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Unformatted text preview: 5.15 {a} The function to be evaluated is {b} Using bisection. the ﬁrst iteration is os+2s
i3. =—=1.s
2 f{5.5)f(1.5) = —32.2552(—5.535945) = 5.995263 Therefore, the root is in the second interval and the lower guess is redeﬁned as I: = 1.5. The
second iteration is 2—1.5 .3 .
i =ﬂ=2 s = 1oo%=25% l' 2 ﬂ f{1.5)f{2) = —5.535946{5.551559) = —5.515524 Therefore, the root is in the ﬁrst interval and the upper guess is redeﬁned as x“ = 2. All the
iterations are displayed in the following table: l x. x. Jr“ Jr“ xr xr It:T 1 5.5 —32.2552 2.5 5.513532 1.5 —5.535545 2 1.5 —5.53555 2.5 5.513532 2 5.551555 25.55%
3 1.5 —5.53555 2 5.551555I 1.?5 5.3?5555 14.25%
4 1.5 —5.53555 1.?5 5.3?5555 1.525 5.25552? ?.55%
5 1.5 —5.53555 1.525 5.25552]Ir 1.5525 5.551555 4.55%
5 1.5 —5.53555 1.5525 5.55?555 1.53125 5.535251 2.54%
? 1.5 —5.53555 1.53125 5.535251 1.515525 5.553353 1.53%
5 1.5 —5.53555 1.515525 5.553353 1.55?5125 —5.513555 5.52% After eight iterations, we obtain a root estimate of 1.5028125 with an approximate error of
0.52%. {c} Using false position, the ﬁrst iteration is _ '3"
x. = 2.5 ——0'81303m'5 "5) = 2.451333 — 32.2582 — 0.81303 f{0.5)f(2.45083) = —32.25321{o.rsssv) = —2s.eo24s Therefore, the root is in the ﬁrst interval and the upper guess is redeﬁned as x” = 2.45083. The
second iteration is 0.2993? 0.5 — 2.45033 2.40363 — 2.45033
.35.. = 2.45083 ( ) = 2.40363 5“. = 100% = 1.96%
—32.2582l—0.79987 2.40353 f{0.5)f(2.40383) = —32.2582(0.?8812)= —2s.ssess The root is in the first interval and the upper guess is redefined as x” = 2.40363. All the iterations
are displayed in the following table: if x; in x" itLl itr Mr Ea 1 0.5 —32.2582 2.50000 0.81303 2.45083 0.?888? 2 0.5 —32.2582 2.45083 0.?888? 2.40383 0.?8812 1.88%
3 0.5 —32.2582 2.40383 0.?8812 2.35834 0.??1?8 1.82%
4 0.5 —32.2582 2.35834 0.??1?8 2.31482 0.?5888 1.88%
5 0.5 —32.2582 2.31482 0.?5888 2.2?331 0.?4145 1.83%
8 0.5 —32.2582 2.2?331 0.?4145 2.2334? 0.?254? 1.?8%
? 0.5 —32.2582 2.2334? 0.?254? 2.18534 0.?0800 1.?4%
8 0.5 —32.2582 2.18534 0.?0800 2.15888 0.88208 1.88%
8 0.5 —32.2582 2.15888 0.88208 2.12404 0.8?488 1.84%
10 0.5 —32.2582 2.12404 0.8?488 2080?? 0.85883 1.58% After ten iterations we obtain a root estimate of 2.0802? with an approximate error of 1.58%.
Thus. after ten iterations. the false position method is converging at a very slow pace and is still far from the root in the vicinity of 1.5 that we detected graphically. Discussion: This is a classic example of a case where false position performs poorly and is
inferior to bisection. Insight into these results can be gained by examining the plot that was
developed in part (a). This ﬁinction violates the premise upon which false position was
based—that is. if ﬁx”) is much closer to zero than ﬁn). then the root is closer to .35., than to so {recall
Figs. 5.8 and 5.9). Because of the shape of the present function. the opposite is true. 5.13 This problem can be solved by determming the root of the derivative of the elastic curve _= = ' (—5.¥4+6L212—L4) Therefore, after substituting the parameter 1values, we must determine the root of
ﬁx) = —5x4 + 2,15I:1.1:ro:1:c3 —1.2s5 ><1C111 = 5 A plot of the function indicates a root at about .1; = ETD. 2E+11
1E+11
ﬂ
1E+11
2E+11 Bisectiou can be used to determine the root. Here are the ﬁrst few iterations: i x; Jr" Mr Exit ﬂint flxdxﬂxr! 5,;
1 a 555 255 4.3215“ —1.4x1a‘“ 1.532152‘ 2 255 555 325 —1.4x15‘° 2.53515‘ﬂ —1.1x152‘ 33.33%
3 255 375 312.5 —1.4x1o‘° 3.3mm” 4.55152” 25.5551.
4 251] 312.5 251.25 —1.4x15‘° 5.52‘2113’*'I 4.42152” 11.1151.
5 251] 251.25 255.525 —1.4x15‘° 21211]9 2.5521519 5.55% After 20 iterations, the root is determined as x = 268.323. This 1aalue can be substituted into Eq.
(£35.13) to compute the maxiimun deﬂection as 2.5 1'=—(—I[2158.328)S + 22o,555(255.325)3 —1.295><15“{255325))= 43.51519
' 12a(55.505)3a,555(550) 6.2 (a) Fixed point. The equation can be solved in two ways. The 3'33: that converges is
1H = ‘“11.8th + 2.5 The resulting iterations are i x. .53 II] 5 ’1 3.391165 4?.4493
2 2.9332?4 ’15. 61 E33
3 2199246 5.16%
4 2.?423?9 ’1 .1’1 93
5 2.?26955 11.51%
6 2.?21959 11.19%
'1' 2.?201?4 11.06% 8 2.?19616 [1.02% Thus. after 3 iterations. the root estimate is 2.719616 with an approximate error of 0.92%. The
result can be checked by substituting it back into the original function. ft2.7’19616}=—(2.7'196l6)1 +1.3(2.T19616}+ 2.5 =—O.CDl
(I3) Fen‘ton—Raphson —.1f+1.33.I.+2.5 111 = '11— .
—le+1.8 1' X. ﬂxl ﬁx] £3 3 3 133 3.2 1 3.333333 221344 433732 43.3333
2 2.331332 3.33333 3.33233 13.2213
3 2.221133 3.33344 3.34222 2.3533
4 2.213341 3.1333 3.33333 3.3333
3 2.213341 3.4313 3.33333 3.333 After 5 iterations. the root estimate is 2.119341 with an approximate error of 0.00%. The result
can be checked by substituting it back into the oliginal function. J1I“(1.7’191341)=—[1.’.’191341}2 +1.3(2.213341}+ 2.5 = —'.*.33><13‘13 6.115 (a) The ﬁmetion to be evaluated is 'w
ﬂrﬂzim ﬂ +5_£ *1 —1s
12 11,1 J 12 The selutieu can be obtained with the £2 arc function as a} format lcng
a? fTa = @{Ta} TaflﬂecmshﬁECCfTa]+ETaf1215;
a} Ta = fzercﬁfTa,lUUC} Ta =
l.63436503£81737£e+003 (II) A plot of the cable can be generated as a} x = linspaEE{SD,1CC];
}} w = 12;yD = E;
a} v = Ta£w*:Dsh{w*xfTa] + y:  Taiw; ﬁb ﬁlctﬁx,y],grid 5
E] III E] 12 (a) The ﬁlnetien can be diffﬂentiated to give 1EE f'Ex)=—2x+8 This functi on can be set equal to zero and solved for x = 3.33 = 4. The derivative can be
differentiated to give the second derivative f" (I) = —2
Because this is negative, it indicates that the ﬁmction has a maximum atx = 4. (l1) Using Eq. 11o _i(2—o)"[o—o]—(2—s)1[o+12]_
a {a—o)[o—o]—(2—s)[o+12] _ 7.4 (a) The function can be plotted
4D 4D 3D 420 (la) The function can be diﬁei'entiated twice to give
ﬂu) = 451‘ — 24x3 Thus; the second derivative will always be negative and hence the ﬁinction is concave for all
values of .v. {c} Differentiating the function and setting the result equal to zero results in the following roots
problem to locate the maximum f'tx) :5 = —s.:.5 — 5x3 +12 A plot of this function can be developed 455 255 2 55 4155 A technith such as bisection can he employed to determine the root. Here are the ﬁrst few
iterations: iteration Jr; Jr" x, x. x, x; x Xr a,
1 5.55555 2.55555 1.55555 12 —5 —55.5555
2 5.55555 1.55555 5.55555 12 15.11 B?5 125.5255 155.55%
3 5.55555 1.55555 5.?5555 15.?15T5 5.455255 55.555? 33.33%
4 5.?5555 1.55555 5.5T555 5.455255 2.524445 13.131'1 14.25%
5 5.5T555 1.55555 5.53?55 2.524445 —1 .15555 —2.2453 5.5T% The approach can he continued to yield a result of I = 5.91592. 2.5 First, the golden ratio can he used to create the interior points. F_
a: “‘52 1(2—a)=12351 I1 = 5 +1.2351=1.2351
I: = 2 —1.2351 2 5.?539 The function can be evaluated at the interior points flﬁxz) = f{5.2639) = 3.1329
f{.r1)=f{1.2351)=4.5142 Because ﬁn) ﬁﬂxl). the maximum is in the mtei'val deﬁned 11v 9.]. x}. and x1 .where x; is the
optimum. The error at this point can be computed as 2 — 0
0.31639 x100% =100% as = (1— 0.61803)‘ For the second iteration, .1]: ID and x“ = 1.2361. The fmmei' I} value becomes the new x1. that is.
1'1 = 0.?639 and ﬂxl) = 8.18?9. The new values of d and 1'; can be computed as E d = 1 _1(1.2351—0) = 03539 2
.153 = 1.2361 — 03639 = 0.4721 The function evaluation at ﬂxg) = 5.5496. Since this value is less than the fmetion value at 11. the
maximum is in the interval prescribed by :53, 1'1 and x”. The process can be repeated and all three
iterations summarized as 1 X1 X1 X2 X2 X1 JI'1 Jru Jl'u 4' Na: «Ea
1 0.0000 0.0000 0.7039 3.1379 1.2351 4.8142 2.0000 —104.0000 1.2351 0.?539 100.00% 2 [1.80138 0.0001] 0.4?21 5.5498 0.?638 3.13.?9 1.2361 4.3142 0.?638 0.?638 61.88%
3 0.4?21 5.5486 0153.9 3.1378 [1.9443 3.5773 1.2351 4.3142 13.4721 118443 30.80% 16 First, the function values at the initial values can be evaluated ﬁll) 2110) = 0
ﬂxﬂ =ﬂ1) = 35
51913) :11?) = 4'34 and substituted into Eq. (71.10) to give. 3; _l_i(1—0)1[8.5 +104]—(1 — 2)3[8.5 —o)] _
' 4 _ 2 (1 0)[8.5 : 104] (1 2)[a.5 0)] _
which has a function value of ﬂD.5TDE48) = 6.5199. Because the ﬁinction value for the new point is lower than for the intermediate point {3(3) and the new x value is to the left of the mteimediate
point. the lower guess {x1} is discarded. Therefore. for the next iteration. 0.520348 ﬂxl) =ﬁo.5roz4a} = 5.45799
ﬂxr) =ﬂ1) = 35
ﬂxg) =ﬂ2) = —104 which can be substituted into Eq. {1.18) to give .154 = 11.813431, which has a function value of
ﬂ0812431) = 8.446523. At this point, an approxmtate error can be computed as _ a
Ea = 0.81243 tit510443 ﬂat1943:3318;
0.81243 The process can be repeated: with the results tabulated below: if x1 {liq} x2 5le x3 [1x31 14 3x4} 5...
0.00000 0.00000 1.00000 8.50000 2.0000 —104 0.5025 0.5?091 1
2 0.5T025 6.5?801 1.00000 8.50000 2.0000 104 0.31243 8.44052 20.81%
3 0.81243 8.44652 1.00000 8.50000 2.0000 104 0.00??? 8.805T5 10.50% Thus. after 3 iterations, the result is converging on the true value of ﬁx) = 8.69193 at x = 0.91692. 2.28 This is a trick question. Because of the presence of (1 — a) in the denominator. the function
Ta‘ill experience a division by zero at the maximum. This can be rectiﬁed by merely canceling the
(1— s) teims in the numerator and denominator to give 15.; 1"=——1———————
45' —33+4 Any of the optimizers described in this section can then he used to determine that the maximum
of T: 3 occurs ats =1. }} format lcng
F} T=@[S] —[:515f[413A2—313+4]];
}} [smin,Tmin]=fminbnd{C,0,4}
3min = 0.99338407348109
Cmin = 2.98338998339122 2.32 Here is a diagram for this problem: 0'1 £1'1 5' = tan — (1) P=a+2aw1+a1 (2 .«1=(l11 +sat}oT (3}
bl = bl + End (4)
We can solve Eq. 3 for :51. reﬁne (s) and substitute the result into Eq. 2 to give. A u 3 '
P=E+d[2vl+s —.r] (5) Given a value for A. we can determine the values of al and .s for this two—dimensional ﬂinction, F} P=E[x] 50fx(1}+x(1}*[2*3qrt[1+x{2].A2}x(2}};
e} [xmin,fmin]=fminsear:h(P,[5,1]} xmin = 5.3725 ChETTB
fmin = 18.61:]. The optimal parameters are d = 5.339 and .s = 13.5??3. Using Eq. 1 gives 1 . 130°
1 = inane radians x
users fr H = tan‘ = 60° Thus. this speciﬁc application indicates that a 60° angle yields the minimum wetted perimeter.
The other dimensions can be computed as (Eqs. 5 and 4). _ so
_ asses bl —D.5?T3(5.3T29) = 5.2041 323 = 5.21341 + 2(ﬂ.5??3)(5.3?29) 211413313? The verification of wheﬂier this result is universal can be attained mductivelv or deductivelv. The
inductive approach involves trying several different desired areas in conjunction with our
MTLAB solution. As long as the desired area is greater than D, the result for the optimal design
will be fit)". The deductive veriﬁcation involves calculus. If both A and d are constants and sis a variable the
condition for the minimum perimeter is dPMs = D. Differentiating Eq. 6 with respect to s and
setting the resulting equation to zero, d_P_
d5— —1]=u (4) The term in parentheses can be solved for S 21.51,"; Using Eq. l= this conespends to 5': 6D”. ...
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 Spring '09
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