Multicriteria Decision Making Notes

Sándor Zoltán NémethSchool of Mathematics, University of Birmingham

Watson Building, Room 324Email: s.nemeth@bham.ac.uk

Decisions, decisions . . .

Decision are everywhere to make. . .

. . . and some of them are pretty hard!

Contents

1 Some organisational stuff 2

2 What is Multicriteria Decision Making? 32.1 Decision Making in the Real World . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

3 How to model a Multicriteria Decision Making problem? 63.1 Some general ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 How to order vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73.3 Classical examples of convex cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.3.1 Polyhedral cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123.3.2 Simplicial cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.3.3 The Lorentz cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3.4 Cone generated by a set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3.4 The set of efficient points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.5 Topological properties of the set of efficient points . . . . . . . . . . . . . . . . . . . . . . . 17

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4 Examples for Multicriteria Decision Making Problems 19

5 Computing efficient points 245.1 The dual cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245.2 Examples of classical dual cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

5.2.1 Geometrical interpretation of dual cones . . . . . . . . . . . . . . . . . . . . . . . . 245.2.2 Dual of closed convex cones in RRR2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255.2.3 Dual of the nonnegative orthant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.4 Dual of the monotone nonnegative cone . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.5 Dual of the Lorentz cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.6 Dual of polyhedral cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265.2.7 Any dual of a set is a dual of a closed convex cone . . . . . . . . . . . . . . . . . . . 28

5.3 Conditions for the efficiency of a point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295.3.1 Finding efficient points by using K-monotone functions . . . . . . . . . . . . . . . . 33

5.4 The set of properly efficient points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355.4.1 Geometical interpretation of the set of properly efficient points . . . . . . . . . . . . 37

5.5 Finding efficient points by minimising norms . . . . . . . . . . . . . . . . . . . . . . . . . . 395.5.1 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.5.2 Examples of norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 395.5.3 Any Minkowski functional is a norm . . . . . . . . . . . . . . . . . . . . . . . . . . 405.5.4 Cone-monotone norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415.5.5 A necessary and sufficient condition for efficiency . . . . . . . . . . . . . . . . . . . 425.5.6 How to choose u when K = Rn+? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

1 Some organisational stuff

• 2 class tests of 25% each for 3GTMDM and 22.5% each for 4GTMDM (i.e., fourth year version ofthe module).

• 50% examination for 3GTMDM and 45% for 4GTMDM.

• 10% extra task for 4GTMDM.

• More details later.

• Assessment arrangements are subject to possible change.

Recommended Literature

• M. Ehrgott: Multicriteria Optimization, Springer, 2005.

• P. C. Fishburn: Mathematics of decision theory. Mouton, 1972.

• A. Goepfert, C. Tammer, C. Zalinescu, H. Riahi: Variational methods in partiallyordered spaces. Springer-Verlag, 2003.

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• C. Hillermeier: Nonlinear Multiobjective Optimization: a Generalized HomotopyApproach. Birkhaeuser Verlag, 2001.

• J. Jahn: Vector Optimization: Theory, Applications, and Extensions. Springer-Verlag, 2004.

• K. Miettinen: Nonlinear Multiobjective Optimization. Kluwer, 1999.

• Y. Sawaragi, H. Nakayama, T. Tanino: Theory of Multiobjective Optimization.Academic Press, 1985.

• R. E. Steuer: Multiple Criteria Optimization: Theory, Computations, andApplication. Wiley, 1986.

• http://www.convexoptimization.com/wikimization/index.php/Farkas’_lemma#Extended_Farkas.27_lemma

2 What is Multicriteria Decision Making?

Multicriteria Decision Making is the mathematical theory of decision making with conflicting objectivesand/or decision makers which conflict each other.

Typical examples include

• portfolio optimization (return ↔ risk)

• power plant control (cost ↔ wear & tear)

• hazardous material deposing/routing (cost ↔ risk)

• health care management (efficiency ↔ cost)

• material optimization (stability ↔ cost)

• parameter estimation (each equation one objective function)

• any model including at least two of the following: cost, risk, stability, etc. . . .

What is Multicriteria Decision Making for?

• Understand why the ”decision makers” behave as they do. (Thereby understanding, e. g., economicdecisions better.)

• Help the decision maker (or even replace him with an automated procedure!): find the ”best” decision(if there is a ”best” way).

(The second reason is the more practical one. . . )

Some other names for the topic:

Multicriteria Decision Making, Multiple Criteria Decision Making, Multicriteria Optimisation,Multiobjective Decision Making, Multiobjective Optimisation, Vector Optimisation, Decision Theory, . . .

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2.1 Decision Making in the Real World

When the Naskapi Indians (Canada) had to send a hunting party out, they had to decide upon a searchdirection for the party.

For this, the shoulder bones of a caribou was held over hot coals causing cracks in the bone which are thenused to direct a hunting party.

This procedure worked!

”Worked” means ”a decision has been made”, not ”the best decision has been made”.

The fact that a decision is made is a relief, a comfort; it is satisfying to remove uncertainty and indecision.Any outcome can be satisfying.

Can we do better??

A (very first) example: buying a car

Please help me in buying a car.

There are (only) the criteria prize and style.

The market has been boiled down to the following models with the following prices:

A = Mini: £12K

B = Golf: £14K

C = Alfa Romeo: £21K

D = Mitsubishi: £9K

E = Fiat: £8K

F = Jaguar: £25K

Style: let’s do a poll −→

Question 1: which car would you NOT choose?

Question 2: which car would you choose?

Let us now use some more criteria: price, size, reliability, style (aesthetics).

But let us simplify the market to four models: α, β, γ, δ.

For each criterium, one can rank each car (1 = best, . . . , 4 = worst).α β γ δ

price 1 3 2 4size 1 4 3 2

reliability 2 3 4 1style 4 2 1 3

Which car would you choose ??

This is just a simple example: add criteria comfort, risk of theft, weekly costs, safety, depreciation, . . .

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. . . and replace style by status and colour . . .

Another motivating example

Suppose a decision maker can invest a certain amount of money on the stock market. Let x denote thedecision taken.

It will usually be the case that the decision maker tries to optimise several objectives at once, i. e.

reward (expected return): %(x) = f1(x) −→ maxrisk (variance of return): R(x) = f2(x) −→ min

dividends: f3(x) −→ maxsocial responsibility: f4(x) −→ maxnumber of securities: f5(x) −→ max

short selling: f6(x) −→ minsecurities sold short: f7(x) −→ min

liquidity: f8(x) −→ maxskewness: f9(x) −→ minturnover: f10(x) −→ min

amount invested in R&D: f11(x) −→ maxgrowth in sales: f12(x) −→ max

And another motivating example

(Based on a true story, Illinois, USA, 1977)

A region of undeveloped land (close to a suburban area of Chicago) is under consideration. It is dividedinto several different planning regions. Goal: assign a certain land use (small residential, large residential,commercial, offices, manufacturing, schools, open space) to each region.

Objectives:

Maximise compatibility of assignment. (Not every region is as usable as every other.)

Minimise expected total transportation.

Minimise tax load (operating cost).

Minimise environmental impact.

Minimise facility cost.

So far we have seen:

Multicriteria Decision Making looks more difficult than ”standard” decision making.

We will see:

Multicriteria Decision Making IS more difficult than ”standard” decision making . . .

. . . but it is still within our reach!

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3 How to model a Multicriteria Decision Making problem?

3.1 Some general ideas

General scenario: we have n > 1 functions f1, . . . , fn : X −→ R (”criteria”) to be minimised. (Ex.: X:set of feasible portfolios).

(We deal only with minimisation, just for simplicity. For maximisation, multiply criteria with −1.)

We also write f : X −→ Rn, with f(x) = [f1(x), . . . , fn(x)]>.

Example: above, n = 2, f1 = −%, f2 = R.

Observation: only in very special cases does there exist an x ∈ X which minimises all criteria f1, . . . , fnsimultaneously.=⇒

We need a new concept of solution.

What is a solution to a multicriteria problem??

Example: n = 2, X = [−1, 1 ], f1(x) = −x/2, f2(x) = −(5− x2)1/2.−→

Some solution ideas:

1. Build tradeoff model: choose tradeoff parameter µ and solve minx∈X µf1(x) + f2(x).

2. Bound one function value: choose parameter % and minimise f2(x) subject to x ∈ X, f1(x) ≤ %.3. Bound the other function value: choose parameter % and minimise f1(x) subject to x ∈ X, f2(x) ≤ %.4. Every x ∈ [ 0, 1 ] is a solution: changing x makes at least one fi larger.5. Minimise f1 first, then f2.

6. Minimise f2 first, then f1.

7. Worst-case approach: solve minx∈X maxi fi(x).

8. Weight importance in the worst case approach: choose weights µi and solveminx∈X maxi µifi(x).

9. Solve minx∈X((f1(x))2 + (f2(x))

2)1/2.

10. Solve minx∈X(µ1(f1(x))2 + µ2(f2(x))

2)1/2.

Lots of open questions:

1. Which approach is the right one?

2. Are these different ideas/approaches interconnected with each other?

3. What are the fundamental differences between them?

4. Or are some of them equivalent?

5. What are the advantages and disadvantages of the different approaches?

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6. Why are there so many of them??

7. Are there even more???

8. . . .

So many different approaches are a clear sign for the need of a good mathematical theory!

• There are (literally!) hundreds of different solution methods for multicriteria decision making.

• We can not discuss all or even many of them.

• Emphasis will be on understanding the main concepts and their theoretical underpinnings.

• If you understand this theory, practical applications and new methods will rarely suprise you.

• Many solution methods need a method from Linear Programming & Combinatorial Optimisation orNonlinear Programming as a “subroutine”. (Hint: we can’t cover everything in one lecture. . . )

3.2 How to order vectors

Recall the situation in R1 = R (i. e. n = 1): f : X −→ R has to be minimised. The image set R isequipped with a relation ”≤” with has a meaning to the decision maker:

Let f(x) = v and f(y) = w and v ≤ w. Then v (resp. x) is at least as good as w (resp. y).

(Or, if v 6= w, the value v (resp. x) is preferred to w (resp. y): ≤ is a preference relation. Or, if v 6= w, thevalue v (resp. x) dominates w (resp. y).)

x, y: decisions

v = f(x), w = f(y): values of the decisions

=⇒

We are in search for a relation (order) � on Rn (with n > 1) that has some meaning for the decision-maker.(And that also has nice mathematical and computational properties, see below!)

Decision makers want, first and all, two properties:

1. � should be translation invariant, i. e. for all v, w ∈ Rn:

v � w =⇒ v + u � w + u for all u ∈ Rn

(”compatible with addition”)

2. � should be scale invariant, i. e. for all v, w ∈ Rn:

v � w & λ > 0 =⇒ λv � λw.

(”compatible with positive scalar multiplication”)

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Let K ⊆ Rn be given. Define the relation ≤K by

v ≤K w :⇐⇒ w − v ∈ K (1)

(The order ≤K is induced by K.)

Example: n = 1, K = R+, standard order ≤.(Idea: K simple =⇒ x ≤K y easy to check.)

Example: −→

Theorem 1. A relation � is translation invariant if and only if it is induced by a set K, i. e. if � = ≤Kholds.

Proof. Suppose that � is translation invariant. Then, x � y implies 0 = x + (−x) � y + (−x) = y − x.Hence, y − x ∈ K := {z ∈ Rn | 0 � z} and thus x ≤K y. On the other hand x ≤K y implies y − x ∈ Kand thus 0 � y− x. Hence, x = 0 + x � (y− x) + x = y. In conclusion x � y is equivalent to x ≤K y andtherefore �=≤K .Conversely, suppose that �=≤K , for some set K ⊆ Rn. Hence, x � y implies x ≤K y and thus(y + z) − (x + z) = y − x ∈ K. It follows that x + z ≤K y + z which implies x + z � y + z, for anyz ∈ Rn. Therefore, � is translation invariant. �

Definition. A set K ⊆ Rn is called a cone if for all v ∈ K and all λ ∈ R, λ > 0, we have λv ∈ K. Acone K is called pointed if K ∩ −K ⊆ {0}.

We remark that a cone K is pointed if and only if there is no a 6= 0 such that a,−a ∈ K.

Corollary 1. The relation ≤K is scale invariant if and only if K is a cone.

Proof. Suppose that ≤K is scale invariant. Let x ∈ K and λ ∈ R with λ > 0. Then, 0 ≤K x. Hence,0 = λ0 ≤K λx, which implies λx ∈ K. Hence, K is a cone.Conversely, suppose that K is a cone. Let x, y ∈ Rn with x ≤K y and λ ∈ R with λ > 0. Then,y − x ∈ K and since K is a cone we have λy − λx = λ(y − x) ∈ K. Hence, λx ≤K λy. Therefore, ≤K isscale invariant. �

Example: The nonnegative quadrant

K = R2+ := {x = (x1, x2)> ∈ R2 | x1 ≥ 0 and x2 ≥ 0}

is a cone which is a convex set. Such cones are simply called convex cones. The relation induced by K isdefined by v ≤K w ⇐⇒ w − v ∈ K ⇐⇒ w1 ≥ v1 and w2 ≥ v2. It is easy to check that ≤K is a scaleinvariant relation.

Example: The setK = {x = (x1, x2)> ∈ R2 | x1 ≥ 0 and − x2 ≤ x1}

is a cone which is a convex set, simply called convex cone. The relation induced by K is defined byv ≤K w ⇐⇒ w− v ∈ K ⇐⇒ w1 − v1 ≥ 0 and v2 −w2 ≤ w1 − v1. It is easy to check that this is a scaleinvariant relation. Indeed, w1 − v1 ≥ 0 and v2 − w2 ≤ w1 − v1 implies

λw1 − λv1 = λ(w1 − v1) ≥ 0 and λv2 − λw2 = λ(v2 − w2) ≤ λ(w1 − v1) = λw1 − λv1.

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Hence, λv ≤K λw.Example: The set

K = {(t, 0)> ∈ R2 | t ≥ 0} ∪ {(0, t)> ∈ R2 | t ≥ 0}is a cone, but not a convex cone. It is easy to check that K is a cone. K is not convex because a :=(2, 0)> ∈ K, b := (0, 2)> ∈ K, but (1/2)(a+ b) = (1, 1)> /∈ K.Example: The set

K = {x = (x1, x2)> ∈ R2 | x1 ≥ 0 or x2 ≥ 0}is a cone, but not a convex cone. It is easy to check that K is a cone. K is not convex because a :=(−2, 0)> ∈ K, b := (0,−2)> ∈ K, but (1/2)(a+ b) = (−1,−1)> /∈ K.Example: The set

K = {x = (x1, x2)> ∈ R2 | 0 ≤ x1 ≤ 1 and 0 ≤ x2 ≤ 1}is not a cone. Indeed, (1, 1)> ∈ K, but 2(1, 1)> = (2, 2)> /∈ K. Observe that K is a bounded set. Therelation induced by K is defined by v ≤K w ⇐⇒ w− v ∈ K ⇐⇒ 0 ≤ w1− v1 ≤ 1 and 0 ≤ w2− v2 ≤ 1.As expected this relation is not translation invariant because (0, 0)> ≤K (1, 1)> but (0, 0)> = 2(0, 0)> 6≤K2(1, 1)> = (2, 2)>.

Example: The setK = {x = (x1, x2)> ∈ R2 | x21 + x22 ≤ 1}

is not a cone. Indeed (1/2, 1/2)> ∈ K, but 2(1/2, 1/2)> = (1, 1) /∈ K. Observe that K is a bounded set.Example: The only nonempty bounded cone in Rn is the trivial cone K = {0}. Indeed, {0} is a boundedcone. If K 6= {0} is a nonempty cone than ∃ v ∈ K \ {0}. Hence, vk := kv ∈ K, for any positive integerk. But ‖vk‖ = k‖v‖ → ∞, when k →∞. Thus, K is not bounded.Example: The nonnegative orthant

K = Rn+ := {x = (x1, . . . , xn)> ∈ Rn | x1 ≥ 0, . . . , xn ≥ 0}

is a convex cone.

The relation induced by K is defined by v ≤K w ⇐⇒ w − v ∈ K ⇐⇒ w1 ≥ v1, . . . , wn ≥ vn. It is easyto check that ≤K is a scale invariant relation. This relation is called standard or componentwise order.For A,B ⊂ Rn, v ∈ Rn and λ ∈ R denote

λA := {λa | a ∈ A},

−A = (−1)A = {−a | a ∈ A},A+B := {a+ b | a ∈ A, b ∈ B},

A−B := A+ (−B) = {a− b | a ∈ A, b ∈ B},v +B := {v}+B = {v + b | b ∈ B},v −B := v + (−B) = {v − b | b ∈ B}.

Remark 1. Let K ⊆ Rn. Note that

K = {x ∈ Rn | 0 ≤K x} = {x ∈ Rn | 0 is at least as good as x},−K = {x ∈ Rn | x ≤K 0} = {x ∈ Rn | x is at least as good as 0},

v +K = {w ∈ Rn | v ≤K w} = {w ∈ Rn | v is at least as good as w},v −K = {w ∈ Rn | w ≤K v} = {w ∈ Rn | w is at least as good as v},

v +K \ {0} = {w ∈ Rn | v ≤K w & v 6= w} = {w ∈ Rn | w is dominated by v},v −K \ {0} = {w ∈ Rn | w ≤K v & v 6= w} = {w ∈ Rn | w is dominating v}.

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Reminder. Recall the following properties of the relation � on Rn:

1. The relation � is called reflexive if v � v, ∀v ∈ Rn.

2. The relation � is called transitive if u � v and v � w implies u � w.

3. The relation � is called antisymmetric if v � w and w � v implies v = w

4. The relation � is called total if v � w or w � v, ∀v, w ∈ Rn (in other words any two elements in Rnare comparable).

The properties of reflexivity, transitivity, antisymmetry and totality are also “desirable” properties for ≤K .When do these properties hold? The answer is given by the next theorem.

Theorem 2. Let K ⊆ Rn be a nonempty set (but not necessarily a cone) and ≤K be the binary relationdefined by (1). Then, the following holds:

1. The order ≤K is reflexive if and only if 0 ∈ K.

2. The relation ≤K is transitive if and only if {v + w | v, w ∈ K} =: K +K ⊆ K.

3. The relation ≤K is antisymmetric if and only if K ∩ −K ⊆ {0}. In particular if K is a cone, then≤K is antisymmetric if and only if K is pointed.

4. The order ≤K is total if and only if the equality K ∪ −K = Rn holds.

5. The set K is closed if and only if the relation ≤K is ”continuous at 0” in the following sense: Forall b ∈ Rn and all sequences (wi)i∈N in Rn with limi→∞wi = b and 0 ≤K wi for all i ∈ N it followsthat 0 ≤K b holds.

Proof.

1. Suppose that ≤K is reflexive. Then, 0 ≤K 0, or equivalently 0− 0 ∈ K. Hence, 0 ∈ K.

Conversely, suppose that 0 ∈ K. It follows that v − v ∈ K, ∀v ∈ Rn. Hence, v ≤K v, ∀v ∈ Rn, orequivalently ≤K is reflexive.

2. Suppose that ≤K is transitive. Let v + w ∈ K + K be arbitrary. This means that v, w ∈ K arearbitrary. We have −w ≤K 0 and 0 ≤K v, which by the transitivity of ≤K implies −w ≤K v. Hence,v + w ∈ K. Therefore, K +K ⊆ K.

Conversely, suppose that K +K ⊆ K. Then, u ≤K v and v ≤K w imply v− u ∈ K and w− v ∈ K,respectively. Hence, w − u = (w − v) + (v − u) ∈ K + K ⊆ K. Thus, u ≤K w. Therefore, ≤K istransitive.

3. Suppose that K ∩ −K ⊆ {0}. Then, v ≤K w and w ≤K v imply w − v ∈ K and v − w ∈ K,respectively. Thus, w − v ∈ K and w − v ∈ −K. Hence, w − v ∈ K ∩ −K ⊆ {0}. Therefore,w − v = 0, or equivalently v = w.

Conversely, suppose that ≤K is antisymmetric. If K ∩ −K = ∅, then obviously K ∩ −K ⊆ {0}. IfK ∩ −K 6= ∅, then let v ∈ K ∩ −K be arbitrary. Hence, v ∈ K and v ∈ −K. Thus, v ∈ K and−v ∈ K, which imply 0 ≤K v and v ≤K 0. Since ≤K is antisymmetric, we obtain v = 0 ∈ {0}.Therefore, K ∩ −K ⊆ {0}.

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4. Suppose that ≤K is total. We only need to show that Rn ⊆ K∪−K, because K∪−K ⊆ Rn triviallyholds. Let v ∈ Rn be arbitrary. Since ≤K is total, we have 0 ≤K v or v ≤K 0. Hence, v ∈ K or−v ∈ K. Thus, v ∈ K or v ∈ −K, which implies v ∈ K ∪ −K. Therefore, Rn ⊆ K ∪ −K.

Conversely, suppose that K ∪ −K = Rn. Let v, w ∈ Rn be arbitrary. Then,w − v ∈ Rn = K ∪ −K.

Hence, w−v ∈ K or w−v ∈ −K. Thus, w−v ∈ K or v−w ∈ K, which implies v ≤K w or w ≤K v,respectively. Therefore, ≤K is total.

5. The set K is closed if and only if for all b ∈ Rn and all sequences (wi)i∈N in Rn with limi→∞wi = band wi ∈ K for all i ∈ N it follows that b ∈ K holds. Hence, the equivalence follows by observingthat wi ∈ K and b ∈ K are equivalent to 0 ≤K wi and 0 ≤K b, respectively.

�

Definition. A set S ⊆ Rn is called connected if S = A ∪ B and A ∩ B = ∅ imply that at least one of Aand B is not closed.

Recall that if A ∩B = ∅ and S = A ∪B, then we say that S is the disjoint union of A and B. Hence, Sis connected if and only if it cannot be written as a disjoint union of two closed sets.

Example: It is known that the unit sphere S := {v ∈ Rn | ‖v‖2 = 1} is connected.

One can’t have everything:

Suppose that K is a pointed cone such that ≤K is total.Then: v ∈ K, v 6= 0 =⇒ −v /∈ K.Consider S := {v ∈ Rn | ‖v‖2 = 1}.≤K total =⇒ S = S ∩ Rn = S ∩ (K ∪ −K) = (S ∩K) ∪ (S ∩ −K).But (S ∩K) ∩ (S ∩ −K) = S ∩ (K ∩ −K) ⊆ S ∩ {0} = ∅= ∅= ∅ (because 0 6∈ S).Therefore S = S ∩ (K ∪ −K) = (S ∩K) ∪ (S ∩ −K) is a disjoint union.Consequence: K is not closed (otherwise S would be a disjoint union of two closed sets, which is impossiblebecause S is connected).

=⇒≤K is not continuous at 0.One can not have translation invariance, scale invariance, antisymmetry, totality as well as continuity!

Example: The componentwise order in Rn seems to satisfy all properties listed before. Hang on a minute,maybe not...? If not, then which property is not satisfied?

Example: The lexicographic cone in Rn is is defined by

L := {(x1, . . . , xn)> ∈ Rn | x1 > 0 or(x1 = 0 & x2 > 0) or(x1 = x2 = 0 & x3 > 0) or...(x1 = x2 = . . . xn−1 = 0 & xn > 0) orx = 0}

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u ≤L v ⇐⇒

u1 < v1 or(u1 = v1 & u2 < v2) or(u1 = v1, u2 = v2 & u3 < v3) or...(u1 = v1, u2 = v2, . . . un−1 = vn−1 & un < vn) oru = v

The relation ≤L is called lexicographic order. The relation ≤L seems to satisfy all properties listed before.Hang on a minute, maybe not...? If not, then which property is not satisfied?

Example: Let K = {x ∈ Rn | xn ≥ 0}. The relation ≤L seems to satisfy all properties listed before.Hang on a minute, maybe not...? If not, then which property is not satisfied?

Discussion

• Translation invariance & scale invariance too important: use a cone K.

• Continuity more important than totality: use a closed set K.

• Transitivity is natural: use a set K with K +K ⊆ K.

Theorem 3. A cone K is convex if and only if K +K ⊆ K holds.

Proof. Let K be a cone.

Suppose that K is convex. Take an arbitrary x + y ∈ K + K, which means that x, y ∈ K are arbitrary.Since K is a cone 2x, 2y ∈ K and since K is convex x+ y = (1/2)(2x+ 2y) ∈ K. Therefore, K +K ⊆ K.Conversely, suppose that K + K ⊆ K holds. Let 0 < λ < 1 and x, y ∈ K be arbitrary. Then, λ > 0 and1 − λ > 0. Hence, λx ∈ K and (1 − λ)y ∈ K. Thus, λx+ (1− λ)y ∈ K + K ⊆ K. Therefore, K isconvex. �

Conclusion

From now on, the decision maker should choose (somehow) a closed convex cone K to define his preferencerelation ≤K .

Assumption: Let K 6= ∅ be a closed convex cone with K 6= {0} and K 6= Rn.

3.3 Classical examples of convex cones

3.3.1 Polyhedral cones

Let u1, . . . , um ∈ Rn. Then, the set

K = {λ1u1 + . . . λmum | λ1 ≥ 0, . . . , λm ≥ 0}

is denoted bycone{u1, . . . , um}

and it is a closed convex cone cone, called the polyhedral cone generated by the vectors u1, . . . , um. Thevectors u1, . . . , um are called the generators of the polyhedral cone. It can be proved that each polyhedral

12

cone is the intersection of a finite number of closed half spaces (highly nontrivial to show). This is fromwhere it follows that the cone{u1, . . . , um} is closed. However, the convexity of cone{u1, . . . , um} can beshown easily from its definition as follows. If

x = λ1u1 + . . . λmu

m ∈ K,

y = µ1u1 + . . . µmu

m ∈ K

and λ > 0, thenλx = (λλ1)u

1 + · · ·+ (λλm)um ∈ K,

(because λλ1 ≥ 0 . . . λλm ≥ 0) and

x+ y = (λ1 + µ1)u1 + · · ·+ (λm + µm)um} ∈ K

(because λ1 + µ1 ≥ 0, . . . , λm + µm ≥ 0). Let u1, . . . , um ∈ Rn. Then,

U = {x ∈ Rn | (u1)>x ≥ 0, . . . , (um)>x ≥ 0}

is a closed convex cone. This follows easily because U is an intersection of closed half spaces which arealso closed convex cones. Indeed, it is easy to see from the corresponding definitions that the properties“closed”, “convex” and “cone” of a set are preserved by intersection. In fact U is a polyhedral cone (highlynontrivial to show). If you change some of the inequalities in the definition of U into strict ones you willalso get a convex cone.

3.3.2 Simplicial cones

Let u1, . . . , un ∈ Rn be linearly independent. Then,

K = cone{u1, . . . , un}

is called a simplicial cone which is a particular polyhedral cone. The name follows from the observationthat cutting K with a hyperplane which does not contain the origin one obtains a simplex: triangle in3 dimensions, tethraedron in 4 dimensions,.... The convex hull of w1, w2, . . . , wn, wn+1 ∈ Rn such thatw2 − w1, . . . , wn − w1, wn+1 − w1 are linearly independent is called a simplex. It can be shown thatsimplicial cones are of the form

K = {x ∈ Rn | (v1)>x ≥ 0, . . . , (vn)>x ≥ 0},

where v1, . . . , vn ∈ Rn are linearly independent.

Classical simplicial cones

• Nonnegative orthant:

Rn+ = {x = (x1, . . . , xn) ∈ Rn | x1 ≥ 0, . . . , xn ≥ 0}.

• Monotone nonnegative cone:

Rn≥+ = {x = (x1, . . . , xn) ∈ Rn | x1 ≥ x2 ≥ · · · ≥ xn ≥ 0}.

13

3.3.3 The Lorentz cone

The Lorentz cone is also called second order cone or icecream cone. It is defined by

L = {( xxm+1 ) ∈ Rm+1 | xm+1 ≥ ‖x‖},

where ‖ · ‖ is the Euclidean norm in Rm. From the definition it can be easily seen that L is closed. The setL is a convex cone. Indeed, if ( xxm+1 ) ∈ L and λ > 0, then λ ( xxm+1 ) =

(λx

λxm+1

)∈ L because xm+1 ≥ ‖x‖

implies λxm+1 ≥ λ‖x‖ = ‖λx‖. Hence L is a cone. Moreover, if ( xxm+1 ) ∈ L and ( yym+1 ) ∈ L, then(x+y

xm+1+ym+1

)= ( xxm+1 ) + (

yym+1 ) ∈ L because xm+1 + ym+1 ≥ ‖x‖ + ‖y‖ ≥ ‖x + y‖, where we used the

triangle inequality. Hence, L is a convex cone. The Lorentz cone is an example of a convex cone which isnot polyhedral.

x

x

x1

2

3

O

a

Figure: Lorentz cone in 3 dimensions. The angle between Oa and Ox3 is 45 degrees.

3.3.4 Cone generated by a set

If A ⊂ Rn is a nonempty set, then denote by closure(A) the smallest closed set (with respect to theinclusion of sets) containing A. Then,

cone(A) := closure{λ1u1 + . . . λmum | m ∈ N and u1, . . . , um ∈ A and λ1, . . . , λm > 0}

is a convex cone, called the cone generated by A. It can be shown that it is the smallest closed convexcone (with respect to the inclusion of sets) containing A.

3.4 The set of efficient points

Remark For v, w ∈ Rn we might have v 6≤K w and w 6≤K v: no totality!If n = 1 and K = R+, we know what ”minimal” means.

What about the more general case here??

Definition. Let M ⊆ Rn be a set. The set of minimal elements of M with respect to ≤K (resp. K) is

E(M,K) := {v ∈M |6 ∃w ∈M : w ≤K v & w 6= v}.

14

Remark Minimal elements are also called ”efficient”, ”preferred”, ”dominating”, ”Pareto-optimal”,”Pareto-minimal”, ”Edgeworth-Pareto-optimal”, etc.

We have

E(M,K) = {v ∈M |6 ∃w ∈M : w ≤K v & w 6= v}= {v ∈M | ∀w ∈M,w 6= v : w 6≤K v}= {v ∈M | ∀w ∈M,w 6= v : v − w /∈ K}= {v ∈M | ∀w ∈M : v − w /∈ K \ {0}}= {v ∈M | (v −M) ∩ (K \ {0}) = ∅}= {v ∈M |M ∩ (v −K \ {0}) = ∅}.

To show this denote the sets on the right hand sides of the equalities above from top to bottom byE1, E2, E3, E4, E5, E6. E1 is the original definition of E(M,K). Hence, we only need to show thatE1 = E2 = E3 = E4 = E5 = E6.

E1 and E2 logically mean the same thing. Indeed, saying that 6 ∃w ∈ M : w ≤K v & w 6= v (i.e., v ∈ E1)it is the same as saying that ∀w ∈ M : w 6≤K v or w = v, which means that ∀w ∈ M,w 6= v : w 6≤K v(i.e., v ∈ E2). Hence, E1 = E2.The relation w 6≤K v is equivalent to v − w 6∈ K, hence E2 = E3 (because the remaining parts of E2 andE3 are identical).

E3 and E4 logically mean the same thing. Indeed saying that ∀w ∈M,w 6= v : v−w /∈ K (i.e., v ∈ E3) itis the same as saying that ∀w ∈M : v−w = 0 or v−w /∈ K, which means that ∀w ∈M : v−w /∈ K \{0}(i.e., v ∈ E4). We used that v − w ∈ K \ {0} is equivalent to v − w ∈ K and v − w 6= 0, which negatedgives that v − w /∈ K \ {0} is equivalent to v − w = 0 or v − w /∈ K. Hence, E3 = E4.The relation v /∈ E6 is equivalent to ∃w ∈M ∩ (v−K \{0}), which means that ∃w ∈M : w ∈ v−K \{0}.Hence, v /∈ E6 if and only if ∃w ∈M : v−w ∈ K\{0}, which is equivalent to ∃w ∈M : v−w ∈ K & w 6= v,or to v /∈ E1. Thus, E1 = E6. It can similarly be shown that E1 = E5.In conclusion we have E1 = E2 = E3 = E4 = E5 = E6.

Example: Let M = [−1, 1]× [−1, 1] ⊆ R2 and K = R2+. Then, E(M,K) = {(−1,−1)>}.Example: Let M = {v ∈ R2 | v21 + v22 ≤ 1} and K = R2+. Then,

E(M,K) = {v ∈ R2 | v1 ≤ 0, v2 ≤ 0, v21 + v22 = 1}

Example: Let M = M1 ∪M2 ∪M3, where

M1 = {(x1, x2)> ∈ R2 | x1 ≥ x2 and x2 ≥ 0 and x2 ≥ 2x1 − 2},

M2 = {(0, λ)> ∈ R2 | −2 ≤ λ ≤ 0},

M3 = {(λ,−2)> ∈ R2 | 0 ≤ λ ≤ 1}

and

K1 = {(x1, x2)> ∈ R2 | x1 > x2 and x2 ≥ 0},

K2 = {(x1, x2)> ∈ R2 | x1 ≥ x2 and x2 > 0},

15

K3 = {(λ, 0)> ∈ R2 | λ ≥ 0} ∪ {(0, λ)> ∈ R2 | λ ≥ 0},

K4 = {(x1, x2)> ∈ R2 | x1 > 0 or (x1 = 0 & x2 > 0). or x1 = x2 = 0}(lexicographic cone)

Determine the set of efficient points E(M,Ki) for all i ∈ {1, 2, 3, 4}.Help me sketch the set set of efficient points on the board!

Note that M is usually the set of possible values of decisions: v, w ∈ M , v = f(x), w = f(y). Hence,solving a decision making problem means that we have a set X of decisions and we want to choose theoptimal decisions. In order to do so, we need to be given (or for a practical problem often identify byourselves) a set of objective functions f1, · · · , fn : X → R. Let f := (f1, · · · , fn) : X → Rn. UsuallyX is a subset of another finite dimensional space Rm, or it can be identified with such a subset. Solvingour problem means that a cone K ⊆ Rn is given (if no cone is given we assume that the cone is Rn+; forpractical problems we need to decide which cone to use), we need to find the image set

M = f(X) := {f(x) | x ∈ X},

the set of efficient points E(M,K) of M with respect to the order ≤K induced by K (or simply withrespect to K) and the decisions

f−1(E(M,K)) = {x ∈ X | f(x) ∈ E(M,K)}

corresponding to E(M,K), that is, the preimage set. A part of this process is illustrated in the next figure.

K

f

X M=f(X)

Remark Let v, w ∈ E(M,K), v 6= w. Then v 6≤K w and w 6≤K v: efficient points are not comparablewith each other. (This feature is called ”inner stability”.)

=⇒The decision maker can not compare efficient elements with each other without resorting to measuresoutside the mathematical model!

RemarkK1 ⊆ K2 =⇒ E(M,K2) ⊆ E(M,K1)

(”larger cone =⇒ weaker ordering =⇒ less minimal elements”)

16

3.5 Topological properties of the set of efficient points

Theorem 4. The set E(M,K) is a subset of the boundary of M .

Proof. Denote the boundary of M by ∂M and the interior of M by int(M). We need to show thatE(M,K) ⊆ ∂M . Suppose to the contrary that there is a v ∈ E(M,K) such that v /∈ ∂M . Then,v ∈M \ ∂M = intM . Since, K 6= ∅ and K 6= {0} there exists u ∈ K with u 6= 0. Let w = v − λu, whereλ is a positive real number. Since v ∈ int(M), u 6= 0 and K is a cone, it follows thatv 6= w = v − λu ∈M ∩ (v −K \ {0}) for λ sufficiently small. Thus, w ∈M and it is dominating v, whichcontradicts v ∈ E(M,K). Hence, E(M,K) ⊆ ∂M . �

Example: −→Example: −→

Reminder: A set S ⊆ Rn is called connected if there do not exist open sets U1, U2 ⊂ Rn such thatS ⊆ U1 ∪ U2 and

S ∩ U1 6= ∅, S ∩ U2 6= ∅, S ∩ U1 ∩ U2 = ∅.Examples: −→

Is E(M,K) connected ?

Not always, but:

Theorem 5. Let M ⊆ Rn be convex, and closed. If K is a pointed closed convex cone, then E(M,K) isconnected.

(No Proof.)

Example: −→

Reminder (not examinable): A total order on a set X is any binary relation � on X that isantisymmetric, transitive, and total, i.e., for all a, b, c ∈ X

if a � b and b � c then a � c (transitivity);

if a � b and b � a then a = b (antisymmetry);

a � b or b � a (totality).

A set paired with an associated total order on it is called a totally ordered set.

Notice that the totality condition implies reflexivity, that is, a � a. Thus a total order is also a partialorder, that is, a binary relation which is reflexive, antisymmetric and transitive.

Lemma 1 (Zorn’s lemma). Every non-empty partially ordered set, in which every totally ordered subsethas a lower bound, contains at least one minimal element.

(No Proof.)

Let X be a set with A = {Ai}i∈I a family of subsets of X. Then the collection A has the finite intersectionproperty, if any finite subcollection J ⊂ I has non-empty intersection

⋂i∈J Ai. A set X is compact if

and only if every collection of closed subsets of X satisfying the finite intersection property has nonemptyintersection itself (no proof).

17

End of Reminder

In which situations do minimal elements exist?

Theorem 6. If K is pointed and there exists a v ∈M such that (v −K) ∩M is compact and nonempty,then

E(M,K) 6= ∅.

Proof (not examinable):

Below only a sketch of the proof is presented. You are challenged to fill in the missing details of the proof.This is a real research mini-project for you. Please let me know if you would like to get the missing detailsfrom me.

Suppose that Mv = (v − K) ∩ M is compact. Let {ui}i∈I be any totally ordered subset of Mv. Thefamily of closed subsets Mui (i ∈ I) has the finite intersection property; that is, every finite subfamilyhas a nonempty intersection. Since Mv is compact, the family of subsets Mui (i ∈ I) has a nonemptyintersection; that is, there is an element

u ∈ ∩i∈IMui = ∩i∈I(ui −K) ∩Mv.

Hence, u is a lower bound of the subset {ui}i∈I and belongs to Mv. Then by Zorn’s lemma Mv has at leastone minimal element which is also a minimal element of M . �

Example: −→

Recall the following very important phrase from before:

“Note again that M is usually the set of possible values of decisions: v, w ∈ M , v = f(x), w = f(y).Hence, solving a decision making problem means that we have a set X of decisions and we want to choosethe optimal decisions. In order to do so, we need to be given (or for a practical problem often identifyby ourselves) a set of objective functions f1, · · · , fn : X → R. Let f := (f1, · · · , fn) : X → Rn. UsuallyX is a subset of another finite dimensional space Rm, or it can be identified with such a subset. Solvingour problem means that a cone K ⊆ Rn is given (if no cone is given we assume that the cone is Rn+; forpractical problems we need to decide which cone to use), we need to find the image set

M = f(X) := {f(x) | x ∈ X},

the set of efficient points E(M,K) of M with respect to the order ≤K induced by K (or simply withrespect to K) and the decisions

f−1(E(M,K)) = {x ∈ X | f(x) ∈ E(M,K)}

corresponding to E(M,K), that is, the preimage set. A part of this process is illustrated in the next figure.

18

K

f

X M=f(X)

”

4 Examples for Multicriteria Decision Making Problems

1. Solve the following decision making problem:Minimise x2Minimise x1 + x2Subject to x1 + x2 ≥ 0

x42 − x22 − 2x1x2 − x21 ≤ 0

with respect to the cone K = R2+ := {(x1, x2)> ∈ R2 : x1 ≥ 0, x2 ≥ 0}.

Solution: The decision variables are x1, x2. The decision set is

{(x1, x2)> ∈ R2 : x1 + x2 ≥ 0, x42 − x22 − 2x1x2 − x21 ≤ 0}.

The objective functions are y1 = x2, y2 = x1 + x2. The image set is

M = {(y1, y2)> ∈ R2 : y2 ≥ y21}.

The efficient set isE(M,K) = {(y1, y2)> ∈ R2 : y2 = y21, y1 ≤ 0}.

Indeed, E(M,K) ⊆ ∂M , where ∂M is the boundary of M . Let (y1, y2)> ∈ ∂M . Then, y2 = y21.Suppose y1 > 0. Then, y2 > 0, and hence (0, y2)

> ≤K (y1, y2)> (componentwise ordering) and(y1, y2)

> 6= (0, y2)> ∈ M . Hence (y1, y2)> /∈ E(M,K). Suppose y1 ≤ 0 and let (z1, z2)> ∈ Msuch that (z1, z2)

> ≤K (y1, y2)>. Hence, y21 = y2 ≥ z2 ≥ z21 and z1 ≤ y1 ≤ 0 which implies−y1 = |y1| ≥ |z1| = −z1, or equivalently z1 ≥ y1. Thus, y1 = z1, which by y21 = y2 ≥ z2 ≥ z21implies y2 = z2. Thus, (z1, z2)

> = (y1, y2)>. So, no point in M is dominating (y1, y2)

>. Thus,(y1, y2)

> ∈ E(M,K). In conclusion,

E(M,K) = {(y1, y2)> ∈ R2 : y2 = y21, y1 ≤ 0}.

The preimage set of E(M,K) is

{(x1, x2)> ∈ R2 : x1 + x2 = x22, x2 ≤ 0}.

19

2. Solve the following decision making problem:Minimise x1 cosx2Minimise x1 sinx2Subject to x21 ≤ 1

with respect to the cone K = R2+.

Solution: Let y1 = x1 cosx2 and y2 = x1 sinx2. The image set is

M = {(y1, y2)> ∈ R2 : y21 + y22 ≤ 1}.

a

b

c

a−K

c−K

b−K

x−K

x

y

y−K

d−K

d

For any point v on the arc a, x and b we have (v −K) ∩M ⊆ {v} (such points are for example a, band x). For any other point u on the boundary of M but not on this arc we have (u−K)∩M 6⊆ {u}.Hence, the efficient set is the arc joining a, x and b. This can be expressed as:

E(M,K) = {(y1, y2)> ∈ R2 : y21 + y22 = 1, y1 ≤ 0, y2 ≤ 0}.

The preimage set of E(M,K) is(∪k∈ZAk) ∪ (∪k∈ZBk),

where

Ak =

{(x1, x2)

> ∈ R2 : x1 = 1, x2 ∈[π + 2kπ,

3π

2+ 2kπ

]}and

Bk ={

(x1, x2)> ∈ R2 : x1 = −1, x2 ∈

[2kπ,

π

2+ 2kπ

]}.

3. Solve the following decision making problem:

Minimise x1 + x2Minimise x1 − x2Subject to x1 − 2 ≥ 0

3x1 − 5x2 + 4 ≥ 02x2 − 7 ≤ 03x1 + 5x2 + 4 ≥ 02x2 + 7 ≥ 0

20

with respect to the cone K = R2+.

Solution: Let y1 = x1 +x2 and y2 = x1−x2. Then, x1 = (y1 +y2)/2 and x2 = (y1−y2)/2. Insertingthese expressions for x1 and x2 into the constraints we obtain the image set of the decision set:

M = {(y1, y2)> ∈ R2 : y1 + y2 ≥ 4, y1 − 4 ≤ 4y2, y1 − 7 ≤ y2, 4y1 + 4 ≥ y2,y1 + 7 ≥ y2, y1 ≥ 0, y2 ≥ 0}.

By drawing the image set and inspecting it we get

E(M,K) = {(y1, y2)> ∈ R2 : y1 + y2 = 4, y1 ≥ 0, y2 ≥ 0}.

Indeed, the image set is drawn below

(8,1)

(1,8)

4y +4=y2

2

T

y +7=y

1

y − 4=4y1

21

y

y

1

2

O

T

a=(0,4)

b=(4,0)

(0,7)

(7,0)

T

TT

T

c

c−K

d−K

d

e

e−K

f

f−K

gg−K

y −7=y1 2

We can observe that for any point v on the straight line segment [(0, 4)>, (4, 0)>] we have (v −K) ∩M ⊆ {v} (for example such points are a = (0, 4)>, b = (4, 0)> and c) and for any point u onthe boundary of M which is not on this straight line segment we have that (u − K) ∩M 6⊆ {u}(for example such points are d, e, f and g, one for each edge of M different from [a, b]). Hence,E(M,K) = [(0, 4)>, (4, 0)>] which can be written as

E(M,K) = {(y1, y2)> ∈ R2 : y1 + y2 = 4, y1 ≥ 0, y2 ≥ 0}.

21

The set of the decision variables correponding to the efficients set (the preimage set of E(M,K)) is

{(x1, x2)> ∈ R2 : x1 = 2, x1 + x2 ≥ 0, x1 − x2 ≥ 0},

or equivalently{(2, x2)> ∈ R2 : −2 ≤ x2 ≤ 2},

4. Solve the following decision making problem:Minimise 2x1 + x2Minimise x1 − 2x2Subject to (2a+ b)x1 + (a− 2b)x2 + c ≥ 0

2x1 + x2 ≥ 0x1 − 2x2 ≥ 0

with respect to the cone K = R2+, where a, b, c are real constants with a > 0, b > 0 and c < 0.

Solution: Let y1 = 2x1 + x2 and y2 = x1 − 2x2. The image set is

M = {(y1, y2)> ∈ R2 : ay1 + by2 + c ≥ 0, y1 ≥ 0, y2 ≥ 0}.

The efficient set is

E(M,K) = {(y1, y2)> ∈ R2 : ay1 + by2 + c = 0, y1 ≥ 0, y2 ≥ 0}.

Indeed, the image set M is drawn below

y

y2

1

p

r

p−K

x−K

q−K

r−Ky−K

O

q

z=(0,−c/b)

w=(−c/a,0)

If z = (0,−c/b) and w = (−c/a, 0), then E(M,K) is the straight line segment [z, w]. Indeed, for anypoint v on this segment (for example such points are z, w and p) we have (v−K)∩M ⊆ {v} and for

22

any other point u on the boundary (for example such points are q and r) we have (u−K)∩M 6⊆ {u}.The efficient set E(M,K) can be also written as

E(M,K) = {(y1, y2)> ∈ R2 : ay1 + by2 + c = 0, y1 ≥ 0, y2 ≥ 0}.The preimage set of E(M,K) is

{(x1, x2)> ∈ R2 : (2a+ b)x1 + (a− 2b)x2 + c = 0, 2x1 + x2 ≥ 0, x1 − 2x2 ≥ 0}.

5. Solve the following decision making problem:

Minimise x31Minimise x52Subject to x1 ≥ 0

x1 ≤ 3x2 ≥ 0x2 ≤ 2

with respect to the cone K = {(x1, x2)> ∈ R2 : x1 ≥ x2, x2 > 0}.

Solution: Let y1 = x31 and y2 = x

52. The image set is

M = {(y1, y2)> ∈ R2 : y1 ≥ 0, y1 ≤ 27, y2 ≥ 0, y2 ≤ 32}.The set M is a box. For any point v on the left vertical edge of M we have (v − K) ∩M ⊆ {v}and for any point u on the bottom horizontal edge of M we have (u−K) ∩M ⊆ {u} (because thehorizontal line {(x1, x2)> : x2 = 0, x1 > 0} is not part of K). For all other points w of the boundaryof M we have (w −K) ∩M 6⊆ {w}. In fact these ideas are true regardless of the size of the box M ,as illustrated below.

M

a

b

c

d

b−K

d−K

c−K

a−K

Hence, the efficient set is

E(M,K) = {(y1, y2)> ∈ R2 : y1 = 0, y2 ≥ 0, y2 ≤ 32} ∪ {(y1, y2)> ∈ R2 : y2 = 0,y1 ≥ 0, y1 ≤ 27}.

The preimage set of E(M,K) is

{(x1, x2)> ∈ R2 : x1 = 0, x2 ≥ 0, x2 ≤ 2} ∪ {(x1, x2)> ∈ R2 : x2 = 0, x1 ≥ 0, x1 ≤ 3}.

23

5 Computing efficient points

5.1 The dual cone

Definition. Define the dual cone K∗ by

K∗ := {z ∈ Rn | ∀ v ∈ K : z>v ≥ 0}.

We have K∗∗ = K (no proof).

Lemma 2. If x ∈ int(K) and y ∈ K∗ \ {0}, then y>x > 0. If x ∈ K \ {0} and y ∈ int(K∗), then y>x > 0.

Proof. Let x ∈ int(K) and y ∈ K∗ \ {0}. Since x ∈ int(K), x − λy ∈ K for λ > 0 sufficiently small.Since y ∈ K∗ and x − λy ∈ K, it follows that y>(x − λy) ≥ 0. Thus, y>x ≥ λ‖y‖2 > 0, because y 6= 0.The second statement follows from the first one by using that K = (K∗)∗. �

A converse of Lemma 2. If int(K) 6= ∅ and y>x > 0 for all x ∈ int(K), then y ∈ K∗ \ {0}.

Example: −→Example: −→

5.2 Examples of classical dual cones

5.2.1 Geometrical interpretation of dual cones

Note that the scalar product of z and v is z>v = z1v1 + · · · + znvn, where z = (z1, . . . , zn)> ∈ Rn andv = (v1, . . . , vn) ∈ Rn.Define the dual of the set C ⊆ Rn by

C∗ = {z ∈ Rn | ∀v ∈ C : z>v ≥ 0}.

24

The dual of any set C is a closed convex cone. Indeed the figure below shows that C∗ = ∩v∈C{z ∈ Rn |z>v ≥ 0} which is an intersection of closed half spaces which are closed convex cones.

Tz v>0

half plane

z v=0T

straight line

O

v

Convexity, closedness and “being a cone” are all properties preserved by intersection of any number ofsets. Hence C∗ is a closed convex cone. It is easy to see from definition as well that C∗ is a convex cone.Indeed, z ∈ C∗ and λ > 0 implies (λz)>v = λ(z>v) ≥ 0, for all v ∈ C. Hence, λz ∈ C, which shows thatC∗ is a cone. If x, y ∈ C∗, then (x+ y)>v = x>v + y>v ≥ 0, for all v ∈ C. Hence, x+ y ∈ C∗.Note that if K is a closed convex cone and L = K∗, then L∗ = K∗∗ = K. Therefore, K and L are calledmutually dual cones. The cones K ⊆ Rn with the property K ⊆ K∗ are called subdual and the cones withthe property K∗ ⊆ K are called superdual. Cones K ⊆ Rn which are both supdual and superdual, that isK = K∗, are called self-dual. Note that if n > 2, then there are many cones K ⊆ Rn which are neithersubdual nor superdual. However, if K is a subdual closed convex cone, then L := K∗ is superdual. Indeed,L∗ = K and thus L∗ = K ⊆ K∗ = L.

5.2.2 Dual of closed convex cones in RRR2

If K ⊆ R2 is a closed convex cone, then either K ⊆ K∗ or K∗ ⊆ K. So any closed convex cones in R2 areeither subdual or superdual (see the figure below).

O

LK

a

b

c

d

In this figure K = cone{a, b} and L = cone{c, d}, where a ⊥ c and b ⊥ d. It can be seen that K∗ = L,L = K∗, K is subdual and L is superdual.

25

5.2.3 Dual of the nonnegative orthant

If K = Rn+, then K∗ = K, that is, K is self-dual. Indeed, if z ∈ K = Rn+, then z>v ≥ 0 for any v ∈ K = Rn+because z>v = z1v1 + · · ·+ znvn ≥ 0, where z = (z1, . . . , zn)> and v = (v1, . . . , vn)>. Hence, z ∈ K∗ whichimplies K ⊆ K∗, that is, K is subdual. Conversely, let z = (z1, . . . , zn)> ∈ K∗. Denote ei the vector whichhas all components 0 except the i-th component which is 1. Then, ei ∈ K = Rn+ for any i ∈ {1, . . . , n}.Hence, z>ei = zi ≥ 0 for any i ∈ {1, . . . , n}. Thus, z ∈ Rn+ = K. Therefore, K∗ ⊆ K, that is, K issuperdual. It follows that K is self-dual, that is, K∗ = K.

5.2.4 Dual of the monotone nonnegative cone

It can be seen that if C and D are two cones, such that C ⊆ D, then D∗ ⊆ C∗. Indeed, let x ∈ D∗. Then,for any p ∈ D we have x>p ≥ 0. Thus, for any q ∈ C ⊆ D we have x>q ≥ 0. Hence, x ∈ C∗ which impliesD∗ ⊆ C∗. In particular if D is self-dual, then C is subdual, because C ⊆ D = D∗ ⊆ C∗. Let

K = Rn≥+ = {x = (x1, . . . , xn) ∈ Rn | x1 ≥ x2 ≥ · · · ≥ xn ≥ 0}.

be the monotone nonnegative cone. Hence, since K = Rn≥+ ⊆ Rn+ and Rn+ is self-dual, it follows thatK = Rn≥+ is subdual. So, K∗ is superdual. Let y = (y1, . . . , yn) ∈ K∗. Note that (1, 0, . . . , 0)> ∈ K,(1, 1, 0, . . . , 0)> ∈ K,...,(1, 1, . . . , 1, 0)> ∈ K and (1, 1, . . . , 1, 1)> ∈ K. Hence, y>(1, 0, . . . , 0)> = y1 ≥ 0,y>(1, 1, 0, . . . , 0)> = y1 + y2 ≥ 0, ... , y>(1, 1, . . . , 1, 0)> = y1 + y2 + ... + yn−1 ≥ 0, y>(1, 1, . . . , 1, 1)> =y1 + y2 + ...+ yn−1 + yn ≥ 0. Thus K∗ ⊆ L, where

L = {y = (y1, y2, . . . , yn−1, yn)> ∈ Rn | y1 ≥ 0,y1 + y2 ≥ 0, . . . , y1 + y2 + ...+ yn−1 ≥ 0, y1 + y2 + ...+ yn−1 + yn ≥ 0}.

Now let any y = (y1, y2, . . . , yn−1, yn)> ∈ L. Then, the identity

y1(x1 − x2) + (y1 + y2)(x2 − x3) + · · ·+ (y1 + y2 + · · ·+ yn−1)(xn−1 − xn)+(y1 + y2 + · · ·+ yn−1 + yn)xn = x1y1 + x2y2 + · · ·+ xn−1yn−1 + xnyn = y>x

shows that for for any x ∈ K and any y ∈ L we have y>x ≥ 0. Hence, L ⊆ K∗. It follows that K∗ = L.

5.2.5 Dual of the Lorentz cone

LetL = {( xxm+1 ) ∈ Rm+1 | xm+1 ≥ ‖x‖},

be the Lorentz cone. Then, L∗ = L. Indeed, let ( xxm+1 ) ∈ L∗. Suppose x = 0. We have ( 01 ) ∈ L.Hence, 0 ≤ (0>, xm+1) ( 01 ) = xm+1. Thus, xm+1 ≥ 0 = ‖x‖ and therefore (

xxm+1 ) ∈ L. Suppose x 6= 0.

Then,( −x‖x‖)∈ L. Hence, 0 ≤ (x>, xm+1)

( −x‖x‖)

= −‖x‖2 + xm+1‖x‖. It follows that xm+1‖x‖ ≥ ‖x‖2, orequivalently, xm+1 ≥ ‖x‖. Thus, ( xxm+1 ) ∈ L and therefore L∗ ⊆ L.Conversely, let ( xxm+1 ) ∈ L. Then, for any ( yym+1 ) ∈ L we have (x>, xm+1) ( yym+1 ) = x>y + xm+1ym+1 ≥x>y + ‖x‖‖y‖ ≥ 0 (by using the Cauchy inequality ‖x‖‖y‖ ≥ |x>y|). Hence, ( xxm+1 ) ∈ L∗ and thereforeL ⊆ L∗. In conclusion, L∗ = L, that is, the Lorentz cone is self-dual.

5.2.6 Dual of polyhedral cones

Theorem A. Let u1, · · · , um ∈ Rn,K = cone{u1, . . . , um}

26

andL = {x ∈ Rn | (u1)>x ≥ 0, . . . , (um)>x ≥ 0}.

Then, K and L are mutually dual cones, that is, K∗ = L and L∗ = K.

Proof. Since, K is a closed convex cone, it is enough to show that K∗ = L. Let x ∈ K∗. Then, sinceu1, · · · , um ∈ K, we obtain (u1)>x ≥ 0, · · · , (um)>x ≥ 0. Hence, K∗ ⊆ L. Suppose that x ∈ L. Then, forall y = λ1u

1 + · · ·+ λmum ∈ K we have

y>x = (λ1u1 + · · ·+ λmum)>x = λ1(u1)>x+ · · ·+ λm(um)>x ≥ 0.

It follows that x ∈ K∗. Hence, L ⊆ K∗. In conclusion, K∗ = L. �

Numerical examples

1. Determine the dual cone of the following cones

(a) K1 = {(x1, x2)> ∈ R2 | 2x1 + 5x2 ≥ 0, 3x1 − 4x2 ≤ 0},(b) K2 = cone{(1, 4)>, (3, 2)>} ⊆ R2,(c) K3 = {(x1, · · · , x7)> ∈ R7 | x1 + x7 ≤ 0, x2 + x5 ≤ 0},(d) K4 = cone{(1, 0, 0,−1)>, (0,−1, 1, 0)>} ⊆ R4.

Solution

(a) K∗1 = cone{(2, 5)>, (−3, 4)>}(b) K∗2 = {(x1, x2)> ∈ R2 | x1 + 4x2 ≥ 0, 3x1 + 2x2 ≥ 0}(c) K∗3 = cone{(−1, 0, 0, 0, 0, 0,−1)>, (0,−1, 0, 0,−1, 0, 0)>}(d) K∗4 = {(x1, x2, x3, x4)> ∈ R2 | x1 − x4 ≥ 0, −x2 + x3 ≥ 0}

2. LetK5 = cone{(2, 7)>, (7, 2)>} ⊆ R2. Determine the vectors u, v ∈ R2 such that u>u = 53, v>v = 53,and K∗5 = cone{u, v}.

3. Write the dual cone of K6 = {(x1, x2)> ∈ R2 | x1 + 5x2 ≥ 0, 2x1 − x2 ≤ 0} as an intersection of halfspaces.

Solution

2. We haveK∗5 = {(x1, x2)> ∈ R2 | 2x1 + 7x2 ≥ 0, 7x1 + 2x2 ≥ 0}.

The generators of K∗5 can be obtained by solving the equations 2x1 + 7x2 = 0, 7x1 + 2x2 =0. Hence, the generators of K∗5 are of the form u = (7λ,−2λ)> and v = (2µ,−7µ)> withλ, µ ∈ R \ {0}. Since u ∈ K∗5 and (7, 2)> ∈ K5, we have 0 ≤ 7(7λ) + 2(−2λ) = 45λ. Hence,λ > 0. Since v ∈ K∗5 and (2, 7)> ∈ K5, we have 0 ≤ 2(2µ) + 7(−7λ) = −45µ. Hence, µ < 0.Since 53 = u>u = 53λ2 and λ > 0 it follows that λ = 1. Therefore, u = (7,−2)>. Since53 = v>v = 53µ2 and µ < 0, it follows that µ = −1. Hence, v = (−2, 7)>.

27

3. A set of generators of K6 can be obtained by solving the equations x1 + 5x2 = 0, 2x1 − x2 = 0.Hence a set of generators of K6 can be searched in the form u = (5λ,−λ)> and v = (µ, 2µ)>with λ, µ ∈ {−1, 1}. Since u ∈ K6 and (−2, 1)> ∈ K∗6 , we have 0 ≤ (−2)5λ−λ = −11λ. Henceλ = −1 and u = (−5, 1)>. Since v ∈ K6 and (1, 5)> ∈ K∗6 , we have 0 ≤ µ + 5(2µ) = 11µ.Hence, µ = 1 and v = (1, 2)>. Therefore,

K6 = cone{(−5, 1)>, (1, 2)>}

andK∗6 = {(x1, x2)> ∈ R2 | −5x1 + x2 ≥ 0, x1 + 2x2 ≥ 0}.

Another way to find this is as follows: We have

K∗6 = cone{(1, 5)>, (−2, 1)>}.

Hence,K∗6 = {(x1, x2)> ∈ R2 : x1 = λ− 2µ, x2 = 5λ+ µ where λ, µ ≥ 0}.

So, we get −5x1 + x2 = 11µ ≥ 0 and x1 + 2x2 = 11λ ≥ 0. Therefore,

K∗6 = {(x1, x2)> ∈ R2 | −5x1 + x2 ≥ 0, x1 + 2x2 ≥ 0}.

5.2.7 Any dual of a set is a dual of a closed convex cone

If C is not a cone, thenC∗ = (cone(C))∗.

For example see the figure below:

A

C

B

b

a

x1

2x

O

D

In this figure we have 3 circular disks. A is tangent to the horizontal axis in the origin. B is tangentto both the horizontal and vertical axis and C is in the interior of the nonnegative quadrant. We havecone(A) = {(x1, x2)> ∈ R2 | x2 ≥ 0} and cone(B) = R2+ and in general cone(C) is the conic hull of thetangents from the origin to C, that is, cone(C) = cone{a, b}. If the origin O is in the interior of a circulardisk D, then cone(D) = R2.

28

Dual of the lexicographic cone

Let L ⊆ Rn be the lexicographic cone.

cone{L} = closure{L} = {x ∈ Rn | x1 ≥ 0} =⇒ L∗ = (cone(L))∗ = cone{(1,n−1︷ ︸︸ ︷

0, . . . , 0)>}.

5.3 Conditions for the efficiency of a point

Separation Theorem. Let K ⊆ Rn be a closed convex cone and S ⊆ Rn a convex set with int(K)∩S = ∅.Then, there exists a z ∈ Rn with z>v > 0, ∀v ∈ int(K) and z>w ≤ 0, ∀w ∈ S.

Theorem 7. Let int(K) 6= ∅, M be convex and v ∈M . Then v ∈ E(M, int(K)∪{0}) if and only if thereexists an z ∈ K∗ \ {0} such that for all w ∈M we have z>v ≤ z>w.

Illustration of minimisation of linear functionals on a set:

TT

z v = 100T

T

M

T

z v < z w, for any w in M

z v = 0

z v = 50

z

=

Example for Theorem 7 and comments

Let K := R2+, C := intK ∪ {0} and M be a square with sides parallel to the coordinate axes.

29

int(K)

E(M,C)

only point in E(M,K), because the boundary of K is included

E(M,K) = {v ∈M |6 ∃w ∈M : w ≤K v︸ ︷︷ ︸wi≤vi,∀i

, w 6= v}

E(M,C) = E(M, int(K) ∪ {0}) = {v ∈M |6 ∃w ∈M : w ≤int(K) v︸ ︷︷ ︸wiw (a standard optimisation problem considerably simpler to solve),

3. result: v ∈M ,

4. Use Theorem 7 to conclude that v is efficient,

5. GOTO 1.

30

Recall the following very important phrase from before:

“Note again that M is usually the set of possible values of decisions: v, w ∈ M , v = f(x), w = f(y).Hence, solving a decision making problem means that we have a set X of decisions and we want to choosethe optimal decisions. In order to do so, we need to be given (or for a practical problem often identifyby ourselves) a set of objective functions f1, · · · , fn : X → R. Let f := (f1, · · · , fn) : X → Rn. UsuallyX is a subset of another finite dimensional space Rm, or it can be identified with such a subset. Solvingour problem means that a cone K ⊆ Rn is given (if no cone is given we assume that the cone is Rn+; forpractical problems we need to decide which cone to use), we need to find the image set

M = f(X) := {f(x) | x ∈ X},

the set of efficient points E(M,K) of M with respect to the order ≤K induced by K (or simply withrespect to K) and the decisions

f−1(E(M,K)) = {x ∈ X | f(x) ∈ E(M,K)}

corresponding to E(M,K), that is, the preimage set. A part of this process is illustrated in the next figure.

K

f

X M=f(X)

”Hence, the standard procedure described before for finding an efficient point by minimising a linearfunctional has to be rephrased as follows:

...

2’. Solve miny∈X z>f(y),

3’. result: x ∈ X, f(x) = v.

...

31

Remark:

Be careful z might not be unique. See the next figure which should be self-explanatory.

z

z’

M

K

Proof of Theorem 7.

“⇒” Let v ∈ E(M, int(K) ∪ {0}), i.e., (v − M) ∩ int(K) = ∅. Use the Separation Theorem withS = v − M . Then, ∃z ∈ Rn : z>u > 0, ∀u ∈ int(K), i.e., z ∈ K∗ \ {0} (by Lemma 2 and itsconverse); and z>(v − w) ≤ 0, ∀w ∈M , i.e., z>v ≤ z>w, ∀w ∈M .

“⇐” Conversely, suppose that∃z ∈ K∗ \ {0} : z>v ≤ z>w, ∀w ∈M.

We want to prove that v ∈ E(M, int(K) ∪ {0}). Suppose to the contrary, that

v /∈ E(M, int(K) ∪ {0}).

Then, ∃w ∈ (v − int(K)) ∩ M . It follows that v − w ∈ int(K). Hence, Lemma 2 implies thatz>(v − w) > 0. It follows that

z>v > z>w,

which is a contradiction. Therefore, v ∈ E(M, int(K) ∪ {0}).

�

Other examples for Theorem 7

Example:

Let M ⊆ R2 be a circular disk around 0 with radius 1 and K = R2+. Then,

E(M, int(K) ∪ {0}) = {v ∈ R2 | v1 ≤ 0, v2 ≤ 0, v21 + v22 = 1}.

Let v ∈ E(M, int(K) ∪ {0}). In order to use Thorem 7 it is enough to choose z = −v ∈ K∗, as illustratedin the next figure.

32

z

v

Example:

Let M = {x ∈ R2 | −1 ≤ x1 ≤ 1,−1 ≤ x2 ≤ 1} and K = R2+. Then, it is easy to see that

E(M, int(K) ∪ {0}) = {v ∈M | v1 = −1 or v2 = −1}.

(a) To find the solutions with v1 = −1 by using Theorem 7, choose z = (1, 0)> ∈ K∗.

(b) To find the solutions with v2 = −1 by using Theorem 7, choose z = (0, 1)> ∈ K∗.

(c) To find the solution with v = (−1,−1)> by using Theorem 7, choose any z ∈ K∗ \ {0}.

In (a) one solves minw∈M z>w, i.e.,

Minimise w1Subject to −1 ≤ w1 ≤ 1,

−1 ≤ w2 ≤ 1.

In (b) one solves minw∈M z>w, i.e.,

Minimise w2Subject to −1 ≤ w1 ≤ 1,

−1 ≤ w2 ≤ 1.

5.3.1 Finding efficient points by using K-monotone functions

Definition. A function s : Rn −→ R is called K-monotone, if

v ≤K w =⇒ s(v) ≤ s(w).

The function s is called strictly K-monotone, if

v ≤K w & v 6= w =⇒ s(v) < s(w).

33

Example: Let z ∈ K∗ \ {0} and define s(v) := z>v. Then s is K-monotone.Proof. Indeed, v ≤K w implies w − v ∈ K. Hence, z>(w − v) ≥ 0 (because z ∈ K∗ and w − v ∈ K), orequivalently

s(v) = z>v ≤ z>w = s(v).�

Example: Let z ∈ int(K∗) and define s(v) := z>v. Then s is strictly K-monotone.Proof. Indeed, v ≤K w & v 6= w implies w − v ∈ K \ {0}. Since z ∈ int(K∗) and w − v ∈ K \ {0},Lemma 2 implies z>(w − v) > 0, or equivalently

s(v) = z>v < z>w = s(v).

�

Example: Let K = R2+, z = (1, 1)> and s(u) = z>u. Then, z ∈ int(K) = int(K∗). One has s(u) = z>u =u1 + u2. Let v ≤K w, i.e., w − v ∈ K, i.e., v1 ≤ w1 & v2 ≤ w2. It follows easily that s(v) = v1 + v2 ≤w1 + w2 = s(w). Moreover, if v 6= w, then at least one of the inequalities v1 ≤ w1 & v2 ≤ w2 is strict,hence s(v) = v1 + v2 < w1 + w2 = s(w). Hence, s is both K-monotone and strictly K-monotone. Infact any strictly K-monotone function is K-monotone, but the converse is not true, as shown by the nextexample.

Example: Let K = R2+, z = (1, 0)> and s(u) = z>u = u1. Then, z ∈ K \ {0} = K∗ \ {0}. Let v ≤K w,i.e., w − v ∈ K, i.e., v1 ≤ w1 and v2 ≤ w2. It follows easily that s(v) = v1 ≤ w1 = s(w). Hence, s isK-monotone. However, s is not strictly K-monotone, because a := (0, 0)> ≤K (0, 1) =: b> and a 6= b>,but s(a) = s(b).

Theorem 8. Let s : Rn −→ R be a function and let v ∈ M be a minimum of s over M . Then, thefollowing hold.

1. If s is K-monotone and v is the unique minimum of s over M , then v ∈ E(M,K).

2. If s is strictly K-monotone, then v ∈ E(M,K).

Proof. Let s : Rn → R be K-monotone and v be a solution of{minimise s(w)subject to w ∈M

Assume v /∈ E(M,K), i.e., ∃w ∈M , w 6= v: w ≤K v. Since s is K-monotone, s(w) ≤ s(v).

case 1 v is unique: But, w 6= v is also a minimum; contradiction.

case 2 s is strictly K-monotone: Then, w ≤K v and w 6= v implies s(w) < s(v). So, v is not a minimum;contradiction.

�

Corollary 2. Let z ∈ int(K∗). Define s(v) = z>v. Solve{minimise s(w)subject to w ∈M

If v is a solution, then v ∈ E(M,K).

34

Theorem 9. Let int(K∗) 6= ∅ and v ∈ M . Then v ∈ E(M,K) if and only if there exists a z ∈ int(K∗)and a % ∈ Rn such that v solves the following optimisation problem:

min z>w

subject to w ∈M,w ≤K %.

Proof.

“⇐” Let v ∈M be a solution of

(P )

minimise z>wsubject to w ∈M

w ≤K ρSuppose v /∈ E(M,K), i.e., ∃w ∈ M , w 6= v: w ≤K v. By transitivity of “≤K”, w ≤K ρ, i.e., w isfeasible for (P ). Since z ∈ int(K∗), s(w) = z>w is strictly K-monotone. It follows that z>w < z>v.Hence v does not solve (P ); contradiction.

“⇒” Let v ∈ E(M,K). Let ρ := v and z ∈ int(K∗) arbitrary. Then, v is the only feasible point for (P ).Hence v solves (P ).

�

Recall that in general M = f(X) and a cone K is given in the image space of f as illustrated in thenext figure. Finding an efficient point v ∈ E(M,K) and its corresponding decision x ∈ X with f(x) = vby solving the problem minx∈X s(f(x)), where s : Rn → R is called scalarisation. If s(u) = z>u, thenthe latter problem is called weighted scalarisation and it can be written in the form minx∈X

∑ni=1 zifi(x).

Note that if the latter problem provides an efficient point (or efficient points), then we will find the sameefficient point(s) if we replace z with λz, for any λ > 0.

K

f

X M=f(X)

5.4 The set of properly efficient points

Definition. The set of properly efficient points P (M,K) is defined by

P (M,K) :=⋃

z∈int(K∗)

{v ∈M | ∀w ∈M : z>v ≤ z>w}.

35

Example: Let M = [−1, 1]× [−1, 1] ⊆ R2 and K = R2+. Then, E(M,K) = P (M,K) = {(−1,−1)>}. Tofind (−1,−1)> ∈ P (M,K) choose any z ∈ int(K∗).Example: Let M = {v ∈ R2 | v21 + v22 ≤ 1} and K = R2+. Then,

E(M,K) = {v ∈ R2 | v1 ≤ 0, v2 ≤ 0, v21 + v22 = 1}

and P (M,K) = E(M,K) \ {(−1, 0)>, (0,−1)>}. To find v ∈ P (M,K) choose z = −v.

Theorem 10. The following inclusions hold:

1. P (M,K) ⊆ E(M,K).

2. If K is pointed and M is convex and closed, then E(M,K) ⊆ closure(P (M,K)).

Proof. We prove 1. only. It follows from the definition of P (M,K) and Corollary 2.

Denote the components of any vector a ∈ Rn by a1, . . . , an, that is, a = (a1, . . . , an)>.

Remark 2. If K is pointed and M is convex and closed, then P (M,K) ⊆ E(M,K) ⊆ closure(P (M,K)).This means that a computer cannot distinguish between E(M,K) and P (M,K).

Theorem 11. Suppose K = Rn+ and M is polyhedral, i. e. there exists a matrix A ∈ Rk×n and a vectorb ∈ Rk such that

M = {v ∈ Rn | (Av)i ≤ bi (i = 1, . . . , k)}.

Then P (M,K) = E(M,K).

(No Proof.)

Extension of Theorem 11. Suppose K is a simplicial cone in Rn, that is, K = cone{u1, . . . , un}, whereu1, . . . , un ∈ Rn are linearly independent. Also suppose that M is polyhedral, that is, there exists a matrixA ∈ Rk×N and a vector b ∈ Rk such that

M = {v ∈ Rn | (Av)i ≤ bi (i = 1, . . . , k)}.

Then P (M,K) = E(M,K).

(No Proof.)Idea:

• Choose (several) z ∈ int(K∗).• Solve (several) optimization problems of the form minv∈M z>v.

• Result: (several??) points in P (M,K).

Caution: P (M,K) = E(M,K) does not mean that this idea works! Consider K = R2+,

M :={

(v1, v2)> | v1 + v2 ≥ 4, v1 ≥ 1, v2 ≥ 1

}.

Indeed, P (M,K) = E(M,K) ={

(v1, v2)> | v1 + v2 = 4, v1 ≥ 1, v2 ≥ 1

}and a point v ∈ P (M,K) with

v1 > 1, v2 > 1 can be obtained by choosing a positive scalar multiple of z = (1, 1)> only. It is unlikely

that a computer will find this unique direction by randomly generating several z ∈ int(K∗).

36

5.4.1 Geometical interpretation of the set of properly efficient points

Recall that the scalar product of z and v is z>v = z1v1 + · · · + znvn, where z = (z1, . . . , zn)> ∈ Rn andv = (v1, . . . , vn) ∈ Rn.

Definition. A hyperplane (with normal vector z 6= 0 and through v ∈ Rn) is a set of form

H(z, v) = {w ∈ Rn | z>w = z>v}.

A hyperplane H(z, v) determines two closed halfspaces H−(z, v) and H+(z, v) of Rn, defined by

H−(z, v) = {w ∈ Rn | z>w ≤ z>v},

andH+(z, v) = {w ∈ Rn | z>w ≥ z>v},

A hyperplane H(z, v) is a supporting hyperplane of M at v ∈M if M ⊆ H+(z, v).

See the following figure for the geometric meaning of the supporting hyperplane.

halfspace

O

T

v

w

z

M

Tz w=z v

supporting hyperplane

z w>z vT

T

From Definition 3 it follows that v ∈ P (M,K) if and only if there exists z ∈ intK∗ such that H(z, v) is asupporting hyperplane of M at v.

Example: Let M = M1 ∪M2, whereM1 = {(x1, x2)> ∈ R2 | 0 ≤ x1 ≤ 1 and 1 ≤ x2 ≤ 2},M2 = {(x1, x2)> ∈ R2 | 1 ≤ x1 ≤ 2 and 0 ≤ x2 ≤ 2 and x1 + x2 ≥ 2},and

K = {(x1, x2)> ∈ R2 | x1 ≥ x2 and x2 ≥ 0}. Determine the set of properly efficient points P (M,K).Solution: See the next figure.

37

x

x

x

x

M −K

T

T

T

T

1

2

1

2

*K

(0,2)

(0,1)

(2,0)

(2,2)

T

(1,1)

Let us first determine the set of efficient points E(M,K). It is easy to see that v ∈ E(M,K) if and onlyif (v −K) ∩M ⊆ {v}. The boundary ∂M of M is formed by the straight line segments

[(0, 2)>, (0, 1)>], [(0, 1)>, (1, 1)>], [(1, 1)>, (2, 0)>], [(2, 0)>, (2, 2)>], [(2, 2)>, (0, 2)>].

We know that E(M,K) ⊆ ∂M . For any point v belonging to one of the closed straight line segments

[(2, 0)>, (2, 2)>], [(0, 1)>, (1, 1)>], [(0, 2)>, (2, 2)>]

we have (v −K) ∩M 6⊆ {v}. However, for any point v belonging to one of the straight line segments

[(0, 2)>, (0, 1)>], ((1, 1)>, (2, 0)>]

we have that (v −K) ∩M = {v} (note that the second segment does not contain the point (1, 1)>). Forthis observe that the set v −K is the translation of the set −K into the point v. Therefore,

E(M,K) = [(0, 2)>, (0, 1)>] ∪ ((1, 1)>, (2, 0)>]= {(0, λ)> ∈ R2 | 1 ≤ λ ≤ 2} ∪ {(1 + λ, 1− λ)> ∈ R2 | 0 < λ ≤ 1}.

We know that P (M,K) ⊆ E(M,K). Observe that p = (1, 0)> ∈ int(K∗). Indeed (1, 0)(x1, x2)> = x1 > 0for any (x1, x2)

> ∈ K \ {0}, because x1 ≥ x2 ≥ 0 and x1 = 0 implies x2 = 0. Minimising s(v) := p>v onthe set M we obtain the segment [(0, 2)>, (0, 1)>]. Hence, [(0, 2)>, (0, 1)>] ⊆ P (M,K). It is easy to seethat except the point (2, 0)> there is no point on the segment ((1, 1)>, (2, 0)>] through which there exista supporting hyperplane for M (that is, there is no line which leaves M on one of its sides or half-planes).The straight line through (0, 1)> and (2, 0)> is a supporting hyperplane of M . The vector z = (1, 2)> is anormal vector of the supporting hyperplane which is directed towards its half-plane containing M . It canbe seen that z ∈ int(K∗). Indeed, K ⊆ R2+ and hence (R2+)∗ = R2+ ⊆ K∗. Thus, z ∈ int(R2+)∗ ⊆ int(K∗).Therefore (2, 0)> ∈ P (M,K). In conclusion

P (M,K) = [(0, 2)>, (0, 1)>] ∪ {(2, 0)>} = {(0, λ)> ∈ R2 | 1 ≤ λ ≤ 2} ∪ {(2, 0)>}.

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5.5 Finding efficient points by minimising norms

5.5.1 Norms

Reminder: A function η : Rn −→ R is called a norm if

1. η(v) ≥ 0 for all v ∈ Rn.

2. η(v) = 0 implies v = 0.

3. η(v + w) ≤ η(v) + η(w) for all v, w ∈ Rn (subadditivity of η or triangle inequality)

4. η(λv) = |λ|η(v) for all v ∈ Rn, λ ∈ R (homogeneity of η).

5.5.2 Examples of norms

Euclidean norm

• In R2: η(x) = ‖x‖2 :=√x21 + x

22.

• In Rn: η(x) = ‖x‖2 :=√∑n

i=1 x2i .

1-norm

• In R2: η(x) = ‖x‖1 := |x1|+ |x2|.

• In Rn: η(x) = ‖x‖1 :=∑n

i=1 |xi|.

∞-norm

• In R2: η(x) = ‖x‖∞ := max(|x1|, |x2|).

• In Rn: η(x) = ‖x‖∞ := maxi=1,...,n |xi|.

p-norm

Let 1 ≤ p 0 given weights

• In R2: η(x) = ‖x‖p,ω := (|ω1x1|p + |ω2x2|p)1/p.

• In Rn: η(x) = ‖x‖p,ω := (∑n

i=1 |ωixi|p)1/p.

• In Rn: η(x) = ‖x‖∞,ω := maxni=1 |ωixi|.

Example: If ω1 = 1/2, ω2 = 1 and p = 2, then ‖(1, 0)>‖p,ω = ((1/2 · 1)2 + 02)1/2 = 1/2.

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Rotated weighted p-norm

In R2: Let 1 ≤ p 0 given weights and R = ( cosα sinα− sinα cosα ).‖x‖p,ω,R = ‖Rx‖p,ω.

5.5.3 Any Minkowski functional is a norm

Definition. Let A ⊆ Rn. Then, A is called symmetric if x ∈ A implies −x ∈ A; or equivalently A = −A.

Definition. Let A ⊆ Rn be a compact convex symmetric set such that 0 ∈ intA. Then, the functionρA : Rn → R defined by

ρA(x) =

{min{r > 0 : x ∈ rA} if x 6= 00 if x = 0

is called the Minkowski functional.

Lemma 3. Let A ⊆ Rn be a convex set and r, s > 0. Then rA+ sA ⊆ (r + s)A.

Proof. Since A is convex, λA+ (1− λ)A ⊆ A, for any λ ∈ [0, 1]. Take λ = r/(r + s). � �

Proposition 1. Let A ⊆ Rn be a compact convex symmetric set such that 0 ∈ int(A). Then, the Minkowskifunctional is well defined and it is a norm.

Proof. Let x ∈ Rn \ {0}. DenoteI = {r > 0 : x ∈ rA} ⊂ R.

The set I is nonempty: Indeed, if λ > 0 is sufficiently large, then x/λ is sufficiently close to the origin,and hence x/λ ∈ A; that is λ ∈ I.The set I is closed: Indeed, let rn ∈ I with rn → r∗ as n → ∞. Then, x/rn ∈ A. Since A is closed,x/r∗ ∈ A too; that is, r∗ ∈ I.The set I is convex: Indeed, if p, q ∈ I and 0 < t < 1, then x ∈ pA and x ∈ qA. Hence, by using theconvexity of A and Lemma 3, we get

x = tx+ (1− t)x ∈ tpA+ (1− t)qA ⊆ [tp+ (1− t)q]A,

that is, tp+ (1− t)q ∈ I.Since the set I is nonempty, closed, convex and bounded from below, it is a closed interval, bounded frombelow. Hence, it has a minimal element. This means that the Minkowski functional is well defined.

The conditions 1. and 2. of a norm are trivially satisfied.

Subadditivity of ρA: Consider v, w ∈ Rn \ {0}. Then, v ∈ ρA(v)A and w ∈ ρA(w)A. Thus, since A isconvex, by Lemma 3 we have

v + w ∈ ρA(v)A+ ρA(w)A ∈ [ρA(v) + ρA(w)]A.

Hence, ρA(v + w) ≤ ρA(v) + ρA(w).Homogeneity of ρA: Consider v ∈ Rn \ {0} and λ ∈ R \ {0}. By the symmetry of A it follows thatλv ∈ rA if and only if v ∈ r|λ|A. Therefore,

ρA(λv) = min {r > 0 : λv ∈ rA} = min{r > 0 : v ∈ r

|λ|A

}= min

{|λ| r|λ|

> 0 : v ∈ r|λ|A

}= |λ|ρA(v).

40

�

Miscellaneous remarks related to the Minkowski functional

Let A ⊆ Rn be a compact convex symmetric set such that 0 ∈ int(A).

• rA = {rx : x ∈ A}.

• Note that for any x ∈ Rn we have x ∈ ρA(x)A.

• Let t ∈ [0, 1]. If x ∈ A, then tx ∈ A. Indeed x ∈ A implies tx = tx + (1 − t)0 ∈ A because A isconvex. This implies tA ⊆ A and hence A = {x ∈ Rn : ρA(x) ≤ 1}. Indeed, ρA(x) ≤ 1 impliesx ∈ ρA(x)A ⊆ A and x ∈ A implies ρA(x) ≤ 1.

• In the proof of Proposition 1, when showing that I is closed, we can assume r∗ 6= 0, otherwisex/rn ∈ A and ‖x/rn‖ = ‖x‖/rn → ∞ which contradict the boundedness of A. Here ‖ · ‖ is theEuclidean norm.

• In the proof of Proposition 1, when showing that ρA is subadditive, we assumed that v 6= 0 andw 6= 0. If one of them is zero, then the subadditivity is trivial.

• In the proof of Proposition 1, when showing the homogeneity of ρA, we assumed v 6= 0 and λ 6= 0.Otherwise the homogeneity is trivial.

• It can be shown that any norm in Rn is a Minkowski functional. More specifically, if η is a norm inRn, then η = ρA, where A = {x ∈ Rn : η(x) ≤ 1}.

5.5.4 Cone-monotone norms

Let K be a pointed closed convex cone, ∅ 6= K 6= {0}, K 6= Rn.

Definition. The norm η : Rn → R is called K-monotone if 0 ≤K x ≤K y implies η(x) ≤ η(y).

Although this definition seems very similar to the definition of a K-monotone function, the two notionsare completely different: Please note that there is no K-monotone norm which is a K-monotone function.Moreover, there is no norm which is a K-monotone function. Indeed, since ∅ 6= K 6= {0}, there exists anx ∈ K \ {0}. Then, −2x ≤K −x. If we suppose that the norm η : Rn → R is a K-monotone function,then we get

0 ≤ 2η(x) = | − 2|η(x) = η(−2x) ≤ η(−x) = η((−1)x) = | − 1|η(x) = η(x),

which implies 0 ≤ η(x) ≤ 0. Thus, η(x) = 0 and therefore x = 0, which is a contradiction. Hence, ηcannot be a K-monotone function.

Examples of cone-monotone norms

1. Suppose K ⊂ Rn is subdual. Then, the Euclidean norm ‖ · ‖ := ‖ · ‖2 is K-monotone norm. Indeed,let 0 ≤K x ≤K y. Then, x ∈ K ⊆ K∗ and y − x ∈ K imply

0 ≤ x>(y − x) = x>y − ‖x‖2. (2)

Similarly, y − x ∈ K ⊆ K∗ and y ∈ K implies

0 ≤ (y − x)>y = ‖y‖2 − x>y (3)

Hence, formulas (2) and (3) imply that ‖x‖2 ≤ x>y ≤ ‖y‖2. Therefore, ‖x‖ ≤ ‖y‖.

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2. Let K = Rn+, 1 ≤ p < ∞ and ω = (ω1, . . . , ωn)>, where ωi > 0 for any i ∈ {1, . . . , N}. Then, theweighted p-norm ‖ · ‖p,ω is a K-monotone norm. Indeed, 0 ≤K x ≤K y implies 0 ≤ xi ≤ yi for anyi ∈ {1, . . . , N}, which implies

‖x‖p,ω =

(n∑i=1

|ωixi|p) 1

p

=

(n∑i=1

(ωixi)p

) 1p

≤

(n∑i=1

(ωiyi)p

) 1p

=

(n∑i=1

|ωiyi|p) 1

p

= ‖y‖p,ω.

Therefore, ‖x‖p,ω ≤ ‖y‖p,ω.

3. Let K = Rn+, and ω = (ω1, . . . , ωn)>, where ωi > 0 for any i ∈ {1, . . . , N}. Then, the weighted∞-norm ‖ · ‖∞,ω is a K-monotone norm. Indeed, 0 ≤K x ≤K y implies 0 ≤ xi ≤ yi for anyi ∈ {1, . . . , N}, which implies

‖x‖∞,ω = maxi=1,...,N

|ωixi| = maxi=1,...,N

(ωixi) ≤ maxi=1,...,N

(ωiyi) = maxi=1,...,N

|ωiyi| = ‖y‖∞,ω.

Therefore, ‖x‖∞,ω ≤ ‖y‖∞,ω.

Example of a norm which is not an R2+R2+R2+-monotone norm

Consider the rotated weighted 2-norm ‖ · ‖2,ω,R, where R = ( cosα sinα− sinα cosα ) with α = π4 , ω1 =√22

and ω2 = 1,that is,

‖ · ‖2,ω,R(x) = ‖Rx‖2,ω =

√√√√12

(√2

2x1 +

√2

2x2

)2+

(−√

2

2x1 +

√2

2x2

)2

=

√1

2

(x212

+x222

+ x1x2

)+

(x212

+x222− x1x2

)=

√3

2(x21 + x

22)−

1

2x1x2,

for any x = (x1, x2)> ∈ R2. A simple calculation yields that

∥∥∥∥(0,√23 )>∥∥∥∥2,ω,A

= 1 and

∥∥∥∥(14 ,√23 )>∥∥∥∥2,ω,R

=

0.9916879 < 1. Hence, (0, 0)> ≤R2+(

0,√

23

)>≤R2+

(14,√

23

)>and∥∥∥∥∥∥

(0,

√2

3

)>∥∥∥∥∥∥2,ω,R

>

∥∥∥∥∥∥(

1

4,

√2

3

)>∥∥∥∥∥∥2,ω,R

.

Therefore, ‖ ·‖2,ω,R is not an R2+-monotone norm. I will not show this example on the whiteboard. InsteadI will show you on a whiteboard-figure a more general idea behind this example.

5.5.5 A necessary and sufficient condition for efficiency

Theorem 12. Let there exist a u ∈ Rn with M ⊆ u + K. Then, v ∈ E(M,K) if and only if there existsa K-monotone norm η : Rn −→ R such that v is the only solution to the optimisation problem

minimise η(w − u)subject to w ∈M.

(No proof.)

Example −→Problem: finding an appropriate η is, in general, not easy.

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5.5.6 How to choose u when K = Rn+?

For f = (f1, . . . , fn) : X →M do the following:

Solve minx∈X fi(x), i = 1, . . . , n. The minimisers of these problems are xi, i = 1, . . . , n, respectively. The

corresponding minimal values are fi(xi), i = 1, . . . , n, respectively.

The point u = (f1(x1), . . . , fn(x

m))> is a suitable point and it is called utopia point (why?).

Theorem 12 is called goal programming in the literature.

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