Assessing investments: Accounting approach
APP, Accounting payback period
Accounting payback period
The time required to recoup the costs of a project is called the accounting payback period and is abbreviated as APP.
The APP is calculated using the net return up to and including a specified period; that is the sum of the cash flows up to and including that period. The net return on the investment is the return up to and including the last period #n#.
The accounting payback period is therefore the first period in which the net return up to and including that period are not negative.
The accounting payback period in a project of the duration #n# is therefore a natural number less than or equal to #n#, unless the net return on the investment is negative. In the latter case, the APP is indefinite. Because we are talking about plans to make a profit (or at least not to make a loss), this is generally not an issue.
We now convert the concept of APP into formulas.
Accounting payback period
Given an investment with the duration #n# and cash flows #C_0#, #C_1 ,\ldots, C_n#, the accounting payback period is the period #j# in which the net return up to and including period #j# is non-negative for the first time:
\[\sum_{i=0}^{k} C_i \lt 0\quad \text{voor alle }k\lt j\quad \text{ en }\quad \sum_{i=0}^{j} C_i \geq 0\]
So we calculate the net returns for the periods #1,2,\ldots# until we get a non-negative result. The first period for which this requirement is met is the accounting payback period.
When comparing two or more projects with the same accounting payback period, preference is given to the investment with the highest net return in the period in which the APP is achieved.
To determine the accounting payback period, we first draw up the table with cumulative cash flows.
\[\begin{array}{c|c} j&\displaystyle \sum_{i=0}^{j} C_i\\ \hline\ 0 & -305\\\ 1 & -155 \\ \ 2 & -5 \\ \ 3 & 145 \\ \ 4 & 295 \\ \ 5 & 445 \\ \end{array}\]
The accounting payback period is the period in which the sum of cash flows is greater than or equal to #0# for the first time. Since
- #\sum_{i=0}^{2}C_i =-5 \lt 0# and
- #\sum_{i=0}^{3}C_i =145 \geq 0#,
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