The three workers are equally efficient and work independently. In order to minimize economic losses, those with less repair time will be arranged first, and the fourth one will be arranged if the first one is completed first, and so on:

A (8, 17,18) = 8+(8+17)+(8+18) = 76 minutes.

B (12,23) =12+(12+23) = 47 minutes.

C (14,30) =14+(14+30) = 58 minutes.

Therefore, the minimum loss is11× (76+47+58) =1×181=199/yuan.

The first question comes from the analysis of other places. Corresponding to your topic, ABC corresponds to three workers, namely 8×3, 12×2, 14×2, which corresponds to the number of times the three numbers in the above formula appear, that is, the time for the three workers to repair their own cars. Let's arrange three cars that can be repaired in the shortest time, which are 8 minutes, 14 minutes and 12 minutes respectively. Party A will repair the car that can be repaired in 8 minutes before repairing the car that can be repaired in 17 minutes. This needs attention. This 17 minute car has stopped during the maintenance period. So you need to add 8 minutes to repair the car on the basis of 17 minutes, and so on. When Party A repairs the third car for 18 minutes, it needs to add the delay time of the first two cars for 8 minutes and 17 minutes before it can be repaired. The same is true for two mechanics B and C to repair vehicles. After simplification, the three formulas become 8× 3, 12× 2, 14× 2.

The second problem is that the cheapest thick pipe is paved with thin pipes, and the thick pipe is paved first, and then the thin pipe is paved. Because thick pipes are expensive, reducing one kilometer from thick pipes can save thin pipes, and increasing money by three kilometers, so let the thick pipes be as few as possible, starting from H point, and then the thin pipes will start from H point to the next three points. What is it in this case, and then push it back and forth, from G to I?