Integer Programming (IP)

Transcription

1 Iteger Programmig (IP) The geeral liear mathematical programmig problem where Mied IP Problem - MIP ma c T + h Z T y A + G y + y b R p + vector of positive iteger variables y vector of positive real variables

2 Iteger Programmig (IP) Problems with iteger variables are said Combiatorial Problems Eamples of applicatios: Productio Schedulig Assig tasks to machies or productio lies ad determie their sequece o the machies Logistics Determie the optimal routig for a fleet of vehicles i order to deliver or pickup goods at customer locatios Facility locatio Determie the best locatio for oe or more distributio cetres i order to optimally reach the customers Network desig (trasportatio ad telecommuicatio) Plaig of ivestmets

3 Iteger Programmig (IP) Use iteger values if fractioal values are ot acceptable approimatios (e.g., determie the yearly productio of airplaes vs ail) Iteger variables ca model the use of idivisible resources or the selectio amog discrete alteratives Whe iteger variables are restricted to biary values 0- LP Biary variables model the choice betwee two alteratives or the occurrece or ot of evets = 0 the the evet evet occurs does ot occur Classical problems modelled with iteger variables Kapsack Problem Matchig/Assigmet Problem Fied Charge Problem Sequecig Problem

4 IP: kapsack problem Possible defiitios: Choose which obects carry i a kapsack for a campsite Select which goods to trasport i a cotaier Select which activities activate amog a set of alteratives (e.g., proects to be fuded): possible proects to be fuded with a maimum available budget b If proect, =,...,, is fuded the required ivestmet is a Proect provides a reveues equal to c Each proect caot be partially fuded: it may be completely fuded or ot fuded The problem Determie the best proects to be fuded i order to maimize the total reveue without eceedig the available budget

5 IP: kapsack problem Kapsack is differet from Product Mi (o fractioal solutio is allowed) Biary variables are eeded = =,..., 0 = -th proect is fuded =0 -th proect is ot fuded Obective fuctio: the total reveue for the fuded proect ma = c

6 IP: kapsack problem Costraits impose that variables assume values cosistet with their meaig The available budget is ot eceed (kapsack costrait) Variables assume biary values The Kapsack problem formulatio ma = = = c a a b b { 0,} = B =,..., { 0,} = B =,...,

7 IP: kapsack problem The multi-dimesioal KP The realizatio of the proects require fudig over m periods (moths) a the fud eeded for proect i period i, =,...,, i=,...,m b i available budget for period i, i=,...,m ma = = c a b i i =,..., m { 0,} = B =,...,

8 IP: assigmet problem Defiitio Assig m activities to workers ( m) Each worker ca at most perform a sigle activity The cost for assigig to worker, =,...,, activity i, i=,...,m, is c The problem Determie the assigmet of all the activities to the workers such that the total cost is miimized

9 IP: assigmet problem Graph represetatio Activities (i) c Workers () The problem: Determie which liks to itroduce betwee activity ad worker odes such that each activity ode is eactly paired to a worker ode

10 IP: assigmet problem Variables: m biary assigmet variables (associated with liks) = 0 Activity i assiged to worker Activity i ot assiged to worker i =,..., m =,..., Obective: Miimize the assigmet cost mi m i= = c

11 IP: assigmet problem Costraits: Each activity must be assiged = = i =,..., m Each worker ca at most perform a sigle activity m i= =,..., Variables are biary to esure that a activity is etirely assiged to a sigle worker { 0,} = B i =,..., m =,...,

12 IP: assigmet problem The assigmet problem formulatio mi m i= = = m i= i c = i =,..., m =,..., { 0, } = B i =,..., m =,...,

13 Trasportatio problem with fied charge A modified versio of the trasportatio problem: Trasportatio of a good (e.g., gas or oil) betwee m suppliers ad customers Each supplier coected to each customer by a trasportatio chael (e.g., pipelie) A flow betwee a supplier i, i=,...,m, ad a customer =,...,, ca be established if: A fied charge f is paid for retig the chael betwee i ad A trasportatio cost c is paid for each uit of good trasported from i to The maimum availability of supplier i is s i, i=,...,m. The demad of customer is r, =,...,. The problem Determie which chaels must be reted ad the flow i them so that the customers demad is satisfied at the miimum total cost

14 Trasportatio problem with fied charge The formulatio eteds that of trasportatio problem m cotiuous variables for the flow betwee i ad R, i=,...,m=,..., Biary variables are itroduced associated with each chael (i,) y = 0 (i,) is used (i,) is ot used i =,..., m =,...,

15 Trasportatio problem with fied charge The cost for each chael has a discotiuity Total cost c f The obective fuctio is o liear but it ca be still liearly modelled

16 Trasportatio problem with fied charge Obective fuctio It icludes both margial ad fied cost Trasportatio problem costraits Supplier capacities mi Customer demads m i = = m i = = ( c + Variables ad y must be liked: trasportatio is allowed oly o active (reted) chaels s i i =,..., f y m = r =,..., ) My i, (fied charge costraits) M is a Big-M

17 Trasportatio problem with fied charge If a maimum chael capacity q is give for the chaels we ca use q istead of M q y i, Defiitios of variables R + y B i =,..., m =,...,

18 Trasportatio problem with fied charge The problem formulatio mi m i= = = m i= ( c = s r My R + i + y f y ) i =,..., m =,..., i, B i =,..., m =,...,

19 Sequecig problem A commo problem i detailed plaig of productio activities: Determie the optimal sequece for a set of operatios (obs or tasks) of a productive resource (machie): idepedet obs must be sequeced o a sigle machie The machie ca process a sigle ob at a time Job processig caot be suspeded ad restarted (o preemptio is allowed) The processig time of ob i, i=,...,, is p i The obective is a cost usually associated with ob completio time, e.g.: Miimize the average ob completio time Miimize the maimum ob completio time (makespa) Miimize the sum of delays w.r.t. obs due dates No assumptio o the cost: we focus o costraits oly

20 Sequecig problem Variables: Cotiuous variables for the start time of obs t i R, i=,..., t i 0 We must model the fact that two obs caot overlap: Give two obs i ad oe of the followig alterative costraits must hold p i. if i precedes t t i + p i ob i ob t t i t p 2. if precedes i t i t + p ob ob i t t t i

21 Sequecig problem Oly oe costrait ca hold: disuctive costraits Accordig to the sequece oly oe of the two disuctive costraits must be active (ad the other oe must be ot bidig) Disuctive costraits are modelled with biary (sequecig) variables y B i =,..., =,..., = 0 For each uordered pair of ob, i,, (i<) the disuctive costraits are y i precedes i ot precede t t i t + p M ( y ) t i i + p My where M >> i= p i Variables y give the sequecig order of obs

22 IP methods - itroductio Solvig a IP problem could seem easier tha a LP oe: IP polyhedra iclude a fiite umber of solutios However, let cosider a 2D eample... 2 LP 2 IP X X X = { R : A b} LP optimum is a verte X = { Z : A b} Ofte IP optimum is a iteral poit

23 IP methods - itroductio Very ofte Simple Alg. does ot geerate a feasible optimal solutio to IP problems From the 2D eample it could seem easy to fid the IP optimum from the LP oe: By approimatig the LP optimum Fidig the iteger poit closest to the LP optimum Both such strategies ca fail.

24 IP methods - itroductio Approimated solutios may be ot iteger feasible

25 IP methods - itroductio The closest iteger solutio may be ot optimal

26 IP methods - itroductio A serious difficulty: whe the problem dimesio (umber of variables) is ot small, eve if fiite, the umber of solutios ca be huge Biary problems (0- LP): the umber of possible solutios double for each added variable It grows as 2 where is the umber of variables eample =4 6 possible combiatios (solutios) =0 024 = Iteger problems: eve worse situatio Eample: 0 variables that ca assume 0 iteger values: the umber of combiatios is