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EGR 599 Advanced Engineering Math II _____________________ LAST NAME, FIRST Problem set #4
1. (P. 15.1 Chapra1) A company makes two types of products, A and B. These products are
produced during a 40-hour work week and then shipped out at the end of the week. They require
20 and 5 kg of raw material per kg of product, respectively, and the company has access to
10,000 kg of raw material per week. Only one product can be created at a time with a production
times for each of 0.05 and 0.15 hrs, respectively. The plant can only store 550 kg of total
product per week. Finally, the company makes profits of $45 and $30 on each unit of A and B,
respectively.
(a) Set up the linear programming problem to maximize profit.
(b) Solve the linear programming problem graphically.
(c) Solve the linear programming problem with the simplex method.
(d) Solve the linear programming problem with Matlab.
Solution
(a) Set up the linear programming problem to maximize profit.
Define xa = amount of product A produced, and xb = amount of product B produced. The objective function is to maximize profit,
P x xa b 45 30
Subject to the following constraints
20 5 10000x xa b {raw materials}
0 05 015 40. .x xa b {production time}
x xa b, 0 {positivity}
(b) To solve graphically, the constraints can be reformulated as the following straight lines
x xb a 2000 4 {raw materials}
x xb a 266 667 0 3333. . {production time}
x xb a 550 {storage}
The objective function can be reformulated as
x P xb a ( / ) .1 30 15
1 Chapra and Canale, Numerical Methods for Engineers , Mc-Graw Hill, 4 th Edition, 2002
The constraint lines can be plotted on the xb-xa plane to define the feasible space. Then the objective function line can be superimposed for various values of P until it reaches the boundary. The result is P 23700 with xa 483 and xb 67. Notice also that material and storage are the binding constraints and that there is some slack in the time constraint.
0
100
200
300
0 200 400 600P
= 15000
P = 5000
P = 23700
xb
xa
time
storage
material
optimum
(c) The simplex tableau for the problem can be set up and solved as
Basis P xa xb S1 S2 S3 Solution InterceptP 1 -45 -30 0 0 0 0
material S1 0 20 5 1 0 0 10000 500time S2 0 0.05 0.15 0 1 0 40 800storage S3 0 1 1 0 0 1 550 550
Basis P xa xb S1 S2 S3 Solution InterceptP 1 0 -18.75 2.25 0 0 22500
xa xa 0 1 0.25 0.05 0 0 500 2000time S2 0 0 0.1375 -0 1 0 15 109.0909storage S3 0 0 0.75 -0.05 0 1 50 66.66667
Basis P xa xb S1 S2 S3 Solution InterceptP 1 0 0 1 0 25 23750
xa xa 0 1 0 0.067 0 -0.333 483.33333time S2 0 0 0 0.007 1 -0.183 5.8333333xb xb 0 0 1 -0.07 0 1.333 66.666667
(d) Solve the linear programming problem with Matlab.
The problem is reformulated using slack variable S1, S2, and S3
Minimize y = 45xa 30xb
Subject to
20xa + 5xb + S1 = 10000 0.05xa + 0.15xb + S2 = 40xa + xb + S3 = 550
xa, xb, S1, S2, and S3 0
The arguments for the Matlab command X=LINPROG(f, A, b, Aeq, beq, LB, UB) are given as
f = , A = 0, b = 0, Aeq = , beq = , LB = , UB =
--------------------------------------------------------------------
% Set 4, Problem 1
f=[-45;-30;0;0;0];
A=zeros(5,5);
b=zeros(5,1);
Aeq=[20 5 1 0 0 ;0.05 0.15 0 1 0;1 1 0 0 1;0 0 0 0 0;0 0 0 0 0];
beq=[10000;40;550;0;0];
LB=[0;0;0;0;0];UB=[inf;inf;inf;inf;inf];
x=linprog(f,A,b,Aeq,beq,LB,UB)
>> s4p1
Optimization terminated successfully.
x =
483.3333
66.6667
0.0000
5.8333
0.0000
------------------------------------------------------The solution from the Matlab program is
xa = 483.3333
x2 = 66.6667
S1 = 0
S2 = 5.8333
S3 = 0
2. (P. 15.2 Chapra)A gas processing plant receives a fixed amount of raw gas each week. The raw gas is processed into three grades of heating gas, regular, premium, and supreme quality. Their production involves both time and on-site storage constraints. For example, only one of the grades can be produced at a time, and the facility is open for only 80 hrs/week. There is limited on-site storage for each of the products. All these factors are listed in Table 4-1 with a metric ton denoted tonne equal to 1000 kg.
Table 4-1Product
Resource Regular Premium Supreme AvailabilityRaw gasProduction timeStorage
7 m3/tonne10 hr/tonne9 tonnes
11 m3/tonne8 hr/tonne6 tonnes
15 m3/tonne12 hr/tonne5 tonnes
154 m3/week80 hr/week
Profit 150/tonne 175/tonne 250/tonne
(a) Set up the linear programming problem to maximize profit.
(b) Solve the linear programming problem with the simplex method.
(c) Solve the linear programming problem with Matlab.
Solution
(a) Set up the linear programming problem to maximize profit.
The total LP formulation is given by
Maximize Z x x x 150 175 2501 2 3 {Maximize profit}
subject to
7 11 15 1541 2 3x x x {Material constraint}10 8 12 801 2 3x x x {Time constraint}x1 9 {“Regular” storage constraint}
{“Premium” storage constraint}{“Supreme” storage constraint}
x x x1 2 3 0, , {Positivity constraints}
(b) The simplex tableau for the problem can be set up and solved as
Basis P x1 x2 x3 S1 S2 S3 S4 S5 Solution InterceptP 1 -150 -175 -250 0 0 0 0 0 0S1 0 7 11 15 1 0 0 0 0 154 10.2667S2 0 10 8 12 0 1 0 0 0 80 6.66667S3 0 1 0 0 0 0 1 0 0 9 S4 0 0 1 0 0 0 0 1 0 6 S5 0 0 0 1 0 0 0 0 1 5 5
Basis P x1 x2 x3 S1 S2 S3 S4 S5 Solution InterceptP 1 -150 -175 0 0 0 0 0 250 1250S1 0 7 11 0 1 0 0 0 -15 79 7.18182
S2 0 10 8 0 0 1 0 0 -12 20 2.5S3 0 1 0 0 0 0 1 0 0 9 S4 0 0 1 0 0 0 0 1 0 6 6x3 0 0 0 1 0 0 0 0 1 5
Basis P x1 x2 x3 S1 S2 S3 S4 S5 Solution InterceptP 1 68.75 0 0 0 21.88 0 0 -12.5 1687.5S1 0 -6.75 0 0 1 -1.375 0 0 1.5 51.5 34.3333x2 0 1.25 1 0 0 0.125 0 0 -1.5 2.5 -1.66667S3 0 1 0 0 0 0 1 0 0 9 S4 0 -1.25 0 0 0 -0.125 0 1 1.5 3.5 2.33333x3 0 0 0 1 0 0 0 0 1 5 5
Basis P x1 x2 x3 S1 S2 S3 S4 S5 SolutionP 1 58.3333 0 0 0 20.83 0 8.33 0 1716.7S1 0 -5.5 0 0 1 -1.25 0 -1 0 48x2 0 0 1 0 0 0 0 1 0 6S3 0 1 0 0 0 0 1 0 0 9S5 0 -0.8333 0 0 0 -0.083 0 0.67 1 2.3333x3 0 0.83333 0 1 0 0.083 0 -0.67 0 2.6667
(c) Solve the linear programming problem with Matlab.
The problem is reformulated as
Minimize y =7 11 15 1541 2 3x x x {Material constraint}10 8 12 801 2 3x x x {Time constraint}x1 9 {“Regular” storage constraint}
{“Premium” storage constraint}{“Supreme” storage constraint}
x x x1 2 3 0, , {Positivity constraints}
We use the Matlab command X=LINPROG(f, A, b, Aeq, beq, LB, UB)
The coefficient vector f for the objective function is
f =
The matrix coefficient A for the inequality constraints is
A =
The right hand vector b for the inequality constraints is
b =
Since there are no equality constraints in this example, Aeq and beq are zeros.
Aeq = 0 and beq = 0
The lower and upper bounds vectors are given by
LB = and UB =
The following Matlab statements are used to solve this linear programming problem.
Matlab Program -----------------------------------------------------
% Set 4, Problem 2f=[-150;-175;-250];A=[7 11 15;10 8 12;0 0 0];b=[154;80;0];Aeq=zeros(3,3);beq=zeros(3,1);LB=[0;0;0];UB=[9;6;5];x=linprog(f,A,b,Aeq,beq,LB,UB)
>> s4p2Optimization terminated successfully.x = 0.0000 6.0000 2.6667-------------------------------------------------------------------------------------------------------------The solution from the Matlab program is
x1 = 0
x2 = 6
x3 = 2.6667
3. (P. 15.3 Chapra)
Consider the linear programming problem:
Maximize f(x, y) = x + y, subject to
x + 2.5y 15
x + y 7
2x + y 9
x 0 and y 0
(a) Solve the linear programming problem graphically.
(b) Solve the linear programming problem with the simplex method.
(c) Solve the linear programming problem with Matlab.
Solution
(a) Solve the linear programming problem graphically.
To solve graphically, the constraints can be reformulated as the following straight lines
y = 6 0.4x
y = 7 x
y = 9 2x
The objective function can be reformulated as
y = x +f(x, y)
The constraint lines can be plotted on the x-y plane to define the feasible space. Then the
objective function line can be superimposed for various values of f(x, y) until it reaches the
boundary. The result is f(x, y) 8.33 with x = 23 and y = 5.
(b) Solve the linear programming problem with the simplex method.
Maximize f(x, y) = x + y, subject to
x + 2.5y + S1 = 15
x + y + S2 = 7
2x + y + S3 = 9
x, y, S1, S2, S3 0
at A, x = 0 y = 0
S1 = 15, S2 = 7, S3 = 9, and f(x, y) = 0
at B, y = 0 S3 = 0
x + S1 = 15
x + S2 = 7
2x = 9
x = 4.5, S1 = 10.5, S2 = 2.5, and f(x, y) = 7.5
at C, S3 = 0 S2 = 0
x + 2.5y + S1 = 15
x + y = 7
2x + y = 9
x = 2, y = 0, S1 = 0.5, and f(x, y) = 8.33
(c) Solve the linear programming problem with Matlab.
The problem is reformulated as
Minimize y = x y
x + 2.5y + S = 15
x + y 7
2x + y 9
x, y, and S 0
We use the Matlab command X=LINPROG(f, A, b, Aeq, beq, LB, UB).
The coefficient vector f for the objective function is
f =
The matrix coefficient A for the inequality constraints is
A =
The right hand vector b for the inequality constraints is
b =
The equality constraints Aeq and beq are given as
Aeq = and beq =
The lower and upper bounds vectors are given by
LB = and UB =
The following Matlab statements are used to solve this linear programming problem.
Matlab Program -----------------------------------------------------
% Set 4, Problem 3f=[-5/3;-1;0];A=[1 1 0;2 1 0;0 0 0];b=[7;9;0];Aeq=[1 2.5 1; 0 0 0;0 0 0];beq=[15;0;0];LB=[0;0;0];UB=[inf;inf;inf];x=linprog(f,A,b,Aeq,beq,LB,UB)
>> s4p3Optimization terminated successfully.x = 2.0000 5.0000 0.5000-------------------------------------------------------------------------------------------------------------The solution from the Matlab program is
x1 = 2
x2 = 5
S = 0.5
4. (P. 13.7 Chapra) Employ the following methods to find the maximum of
f(x) = 2x 1.75x2 + 1.1x3 0.25x4
a) Quadratic interpolation (x0 = 1.75, x1 = 2, x2 = 2.25, iterations = 5)b) Newton method (x0 = 2.5, |xi+1 xi| < 0.01)
Solution
f(x) = 2x 1.75x2 + 1.1x3 0.25x4
f’(x) = 2 3.5x + 3.3x2 x3
f”(x) = 3.5 + 6.6x 3x2
% Set 4, problem 4
fx = '2*x - 1.75*x^2 + 1.1*x^3 - 0.25*x^4';
x0 = 1.75; x = x0; f0 = eval(fx);
x1 = 2; x = x1; f1 = eval(fx);
x2 = 2.25; x = x2; f2 = eval(fx);
f3save=f0;
disp('Quadratic interpolation')
for i=1:5
top = f0*(x1*x1-x2*x2)+f1*(x2*x2-x0*x0)+f2*(x0*x0-x1*x1);
bot = 2*f0*(x1-x2)+2*f1*(x2-x0)+2*f2*(x0-x1);
x3 = top/bot;
x = x3; f3 = eval(fx);
fprintf('x3 = %10.5f, f(x3) = %10.5f\n',x,f3)
% if abs(f3save-f3)<.00001, break, end
f3save=f3;
if x3<x0
x0=x3;x2=x1;x1=x0;
f0=f3;f2=f1;f1=f0;
elseif x3<x1
x2=x1;x1=x3;
f2=f1;f1=f3;
elseif x3<x2
x0=x1;x1=x3;
f0=f1;f1=f3;
else
x0=x1;x1=x2;x2=x3;
f0=f1;f1=f2;f2=f3;
end
end
disp('Newton method')
x = 2.5;
for i=1:10
fx = 2*x - 1.75*x^2 + 1.1*x^3 - 0.25*x^4;
fprintf('x = %8.5f, f(x) = %8.5f\n',x, fx)
dx = (2 - 3.5*x + 3.3*x^2 - x^3)/(- 3.5 + 6.6*x - 3*x^2 );
x= x - dx;
if abs(dx)<0.01, break, end
end
fx = 2*x - 1.75*x^2 + 1.1*x^3 - 0.25*x^4;fprintf('x = %8.5f, f(x) = %8.5f\n',x, fx)
>> s4p4Quadratic interpolationx3 = 2.06166, f(x3) = 1.80775x3 = 2.07412, f(x3) = 1.80813x3 = 2.07807, f(x3) = 1.80817x3 = 2.07898, f(x3) = 1.80817x3 = 2.07926, f(x3) = 1.80817Newton methodx = 2.50000, f(x) = 1.48438x = 2.19565, f(x) = 1.78800x = 2.09170, f(x) = 1.80796x = 2.07951, f(x) = 1.80817x = 2.07935, f(x) = 1.80817
5. (P. 13.9 Chapra) Employ the following methods to find the optimum of
f(x) = 6x + 7.5x2 + 3x3 + x4
a) Quadratic interpolation (x0 = 2, x1 = 1, x2 = 1, iterations = 5)b) Newton method (x0 = 1, |xi+1 xi| < 0.01)
Solution
f(x) = 6x + 7.5x2 + 3x3 + x4
f’(x) = 6 + 15x + 9x2 + 4x3
f”(x) = 15 + 18x + 12x2
% Set 4, problem 5
fx = '6*x + 7.5*x^2 + 3*x^3 + x^4';
x0 = -2; x = x0; f0 = eval(fx);
x1 = -12; x = x1; f1 = eval(fx);
x2 = 1; x = x2; f2 = eval(fx);
f3save=f0;
disp('Quadratic interpolation')
for i=1:5
top = f0*(x1*x1-x2*x2)+f1*(x2*x2-x0*x0)+f2*(x0*x0-x1*x1);
bot = 2*f0*(x1-x2)+2*f1*(x2-x0)+2*f2*(x0-x1);
x3 = top/bot;
x = x3; f3 = eval(fx);
fprintf('x3 = %10.5f, f(x3) = %10.5f\n',x,f3)
% if abs(f3save-f3)<.00001, break, end
f3save=f3;
if x3<x0
x0=x3;x2=x1;x1=x0;
f0=f3;f2=f1;f1=f0;
elseif x3<x1
x2=x1;x1=x3;
f2=f1;f1=f3;
elseif x3<x2
x0=x1;x1=x3;
f0=f1;f1=f3;
else
x0=x1;x1=x2;x2=x3;
f0=f1;f1=f2;f2=f3;
end
end
disp('Newton method')
x = -1;
for i=1:10
fx = 6*x + 7.5*x^2 + 3*x^3 + x^4;
fprintf('x = %8.5f, f(x) = %8.5f\n',x, fx)
dx = (6 + 15*x + 9*x^2 + 4*x^3)/(15 + 18*x + 12*x^2);
x= x - dx;
if abs(dx)<0.01, break, end
end
fx = 6*x + 7.5*x^2 + 3*x^3 + x^4;
fprintf('x = %8.5f, f(x) = %8.5f\n',x, fx)
>> s4p5
Quadratic interpolation
x3 = -0.50980, f(x3) = -1.43952
x3 = 0.18902, f(x3) = 1.42358
x3 = -0.35706, f(x3) = -1.30649
x3 = -0.50715, f(x3) = -1.43906
x3 = -0.52720, f(x3) = -1.44099
Newton method
x = -1.00000, f(x) = -0.50000
x = -0.55556, f(x) = -1.43766
x = -0.52782, f(x) = -1.44099
x = -0.52803, f(x) = -1.44099
6. Minimize the function f(x, y) = 4x2 + y2 2xy, using Newton method with x0 = [1 1]T.
Solution
x i+1 = x i Hi-1f(x i) (1)
f(x i) =
Hi =
Let x i = x i+1 x i, equation (1) becomes
Hix i = f(x i)
For i = 0, f(x i) = =
x = 1, y = 1
x1 = x0 + x = 1 1 = 0
y1 = y0 + y = 1 1 = 0
f(x = 0, y = 0) = 0 is the solution
a company makes two products a and b with two processing departments and wishes to add one
new product line what must be product C’s unit profit in order to profitably add it to the product
line