In this case, the financial equivalence between the value F of the loan and the sequence of n periodic payments implies that:

$F={\displaystyle \underset{k=1}{\overset{n}{\sum }}\frac{P_k}{1+i\times k}}$ (12)

It follows that, in the case of a constant payment denoted by P, we have:

$P=F\times {\displaystyle \underset{k=1}{\overset{n}{\sum }}{\left(1+i\times k\right)}^{-1}}$ (13)

Therefore, in the case of our numerical example, we will have $P=10638.80$. Thus, making use of relations (6) and (11), it follows from the methodology presented in Lachtermacher and de Faro (2023) , that the corresponding value of f is 0.92277415.

We will have the evolution of the debt as shown in Table 1 .

In this case, the financial equivalence between the value F of the loan and the sequence of periodic payments, with the k^th now being denoted by ${\hat{P}}_k$, implies that we have:

$F\times \left(1+i\times n\right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\hat{P}}_k}\times \left\{1+i\times \left(n-k\right)\right\}$ (14)

Therefore, in the case of a constant payment equal to $\hat{P}$, it follows that:

$\hat{P}=\frac{F\times \left(1+i\times n\right)}{n\times \left\{1+\left[i\times \left(n-1\right)\text{}/\text{}2\right]\right\}}$ (14’)

Therefore, since it is assumed that the constant payment is subdivided into components $P^C$ and $P^N$, it follows from relations (6), (11) and (14’), that:

$\frac{F\times \left(1+i\times n\right)}{n\times \left[1+i\times \left(n-1\right)\text{}/\text{}2\right]}=\frac{F\times f}{n}+\frac{F}{n}\times \left\{1-f+\frac{f\times i\times \left(n+1\right)}{2}\right\}$ (15)

In other words, after some algebras, we have that the weighting factor f is such that:

$f=\frac{1}{1+i\times \left(n-1\right)\text{}/\text{}2}$ (16)

An expression that is also presented by Forger (2009) .

Thus, in the case of our numerical example, observing that $\hat{P}=10616.61$ and f = 0.947867299, Table 2 presents the evolution of the debt.

Denoting, now, to follow the presentation of Mari and Aretusi (2018) , by $M_k$ the outstanding balance at time k, and focusing attention on the case of a constant payment equal to P, as given by (13), Mari and Aretusi (2018) postulate that:

$P=M_{k-1}-M_k+i\times M_{k-1}\text{}/\text{}\left\{1+i\times \left(k-1\right)\right\}$ (17)

where

$M_k=\left\{F-P\times {\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{\left(1+i\times \ell \right)}^{-1}}\right\}\times \left(1+i\times k\right)$ (18)

for $k=1,2,\cdots ,n$.

In other words, it is stipulated that the payment due at time k has the following two components:

$C_k=M_{k-1}-M_k$ (19)

and

$I_k=i\times M_{k-1}\text{}/\text{}\left\{1+i\times \left(k-1\right)\right\}$ (20)

for $k=1,2,\cdots ,n$. Respectively interpreted as the amortization component C_k and interest I_k component of the k^th payment.

With this notation, it follows that, in the case of our example, the corresponding evolution of the debt will evolve as shown in Table 3 .

Comparing the results respectively presented in Table 1 and Table 3 , two aspects must be highlighted.

The first concerns the fact that, effectively, the debt is paid off when the last payment is paid. And that the total of interest is the same in both cases.

The second, which is crucial for the analysis that will be done in Section 5, is that the sequence of differences between the corresponding interest sequences.

$d_k=J_k-I_k\text{,}k=1,2,\cdots ,n$ (21)

presented in the last column of Table 3 , has a single signal variation.

Although also considered in Mari and Aretusi (2018, 2019) , we are going to follow the presentation in Annibali et al. (2016, 2020) .

In this case, the value of the constant installment $\hat{P}$ is the same as given by (14’). That is:

$\hat{P}=\frac{F\left(1+i\times n\right)}{n\times \left(1+i\times \frac{n-1}{2}\right)}$ (22)

with

${\hat{M}}_k={\hat{P}}_k\times \left(n-k\right)\times \frac{1+i\times \frac{n-k-1}{2}}{1+i\left(n-k\right)}$ (23)

where ${\hat{M}}_k$ denotes the outstanding balance at time k; for $k=1,2,\cdots ,n$.

With the amortization, ${\hat{C}}_k$, and interest ${\hat{I}}_k$ components being, respectively:

${\hat{C}}_k={\hat{M}}_{k-1}-{\hat{M}}_k$ (24)

and

${\hat{I}}_k=\frac{i\times M_{k-1}}{1+i\left(n-k\right)}$ (25)

for $k=1,2,\cdots ,n$.

In Table 4 , still relating to our numerical example, we have the corresponding evolution of the debt.

Similarly to the previous case, the debt is effectively redeemed with the payment of the last installment. And, also, that the total interest and installments are the same as those presented in Table 2 .

Additionally, considering the corresponding interest sequences, we have the respective differences, given by the relationship:

${\hat{d}}_k={\hat{J}}_k-{\hat{I}}_k\text{,}k=1,2,\cdots ,n$ (26)

also have a single signal variation.

$\begin{array}{l}{S}_k=F-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{A}_{\ell }}\\ S_6=120000-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{A}_{\ell }}=120000-A_1-A_2-\cdots -A_6\\ S_6=120000-\left[9459.48+9557.75+9656.03+9754.31+9852.58+9950.86\right]\\ =120000-58231.01=61768.99\end{array}$

$\begin{array}{l}{S}_k={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}\left(P_{\ell }-J_{\ell }\right)}={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{P}_{\ell }}-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ S_6=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ =\left(12-6\right)\times 10638.80-\left[589.66+491.39+393.11+294.83+196.55+98.28\right]\\ S_6=63832.80-2063.82=61768.98\end{array}$

$\begin{array}{l}{S}_k=S_{k-1}+J_k-P_k\Rightarrow S_k=S_{k-2}+J_{k-1}-P_{k-1}+J_k-P_k\text{, etc}\text{.}\\ S_k=S_0+{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{P}_{\ell }}\\ S_6=S_0+{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{P}_{\ell }}=120000+5601.80-63832.81=61768.99\end{array}$

$\begin{array}{l}{S}_k=F-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{A}_{\ell }}\\ S_6=120000-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{A}_{\ell }}=120000-A_1-A_2-\cdots -A_6\\ S_6=120000-\left[9478.67+9573.46+9668.25+9763.03+9857.82+9952.61\right]\\ =120000-58293.84=61706.16\end{array}$

$\begin{array}{l}{S}_k={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}\left(P_{\ell }-J_{\ell }\right)}={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{P}_{\ell }}-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ S_6=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ =\left(12-6\right)\times 10616.114-\left[568.72+473.93+379.15+284.36+189.57+94.79\right]\\ S_6=63696.68-1990.52=61706.16\end{array}$

$\begin{array}{l}{S}_k=S_{k-1}+J_k-P_k\Rightarrow S_k=S_{k-2}+J_{k-1}-P_{k-1}+J_k-P_k\\ \Rightarrow S_k=S_0+{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{P}_{\ell }}\\ S_6=S_0+{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{P}_{\ell }}=120000+5402.84-63696.68=61706.16\end{array}$

$\begin{array}{l}{S}_k=F-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{A}_{\ell }}\\ S_6=120000-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{A}_{\ell }}=120000-A_1-A_2-\cdots -A_6\\ S_6=120000-\left[9438.80+9544.14+9648.44+9751.73+9854.02+9955.35\right]\\ =120000-58192.48=61807.52\end{array}$

$\begin{array}{l}{S}_k={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}\left(P_{\ell }-J_{\ell }\right)}={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{P}_{\ell }}-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ S_6=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ =\left(12-6\right)\times 10638.80-\left[583.09+483.66+385.15+287.55+190.93+94.99\right]\\ S_6=63832.81-2025.28=61807.53\end{array}$

$\begin{array}{l}{S}_k=S_{k-1}+J_k-P_k\Rightarrow S_k=S_{k-2}+J_{k-1}-P_{k-1}+J_k-P_k\\ \Rightarrow S_k=S_0+{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{P}_{\ell }}\\ S_6=S_0+{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{P}_{\ell }}=120000+5640.34-63832.81=61807.53\end{array}$

$\begin{array}{l}{S}_k=C-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{A}_{\ell }}\\ S_6=120000-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{A}_{\ell }}=120000-A_1-A_2-\cdots -A_6\\ S_6=120000-\left[9535.03+9611.89+9690.86+9772.02+9855.46+9941.26\right]\\ =120000-58406.52=\mathrm{61.593.48}\end{array}$

$\begin{array}{l}{S}_k={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}\left(P_{\ell }-J_{\ell }\right)}={\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{P}_{\ell }}-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ S_6=\left(n-k\right)\times P-{\displaystyle \underset{\ell =k+1}{\overset{n}{\sum }}{J}_{\ell }}\\ =\left(12-6\right)\times 10616.114-\left[586.60+495.81+402.37+306.18+207.13+105.11\right]\\ S_6=63696.68-2103.20=61593.48\end{array}$

$\begin{array}{l}{S}_k=S_{k-1}+J_k-P_k\Rightarrow S_k=S_{k-2}+J_{k-1}-P_{k-1}+J_k-P_k\\ \Rightarrow S_k=S_0+{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{k}{\sum }}{P}_{\ell }}\\ S_6=S_0+{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{J}_{\ell }}-{\displaystyle \underset{\ell =1}{\overset{6}{\sum }}{P}_{\ell }}=120000+5290.17-63696.68=61593.49\end{array}$

Denoting:

$V_1\left(\rho \right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\left(1+\rho \right)}^{-k}}\times J_k$ Forger Method (27)

and

$V_2\left(\rho \right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\left(1+\rho \right)}^{-k}}\times I_k$ Mari & Aretusi Method (28)

the respective present values, at the periodic rate, ρ, of the corresponding interest parcels, the financing institution must choose the proposition that presents the lowest present value of the interest sequence.

Now, as can be seen from Table 3 , considering the sequence $d_k=J_k-I_k$, we see that it presents a single signal variation. In other words, the sequence in question identifies a so-called conventional investment project. Which, according to de Faro (1974) , has a single internal rate of return. Which, in this case, as ${\displaystyle \underset{k=1}{\overset{n}{\sum }}{J}_k}-{\displaystyle \underset{k=1}{\overset{n}{\sum }}{I}_k}=0$, is null.

Consequently, which also happens in the general case of n periods, we have $V_1\left(\rho \right)<V_2\left(\rho \right)$, for $\rho >0$. Therefore, the financial institution must choose to implement the procedure suggested by Forger (2009) .

To assess the relevance of the differences between the two options, Tables 5-8 present the behavior of the percentage tax gain:

$\Delta =\left(\frac{V_1\left(\rho \right)}{V_2\left(\rho \right)}-1\right)\times 100$ (29)

for monthly rates i ranging from 0.5% to 2%, for terms n ranging from 5 to 30 years, with the interest rate that reflects the lender’s opportunity cost, in annual terms and identified as ρ_a, ranging from 5% to 30%.

The values presented are significant and, as they are always negative, they indicate that the financing entity should always opt for the methodology recommended in Forger (2009) adapted by de Faro and Lachtermacher (2023b) since $V_1\left(\rho \right)<V_2\left(\rho \right)$, for $\rho >0$ as shown in Tables 5-8 .

Denoting:

$V_3\left(\rho \right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\left(1+\rho \right)}^{-k}}\times {\hat{J}}_k$ Forger Method (30)

and

$V_4\left(\rho \right)={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\left(1+\rho \right)}^{-k}}\times {\hat{I}}_k$ Annibali et al. Method (31)

the respective present values, at the periodic rate ρ, of the corresponding interest payments, the financial institution must choose the proposition that presents the lowest present value of the interest sequence.

Similarly to the case of focal date at time zero, it appears that, as shown in Table 4 , the sequence ${\hat{d}}_k={\hat{J}}_k-{\hat{I}}_k$ also presents a single signal variation, identifying a conventional financing project. For which, as discussed in de Faro (1974), a single internal rate of return is associated. Which, in the case considered, as we also have ${\displaystyle \underset{k=1}{\overset{n}{\sum }}{\hat{J}}_k}={\displaystyle \underset{k=1}{\overset{n}{\sum }}{\hat{I}}_k}$, the respective internal rate of return is zero.

Therefore, which also happens in the general case of n periods, it follows that $V_3\left(\rho \right)>V_4\left(\rho \right)$ for $\rho >0$. Therefore, the financial institution must choose to implement the methodology proposed by Annibali et al. (2016) .

Tables 9-12 show the behavior of the respective percentage tax gain:

${\Delta }^{\prime }=\left(\frac{V_3\left(\rho \right)}{V_4\left(\rho \right)}-1\right)\times 100$ (32)

also, for monthly rates i ranging from 0.5% to 2%, for annual terms n ranging from 5 to 30 years, with the annual rate ρ_a ranging from 5% to 30%.

The results presented, which are also significant, confirm that the financial entity must always opt for the methodology proposed by Annibali et al. (2016, 2020) , since $V_3\left(\rho \right)>V_4\left(\rho \right)$ for $\rho >0$ as shown in Tables 9-12 .