Linear Programming What s Linear Programming? Often your try is to - - PDF document
Linear Programming What s Linear Programming? Often your try is to maximize or minimize an objective within given constraints A linear programming problem can be expressed as a linear function of certain variables and the constraints can
Often your try is to maximize or minimize an objective within given constraints A linear programming problem can be expressed as a linear function of certain variables and the constraints can be expressed as equalities or inequalities on those variables Linear programming could be a minimization linear program or a maximization linear program
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You’re running for office 100K urban voters, 200K suburban, and 50K rural voters are your target You want to win over at least 50% of voters in each area There’ s a catch. These voters may be persuaded differently by advertisements on different issues
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The effect of policies (and hence advertising on those policies) on thousands of voters who can be won over (or lost) by spending $1,000 on particular issue.
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Policy urban suburban rural building roads
5 3 gun control 8 2
farm subsidies 10 gasoline tax 10
Rather than trial and error, you can formulate the problem Let x1 be the number of thousands of dollars spent on advertising on building roads, x2 ... on gun control, x3 ...
transit improvements) You can formulate the requirement to win half urban voters (50k out of 100k voters) as
...(similarly for other areas) You have to ensure xi >= 0 (as spending can’ t be negative) and minimize x1 + x2 + x3 + x4
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Minimize x1 + x2 + x3 + x4 subject to
5x1 + 2x2 + 0x3 + 0x4 >= 100 3x1 - 5x2 + 10x3 - 2x4 >= 25 x1, x2, x3, x4 >= 0 Solution to the linear program yields the optimal strategy.
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Optimize a linear function subject to a set of linear inequalities Given a set of real numbers a1, a2, ..., an and a set of variables x1, x2, ..., xn, we define a linear function f on those variables: f(x1, x2, ..., xn) = a1x1 + a2x2 + ... + anxn = !1<=i<=n aixi plus linear constraints: either a linear equality f(x1, x2, ..., xn) = b or linear inequalities f(x1, x2, ..., xn) <= b
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Standard Maximization of a linear function subject to linear inequalities Slack Maximization of a linear function subject to linear equalities
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A linear program has n variables subject to a set of m linear inequalities A feasible solution is any values for variables that satisfy all the constraints Feasible region is set of all feasible solutions The maximization function is called the objective function and the value of this function at particular point in the feasible region an objective value There may be infinite number of points in a feasible region, so we need a way to practically find the optimal solution for the objective value
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You can plot the constraints on a cartesian coordinate and find an optimal solution for the objective value
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f1 f2 f3 Objective function Optimal Solution
The optimal solution is the vertex of the feasible region If it is a line, you can use end points of the line segment
Each constrain forms a half-space in n-th dimensional space Feasible region formed by the intersection of these half-spaces is called simplex The objective function is a hyperplane and the optimal solution occurs at the vertex of the simplex The simplex algorithm moves through the exterior of the simplex, determining the vertex with maximum value that satisfy the objective function
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Iterative algorithm Set each nonbasic variable to 0 and compute the values
In each iteration you convert from one slack form to another slack form You improve the objective value in each iteration We increase values (subject to constraints) of nonbasic variables from 0 in order to maximize the objective value You convert the equations into equivalent forms until the
coefficient
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Given the linear program in standard form maximize 3x1 + x2 + 2x3 subject to x1 + x2 + 3x3 <= 30 2x1 + 2x2 + 5x3 <= 24 4x1 + x2 + 2x3 <= 36 x1, x2, x3 >= 0
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z = 3x1 + x2 + 2x3 x4 = 30 - x1 - x2 - 3x3 x5 = 24 - 2x1 - 2x2 - 5x3 x6 = 36 - 4x1 - x2 - 2x3 Nonbasic variables on the right, basic variables on the left
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z = 3x1 + x2 + 2x3 x4 = 30 - x1 - x2 - 3x3 x5 = 24 - 2x1 - 2x2 - 5x3 x6 = 36 - 4x1 - x2 - 2x3 Giving values of 0 to nonbasic variables, we have (0, 0, 0, 30, 24, 36) and z gets a value of 0. Now we want to reformulate so the basic solution has a greater objective value.
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z = 3x1 + x2 + 2x3 x4 = 30 - x1 - x2 - 3x3 x5 = 24 - 2x1 - 2x2 - 5x3 x6 = 36 - 4x1 - x2 - 2x3 Take a nonbasic variable whose coefficient in the
much as possible without violating constraints. This variable becomes basic and some other variable becomes nonbasic.
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z = 3x1 + x2 + 2x3 x4 = 30 - x1 - x2 - 3x3 x5 = 24 - 2x1 - 2x2 - 5x3 x6 = 36 - 4x1 - x2 - 2x3 Increasing x1 will result in decreasing x4, x5, and x6. However, non of them can become negative due to
respectively. The third constraint is the most limiting. We switch roles
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z = 3x1 + x2 + 2x3 x4 = 30 - x1 - x2 - 3x3 x5 = 24 - 2x1 - 2x2 - 5x3 x1 = 9 - x2/ 4 - x3/2 - x6/ 4 You pivot—that is you chose an entering variable (x1) and a leaving variable (x6) and exchange their roles. You now replace all x1
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z = 27 + x2/ 4 + x3/2 - 3x6/ 4 x1 = 9 - x2/ 4 - x3/2 - x6/ 4 x4 = 21 - 3x2/ 4 - 5x3/2 + x6/ 4 x5 = 6 - 3x2/2 - 4x3 + x6/2 You’re ready to pick the next pivot, an entering variable with positive coefficient in the objective function and a corresponding leaving variable. Let’ s chose x3 and that will lead to the tightest function and hence the leaving variable x5.
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z = 111/ 4 + x2/16 - x5/8 - 11x6/16 x1 = 33/ 4 - x2/16 + x5/8 - 5x6/16 x3 = 3/2 - 3x2/8 - x5/ 4 + x6/8 x4 = 69/ 4 + 3x2/16 + 5x5/8 - x6/16 The lone standing positive coefficient nonbasic variable in
Using that as a pivot,...
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z = 28 - x3/ 6 - x5/ 6 - 2x6/3 x1 = 8 + x3/ 6 + x5/ 6 - x6/3 x2 = 4 - 8x3/3 - 2x5/3 + x6/3 x4 = 18 - x3/2 + x5/2 Giving zero values to the nonbasic variables, we get x1 = 8, x2 = 4, x3 = 0, and the optimal solution z = 28.
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maximize 3x1 + x2 + 2x3 x1 + x2 + 3x3 <= 30 12 <= 30, slack is 18 which is x4 2x1 + 2x2 + 5x3 <= 24 24 <= 24, slack is 0 which is x5 4x1 + x2 + 2x3 <= 36 36 <= 36, slack is 0 which is x6 x1, x2, x3 >= 0 Solution x1 = 8, x2 = 4, x3 = 0, and the optimal solution z = 28. x4, x5, and x6 are the slack variable. The slacks are 18, 0, and 0. If you plug in values for x1, x2, and x3 into original inequalities, you can see the slack.
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It is possible that the feasible region is unbounded It is possible that the iteration leaves the objective value unchanged—known as degeneracy You need to look for cycles where the slack forms simply cycle through transformations Since Simplex algorithm is deterministic, it will cycle
You can determine it cycles if it does not terminate in at most ( ) iterations
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n+m m