MTH Operations Research - Wilkes University

MTH %&% Operations Research

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A factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

An interdisciplinary branch of mathematics which uses mathematical models and tools to arrive at optimal or near-optimal decisions in complex decision making problems

Let 𝑥 = The number of regular gadgets manufactured each day.and 𝑦= The number of premium gadgets manufactured each day.

The objective function is

𝑃=20𝑥+30𝑦

We now write the constraints. The company can make at most 7 gadgets a day. This translates as

𝑥+𝑦 ≤ 7

Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get

𝑥+2𝑦≤12

Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint.

MTH %&% Operations ResearchA factory manufactures two types of gadgets, regular and premium. Each gadget requires the use of two operations, assembly and finishing, and there are at most 12 hours available for each operation. A regular gadget requires 1 hour of assembly and 2 hours of finishing, while a premium gadget needs 2 hours of assembly and 1 hour of finishing. Due to other restrictions, the company can make at most 7 gadgets a day. If a profit of $20 is realized for each regular gadget and $30 for a premium gadget, how many of each should be manufactured to maximize profit?

Let 𝑥 = The number of regular gadgets manufactured each day. 𝑦 = The number of premium gadgets manufactured each day.

The objective function is 𝑃 = 20𝑥 + 30𝑦

We now write the constraints. The company can make at most 7 gadgets a day. This translates as 𝑥 + 𝑦 ≤ 7

Since the regular gadget requires one hour of assembly and the premium gadget requires two hours of assembly, and there are at most 12 hours available for this operation, we get 𝑥 + 2𝑦 ≤ 12

Similarly, the regular gadget requires two hours of finishing and the premium gadget one hour. Again, there are at most 12 hours available for finishing. This gives us the following constraint. 2𝑥 + 𝑦 ≤ 12

The fact that 𝑥 and 𝑦 can never be negative is represented by the following two constraints: 𝑥 ≥ 0, and 𝑦 ≥ 0.

We have formulated the problem as follows:

𝐌𝐚𝐱𝐢𝐦𝐢𝐳𝐞 P = 20x + 30y 𝐒𝐮𝐛𝐣𝐞𝐜𝐭 𝐭𝐨: x + y ≤ 7 x + 2y ≤ 12 2x + y ≤ 12 x ≥ 0; y ≥ 0

In order to solve the problem, we next graph the constraints and feasibility region.



Theorems:

• The set of all feasible solutions to a linear programming problem is a convex set.

• An extreme point of a convex set lies on the boundary of the set and is, in fact, a corner point.


A small scale industrialist produces three types of machine components M1, M2, and M3, made of steel and brass. The amounts of steel, brass required for each component and the number of man-weeks of labor required to manufacture and assemble one unit of each component are as follows :

This labor is restricted to 20 man-weeks, steel is restricted to 100 kg. per week and the brass to 75 kg. per week. The industrialist’s profit on each unit of M1, M2,, and M3, is $6 , $4 and $7 respectively. Give its mathematical formulation as a linear programming problem such that the total profit is maximum.


Theorem:

• The objective function of a linear programming problem assumes its optimum at a corner point of the convex set of all feasible solutions.

MTH %&% Operations Researchclc

%% ===== Setup ========S = [[6 5 3 100];[3 4 9 75];[1 2 1 20]];sol = []; c = S(:,4);B = S; B(:,4) = []; [sr sc] = size(B);Axes = eye(sc); %% ===== Find intersections =====A = [B(1,:);B(2,:)];b = [c(1);c(2)];sol = [sol linsolve(A,b)]; A = [B(1,:);B(3,:)];b = [c(1);c(3)];sol = [sol linsolve(A,b)]; A = [B(2,:);B(3,:)];b = [c(2);c(3)];sol = [sol linsolve(A,b)];

%% === Check for axes intersections == for i=1:sc for j=1:sr A = [Axes(i,:);B(j,:)]; b = [0;c(j)]; sol = [sol linsolve(A,b)]; endend %% === Find corner points of feasible %% === region ======[r c] = size(sol); feasible = []; tolerance = 0.000001;for i=1:cv = (sol(:,i));x = v(1); y = v(2); z = v(3);if (6*x + 5*y + 3*z) <= 100 + tolerance && (3*x + 4*y + 9*z) <= 75 + tolerance && (x + 2*y + z) <= 20 + tolerance feasible = [feasible sol(:,i)]; end enddisp(feasible)

%% ====== Find optimum =======

opt = 0;xopt = 0; yopt = 0; zopt = 0;[r c] = size(feasible); for i=1:c v = (feasible(:,i)); x = v(1); y = v(2); z = v(3); p = (6*x + 4*y + 7*z); if (p >= opt) opt = p; xopt = x; yopt = y; zopt = z; end end optimum =[xopt yopt zopt opt];disp(optimum)

MTH %&% Operations ResearchTHE SIMPLEX ALGORITHM









Duality



Ties in choosing the leaving variable and more iterations !


If one of the nonbasic variables is valued zero and we choose to enter into another iteration !


Cannot find theta !


An artificial variable in the solution !


Cycling !

MTH %&% Operations Research • Asymptotics

( note: semi_lg axis )

MTH %&% Operations Research • Time to Execute an Algorithm if One Step Takes 1 Microsecond to Execute


• What about the Simplex Method ?


• Variations of the Simplex Method ?



• Simplex Algorithm vs Karmarkar’s Algorithm

MTH %&% Operations Research • Using MatLab’s linprog

As of Matlab R>?@Ab:

The linprog 'simplex' , ''Revised-Simplex'', and 'Dual-Simplex' algorithms have been removed.

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MTH Operations Research - Wilkes University