Question
Transcribed Text
Using MATLAB.
Problem 1 (reference book Anton & Rorres, chapter 10, page 535, section 10.2, technology exercises).
Using MATLAB, solve the linear programming problem:
Consider the feasible region consisting of 0≤x, 0≤y along with the set of inequalities
x cos((2k+1)π/4n)+y sin((2k+1)π/4n)≤cos(π/4n)
for k=0,1,2,⋯,n-1
Maximize the objective function
z=3x+4y
assuming that (a) n=1, (b) n=2, (c) n=3, (d) n=4, (e) n=5, (f) n=6, (g) n=7, (h) n=8, (i) n=9, (j) n=10, and (k) n=11.
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Solution 1> n=1;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
the dual appears to be infeasible (and the primal unbounded).
(The primal residual < TolFun=1.00e-008.)
x =
1.0e+012 *
-0.0000
-3.3538
fopt =
-1.3415e+013
>> n=2;
>> f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
??? end
|
Error: Illegal use of reserved keyword "end".
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
the dual appears to be infeasible (and the primal unbounded).
(The primal residual < TolFun=1.00e-008.)
x =
1.0e+010 *
-2.3918
0.0001
fopt =
-7.1750e+010
>> n=3;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
the dual appears to be infeasible (and the primal unbounded).
(The primal residual < TolFun=1.00e-008.)
x =
1.0e+008 *
-0.0019
-1.4561
fopt =
-5.8301e+008
>> n=4;
>> for i=0:n-1
f=[3;4];b=[cos(pi/(4*n))];A=[cos((2*i+1)*pi/(4*n)) sin((2*i+1)*pi/(4*n))];
end
>> [x,fopt]=linprog(f,A,b)
Exiting: One or more of the residuals, duality gap, or total relative error
has stalled:
the dual appears to be infeasible (and the primal unbounded).
(The primal residual < TolFun=1.00e-008.)
x =
1.0e+008 *
-3.0227
-0.6135
fopt =
-1.1522e+009...
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