1 / 40

EE/Econ 458 Duality

EE/Econ 458 Duality. J. McCalley. Our example (and HW) problem. After one iteration, we obtained the following Tableau:. Optimality test indicates we need to do another iteration (since coefficient of x 1 is negative in above tableau).

mateja
Download Presentation

EE/Econ 458 Duality

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. EE/Econ 458Duality J. McCalley

  2. Our example (and HW) problem After one iteration, we obtained the following Tableau: Optimality test indicates we need to do another iteration (since coefficient of x1is negative in above tableau).

  3. In performing the second iteration, the only choice for the entering variable is x1 (coefficients of all other variables in the tableau are non-negative). Our example (and HW) problem To determine the leaving variable, we identify the constraints that most limit the increase in x1, as shown below. And so the constraint identifying the leaving variable is the last one, since its ratio is smallest (2<4). And so what is the leaving variable?

  4. The leaving variable is x5, since it is the variable in the last constraint which gets pushed to 0 as x1 increases. Our example (and HW) problem Pivot row Pivot element Make the pivot element “1” by dividing the pivot row by 3. Pivot column This is equivalent to dividing both sides of the following by 3. 3x1+0x2+0x3-x4+x5=6  (1/3)*[3x1+0x2+0x3-x4+x5]=6*(1/3)  x1+0x2+0x3-0.3333x4+0.3333x5=2 Now what is the next step?

  5. Use the last equation to eliminate the -3 in the top equation. Our example (and HW) problem This is equivalent to multiplying the bottom equation by 3 and adding it to the top equation: 3*[x1+0x2+0x3-0.3333x4+0.3333x5]=2*3 -3x1+0x2+0x3 -x4 +x5 =6 +{F+3x1+0x2+0x3+2.5x4+0x5=30} ------------------------------------------ F+0x1+0x2+0x3+1.5x4+x5=36

  6. Use the last equation to eliminate the 1 from the 2nd equation: Our example (and HW) problem This is equivalent to multiplying the bottom equation by -1 and adding it to the 2nd equation: -1*[x1+0x2+0x3-0.3333x4+0.3333x5]=2*-1 -x1+0x2+0x3 +0.3333x4 -0.3333x5 =-2 +{x1+0x2+1x3 +0x4 +0x5 =4} ------------------------------------------------------ 0x1+0x2+1x3+0.3333x4-0.3333x5=2

  7. Assume the problem we have been working on is actually a resource allocation problem: Introduction to duality Resources Question: How to gain the most, in terms of the value of our optimized objective function, by increasing one resource or another? CPLEX yields F1*=36 CPLEX yields F2*=36 (increases by 0)

  8. Summary Introduction to duality CPLEX yields F1*=36 CPLEX yields F3*=37.5 (increases by 1.5) CPLEX yields F1*=36 CPLEX yields F4*=37 (increases by 1.0)

  9. Assume F* represents my optimal profits. Assume also that I have a little extra money to spend. Then this information tells me my optimal objective will improve most if I spend that money to increase a unit of resource 2 (b2), less if I spend it to increase a unit of resource 3 (b3), and it will do me no good at all to spend that money to increase resource 1 (b1) . Introduction to duality (Important note: ΔF*/Δbi is how the optimal value of F changes with bi.) Go back and inspect the Tableau for the solution to Problem 1. The coefficients of the slack variables in this final tableau are exactly the same as the right-hand-sides of the above: 0, 1.5, and 1 !

  10. This always happens for linear programs. Here is a general rule: The coefficients of the slack variables in the objective function expression of the final tableau give the improvement in the objective for a unit increase in the right-hand-sides of the corresponding constraints. Introduction to duality x3 is SV for constr 1. x4 is SV for constr 2. X5 is SV for constr 3. What if your objective function was profits per hr, measured in $/hr? And what if one of your equations expressed a constraint on MW output for a certain generator? Units would be $/MW-hr. Where have we seen this before?

  11. These slack variable coefficients are called dual variables, dual prices, or shadow prices (implicit rather than explicit values associated with obtaining additional resources). We denote them as λicorresponding to the ith constraint. Introduction to duality

  12. But we need to be a little careful… From the left and below, we see that λ3 =1, which we now understand to mean that if we increase b3 by 1 unit, our objective function will increase by 1, from 36 to 37, and this, as we have learned, is the case. Introduction to duality But what if we increase b3 by 6, to 24? Will we see an increase in F* by 6 as well, to 42? Will increasing b3 to 26 see F* increase to 44? CPLEX yields F6*=42 CPLEX yields F1*=36 CPLEX yields F5*=42

  13. There are two observations to be made here: • Changing some bi do not influence F*. • Only a subset of resources limit the objective. • You may have labor hours, trucks, & pipe fittings as resources, each of which are individually constrained. You have a large truck fleet and a whole warehouse of pipe fittings, but you don’t have enough labor. And so you increase labor until you hit the limit on pipe fitting inventory. You then increase pipe fitting inventory, but soon you hit the limit on trucks. Introduction to duality 2. ΔF*/Δbi is not accurate if Δbi gets too large. A better definition of the dual variables follows: This says that the λi = ΔF*/Δbi only for incrementally small Δbi. In an LP, λi = ΔF*/Δbi is no longer true when Δbi becomes so large that its corresponding constraint is no longer binding.

  14. Consider our now very familiar example problem (call it P): P Motivating the dual problem Let’s express linear combinations of multiples of the constraints, where the constraint multipliers are denoted λion the ith constraint. We assume that λi≥0 to avoid having to change the inequality. The composite inequality

  15. Composite inequality: The left-hand-side of the composite inequality, being a linear combination of our original inequalities, must hold at any feasible solution (x1,x2), and in particular, at the optimal solution. That is, it is a necessary condition for satisfying the inequalities from which it came. It is not, however, a sufficient condition (see notes for example) unless we constrain how we choose the λi. Motivating the dual problem The objective function. The composite inequality. Let’s develop criteria for selecting λ1, λ2,λ3. Concept 1: Make sure our choices of λ1, λ2,λ3 are such that each coefficient of xi in the composite inequality is at least as great as the corresponding coefficient in the objective function expression: Any solution (x1,x2)results in a value F less than or equal to the left-hand-side of the composite inequality.

  16. Composite inequality: Concept 2: Because F is less than or equal to the left-hand side of the composite inequality (by concept 1), and because the left-hand-side of the composite inequality is less than or equal to the right hand side of the composite inequality, we can also say that any solution (x1,x2) results in a value F which is less than or equal to the right-hand-side of the composite inequality. Motivating the dual problem Concept 3: Concept 2 implies the right-hand-side of the composite inequality is an upper bound on the value that F may take. This is true for any value of F, even the maximum value F*.

  17. Composite inequality: Concept 4: Now choose λ1, λ2,λ3 to minimize the right-hand-side of the composite inequality, subject to the constraints we imposed in choosing the λi ‘s. Motivating the dual problem This creates a least upper bound to F*, i.e., it pushes right-hand-side of the composite inequality as far left as possible, while guaranteeing right-hand-side remains ≥ F* (due to enforcement of above constraints).

  18. Composite inequality: Concept 5: Given that the right-hand-side of the composite inequality is an upper bound to F*, then finding its minimum, subject to constraints, implies Δ=0. So we have a new problem to solve: Motivating the dual problem What does concept 5 tell us about the above problem, if we solve it? The value of the obtained objective function, at the optimum, will be the same as the value of the original objective function at its optimum, i.e., F*=G*.

  19. In other words, solving problem D is equivalent to solving problem P. Motivating the dual problem Problem P is called the primal problem. Problem D is called the dual problem. They are precisely equivalent. To show this, let’s use CPLEX to solve Problem D.

  20. The CPLEX code to do it is below (note I use y instead of λ in code): minimize 1 y1 + 12 y2 + 18 y3 subject to 1 y1 + 3 y3 >= 3 2 y2 + 2 y3 >=5 y1 >= 0 y2 >= 0 y3 >= 0 end The solution gives G*=36 λ1=0 λ2=1.5 λ3=1 Motivating the dual problem • They are: • Coefficients of the SVs λ 4 , λ 5of the objective function expression in last tableau of dual problem (not the values of the SVs). They are sensitivities. • Values of the DVs at the optimal solution to the dual of this problem – “dual to the dual” which is the primal! So these are the values x1 and x2. Using CPLEX, following solution of the above dual problem, I typed the following command: display solution dual – CPLEX gave: Constraint Name Dual Price c1 2.000000 c2 6.000000 What are these?

  21. The CPLEX code to do it is below (note I use y instead of λ in code): The solution gives G*=36 λ1=0 λ2=1.5 λ3=1 The solution gives G*=30 λ1=0 λ2=1 λ3=1 Motivating the dual problem Using CPLEX, following solution of the above dual problem, I typed the following command: display solution dual – CPLEX gave: Constraint Name Dual Price c1 2.000000 c2 6.000000 c1, c2 are right-hand-sides of the two constraints in the dual

  22. There is a circular relationship here, which can be stated as: The dual of the dual to a primal is the primal. If you called Problem D our primal problem, and took its dual, you would get our original primal problem P back, as illustrated below. Motivating the dual problem

  23. 1. Number of decision variables and constraints: • # of dual decision variables = # of primal constraints. • # of dual constraints = # of primal decision variables. • 2. Coefficients of decision variables in dual objective • are right-hand-sides of primal constraints. Primal-dual relationships 3. Coefficients of decision variables in primal objective are right-hand-sides of dual constraints.

  24. 4. Coefficients of one variable across multiple primal constraints are coefficients of multiple variables in one dual constraint. Primal-dual relationships

  25. 5. Coefficients of one variable across multiple dual constraints are coefficients of multiple variables in one primal constraint. Primal-dual relationships 6. If primal objective is maximization, dual objective is minimization 7. If primal constraints are ≤, dual constraints are ≥.

  26. We can immediately write down the dual given the primal. Example: Let’s return to the example we used to illustrate use of CPLEX in the notes called “Intro_CPLEX.” By inspection: Obtaining the dual Subject to: Subject to: Let’s check it using CPLEX…. (if we got it right above, they should have the same objective function value at the optimal solution).

  27. Primal Dual minimize 5 y1 + 11 y2 + 8 y3 subject to 2 y1 + 4 y2 + 3 y3 >= 5 3 y1 + 1 y2 + 4 y3 >= 4 1 y1 + 2 y2 + 2 y3 >= 3 y1 >= 0 y2 >= 0 y3 >= 0 end maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end Obtaining the dual The solution is (λ1, λ2, λ3)=(1,0,1), G*=13. The dual SV coefficients, i.e., the values of the dual variables are: (x1,x2,x3)=(2,0,1). The solution is (x1,x2,x3)=(2,0,1), F*=13. The primal SV coefficients, i.e., the values of the dual variables are: (λ1, λ2, λ3)=(1,0,1)

  28. Generalize the primal-dual problems: Viewing primal-dual relationship

  29. Viewing primal-dual relationship

  30. Viewing primal-dual relationship

  31. Duality theorem: If the primal problem has an optimal solution x*, then the dual problem has an optimal solution λ* such that The duality theorem But what if the primal does not have an optimal solution? Then what happens in the dual? To answer this, we must first consider what are the alternatives for finding an optimal solution to the primal? There are two: 1. The primal is unbounded. 2. The primal is infeasible.

  32. Unbounded Infeasible CPLEX> read ex.lp Problem 'ex.lp' read. Read time = 0.00 sec. CPLEX> primopt Bound infeasibility column 'x2'. Presolve time = 0.00 sec. Presolve - Unbounded or infeasible. Solution time = 0.00 sec. CPLEX> CPLEX> read ex1.lp Problem 'ex1.lp' read. Read time = 0.00 sec. CPLEX> primopt Dual infeasible due to empty column 'x2'. Presolve time = 0.00 sec. Presolve - Unbounded or infeasible. Solution time = 0.00 sec. CPLEX> Unbounded vs. infeasibility

  33. Let’s inspect the unbounded problem. An unbounded primal Recall that the objective function for the dual establishes an upper bound for the objective function of the primal, i.e.,. for any sets of feasible solutions λ and x. If F(x) is unbounded, then the only possibility for G(λ) is that it must be infeasible.

  34. Write down the dual. This constraint cannot be satisfied, therefore this problem is infeasible. An unbounded primal Likewise, we can show that if the dual is unbounded, the primal must be infeasible. However, it is not necessarily true that if the primal (or dual) is infeasible, then the dual (or primal) is unbounded. It is possible for an infeasible primal to have an infeasible dual and vice-versa, that is, both the primal and the dual may be both be infeasible. Finally, if both primal and dual have feasible solutions, both have optimal solutions.

  35. Primal-Dual relations

  36. maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end CPLEX> primopt …  Primal simplex - Optimal: Objective = 1.3000000000e+01 …  CPLEX> display solution variables - Variable Name Solution Value x1 2.000000 x3 1.000000 All other variables in the range 1-3 are 0. CPLEX> display solution slacks - Constraint Name Slack Value slack c2 1.000000 slack c4 -2.000000 slack c6 -1.000000 All other slacks in the range 1-6 are 0. CPLEX> display solution dual - Constraint Name Dual Price c1 1.000000 c3 1.000000 All other dual prices in the range 1-6 are 0. CPLEX> quit A comment on using CPLEX CPLEX will name the slack variables c1, c2, …, cm, cm+1,…, cm+n where there are m constraints and n decision variables. Therefore the first m slack variables (c1, c2, …, cm) correspond to the explicit inequality constraints, and the last n slack variables (cm+1,…, cm+n) correspond to the nonnegativity constraints on the decision variables.

  37. maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end CPLEX> primopt …  Primal simplex - Optimal: Objective = 1.3000000000e+01 …  CPLEX> display solution variables - Variable Name Solution Value x1 2.000000 x3 1.000000 All other variables in the range 1-3 are 0. CPLEX> display solution slacks - Constraint Name Slack Value slack c2 1.000000 slack c4 -2.000000 slack c6 -1.000000 All other slacks in the range 1-6 are 0. CPLEX> display solution dual - Constraint Name Dual Price c1 1.000000 c3 1.000000 All other dual prices in the range 1-6 are 0. CPLEX> quit A comment on using CPLEX The values of the slack variables at the solution were c1=0, c2=1, c3=0, c4=-2, c5=0, c6=-1. Note that CPLEX does not print the values of slack variables that are zero.

  38. maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end CPLEX> primopt …  Primal simplex - Optimal: Objective = 1.3000000000e+01 …  CPLEX> display solution variables - Variable Name Solution Value x1 2.000000 x3 1.000000 All other variables in the range 1-3 are 0. CPLEX> display solution slacks - Constraint Name Slack Value slack c2 1.000000 slack c4 -2.000000 slack c6 -1.000000 All other slacks in the range 1-6 are 0. CPLEX> display solution dual - Constraint Name Dual Price c1 1.000000 c3 1.000000 All other dual prices in the range 1-6 are 0. CPLEX> quit A comment on using CPLEX The fact that c1=0 and c3=0 indicates that the first and third constraints are binding. That c2=1 indicates the left-hand side of the second constraint is less than the right-hand-side by 1. Let’s check: Constraint 1: Constraint 2: Constraint 3:

  39. maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end CPLEX> primopt …  Primal simplex - Optimal: Objective = 1.3000000000e+01 …  CPLEX> display solution variables - Variable Name Solution Value x1 2.000000 x3 1.000000 All other variables in the range 1-3 are 0. CPLEX> display solution slacks - Constraint Name Slack Value slack c2 1.000000 slack c4 -2.000000 slack c6 -1.000000 All other slacks in the range 1-6 are 0. CPLEX> display solution dual - Constraint Name Dual Price c1 1.000000 c3 1.000000 All other dual prices in the range 1-6 are 0. CPLEX> quit A comment on using CPLEX The fact that c5=0 indicates that the second inequality constraint is binding, i.e., which is consistent with the fact that x2=0.

  40. maximize 5 x1 + 4 x2 + 3 x3 subject to 2 x1 + 3 x2 + x3 <= 5 4 x1 + x2 + 2x3 <= 11 3 x1 + 4 x2 + 2 x3 <= 8 x1 >= 0 x2 >= 0 x3 >= 0 end CPLEX> primopt …  Primal simplex - Optimal: Objective = 1.3000000000e+01 …  CPLEX> display solution variables - Variable Name Solution Value x1 2.000000 x3 1.000000 All other variables in the range 1-3 are 0. CPLEX> display solution slacks - Constraint Name Slack Value slack c2 1.000000 slack c4 -2.000000 slack c6 -1.000000 All other slacks in the range 1-6 are 0. CPLEX> display solution dual - Constraint Name Dual Price c1 1.000000 c3 1.000000 All other dual prices in the range 1-6 are 0. CPLEX> quit A comment on using CPLEX The facts that c4=-2, c6=-1 is interesting because these slacks are negative. This is a result of the fact that the corresponding constraints are actually “greater than or less to” constraints instead of “less than or equal to constraints.” The way they are treated in CPLEX is as follows: so that when x1=2,, c4=-2, and when x3=1, , c6=-1..

More Related