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Solving LP Models Improving Search Special Form of Improving Search
Unimodal Convex feasible region Should be successful! Special Form of Improving Search Simplex method (now) Interior point methods (later) Part 3 IE 312
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Simple Example Top Brass Trophy Company Makes trophies for
football wood base, engraved plaque, brass football on top $12 profit and uses 4’ of wood soccer wood base, engraved plaque, soccer ball on top $9 profit and uses 2’ of wood Current stock 1000 footballs, 1500 soccer balls, 1750 plaques, and 4800 feet of wood Part 3 IE 312
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Formulation Part 3 IE 312
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Graphical Solution Optimal Solution 2000 1500 1000 500
Optimal Solution Part 3 IE 312
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Feasible Solutions Feasible solution is a
boundary point if at least one inequality constraint that can be strict is active interior point if no such constraints are active Extreme points of convex sets do not lie within the line segment of any other points in the set Part 3 IE 312
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Example 2000 1500 1000 500 Part 3 IE 312
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Optimal Solutions Every optimal solution is a boundary point
We can find an improving direction whenever we are at an interior point If optimum unique the it must be an extreme point of the feasible region If optimal solution exist, an optimal extreme point exists Part 3 IE 312
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LP Standard Form Easier if we agree on exactly what a LP should look like Standard form only equality main constraints only nonnegative variables variables appear at most once in left-hand-side and objective function all constants appear on right hand side Part 3 IE 312
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Converting to Standard
Inequality constraints Add nonnegative, zero-cost slack variables Add in inequalities Subtract in inequalities Variables not nonnegative nonpositive - substitute with negatives unrestrictive sign (URS) - substitute difference of two nonnegative variables Part 3 IE 312
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Top Brass Model Part 3 IE 312
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Why? Feasible directions Equality constraints Inequality constraints
Check only if active Keep track of active constraints Equality constraints Always active Inequality constraints May or may not be active Prefer equality constraints! Part 3 IE 312
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Standard Notation Part 3 IE 312
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LP Standard Form In standard notation In matrix notation Part 3 IE 312
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Write in Matrix Form Part 3 IE 312
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Extreme Points Know that an extreme point optimum exists
Will search trough extreme points An extreme point is define by a set of constraints that are active simultaneously Part 3 IE 312
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Improving Search Move from one extreme point to a neighboring extreme point Extreme points are adjacent if they are defined by sets of active constraints that differ by only one element An edge is a line segment determined by a set of active constraints Part 3 IE 312
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Basic Solutions Extreme points are defined by set of active nonnegativity constraints A basic solution is a solution that is obtained by fixing enough variable to be equal to zero, so that the equality constraints have a unique solution Part 3 IE 312
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Example Choose x1, x2, x3, x4 to be basic Part 3 IE 312
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Where is the Basic Solution?
2000 1500 1000 500 Part 3 IE 312
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Example Compute the basic solution for x1 and x2 basis: Solve Part 3
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Existence of Basis Solutions
Remember linear algebra? A basis solution exists if and only if the columns of corresponding equality constraint form a basis (in other words, a largest possible linearly independent collection) Part 3 IE 312
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Checking The determinant of a square matrix D is
A matrix is singular if its determinant = 0 and otherwise nonsingular Need to check that the matrix is nonsingular Part 3 IE 312
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Example Check whether basic solutions exist for Part 3 IE 312
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Basic Feasible Solutions
A basic feasible solution to a LP is a basic solution that satisfies all the nonnegativity contraints The basic feasible solutions correspond exactly to the extreme points of the feasible region Part 3 IE 312
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Example Problem Suppose we have x3, x4, x5 as slack variables in the following LP: Lets plot the original problem, compute the basic solutions and check feasibility Part 3 IE 312
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Solution Algorithm Simplex Algorithm Standard display:
Variant of improving search Standard display: Part 3 IE 312
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Simplex Algorithm Starting point Direction
A basic feasible solution (extreme point) Direction Follow an edge to adjacent extreme point: Increase one nonbasic variable Compute changes needed to preserve equality constraints One direction for each nonbasic variable Part 3 IE 312
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Top Brass Example Basic variables Initial solution Part 3 IE 312
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Looking in All Directions …
Can increase either one of those Must adjust these! Part 3 IE 312
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So Many Choices ... Want to try to improve the objective
The reduced cost of a nonbasic variable: Want Defines improving direction Part 3 IE 312
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Top Brass Example Improving x1 gives Improving x2 gives
Both directions are improving directions! Part 3 IE 312
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Where and How Far? Any improving direction will do
If no component is negative Improve forever - unbounded! Otherwise, compute the minimum ratio Part 3 IE 312
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Computing Minimum Ratio
Part 3 IE 312
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Moving to New Solution Part 3 IE 312
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Updating Basis New basic variable New nonbasic variable(s)
Nonbasic variable generating direction New nonbasic variable(s) Basic variables fixing the step size Part 3 IE 312
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What Did We Do? 2000 1500 1000 500 Part 3 IE 312
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Where Will We Go? Optimum in three steps! Why is this guaranteed? 2000
1500 1000 500 Optimum in three steps! Why is this guaranteed? Part 3 IE 312
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Simplex Algorithm (Simple)
Step 0: Initialization. Choose starting feasible basis, construct basic solution x(0), and set t=0 Step 1: Simplex Directions. Construct directions Dx associated with increasing each nonbasic variable xj and compute the reduced cost cj =c ·Dx. Step 2: Optimality. If no direction is improving, then stop; otherwise choose any direction Dx(t+1) corresponding to some basic variable xp. Step 3: Step Size. If no limit on move in direction Dx(t+1) then stop; otherwise choose variable xr such that Step 4: New Point and Basis. Compute the new solution and replace xr in the basis with xp. Let t = t+1 and go to Step 1. Part 3 IE 312
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Stopping The algorithm stop when one of two criteria is met:
In Step 2 if no improving direction exists, which implies local optimum, which implied global optimum In Step 3 if no limit on improvement, which implies problem is unbounded Part 3 IE 312
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Optimization Software
Spreadsheet (e.g, MS Excel with What’s Best!) Optimizers (e.g., LINDO) Combination Modeling Language Solvers Either together (e.g., LINGO) or separate (e.g., GAMS with CPLEX) LINDO and LINGO are in Room 0010 (OR Lab) Also on disk with your book Part 3 IE 312
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LINDO The main software that I’ll ask you to use is called LINDO
Solves linear programs (LP), integer programs (IP), and quadratic programs (QP) We will look at many of its more advanced features later on, but as of yet we haven’t learned many of the concepts that we need Part 3 IE 312
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Example Part 3 IE 312
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LINDO Program Part 3 IE 312 MAX 12 x1 + 9 x2 ST x1 + x2 = 1000
END Part 3 IE 312
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Part 3 IE 312
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Part 3 IE 312
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Output LP OPTIMUM FOUND AT STEP 4 OBJECTIVE FUNCTION VALUE 1) 12000.00
1) VARIABLE VALUE REDUCED COST X X X X X X Part 3 IE 312
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Output (cont.) ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 12.000000
2) 3) 4) 5) 6) 7) 8) 9) 10) 11) NO. ITERATIONS= Part 3 IE 312
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LINDO: Basic Syntax Objective Function Syntax: Start all models with MAX or MIN Variable Names: Limited to 8 characters Constraint Name: Terminated with a parenthesis Recognized Operators (+, -, >, <, =) Order of Precedence: Parentheses not recognized Part 3 IE 312
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Syntax (cont.) Adding Comment: Start with an exclamation mark
Splitting lines in a model: Permitted in LINDO Case Sensitivity: LINDO has none Right-hand Side Syntax: Only constant values Left-hand Side Syntax: Only variables and their coefficients Part 3 IE 312
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Why Modeling Language? More to learn!
More ‘complicated’ to use than LINDO (at least at first glance) Advantages Natural representations Similar to mathematical notation Can enter many terms simultaneously Much faster and easier to read Part 3 IE 312
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Why Solvers? Best commercial software has modeling language and solvers separated Advantages: Select solver that is best for your application Learn one modeling language use any solver Buy 3rd party solvers or write your own! Part 3 IE 312
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Example Problem Part 3 IE 312
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Problem Formulation Part 3 IE 312
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LINDO Solution max 1.60 x1 + 1.40 x2 + 1.90 x3 + 1.20 x4 st
end Part 3 IE 312
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LINGO Solution Capacity constraint @SUM(REGIONS(I): CASES(I))
<=1200; Minimum/maximum cases @FOR(REGIONS(I): CASES(I) <= UBOUND; CASES(I) >= LBOUND); Part 3 IE 312
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LINGO Solution Objective function MAX = @SUM(REGIONS(I):
PROFIT*CASES(I)); We also need to define REGIONS, CASES, etc, and type in the data. Part 3 IE 312
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LINGO Solution Defining sets SETS: REGIONS / NE SE MW W/: LBOUND,
UBOUND, PROFIT, CASES; ENDSETS Part 3 IE 312
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LINGO Solution Enter the data DATA: LBOUND = 310 245 255 190;
UBOUND = ; PROFIT = ; ENDDATA Part 3 IE 312
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Sensitivity Analysis Basic Question: How does our solution change as the input parameters change? The objective function? More/less profit or cost The optimal values of decision variables? Make different decisions! Why? Only have estimates of input parameters May want to change input parameters Part 3 IE 312
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What We Know Qualitative Answers for All Problems
Quantitative Answers for Linear Programs (LP) Dual program Same input parameters Decision variables give sensitivities Dual prices Easy to set up Theory is somewhat complicated Part 3 IE 312
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Back to Example Problem
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LINDO Formulation max 1.60 x1 + 1.40 x2 + 1.90 x3 + 1.20 x4 st
end Part 3 IE 312
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LINDO Solution (second half)
ROW SLACK OR SURPLUS DUAL PRICES 2) 3) 4) 5) 6) 7) 8) 9) 10) Part 3 IE 312
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Dual Prices The Dual is Automatically Formed Report dual prices
Also in LINGO Also in (all) other optimization software Report dual prices Gives us sensitivities to RHS parameter Know how much objective function will change When will the optimal solution change? Need to select that we want sensitivity analysis Part 3 IE 312
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LINDO Sensitivity Analysis (part)
RANGES IN WHICH THE BASIS IS UNCHANGED: OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X X INFINITY X INFINITY X INFINITY X INFINITY Part 3 IE 312
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Interpretation As long as prices for the NE region are between $1.4 and $1.9, we want to sell the same quantity to each region, etc. Part 3 IE 312
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Example An insurance company is introducing two new product lines: special risk insurance and mortgages. The expected profit is $5 per unit on special risk insurance and $2 per unit on mortgages. Management wishes to establish a sales target for the new product lines to maximize the expected profit. The work requirements are as follows: Part 3 IE 312
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LINDO Formulation max 5 x1 + 2 x2 st 3 x1 + 2 x2 <= 2400
end Part 3 IE 312
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Graphical Solution Part 3 IE 312 800 700 600 500 400 300 200 100
800 700 600 500 400 300 200 100 Part 3 IE 312
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Solution VARIABLE VALUE REDUCED COST X1 600.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 3) 4) 5) 6) Part 3 IE 312
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Sensitivity Analysis RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES VARIABLE CURRENT ALLOWABLE ALLOWABLE COEF INCREASE DECREASE X INFINITY X RIGHTHAND SIDE RANGES ROW CURRENT ALLOWABLE ALLOWABLE RHS INCREASE DECREASE INFINITY INFINITY INFINITY Part 3 IE 312
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New Decisions! Optimum Moves! Part 3 IE 312 800 700 600 500 400 300
800 700 600 500 400 300 200 100 Optimum Moves! Part 3 IE 312
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What-If ? Solve New Problem
max 5 x1 + 2 x2 st 3 x1 + 2 x2 <= 2400 x2 <= 290 2 x1 <= 1200 x1 >=0 x2 >=0 end Part 3 IE 312
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New Solution VARIABLE VALUE REDUCED COST X1 600.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 3) 4) 5) 6) Part 3 IE 312
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Interior Point Methods
Simplex always stays on the boundary Can take short cuts across the interior Interior point methods More effort in each move More improvement in each move Much faster for large problems Part 3 IE 312
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Simple Example Frannie’s Firewood sells up to 3 cords of firewood to two customers One will pay $90 per half-cord Other will pay $150 per full cord Part 3 IE 312
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Graphical Solution 4 3 2 1 Part 3 IE 312
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Improving Directions Which direction improves the objective function the most? The gradient Direction: Part 3 IE 312
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Most Improving Direction?
Part 3 IE 312
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Back to Example 4 3 2 1 Part 3 IE 312
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Maintaining Feasibility
At the initial point all directions are feasible because it is an interior point At the new point we have to make sure that a direction Dx at x(1) satisfies Interior point algorithms begin inside and move through the interior, reaching the boundary only at an optimal solution Part 3 IE 312
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Valid Interior Point Search?
2000 1500 1000 500 Part 3 IE 312
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Valid Interior Point Search?
2000 1500 1000 500 Part 3 IE 312
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Valid Interior Point Search?
2000 1500 1000 500 Part 3 IE 312
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LP Standard Form For Simplex used the form In Frannie’s Firewood
Part 3 IE 312
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Benefits of Standard Form?
In Simplex: Made easy to check which variables are basic, non-basic, etc. Needed to know which solutions are on boundary Here quite similar: Know which are not on boundary Check that nonnegativity constraints are strict! A feasible solution to standard LP is interior point if every component is strictly positive Part 3 IE 312
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Interior Points? Part 3 IE 312
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Projections Must satisfy main equality constraints
Want direction Dx that satisfies this equation and is as nearly d as possible The projection of a vector d onto a system of equalities is the vector that satisfies the constraints and minimizes the total squared difference between the components Part 3 IE 312
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Obtaining Projection The projection of d onto ADx=0 is where
is the projection matrix. Part 3 IE 312
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Example: Frannie’s Firewood
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Example The cost vector is c=( ) Part 3 IE 312
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Improvement The projection matrix is design to make an improving direction feasible with minimum changes Is it still an improving direction? Yes! The projection Dx=Pc of c onto Ax=b is an improving direction at every x Part 3 IE 312
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Sample Exercise Determine the direction d of most rapid improvement
Project it onto the main equality constraints to get Dx Verify that the move direction Dx is feasible Verify that the move direction Dx is improving Part 3 IE 312
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