EE/Econ 458DualityJ. 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 x1is negative in above tableau).

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

-3

0

0

2.5

0

30

x3

1

0

1

0

1

0

0

4

x2

2

0

0

1

0

0.5

0

6

x5

3

0

3

0

0

-1

1

6

Add -2 pivot row

Add 5 pivot row

Our example (and HW) problem*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).And so the constraint identifying the leaving variable is the last one, since its ratio is smallest (2

Our example (and HW) problem*The leaving variable is x5, since it is the variable in the last constraint which gets pushed to 0 as x1 increases. Make the pivot element 1 by dividing the pivot row by 3.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=2Now what is the next step?

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

-3

0

0

2.5

0

30

x3

1

0

1

0

1

0

0

4

x2

2

0

0

1

0

0.5

0

6

x1

3

0

3

0

0

-1

1

6

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

-3

0

0

2.5

0

30

x3

1

0

1

0

1

0

0

4

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

Our example (and HW) problem*Use the last equation to eliminate the -3 in the top equation. 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

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

-3

0

0

2.5

0

30

x3

1

0

1

0

1

0

0

4

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

Our example (and HW) problem*Use the last equation to eliminate the 1 from the 2nd equation: 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

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

-3

0

0

2.5

0

30

x3

1

0

1

0

1

0

0

4

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

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

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

Introduction to duality*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) .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 ! (Important note: F*/bi is how the optimal value of F changes with bi.)

Introduction to duality*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.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?This always happens for linear programs. Here is a general rule:

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

0

0

0

1.5

1

36

x3

1

0

0

0

1

0.333

-0.333

2

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

Introduction to duality*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 i corresponding to the ith constraint.

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

0

0

0

1.5

1

36

x3

1

0

0

0

1

0.333

-0.333

2

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

Introduction to duality*From the left and below, we see that 3 =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.But we need to be a little careful 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 F1*=36CPLEX yields F5*=42CPLEX yields F6*=42

Basic variable

Eq. #

Coefficients of

Right side

F

x1

x2

x3

x4

x5

F

0

1

0

0

0

1.5

1

36

x3

1

0

0

0

1

0.333

-0.333

2

x2

2

0

0

1

0

0.5

0

6

x1

3

0

1

0

0

-0.333

0.333

2

Introduction to duality*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 dont 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. 2. F*/bi is not accurate if bi gets too large.A better definition of the dual variables follows:

Motivating the dual problem*Consider our now very familiar example problem (call it P):Lets express linear combinations of multiples of the constraints, where the constraint multipliers are denoted i on the ith constraint. We assume that i0 to avoid having to change the inequality.The composite inequalityP

Motivating the dual problem*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 i.Lets develop criteria for selecting 1, 2, 3. Concept 1: Make sure our choices of 1, 2, 3 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.

Motivating the dual problem*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.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*.

F*

Right-hand-side of composite inequality

0

Motivating the dual problem*Composite inequality:Concept 4: Now choose 1, 2, 3 to minimize the right-hand-side of the composite inequality, subject to the constraints we imposed in choosing the i s. 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).

F*

Right-hand-side of composite inequality

0

F*

0

F*

0

Motivating the dual problem*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 =0. So we have a new problem to solve: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*.

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

Motivating the dual problem*The CPLEX code to do it is below (note I use y instead of in code):minimize 1 y1 + 12 y2 + 18 y3subject to 1 y1 + 3 y3 >= 3 2 y2 + 2 y3 >=5 y1 >= 0 y2 >= 0 y3 >= 0endThe solution givesG*=36 1=0 2=1.5 3=1Using CPLEX, following solution of the above dual problem, I typed the following command: display solution dual CPLEX gave: Constraint Name Dual Pricec1 2.000000c2 6.000000What are these?They are:Coefficients of the SVs 4 , 5 of 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.

Motivating the dual problem*The CPLEX code to do it is below (note I use y instead of in code):The solution givesG*=36 1=0 2=1.5 3=1The solution givesG*=30 1=0 2=1 3=1Using CPLEX, following solution of the above dual problem, I typed the following command: display solution dual CPLEX gave: Constraint Name Dual Pricec1 2.000000c2 6.000000

Motivating the dual problem*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.

Primal-dual relationships*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.3. Coefficients of decision variables in primal objective are right-hand-sides of dual constraints.

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

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Primal-dual relationships*5. Coefficients of one variable across multiple dual constraints are coefficients of multiple variables in one primal constraint.6. If primal objective is maximization, dual objective is minimization7. If primal constraints are , dual constraints are .

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((

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((

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

Obtaining the dual*Primalmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endThe solution is (x1,x2,x3)=(2,0,1), F*=13.The primal SV coefficients, i.e., the values of the dual variables are: (1, 2, 3)=(1,0,1)minimize 5 y1 + 11 y2 + 8 y3subject 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 >= 0endDualThe solution is (1, 2, 3)=(1,0,1), G*=13.The dual SV coefficients, i.e., the values of the dual variables are: (x1,x2,x3)=(2,0,1).

Viewing primal-dual relationship*Generalize the primal-dual problems:

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Viewing primal-dual relationship*

Viewing primal-dual relationship*

The duality theorem*Duality theorem: If the primal problem has an optimal solution x*, then the dual problem has an optimal solution * such thatBut 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.

Unbounded vs. infeasibility*InfeasibleUnboundedCPLEX> read ex.lpProblem 'ex.lp' read.Read time = 0.00 sec.CPLEX> primoptBound infeasibility column 'x2'.Presolve time = 0.00 sec.Presolve - Unbounded or infeasible.Solution time = 0.00 sec.CPLEX>CPLEX> read ex1.lpProblem 'ex1.lp' read.Read time = 0.00 sec.CPLEX> primoptDual infeasible due to empty column 'x2'.Presolve time = 0.00 sec.Presolve - Unbounded or infeasible.Solution time = 0.00 sec.CPLEX>

An unbounded primal*Lets inspect the unbounded problem.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 and x. If F(x) is unbounded, then the only possibility for G() is that it must be infeasible.

An unbounded primal*Write down the dual.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.

Primal-Dual relations

DUALOptimalInfeasibleUnbounded

PRIMALOptimalPossibleImpossibleImpossibleInfeasibleImpossiblePossiblePossibleUnboundedImpossiblePossibleImpossible

A comment on using CPLEXmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endCPLEX> primoptPrimal simplex - Optimal: Objective = 1.3000000000e+01CPLEX> display solution variables -Variable Name Solution Valuex1 2.000000x3 1.000000All other variables in the range 1-3 are 0.CPLEX> display solution slacks -Constraint Name Slack Valueslack c2 1.000000slack c4 -2.000000slack c6 -1.000000All other slacks in the range 1-6 are 0.CPLEX> display solution dual -Constraint Name Dual Pricec1 1.000000c3 1.000000All other dual prices in the range 1-6 are 0.CPLEX> quitCPLEX will name the slack variables c1, c2, , cm, cm+1,, cm+nwhere 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.

A comment on using CPLEXmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endThe 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.CPLEX> primoptPrimal simplex - Optimal: Objective = 1.3000000000e+01CPLEX> display solution variables -Variable Name Solution Valuex1 2.000000x3 1.000000All other variables in the range 1-3 are 0.CPLEX> display solution slacks -Constraint Name Slack Valueslack c2 1.000000slack c4 -2.000000slack c6 -1.000000All other slacks in the range 1-6 are 0.CPLEX> display solution dual -Constraint Name Dual Pricec1 1.000000c3 1.000000All other dual prices in the range 1-6 are 0.CPLEX> quit

A comment on using CPLEXmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endThe 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.Lets check: CPLEX> primoptPrimal simplex - Optimal: Objective = 1.3000000000e+01CPLEX> display solution variables -Variable Name Solution Valuex1 2.000000x3 1.000000All other variables in the range 1-3 are 0.CPLEX> display solution slacks -Constraint Name Slack Valueslack c2 1.000000slack c4 -2.000000slack c6 -1.000000All other slacks in the range 1-6 are 0.CPLEX> display solution dual -Constraint Name Dual Pricec1 1.000000c3 1.000000All other dual prices in the range 1-6 are 0.CPLEX> quitConstraint 1:Constraint 2:Constraint 3:

A comment on using CPLEXmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endThe fact that c5=0 indicates that the second inequality constraint is binding, i.e., which is consistent with the fact that x2=0.CPLEX> primoptPrimal simplex - Optimal: Objective = 1.3000000000e+01CPLEX> display solution variables -Variable Name Solution Valuex1 2.000000x3 1.000000All other variables in the range 1-3 are 0.CPLEX> display solution slacks -Constraint Name Slack Valueslack c2 1.000000slack c4 -2.000000slack c6 -1.000000All other slacks in the range 1-6 are 0.CPLEX> display solution dual -Constraint Name Dual Pricec1 1.000000c3 1.000000All other dual prices in the range 1-6 are 0.CPLEX> quit

A comment on using CPLEXmaximize 5 x1 + 4 x2 + 3 x3subject to 2 x1 + 3 x2 + x3 = 0endThe 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:CPLEX> primoptPrimal simplex - Optimal: Objective = 1.3000000000e+01CPLEX> display solution variables -Variable Name Solution Valuex1 2.000000x3 1.000000All other variables in the range 1-3 are 0.CPLEX> display solution slacks -Constraint Name Slack Valueslack c2 1.000000slack c4 -2.000000slack c6 -1.000000All other slacks in the range 1-6 are 0.CPLEX> display solution dual -Constraint Name Dual Pricec1 1.000000c3 1.000000All other dual prices in the range 1-6 are 0.CPLEX> quitso that when x1=2,, c4=-2, andwhen x3=1, , c6=-1..