小白新来的，请教大神：为啥每次点击运行命令按钮，程序就无响应？
本人小白，谢谢大家的帮助~程序界面如下：
[attach]71219[/attach]
程序代码如下：
Private Sub Command1_Click()
Dim o, xi As Integer, g As Integer
Dim u11, u12 As Integer, u21, u22, theat1 As Integer
Dim q11, q12, q21, q22, q14, q15, theat2, h1, h2, n, sum1, sum2, sum, e As Integer
Dim t1(1 To 6) As Integer '2种方式的路径时间均值
Dim t2(1 To 6) As Integer
Dim d1(1 To 6) As Integer '2种方式的路径时间标准差
Dim d2(1 To 6) As Integer
Dim C1(1 To 7) As Integer '路段通行能力
Dim C2(1 To 7) As Integer
Dim t01(1 To 7) As Integer '路段自由流时间
Dim t02(1 To 7) As Integer
Dim b1(1 To 6) As Integer '2种方式的路径出行时间预算
Dim b2(1 To 6) As Integer
Dim p1(1 To 6) As Integer '2种方式的需求在每条路径上的分配概率
Dim p2(1 To 6) As Integer
Dim f1(1 To 6) As Integer '2种方式在每条路径上的流量
Dim f2(1 To 6) As Integer
Dim f(1 To 6) As Integer '每条路径上的总流量
Dim x(1 To 7) As Integer '每条路段上的流量
Dim ta1(1 To 7) As Integer '2种方式的路段出行时间均值
Dim ta2(1 To 7) As Integer
Dim da1(1 To 7) As Integer '2种方式的路段出行时间的方差
Dim da2(1 To 7) As Integer
Dim y(1 To 7) As Integer '路段辅助流量
Dim c(1 To 7) As Integer
C1(1) = 3000 '通行能力
C1(2) = 2600
C1(3) = 1200
C1(4) = 1200
C1(5) = 1200
C1(6) = 1200
C1(7) = 1200
C2(1) = 2500
C2(2) = 2500
C2(3) = 1500
C2(4) = 1500
C2(5) = 1500
C2(6) = 1500
C2(7) = 1500
t01(1) = 13 '自由流时间
t01(2) = 12
t01(3) = 9
t01(4) = 15
t01(5) = 10
t01(6) = 18
t01(7) = 14
t02(1) = 10
t02(2) = 16
t02(3) = 11
t02(4) = 13
t02(5) = 17
t02(6) = 14
t02(7) = 12
g = 1.29 '可靠度ρ=0.9时标准正态分布的分位数
theta1 = 1 '路径选择分散参数
theta2 = 0.2 '方式选择分散参数
q14 = 2000 '2个OD对的交通需求
q15 = 2500
h1 = 9 '2种方式的吸引力
h2 = 8
e = Exp(-2) '迭代精度
n = 1 '迭代次数
sum1 = 0
sum2 = 0
If Option1.Value = True Then '判断路段a是否在路径k上
o = 1
End If
If Option2.Value = True Then
o = 0
End If
'初始化
t1(1) = t01(1) + t01(4) '路径时间均值
t1(2) = t01(1) + t01(3) + t01(6)
t1(3) = t01(2) + t01(6)
t1(4) = t01(1) + t01(5)
t1(5) = t01(1) + t01(3) + t01(7)
t1(6) = t01(2) + t01(7)
d1(1) = 0 '路径时间标准差
d1(2) = 0
d1(3) = 0
d1(4) = 0
d1(5) = 0
d1(6) = 0
t2(1) = t02(1) + t02(4) '路径时间均值
t2(2) = t02(1) + t02(3) + t02(6)
t2(3) = t02(2) + t02(6)
t2(4) = t02(1) + t02(5)
t2(5) = t02(1) + t02(3) + t02(7)
t2(6) = t02(2) + t02(7)
d2(1) = 0 '路径时间标准差
d2(2) = 0
d2(3) = 0
d2(4) = 0
d2(5) = 0
d2(6) = 0
For i = 1 To 6 '出行时间预算
b1(i) = t1(i)
b2(i) = t2(i)
Next i
'OD对（r，s）上方式m的负效用
u11 = -1 / theta1 * Log(Exp(-theta1 * b1(1)) + Exp(-theta1 * b1(2)) + Exp(-theta1 * b1(3)))
u12 = -1 / theta1 * Log(Exp(-theta1 * b1(4)) + Exp(-theta1 * b1(5)) + Exp(-theta1 * b1(6)))
u21 = -1 / theta1 * Log(Exp(-theta1 * b2(1)) + Exp(-theta1 * b2(2)) + Exp(-theta1 * b2(3)))
u22 = -1 / theta1 * Log(Exp(-theta1 * b2(4)) + Exp(-theta1 * b2(5)) + Exp(-theta1 * b2(6)))
'OD对（r，s）上方式m的需求
q11 = q14 * Exp(-theta2 * (u11 - h1)) / (Exp(-theta2 * (u11 - h1)) + Exp(-theta2 * (u21 - h2)))
q12 = q15 * Exp(-theta2 * (u12 - h1)) / (Exp(-theta2 * (u12 - h1)) + Exp(-theta2 * (u22 - h2)))
q21 = q14 - q11
q22 = q15 - q12
For i = 1 To 3 '2种方式的需求在每条路径上的分配概率
p1(i) = Exp(-theta1 * b1(i)) / (Exp(-theta1 * b1(1)) + Exp(-theta1 * b1(2)) + Exp(-theta1 * b1(3)))
p2(i) = Exp(-theta1 * b2(i)) / (Exp(-theta1 * b2(1)) + Exp(-theta1 * b2(2)) + Exp(-theta1 * b2(3)))
Next i
For i = 4 To 6
p1(i) = Exp(-theta1 * b1(i)) / (Exp(-theta1 * b1(4)) + Exp(-theta1 * b1(5)) + Exp(-theta1 * b1(6)))
p2(i) = Exp(-theta1 * b2(i)) / (Exp(-theta1 * b2(4)) + Exp(-theta1 * b2(5)) + Exp(-theta1 * b2(6)))
Next i
For i = 1 To 3 '2种方式在每条路径上的流量
f1(i) = p1(i) * q11
f2(i) = p2(i) * q21
Next i
For i = 4 To 6
f1(i) = p1(i) * q12
f2(i) = p2(i) * q22
Next i
For i = 1 To 6 '每条路径上的总流量
f(i) = f1(i) + f2(i)
Next i
x(1) = f(1) + f(2) + f(4) + f(5) '每条路段上的总流量
x(2) = f(3) + f(6)
x(3) = f(2) + f(5)
x(4) = f(1)
x(5) = f(4)
x(6) = f(2) + f(3)
x(7) = f(5) + f(6)
Start:
For i = 1 To 7
ta1(i) = t01(i) + 0.15 * t01(i) * x(i) / C1(i) '路段出行时间的均值
ta2(i) = t02(i) + 0.15 * t02(i) * x(i) / C2(i)
da1(i) = (0.15 * t01(i) / C1(i)) ^ 2 * (t01(i) - ta1(i)) ^ 2 '路段出行时间的方差
da2(i) = (0.15 * t02(i) / C2(i)) ^ 2 * (t02(i) - ta2(i)) ^ 2
Next i
t1(1) = ta1(1) + ta1(4) '方式1在各路径出行时间的均值
t1(2) = ta1(1) + ta1(3) + ta1(6)
t1(3) = ta1(2) + ta1(6)
t1(4) = ta1(1) + ta1(5)
t1(5) = ta1(1) + ta1(3) + ta1(7)
t1(6) = ta1(2) + ta1(7)
d1(1) = Sqr(da1(1) + da1(4)) '方式1在各路径出行时间的标准差
d1(2) = Sqr(da1(1) + da1(3) + da1(6))
d1(3) = Sqr(da1(2) + da1(6))
d1(4) = Sqr(da1(1) + da1(5))
d1(5) = Sqr(da1(1) + da1(3) + da1(7))
d1(6) = Sqr(da1(2) + da1(7))
t2(1) = ta2(1) + ta2(4) '方式2在各路径出行时间的均值
t2(2) = ta2(1) + ta2(3) + ta2(6)
t2(3) = ta2(2) + ta2(6)
t2(4) = ta2(1) + ta2(5)
t2(5) = ta2(1) + ta2(3) + ta2(7)
t2(6) = ta2(2) + ta2(7)
d2(1) = Sqr(da2(1) + da2(4)) '方式2在各路径出行时间的标准差
d2(2) = Sqr(da2(1) + da2(3) + da2(6))
d2(3) = Sqr(da2(2) + da2(6))
d2(4) = Sqr(da2(1) + da2(5))
d2(5) = Sqr(da2(1) + da2(3) + da2(7))
d2(6) = Sqr(da2(2) + da2(7))
For i = 1 To 6 '出行时间预算
b1(i) = t1(i) + d1(i) * o
b2(i) = t2(i) + d2(i) * o
Next i
'OD对（r，s）上方式m的负效用
u11 = -1 / theta1 * Log(Exp(-theta1 * b1(1)) + Exp(-theta1 * b1(2)) + Exp(-theta1 * b1(3)))
u12 = -1 / theta1 * Log(Exp(-theta1 * b1(4)) + Exp(-theta1 * b1(5)) + Exp(-theta1 * b1(6)))
u21 = -1 / theta1 * Log(Exp(-theta1 * b2(1)) + Exp(-theta1 * b2(2)) + Exp(-theta1 * b2(3)))
u22 = -1 / theta1 * Log(Exp(-theta1 * b2(4)) + Exp(-theta1 * b2(5)) + Exp(-theta1 * b2(6)))
'OD对（r，s）上方式m的需求
q11 = q14 * Exp(-theta2 * (u11 - h1)) / (Exp(-theta2 * (u11 - h1)) + Exp(-theta2 * (u21 - h2)))
q12 = q15 * Exp(-theta2 * (u12 - h1)) / (Exp(-theta2 * (u12 - h1)) + Exp(-theta2 * (u22 - h2)))
q21 = q14 - q11
q22 = q15 - q12
For i = 1 To 3 '2种方式的需求在每条路径上的分配概率
p1(i) = Exp(-theta1 * b1(i)) / (Exp(-theta1 * b1(1)) + Exp(-theta1 * b1(2)) + Exp(-theta1 * b1(3)))
p2(i) = Exp(-theta1 * b2(i)) / (Exp(-theta1 * b2(1)) + Exp(-theta1 * b2(2)) + Exp(-theta1 * b2(3)))
Next i
For i = 4 To 6
p1(i) = Exp(-theta1 * b1(i)) / (Exp(-theta1 * b1(4)) + Exp(-theta1 * b1(5)) + Exp(-theta1 * b1(6)))
p2(i) = Exp(-theta1 * b2(i)) / (Exp(-theta1 * b2(4)) + Exp(-theta1 * b2(5)) + Exp(-theta1 * b2(6)))
Next i
For i = 1 To 3 '2种方式在每条路径上的流量
f1(i) = p1(i) * q11
f2(i) = p2(i) * q21
Next i
For i = 4 To 6
f1(i) = p1(i) * q12
f2(i) = p2(i) * q22
Next i
For i = 1 To 6 '每条路径上的总流量
f(i) = f1(i) + f2(i)
Next i
y(1) = f(1) + f(2) + f(4) + f(5) '路段上的辅助流量
y(2) = f(3) + f(6)
y(3) = f(2) + f(5)
y(4) = f(1)
y(5) = f(4)
y(6) = f(2) + f(3)
y(7) = f(5) + f(6)
For i = 1 To 7
c(i) = x(i) '将路段流量赋值给c（i）
x(i) = x(i) + (y(i) - x(i)) / n '加权后的路段流量
Next i
For i = 1 To 7
sum1 = sum1 + c(i)
sum2 = sum2 + (x(i) - c(i)) ^ 2
Next i
sum = Sqr(sum2) / (sum1 + 1)
If sum < e Then '停止迭代检验
GoTo Finish
Else
n = n + 1
End If
GoTo Start '返回第继续迭代
Finish:
Text1.Text = q11
Text2.Text = q21
Text3.Text = q12
Text4.Text = q22
Text5.Text = q11 / q14
Text6.Text = 1 - Text5.Text
Text7.Text = q12 / q15
Text8.Text = 1 - Text7.Text
Text9.Text = x(1)
Text10.Text = x(2)
Text11.Text = x(3)
Text12.Text = x(4)
Text13.Text = x(5)
Text14.Text = x(6)
Text15.Text = x(7)
Text16.Text = n
End Sub
[ 本帖最后由 梵蒂冈f3 于 2013-5-30 23:38 编辑 ]
