& =R\\ x& \le 0\end{array}\end{array}
As it is well known, the solution of this constrained optimization problem can be found as well by using the socalled Lagrangian method. If we define the Lagrangian asL(x,y;\lambda )=\gamma \frac{A}{\mathit{\text{xy}}}+\frac{\nu}{3}(x+2y)+\mathit{\text{By}}\lambda (x+yR), the equations given by this Lagrange multiplier method are
\begin{array}{ll}\phantom{\rule{5pt}{0ex}}\frac{\partial L}{\partial x}& =\frac{\gamma A}{y}(\frac{1}{{x}^{2}})+\frac{\nu}{3}\lambda =0\\ =\\ \phantom{\rule{5pt}{0ex}}\frac{\partial L}{\partial y}& =\frac{\gamma A}{x}(\frac{1}{{y}^{2}})+\frac{2\nu}{3}+B\lambda =0\\ =\\ \phantom{\rule{5pt}{0ex}}\frac{\partial L}{\partial \lambda}& =x+y=R,\end{array}
where the multiplier λ has an economic interpretation. As it is well known,
\lambda =\frac{\partial {\underline{m}}^{\ast}}{\partial R}
represents the rate of change of the optimal return point{\underline{m}}^{\ast}=\epsilon ({C}_{0}^{\ast},{C}_{z}^{\ast}) when R increases one unit.
From the first and second equations, we have that
\lambda =\frac{\gamma A}{{x}^{2}y}+\frac{\nu}{3}=\frac{\gamma A}{x{y}^{2}}+\frac{2\nu}{3}+B.
From algebraic computation, it is clear that λ is increasing function of both x
^{2}
y and x y
^{2}: actually, if x
^{2}
y or x y
^{2} increase, then\frac{1}{{x}^{2}y} and\frac{1}{x{y}^{2}} decrease, so\frac{1}{{x}^{2}y} and\frac{1}{x{y}^{2}} increase as well, even more if both are multiplied by non negative constants. Finally, to determine that both x
^{2}
y and x y
^{2} are increasing functions of R, we simply use the test of first derivative: since x + y = R then
\begin{array}{l}{x}^{2}y=f(R)={x}^{2}(Rx),\\ x{y}^{2}=g(R)=(Ry){y}^{2}\end{array}
Hence,
\begin{array}{l}{f}^{\prime}(R)={x}^{2}>0,\\ {g}^{\prime}(R)={y}^{2}>0.\end{array}
As a result of this, we have the following sequence of implications:
\text{if}\phantom{\rule{1em}{0ex}}R\nearrow \phantom{\rule{0.3em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\lambda \nearrow \phantom{\rule{.3em}{0ex}}\Rightarrow \phantom{\rule{1em}{0ex}}\frac{\partial {\underline{m}}^{\ast}}{\partial R}\nearrow \phantom{\rule{1em}{0ex}}\iff \phantom{\rule{1em}{0ex}}\frac{\partial {m}^{\ast}}{\partial R}\nearrow ,
as\lambda =\frac{\partial {\underline{m}}^{\ast}}{\partial R} and{\underline{m}}^{\ast} and m
^{∗} are linearly related. As a consequence of this,

1.
\underset{R\to +\infty}{lim}\frac{\partial {m}^{\ast}}{\partial R}=+\infty, and

2.
\underset{R\to 0}{lim}\frac{\partial {m}^{\ast}}{\partial R}=0,
The economical interpretation of this result is the following: this outcome states that, as the branch size increases, the rate of changes in the optimum for branch costs increases as well. This means that the situation of branch costs deteriorates at major speed when dealing with branches of big size. All the contrary, the situation of possible changes on optimal costs for the branch of small size diminishes (improves) for small values of R.
The fluctuations of the cash flow
This paragraph is devoted to analyzing how the parameter μ, defined in section “The mathematical model”, influences the behavior of the optimal return point m
^{∗}. Let us start then with some remarks about the meaning of the parameter μ.
Recall that μ appears at the first addend of the objective function
\underset{\text{costs due to cash flow}}{\underset{\u23df}{\gamma \frac{A}{({C}_{0}{C}_{z}){C}_{z}}}},
where A = μ
^{2}
t is the variance of daily changes in the cash balance and μ stated for the amount of euros that the branch cash balance increases or decreases in some small fraction of a working day\frac{1}{t}. Then, as we established in section “The mathematical model”, μ shows the branch users behavior as it indicates how active the branch users are in form of fluctuations of the branch cash balance.
Recall also that, in practice, μ is strongly directly related with the geographic location of the branch since the fluctuations of the cash balance in the major part of the cases are caused by some physical reason (in a similar way that a physical phenomena disturbs the state of rest of any object, changing this initial state of rest with a graphical representation close to a straight line into a motion staterepresented by a periodic wave motion, bigger or smaller depending on the intensity of the disturbs).
In our particular context, this physical phenomena which disturbs the branch cash flow may be identified with the proximity to a cash center, in the wide sense of this expression: that is, understanding that a cash center is an industrial, financial, shopping center or whatever other physical zone where the moves of money are higher than usual.
Thus, for our purposes of calibrating the model, μ will quantify the fluctuations in the branch cash flow and μ ∈ [0,+∞). However, due to the direct relationship exposed before between the fluctuations of the cash flow and the physical reason that causes the fluctuations (cash center), μ could be interpreted as geographic location of the branch in reference to a cash center as well.
As function of x
^{∗}, y
^{∗}, m
^{∗} is
\begin{array}{ll}\phantom{\rule{4pt}{0ex}}{m}^{\ast}& =m({x}^{\ast},{y}^{\ast})=\\ =\gamma \frac{A}{{x}^{\ast}{y}^{\ast}}+\nu \frac{{x}^{\ast}+2{y}^{\ast}}{3}+B{y}^{\ast}\phantom{\rule{1.5em}{0ex}}\text{or}\\ \phantom{\rule{4pt}{0ex}}m({y}^{\ast})& =\gamma \frac{A}{(R{y}^{\ast}){y}^{\ast}}+\nu \frac{R+{y}^{\ast}}{3}+B{y}^{\ast}\end{array}
In a similar way, m
^{∗} may be regarded as a function of μ:
\begin{array}{l}{m}^{\ast}=f(\mu )=\gamma \frac{t}{(R{y}^{\ast}){y}^{\ast}}{\mu}^{2}+(\nu \frac{R+{y}^{\ast}}{3}+B{y}^{\ast}),\end{array}
From this expression it can be derived that
\begin{array}{ll}\phantom{\rule{5pt}{0ex}}\frac{d{m}^{\ast}}{d\mu}& =2\gamma \frac{t}{(R{y}^{\ast}){y}^{\ast}}\mu =\\ =2\gamma \frac{t}{{x}^{\ast}{y}^{\ast}}\mu \\ =2\gamma \frac{t}{{x}^{\ast}{y}^{\ast}}\mu <0,\end{array}
since the factor2\gamma \frac{t}{{x}^{\ast}{y}^{\ast}} is negative as x
^{∗} < 0 while the rest of factor (γ, t, y
^{∗}) are positive. As a consequence of this, m
^{∗} is decreasing on μ in all cases.
In contrast with the results based on the parameter R, where the result clearly depends on the values of R, the case of the parameter μ can be written in absolute terms: for all the values of μ the optimal value for branch costs m
^{∗} is always decreasing on μ. That may be interpreted as, if the fluctuations in the branch cash balance increases, the optimum for branch costs decreases: specifically, the situation of branch costs improves when dealing with branches of big fluctuations in the cash flow, that is, when dealing with branches closely located near to cash centers.