这是一份liverpool利物浦大学ACFI202的成功案例

Given two vectors $d$ and $c$ with $n$ components each, let $\min (d, c)$ be the vector whose $i$-th component is the minimum of the $i$-th components of $d$ and $c$ for $i=1, \ldots$, n. For example:
$$
\min ((3,7,2),(0,5,3))=(0,5,2)
$$
If $d$ and $c$ are nonnegative, then $d-\min (d, c)$ and $c-\min (d, c)$ are both nonnegative, and:
$$
[d / / c]=[d-\min (d, c) / / c-\min (d, c)]
$$
For instance:
$$
\begin{aligned}
{[(3,7,2) / /(0,5,3)] } &=[(3,7,2)-(0,5,2) / /(0,5,3)-(0,5,2)] \
&=[(3,2,0) / /(0,0,1)] .
\end{aligned}
$$

ACFI202 COURSE NOTES ：

where one might note the difference between the $\mathrm{T}$-account $A s s e t s(T+1)$ and the vector $A S S E T S(T+1)$. The liabilities are:
$$
\begin{aligned}
\text { Liabilities }(T+1) &=[(0, \ldots, 0) / /(0, \ldots, 0, D)] \
&=[(0, \ldots, 0) / / \operatorname{DEBTS}(T+1)]
\end{aligned}
$$
Hence the end-of-the-period resultant equation zero-term is:
Assets
$$
[(\ldots, 0, C A S H(T+1), F G(T+1), R M(T+1), 1,0,0) / /(0, \ldots)]
$$
Liabilities
$$
+[(0, \ldots, 0) / /(0, \ldots, 0, D)]
$$
Total Assets and Liabilities
$$
\begin{gathered}
+[(0, \ldots, 0, D) / /(\ldots, 0, C A S H(T+1), F G(T+1) \
R M(T+1), 1,0,0)]
\end{gathered}
$$