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tanksize.gms : Tank Size Design Problem

We discuss a tank design problem for a multi product plant, in which the
optimal cycle time and the optimal campaign size are unknown. A mixed in-
teger nonlinear programming formulation is presented, where non-convexities
are due to the tank investment cost, storage cost, campaign setup cost and
variable production rates. The objective of the optimization model is to
minimize the sum of the production cost per ton per product produced. A
continuous-time mathematical programming formulation for the problem is
implemented with a fixed number of event points.
Reference:
Small Model of Type: MINLP
$Title Tank Size Design Problem - (TANKSIZE,SEQ=350) $Ontext We discuss a tank design problem for a multi product plant, in which the optimal cycle time and the optimal campaign size are unknown. A mixed in- teger nonlinear programming formulation is presented, where non-convexities are due to the tank investment cost, storage cost, campaign setup cost and variable production rates. The objective of the optimization model is to minimize the sum of the production cost per ton per product produced. A continuous-time mathematical programming formulation for the problem is implemented with a fixed number of event points. Rebennack, S, Kallrath, J, and Pardalos, P M, Optimal Storage Design for a Multi-Product Plant: A Non-Convex MINLP Formulation. Tech. rep., University of Florida, 2009. Submitted to Computers and Chemical Engineering $Offtext $eolcom // $stitle Define the model size and data Sets p products / P1*P3 / n event points / N1*N3 /; Parameters PRMIN(p) volume flow of products in m^3 per day PRMAX(p) volume flow of products in m^3 per day SLB(p) lower bound on inventory in m^3 SUB(p) upper bound on inventory in m^3 SI(p) initial inventory in m^3 (10% of the lower bound) DLB(p) lower bound on PRODUCTION length d(n) DUB(p) upper bound on PRODUCTION length d(n) DEMAND(p) volume flow of products in m^3 per year!! TS(p) campaign setup times in days CSTI(p) tank variable cost per ton CSTC(p) campaign setup cost B variable part of the tank investment cost / 0.3271 /; Table pdata(p,*) prmin prmax slb sub si dlb dub demand ts csti cstc P1 15.0 50.0 643.0 4018.36 707.0 1 40 4190 0.4 18.8304 10 P2 15.0 50.0 536.0 3348.63 589.0 1 40 3492 0.2 19.2934 20 P3 7.0 50.0 214.0 1339.45 235.0 1 40 1397 0.1 19.7563 30 ; $onechoV > assignpar.gms $label start %1(p) = pdata(p,'%1'); $shift $if not x%1 == x $goto start $offecho $batinclude assignpar prmin prmax slb sub si dlb dub demand ts csti cstc * Derived data Parameters DPD(p) compute the demand per day per product [tons per day] L compute the demand per day [tons per day] CAL longest campain PRL maximum production length CSTCMin minimum setup cost CSTCMax maximum setup cost; DPD(p) = DEMAND(p) / 365; L = sum(p, DPD(p)); CSTI(p) = CSTI(p) / 365 ; // scale the storage cost CAL = max(0, smax(p, DUB(p) + TS(p))); PRL = max(0, smax(p, DUB(p))); CSTCMin = smin(p, CSTC(p)); CSTCMax = smax(p, CSTC(p)); $stitle Model formulation Alias(p,pp); Positive variables d(n) duration of the campaigns pC(p,n) amount of product p produced in campaign n s(p,n) amount of product p stored at the beginning of campaign n sM(p) size of the product tanks in tons sH(p,n) auxiliary variables cI investment costs cC campaign setup costs cS variable storage costs T cycle time; Binary Variables omega(p,n) binary variable indicating product in campaign; Variables cPT cost per ton: the objective variable to minimize; Equations TIMECAP time capacity UNIQUE(n) at most one product per campaign MATBAL(p,n) material balance constraint TANKCAP(p,n) tank capacity constraint PPN1(p,n) compute the nonlinear products pR(rp)*d(n)*omega PPN2(p,n) compute the nonlinear products pR(rp)*d(n)*omega SCCam1(n) semi-continuous bound on campaigns SCCam2(n) semi-continuous bound on campaigns DEFcC campaign setup costs DEFcI investment cost DEFcS variable storage costs DefsH(p,n) define the auxiliary variables DEFcPT total costs per ton produced NONIDLE(n) force not to be idle; * time balance constraint with unknown cycle time T TIMECAP.. sum(n, d(n) + sum (p, TS(p)*omega(p,n))) =e= T; * at most one product per campaign UNIQUE(n).. sum(p, omega(p,n)) =l= 1; * no idle states are allowed NONIDLE(n).. sum(p, DUB(p)*omega(p,n)) =g= d(n); * material balance equation (steady state): * storage at end of n for product p = storage at start of n+1 for product p * storage at end of n for product p = storage at start of n * + total production of product p in n * - total demand in period n MATBAL(p,n).. s(p,n++1) =e= s(p,n) + pC(p,n) - DPD(p)*(d(n) + sum (pp, TS(pp)*omega(pp,n))); * tank capacity constraint: * this connects the tank desing capacity variable with the storage level TANKCAP(p,n).. s(p,n) =l= sM(p); * compute the nonlinear products: pR(p,n)*d(n)*omega * connects the production of product p in period n with * -> the omega variables * -> the lenght of the PRODUCTION period * -> the production rate * PPN(p,n).. pC(p,nbl(n)) =e= pR(p,n)*d(n)*omega(p,n); PPN1(p,n).. pC(p,n) =l= PRMAX(p)*d(n)*omega(p,n); PPN2(p,n).. pC(p,n) =g= PRMIN(p)*d(n)*omega(p,n); * semi-continuous lower and upper bound on campaigns SCCam2(n).. d(n) =g= sum(p, DLB(p)*omega(p,n)); SCCam1(n).. d(n) =l= sum(p, DUB(p)*omega(p,n)); * define the total costs per ton: cPT DEFcPT.. (cPT*L - cI )*T =e= cC + cS; * define the campaign setup costs DEFcC.. cC =e= sum((p,n), CSTC(p)*omega(p,n)); * define the tank investment costs DEFcI.. cI =e= B*sum(p, sqrt(sM(p))); * define the variable tank costs DEFcS.. cS =e= sum((p,n), CSTI(p)*sH(p,n) *(d(n) + sum(pp, TS(pp)*omega(pp,n)))); * auxiliary variables for the objective DefsH(p,n).. sH(p,n) =e= 0.5*(s(p,n++1) + s(p,n)) - SLB(p); * additional constraints to break symmetry Equations SEQUENCE(p,n) redundant consteraint on the omega SYMMETRY(n) break the symmetry of active campaigns; * if a product is produced during period n, then it cannot be produced during * period n+1 SEQUENCE(p,n).. 1 - omega(p,n) =g= omega(p,n+1); * break symmetry buy shift empty periods to the end SYMMETRY(n).. sum(p, omega(p,n)) =g= sum(p, omega(p,n+1)); Model Sequenz /all/; * lower und upper bound on inventory s.lo(p,n) = SLB(p); s.up(p,n) = SUB(p); * initial storage s.fx('P1','N1') = SLB('P1') ; * the inital storage has some implications omega.fx(p,'N1') = 0; omega.fx('P1','N1') = 1; omega.fx('P1','N2') = 0; * lower and upper bound on tank size sM.lo(p) = SLB(p); sM.up(p) = SUB(p); * Get out of the poor starting point omega.l(p,n) = uniform(0,1); Solve Sequenz using minlp minimizing cPT;