Example 7: A town has five primary schools along a circular road, sequentially named First, Second, Third, Fourth, and Fifth Primary Schools. They have 15, 7, 11, 3, and 14 computers, respectively. To make the number of computers equal in each school, some computers need to be transferred to neighboring schools: First to Second, Second to Third, Third to Fourth, Fourth to Fifth, and Fifth to First. If School A gives -3 computers to School B, it means School B gives 3 computers to School A. To minimize the total number of computers moved, how should the transfers be arranged?
Analysis: Let $A, B, C,$
$D, E$ represent the five primary schools in a clockwise order, and let them sequentially transfer $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ computers to their neighboring schools, as shown in Figure 9. Then,
\[
\begin{array}{c}
7 + x_{1} - x_{2} = 11 + x_{2} \\
- x_{3} = 3 + x_{3} - x_{4} = 14 + x_{4} - x_{5} = 15 \div x_{5} \\
- x_{1} = \frac{1}{5}(15 + 7 + 11 + 3 + 14). \\
\therefore x_{2} = x_{1} - 3, x_{3} = x_{1} - 2, \\
x_{4} = x_{1} - 9, x_{5} = x_{1} - 5.
\end{array}
\]
Since the total number of computers transferred should be minimized, we need to find the minimum value of the function
\[
\begin{aligned}
y = & \left|x_{1}\right| + \left|x_{2}\right| + \left|x_{3}\right| + \left|x_{4}\right| + \left|x_{5}\right| \\
= & \left|x_{1}\right| + \left|x_{1} - 3\right| + \left|x_{1} - 2\right| + \left|x_{1} - 9\right| \\
& + \left|x_{1} - 5\right|
\end{aligned}
\]
From the conclusion in Example 1, we know that when $x_{1} = 3$, the function $y_{\text {min }} = 12$. At this point, we have $x_{2} = 0, x_{3} = 1, x_{4} = -6, x_{5} = -2$.

Therefore, Primary School 1 transfers 3 computers to Primary School 2, Primary School 2 transfers 1 computer to Primary School 4, Primary School 5 transfers 6 computers to Primary School 4, and Primary School 1 transfers 2 computers to Primary School 5. This way, the total number of computers transferred is minimized and is 12.