## 2009年9月12日 星期六

### 等巴士之謎

#### 3 則留言:

Pop 提到...

We can discuss this topic by using simple statistics tech .

Assume:
0)班次不準確但車子的數目不變
1) 乘客到達車站的時間是隨機的
Uniform distribution between the bus service time, that is no rush hour. (discrete or continuous doesn't matter)
2) All waiting passengers can take on the first arrival bus.
3) No passenger come to bus stop after the last bus departure.
4) For easier discussion, assume the last bus is on time.

Let have some examples to get some ideas.
i)

Draw a diagram, Y-axix is waiting time and X-axis is the time line.
We will have continues iso. triangles with height 20 and base 20.
Within each class, every 20 mins, the average is waiting time is 10 mins. So the overall mean waiting time is the average of all classes' mean. that is 10 mins.

ii)

Draw a diagram similar as i)
We will have consecutive iso. triangles with height 17 and 23.
so the classes' average waiting time is 17/2 and 23/2.
To calculate the overall mean waiting time, we need to use weighted mean as each classes width is diff.
For example, 40 分鐘可視為一個循環

the overall mean = (17/40)*(17/2)+(23/40)*(23/2)
=(17^2+23^2) / 2*(17+23)

Let Xi be the time in-between 2 bus arrived.
So the overall mean of waiting time
= Sum (Xi^2) / 2*Sum(Xi)

By using s^2=Sum(Xi-u)^2, u = mean of Xi, s = s.d. of Xi
we have s^2 = Sum Xi^2 -2u*Sum(Xi)+Sum(u^2)
s^2= Sum(Xi^2)-n(u^2)

Thus, overall mean waiting time
= [s^2 + n(u^2)] / 2*[n(u)]
= u/2 + s^2/2nu

As s^2(varanice) is positive, we draw the required conclusion.

if 巴士又永遠準時到站, i.e s^2 =0, and the 巴士又永遠準時到站 = u /2

Pop 提到...

Correction:
By using s^2=(1/n)[Sum(Xi-u)^2], u = mean of Xi, s = s.d. of Xi
we have n*s^2 = Sum Xi^2 -2u*Sum(Xi)+Sum(u^2)
n*s^2= Sum(Xi^2)-n(u^2)

Thus, overall mean waiting time
= [n*s^2 + n(u^2)] / 2*[n(u)]
= u/2 + s^2/2u

Kahoo 提到...

Thanks, Pop. Your idea of using statistical techniques is nice.