CS/ECE/ISyE 524 Introduction to Optimization Spring 2017–18 4. Minimax and planning problems ❼ Optimizing piecewise linear functions ❼ Minimax problems ❼ Example: Chebyshev center ❼ Multi-period planning problems ❼ Example: building a house Laurent Lessard (www.laurentlessard.com)
LPs and polyhedra Linear programs have polyhedral feasible sets: { x | Ax ≤ b } = ⇒ Can every polyhedron be expressed as Ax ≤ b ? Not this one... 4-2
LPs and polyhedra If x , y ∈ R n , then the linear combination w = α x + (1 − α ) y for some 0 ≤ α ≤ 1 is called a convex combination . As we vary α , it traces out the line segment that connects x and y . x α x + (1 − α ) y y 0 4-3
LPs and polyhedra If Ax ≤ b and Ay ≤ b , and w is a convex combination of x and y , then Aw ≤ b . Proof: Suppose w = α x + (1 − α ) y . Aw = A ( α x + (1 − α ) y ) = α Ax + (1 − α ) Ay ≤ α b + (1 − α ) b = b Therefore, Aw ≤ b , which is what we were trying to prove. Question: where did we use the fact that 0 ≤ α ≤ 1 ? 4-4
LPs and polyhedra The previous result implies that every polyhedron describable as Ax ≤ b must contain all convex combinations of its points. ❼ Such polyhedra are called convex . ❼ Informal definition: if you were to “shrink-wrap” it, the entire polyhedron would be covered with no extra space. Convex: Not convex: Goes the other way too: every convex polyhedron can be represented as Ax ≤ b for appropriately chosen A and b . 4-5
Piecewise linear functions ❼ Some problems do not appear to be LPs but can be converted to LPs using a suitable transformation. ❼ An important case: convex piecewise linear functions . Consider the following Where f ( x ) is the function: nonlinear optimization: f ( x ) 4 minimize f ( x ) 3 x 2 subject to: x ≥ 0 1 x - 1 1 2 3 4 4-6
Piecewise linear functions The trick is to convert the problem into epigraph form: add an extra decision variable t and turn the cost into a constraint! minimize t minimize f ( x ) x , t = ⇒ x subject to: t ≥ f ( x ) subject to: x ≥ 0 x ≥ 0 f ( x ) t 4 4 polyhedral 3 3 feasible set! 2 2 1 1 x x - 1 1 2 3 4 - 1 1 2 3 4 4-7
Piecewise linear functions t 4 3 minimize t x , t 2 subject to: t ≥ f ( x ) 1 x ≥ 0 x - 1 1 2 3 4 This feasible set is polyhedral . It is the set of ( x , t ) satisfying: t ≥ − 1 2 x + 3 � � t ≥ − 2 x + 3 , t ≥ 3 x − 9 2 , Equivalent linear program: minimize t x , t t ≥ − 1 2 x + 3 subject to: t ≥ − 2 x + 3 , 2 t ≥ 3 x − 9 , x ≥ 0 4-8
Piecewise linear functions Epigraph trick only works if it’s a convex polyhedron . f ( x ) t 4 4 3 3 = ⇒ 2 2 1 1 x x - 1 1 2 3 4 - 1 1 2 3 4 This epigraph is not a convex polyhedron so it cannot be the feasible set of a linear program. 4-9
Minimax problems ❼ The maximum of several linear functions is always convex. So we can minimize it using the epigraph trick. Example: a T � � f ( x ) = max i x + b i i =1 ,..., k f ( x ) t 4 4 3 3 2 2 1 1 x x - 1 1 2 3 4 - 1 1 2 3 4 - 1 - 1 - 2 - 2 min t = ⇒ a T � � min max i x + b i x , t i =1 ,..., k x t ≥ a T s.t. i x + b i ∀ i 4-10
Maximin problems ❼ The minimum of several linear functions is always concave. So we can maximize it using the epigraph trick. Example: a T � � f ( x ) = min i x + b i i =1 ,..., k f ( x ) t 4 4 3 3 2 2 1 1 x x - 1 1 2 3 4 - 1 1 2 3 4 - 1 - 1 - 2 - 2 max t = ⇒ a T � � max min i x + b i x , t i =1 ,..., k x t ≤ a T s.t. i x + b i ∀ i 4-11
Minimax and Maximin problems ❼ A minimax problem: min t a T � � min max i x + b i x , t = ⇒ x i =1 ,..., k t ≥ a T s.t. i x + b i ∀ i ❼ A maximin problem: max t = ⇒ a T � � max min i x + b i x , t x i =1 ,..., k t ≤ a T s.t. i x + b i ∀ i Note: Sometimes called minmax , min-max , min/max . Of course, minmax � = maxmin! 4-12
Minimax and Maximin problems Practical scenario: ❼ Paintco produces specialized paints and we are planning production for the coming year. They have some flexibility in how they produce the paints, but ultimately they require employees, as well as electricity, water, and certain chemicals. ❼ Nobody knows for sure how much paints will sell for, and the future price of electricity, water, and the chemicals is also unknown. But planning decision must be made now. ❼ Three consulting firms are hired to forecast the costs for the coming year. The three firms return with three different forecasts (cost functions f 1 , f 2 , f 3 ). Which one should be used? ❼ The risk-averse approach is to solve the minimax problem: � � min i =1 , 2 , 3 { f i ( x ) } � = max max min x f i ( x ) x i =1 , 2 , 3 4-13
Absolute values ❼ Absolute values are piecewise linear! min t x , t min | x | s.t. Ax ≤ b x = ⇒ s.t. Ax ≤ b t ≥ x t ≥ − x ❼ So are sums of absolute values: min t + r min | x | + | y | x , y , t , r = ⇒ x , y s.t. t ≥ x , t ≥ − x r ≥ y , r ≥ − y 4-14
Chebyshev center What is the largest sphere you can fit inside a polyhedron? If y is the center, then draw perpendicular lines to each face of the polyhedron. d 2 We want to maximize the d 3 y smallest d i . In other words, d 1 max i =1 ,..., 5 d i ( y ) min y d 4 d 5 (the y shown here is obviously not optimal!) 4-15
Chebyshev center What is the largest sphere you can fit inside a polyhedron? If y is the center, then draw perpendicular lines to each face of the polyhedron. d 2 We want to maximize the d 3 smallest d i . In other words, y d 1 max i =1 ,..., 5 d i ( y ) min y d 5 d 4 The optimal y is the Chebyshev center 4-16
Chebyshev center Finding the Chebyshev center amounts to solving an LP! To compute the distance between y 4 and the hyperplane a T x = b , notice 3 a r that if the distance is r , then y + � a � a 2 belongs to the hyperplane: 1 a T � � y r y + � a � a = b - 1 1 2 3 4 Simplifying, we obtain: a T y + � a � r = b “The distance between y and each hyperplane is at least r ” is equivalent to saying that a T i y + � a i � r ≤ b i for each i . 4-17
Chebyshev center Finding the Chebyshev center amounts to solving an LP! The transformation to an LP is given by: max i =1 ,..., k d i ( y ) min max r = ⇒ y y , r a T a T s.t. i y ≤ b i ∀ i s.t. i y + � a i � r ≤ b i ∀ i 4-18
Chebyshev center Example: find the Chebyshev center of the polyhedron defined by the following inequalities: 2 x − y + 2 z ≤ 2 , − x + 2 y + 4 z ≤ 16 , x + 2 y − 2 z ≤ 8 , x ≥ 0 , y ≥ 0 , z ≥ 0 Chebyshev.ipynb 4-19
Multi-period planning problems ❼ Optimization problems with a temporal component. ❼ Decisions must be made over the course of multiple time periods in order to optimize an overall cost. Examples: ❼ scheduling: classes, tasks, employees, projects,... ❼ sequential decisions: investments, commitments,... The decisions at each time period are coupled and must be jointly optimized. Otherwise we risk making decisions that seem good at the time but end up being very costly later. 4-20
Multi-period planning problems ❼ These problems tend to be tricky to model. It is often not clear what the decision variables should be. ❼ There are often more variables than you expect. Important: Decision variables aren’t always things that you decide directly! We will see several examples of this... 4-21
Example: building a house Several tasks must be completed in order to build a house. ❼ Each task takes a known amount of time to complete. ❼ A task may depend on other tasks, and can only be started once those tasks are complete. ❼ Tasks may be worked on simultaneously as long as they don’t depend on one another. ❼ How fast can the house be built? Source: HBR 1963 4-22
Example: building a house The data can be visualized using a directed graph. ❼ Arrows indicate task dependencies. What are the decision variables? ❼ t i : start time of i th task. ❼ precedence constraints are expressed in terms of t i ’s. ❼ minimize t x . Source: HBR 1963 4-23
Example: building a house A small sample: Let t l , t o , t m , t n , t t , t s be start times of the associated tasks. Now use the graph to write the dependency constraints: ❼ Tasks o , m , and n can’t start until task l is finished, and task l takes 3 days to finish. So the constraints are: t l + 3 ≤ t o , t l + 3 ≤ t m , t l + 3 ≤ t n ❼ Task t can’t start until tasks m and n are finished. Therefore: t m + 1 ≤ t t , t n + 2 ≤ t t , ❼ Task s can’t start until tasks o and t are finished. Therefore: t o + 3 ≤ t s , t t + 3 ≤ t s Source: HBR 1963 4-24
Example: building a house Full implementation in Julia: House.ipynb Follow-up: which tasks in the project are critical to finishing on time? Which tasks can withstand delays? ❼ related to notion of duality we will see later. 4-25
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