1、4-2已知某汽車行駛速度x與每公里耗油量的函數關系為f(x)=x + 20/x,試用0.618法確定速度x在每分鐘0.21公里時的經濟速度 x*。精度£ =0.01.Private Sub Comma nd1_Click() a = 0.2 b = 110: a1 = b - 0.618 * (b - a) a2 = a + 0.618 * (b - a) f1 = a1 + 20 / a1 f2 = a2 + 20 / a2If f1 < =f2 The n b = a2: a2 = a1: f2 = f1: a1 = b - 0.618 * (b - a): f1 = a1

2、 + 20 / a1If f1 > f2 The n a = a1: a1 = a2: f1 = f2: a2 = a + 0.618 * (b - a): f2 = a2 + 20 / a2 If Abs(b - a) >= 0.01 The n GoTo 10Prin t "a*=" (a + b) / 2t = (a + b) / 2Prin t "f*=" t + 20 / tEnd Sub運行結果:. 996749540627092 f*= 21.D6igT0T2B9O15最終結果:a*=0.996749540627092f*=21

3、.061970726901515-1試用變量輪換法求目標函數f(X) = 4 +29x12242、-4X2 + X1 + 2x2 - 2X1X2 + X1 - 2x1 X2 的最優解。初始2 -點 X(0) = -2,2.2 T 精度 & = 0.000001。Private Sub Comma nd1_Click()n = 2: E = 0.000001Dim X(2), X0(2), P(2, 2)Forj = 1 To 2X0(j) = InputBox(X0(j)," 輸入 X0 的 x、y 坐標") X(j) = X0(j)Next jGoSub 630K

4、 = 0Fori = 1 To 2Forj = 1 To 2P(i, j) = 0If i = j The n P(i, j) = 1Next j, i260: F0 = FFori = 1 To 2GoSub 470Next iR = 0: K = K + 1: Print FForj = 1 To 2R = R + (X(j) - X0(j) A 2Next jR = Sqr(R)If R < E The n GoTo 420i = nGoSub 470Forj = 1 To 2X0(j) = X(j)Next jGoTo 260470: H = 1: FX = 0 480: GoS

5、ub 600F1 = F: FX = FX + 1If F1 > F0 And FX > 1 The n GoTo 540If F1 > F0 And FX = 1 The n H = -HF0 = F1: H = 2 * HGoTo 480540: H = -0.5 * HIf 10000 * Abs(H) < E The n Return560: GoSub 600F0 = F1: F1 = FIf F1 < F0 The n GoTo 560GoTo 540600: For j = 1 To 2 X(j) = X(j) + H * P(i, j) Next

6、j 630: F = 4 + 2 / 9 * X(1) - 4 * X(2) + X(1) A 2 + 2 * X(2) A 2 - 2 * X(1) * X(2) + X(1) A 4 - 2 * X(1) A 2 * X(2)Return420: For j = 1 To 2Prin t "x*(" j; ")=" X(j)Next jPrint K; "f*=" FEnd Sub運行結果如下：-5 146848195091.21 E5T2T72755033-7. 44934968051013-7.8431&4622B51

7、43-6 0323078783148-8.12261795B5S9S5-& 16448497469126-8.1S3ST24039145-8 192802730968-8. 1K901T9229461-6 1987760776023-8 19963709547631-0.20002906424000-B.20D20T9597DTSB-6. 2COZ3956T54O42-8 20032678315644-6.2QO343750TTE&S-B 200351485B7215 2C03550U561&-8 20D356&1361345-6.20035735104019-

8、6 £0035763437559-0.2OO35703TO22T7-8 £0035790634650-6.20035793T961&1-8 2C035796236501-8.20035795892702-8 2003579619191-6.2OO3579632T0&3-8 2003579639062-8.2C03579641B4充-6.20035756431165-8.20035796437049-8 20035796439656-8.20035796440939-8.2003579644145-8.20035796441712x*( 1 )= 2.3141

9、631630715 x*( 2 )= 4. 33475T197E112T3720035796441712最優解：X1*= 2.314163X2*=4.834757F* = -8.200358最小體積二級圓柱齒輪減速器的最優設計如圖所示的二級斜圓柱齒輪減速器，咼速軸輸入功率Pi = 4.5kw，高速軸轉速ni = 1450rpm ,總傳動比i 2= 31.5，此輪的齒寬系數 “ a= 0.4 ;齒輪材料和熱處理大齒輪45號鋼正火HB = 187207 ,小齒輪45號鋼調質HB = 228255.總工作時間不少于 10年。要求按總中心距 a三最小來確定總體 方案中的各個主要參數。減速器的總中心距計

10、算公式為=ai + a2 =12cos 3mnlZi(1 + ii) + mn2Z3(1 + i2)式中mn1, mn2 高速級與低速級的齒輪法面模數，mmi1, i2高速級與低速級傳動比Z1 , Z3高速級與低速級小齒輪齒數3齒輪的螺旋角1. 選取設計變量計算總中心距涉及的獨立參數有，故取X = mn1, mn2, Z1 , Z3, i1,3 = X1,X2,X3,X4,X5,X6】T2. 建立目標函數f(X) = X1X3 (1+X5) + X2X4 (1 + 31.5/ X5)/(2COS X6)3. 確定約束條件(1)確定約束條件的上下界限 從傳遞功率與轉速可估計2< mn1&l

11、t; 5標準值(2, 2.5, 3, 4, 5)2< mn2 6標準值(3.5, 4, 5, 6)綜合考慮傳動平穩、軸向力不可太大，能滿足短期過載，高速級與低速級大齒輪浸油深度大 致相近，軸齒輪的分度圓尺寸不能太小等因素，?。?4W Z1< 2216W Z3< 225 -5.8< il< 78°w B 三 15°6 -由此建立12個不等式約束條件式# -gi（X） = xi -2 > 0g2（X） = 5 -xi > 0g3（ X）= X2 3.50g4（ X） = 6 -X2 > 0g5（X） = X3 -14> 0g

12、6（X） = 22 -X3> 0g7（X） = X4 T60g8（X） = 22 -X4> 0g9（ X ）= X5 -5.80gi0（X） = 7 -X50 gii（X） = X6 -0.13960gi2（X） = 0.2618 -冷0 （已將角度化成弧度）# -# -(2)按齒面接觸強度公式# -# -925& = T3心皿 <SH , N/mm2bi得到高速級和低速級齒面接觸強度條件分別為233.和mni Zi ii-cos3 f 028(925) K訂i%2mn23Z33i2 da8(925) 2K2T2式中，S許用接觸應力，MpaTi,T2分別為高速軸I和中

13、間軸II的轉矩，N mmKi,K2分別為高速級和低速級載荷系數.(3)按輪齒彎曲強度計算公式-cos301.5 KiTibdi mniyi2< Sf i, N mm# -yi2Si W S 2, N mmy2得到高速級和低速級大小齒輪的彎曲強度條件分別為7 -# -S 1 帕 yi3 KiTiS 2 % y23 KiTi卡3幅y33 K2T232（1 + ii） mni Zi32（1 + ii） mni Zi32（1 + i2）mn2 Z32-cos f02-cos f02-cos0S 4 ©a y43 K2T2其中Si, S2, S3, S4分別為齒輪 1, 2, 3, 4的

14、許用彎曲應力， N/mm2;分別為齒輪1, 2, 3, 4的齒形系數.32（1 + i2）mn2 Z32-cos B02,yi , y2, y3, y4# -(4)按高速級大齒輪與低速軸不干涉相碰的條件a2 -E -de/2 >0得mn2Z3(1 + i2)-2 COS gE + mni) TlnlZiii0式中E低速軸軸線與高速級大齒輪齒頂圓之間的距離，mm;de2高速級大齒輪齒的齒頂圓直徑，mm.對式至代入有關數據：雷=518.75 N mm22& 1=劉 3= 153.5 N mm ,沖2= Sf 4= 141.6 N mmTi =29638 N mm, T2 = 2874

15、9i1 N mmK1 = 1.225 , K2 = 1.204y1=0.248, y2=0.302 , y3=0.256, y4=0.302E = 50mm得g13(X) = 4.331 10X13X33X5 -COS3X6 > 0g14(X) = 1.431 10X23X43 -X5COS3X6 > 0g15(X) = 1.3981農(1+X5)X13X32-cos2X6> 0g18(X) = 1.57010(1+X5)X13X32-COS2X6> 0-43 222g16(X) = 1.514 10X(31.5 + X5)X2 X4 -X5 COS X6 > 0

16、-43222g19(X) = 1.647 1CK(31.5 + X5)X2X4 -X5 COS X6 > 02g17( X ) =X2X4 (31.5 + X5) -2X5COSX6(X1+5O) -C1X3X50g18(X)、g19(X)和g15(X卜g16(X)相比為明顯的消極約束，可省略。共取g1(X)至g17(X)的17個約束條件。4. 選用合適的算法求解這一約束問題采用復合形法求解。在進行優化的過程中，6個變量都是作為連續變量處理的，因為齒輪的齒數應為整數，模數應取標準模數，所以最后對結果進行適當調整。Private Sub Comma nd1_Click()N = 6: E

17、= 0.01Dim X(13, 6), F(13), A(6), B(6)Forj = 1 To 6A(j) = In putBox(A(j),"輸入估計邊界的下界”)B(j) = InputB0X(B(j),"輸入估計邊界的上界 ”)Next j180: For I = 1 To 12190: GoSub 730If AA = 0 The n GoTo 190NeXt II0 = 1: I1 = 1: I2 = 1: K = 0240: For I = 1 To 12If F(I) > F(I0) The n I1 = I0: I0 = I: GoTo 280If

18、F(I) > F(I1) The n I1 = IIf F(I) < F(I2) The n I2 = I280: Next IK = K + 1: R = 0Forj = 1 To 6R = R + (X(IO, j) - X(I2, j)人 2Next jIf Sqr(R) < E The n ZZ = 0: GoTo 670340: For j = 1 To 6X(0, j) = 0For I = 1 To 12X(0, j) = X(0, j) + X(I, j)Next IX(0, j) = (X(0, j) - X(I0, j) / 11Next jI = 0Go

19、Sub 760If AA = 1 The n GoTo 480Forj = 1 To 6A(j) = X(I2, j): B(j) = X(0, j): X(1, j) = X(I2, j)Next jGoTo180480: H = 1.3: I = 13490: For j = 1 To 6X(13, j) = X(0, j) + H * (X(0, j) - X(10, j)Next jGoSub 760H = H / 2If AA = 0 The n GoTo 490If F(13) >= F(I1) The n GoTo 610Forj = 1 To 6X(I0, j) = X(

20、13, j)Next jF(I0) = F(13)GoTo 240610: If H > E The n GoTo 490730: For j = 1 To 6X(I, j) = A(j) + Rn d(1) * (B(j) - A(j)Next j760: AA = 0g1 = X(I, 1) - 2g2 = 5 - X(I, 1)g3 = X(I, 2) - 3.5g4 = 6 - X(I, 2)g5 = X(I, 3) - 14g6 = 22 - X(I, 3)g7 = X(I, 4) - 16g8 = 22 - X(I, 4)g9 = X(l, 5) - 5.8g10 = 7 -

21、 X(l, 5)g11 = X(l, 6) - 0.1396g12 = 0.2618 - X(l, 6)g13 = 4.331 *10 A-7 * X(l, 1) A 3 *X(l, 3) A 3 * X(l, 5) - Cos(X(l, 6) A 3g13 = 1.431 *10 a-5 * X(l, 2) a 3 *X(l, 4)卜 3 - X(l, 5) * Cos(X(l, 6)卜 3g15 = 1.398 * 10 a -4 * (1 + X(l, 5) * X(l, 1) a 3 * X(l, 3) a 2 - Cos(X(l, 6) a 2g16 = 1.514 *10 a-4

22、* (31.5 + X(l,5) * X(l, 2) a 3 * X(l, 4) a 2 - X(l, 5) a 2*Cos(X(l,6) a 2g17 = X(l, 2) *X(l,4) * (31.5 + X(l,5) - 2 * X(l, 5) * Cos(X(l, 6) * (X(l, 1)+ 50) -X(l,1) * X(l,3) * X(l, 5) a 2lf g1 >= 0 And g2 >= 0 And g3 >= 0 And g4 >= 0 And g5 >= 0 And g6 >= 0 And g7 >= 0 And g8 >= 0 And g9 >= 0 And g10 >= 0 And g11 >= 0 And g12 >= 0 And g13 &