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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 37 "Fitting Data Using the Si
mplex Method" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "?simplex" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Question mark command in MAPLE gi
ves you a help window for the command.  This is VERY " }}{PARA 0 "" 0 
"" {TEXT -1 49 "useful assuming you know the name of the command." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(simplex):" }}{PARA 7 "
" 1 "" {TEXT -1 87 "Warning, the protected names maximize and minimize
 have been redefined and unprotected\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 95 "You must load the package simplex in order to access the \+
routines necessary to use this method." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 17 "Sample Proble
m #1" }}{PARA 0 "" 0 "" {TEXT -1 101 "Suppose we want to find the best
 linear fit  y = mx + b to the data \{ (1,2), (2,3.5), (3,4), (4,6) \}
 " }}{PARA 0 "" 0 "" {TEXT -1 97 "using the Chebyshev Approximation Cr
iterion (ie. minimize the largest absolute deviation between " }}
{PARA 0 "" 0 "" {TEXT -1 102 "the data and the line.)  The 4 absolute \+
deviations are |2 - (m + b)|, |3.5 - (2m + b)|, |4 - (3m + b)|" }}
{PARA 0 "" 0 "" {TEXT -1 104 "and |6 - (4m + b)|.  We want to find the
 values of m and b which make the largest of these 4 deviations " }}
{PARA 0 "" 0 "" {TEXT -1 21 "as small as possible." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
19 "m := 1.25:  b := 1:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 78 "r := max
(abs(2 - m - b),abs(3.5 - 2*m - b),abs(4 - 3*m - b),abs(6 - 4*m - b));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rG$\"#v!\"#" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 96 "The command above \"max\" is used to calculate \+
the maximum of the absolute deviations.  Note that " }}{PARA 0 "" 0 "
" {TEXT -1 98 "\"abs\" is the command for the absolute value.  Given a
 particular m and b value, the above command " }}{PARA 0 "" 0 "" 
{TEXT -1 103 "quickly calculates the maximum absolute deviation.  This
 is useful for checking results and estimating " }}{PARA 0 "" 0 "" 
{TEXT -1 10 "best fits." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 101 "This command is necessary to clear the v
alues of m, b and r so you can use them as variables below.  " }}
{PARA 0 "" 0 "" {TEXT -1 101 "This command basically is like restartin
g MAPLE.  Be careful when you use it.  You will need to load " }}
{PARA 0 "" 0 "" {TEXT -1 41 "the simplex package again if you restart.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 101 "We  \+
let  r = maximum absolute deviation.  We wish to minimize r subject to
 the constraints that each " }}{PARA 0 "" 0 "" {TEXT -1 104 "absolute \+
deviation is less than or equal to r.  This is a clever way of avoidin
g the calculation of the " }}{PARA 0 "" 0 "" {TEXT -1 100 "maximum dev
iation.  In other words, if one of the absolute deviations is bigger t
han r, then that r " }}{PARA 0 "" 0 "" {TEXT -1 76 "is inadmissible be
cause r is supposed to be the largest of the deviations.  " }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 100 "There will be 4 \+
constraints in this case, which after removing the absolute value lead
s to 8 linear " }}{PARA 0 "" 0 "" {TEXT -1 107 "constraints.  For exam
ple,  |2 - m - b| <= r  can be written as the two inequalities  r - 2 \+
+ m + b >= 0   " }}{PARA 0 "" 0 "" {TEXT -1 106 "and r + 2 - m - b >= \+
0.  Both of these must be satisfied simultaneously.  We list the 8 con
straints below:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 126 "cnsts :
= \{r+m+b >= 2, r-m-b >= -2, r+2*m+b >= 3.5, r-2*m-b >= -3.5, r+3*m+b \+
>= 4, r-3*m-b >= -4, r+4*m+b >= 6, r-4*m-b >= -6\};" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%&cnstsG<*1\"\"#,(%\"rG\"\"\"%\"mGF*%\"bGF*1!\"#,(F)
F*F+!\"\"F,F01$\"#NF0,(F)F**&F'F*F+F*F*F,F*1$!#NF0,(F)F**&F'F*F+F*F0F,
F01\"\"%,(F)F**&\"\"$F*F+F*F*F,F*1!\"%,(F)F**&F?F*F+F*F0F,F01\"\"',(F)
F**&F<F*F+F*F*F,F*1!\"',(F)F**&F<F*F+F*F0F,F0" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "obj
 := r;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$objG%\"rG" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 62 "The objective function is simply r, which
 we want to minimize." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "minimize(obj,cnsts);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#<%/%\"rG$\"++++]P!#5/%\"mG$\"++++]7!\"*/%\"bG$\"
++++]iF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 99 "Thus we see that a slope of m = 1.25 and \+
a y-int. of b = 0.625 will give us the best fit using the " }}{PARA 0 
"" 0 "" {TEXT -1 95 "Chebyshev criterion.  In this case the maximum ab
solute deviation is only 0.375.  Now that you " }}{PARA 0 "" 0 "" 
{TEXT -1 96 "have an answer, you can go back above and plug in values \+
near these m and b values and see that " }}{PARA 0 "" 0 "" {TEXT -1 
44 "the maximum deviation does in fact increase." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 5 "" 0 "" {TEXT -1 28 "Textbook Example pp. 108-109
" }}{PARA 0 "" 0 "" {TEXT -1 108 "Here we use the constraints and vari
ables from the textbook example and obtain the answer given in the tex
t." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 98 "cnsts := \{r-x1+13 >= 0, r+x1-13 >= 0, r-x2+7
 >= 0, r+x2-7 >= 0, r-x1-x2+19 >= 0, r+x1+x2-19 >= 0\};" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%&cnstsG<(1\"\"!,(%\"rG\"\"\"%#x1G!\"\"\"#8F*1F
',(F)F*F+F*F-F,1F',(F)F*%#x2GF,\"\"(F*1F',(F)F*F2F*F3F,1F',*F)F*F+F,F2
F,\"#>F*1F',*F)F*F+F*F2F*F8F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "obj := r;" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%$objG%\"rG" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 32 "minimize(obj,cnsts,NONNEGATIVE);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#<%/%#x2G#\"#?\"\"$/%#x1G#\"#QF(/%\"rG#\"\"\"F(" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 136 "The option NONNEGATIVE is used to
 restrict the variables to being positive, which according to the exam
ple in question is the case here." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 5 "" 0 "" {TEXT -1 17 "Classroom Example" }}{PARA 0 "" 0 
"" {TEXT -1 93 "Here is the example presented in class about fitting a
 line of the form y = mx to the points " }}{PARA 0 "" 0 "" {TEXT -1 
98 "(1,0), (2,3), (3,6).  For this problem, it is straight-forward to \+
see geometrically that the best " }}{PARA 0 "" 0 "" {TEXT -1 24 "slope
 should be m = 3/2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 91 "cnsts := \{r + m >= 0, r - m >=0, r + 2*m
 >= 3, r - 2*m >= -3, r + 3*m >= 6, r - 3*m >= -6\};" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#>%&cnstsG<(1\"\"!,&%\"rG\"\"\"%\"mGF*1F',&F)F*F+!\"
\"1\"\"$,&F)F**&\"\"#F*F+F*F*1!\"$,&F)F**&F3F*F+F*F.1\"\"',&F)F**&F0F*
F+F*F*1!\"',&F)F**&F0F*F+F*F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "obj := r;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$objG%\"rG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "minimize(obj,cnsts);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#<$/%\"rG#\"\"$\"\"#/%\"mGF&" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 9 "m := 1.5:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "max(
abs(-m),abs(3-2*m),abs(6-3*m));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"#:!\"\"" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 80 "Our answer agrees with what we know to be true from \+
the geometry of the problem." }}}}{MARK "31 1 0" 78 }{VIEWOPTS 1 1 0 
1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }