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Based on customer demand, estimate: ω s : minimum required amount of material s For final products, ω s is the customer demand Calculated once µ i is known for all tasks consuming material s μ i : minimum cumulative production of task i Calculated once ω s is known for all materials produced by task i Constraint 1 Number of times a task is run Number of batches producing a state Constraint 2 Cumulative amount produced by a task. The RHS is increased so it is an integer multiple of β j max Cumulative amount of a state produced Valid Inequalities for Chemical Production Scheduling Sara Zenner and Christos T. Maravelias Department of Chemical and Biological Engineering University of Wisconsin – Madison - WI P ROBLEM S TATEMENT Given are a set of tasks i I, processing units j J, and materials s S A processing unit j can be used to carry out tasks i I j. Task i in unit j has processing time τ ij Unit j has variable batchsize in [β j min, β j max ] A material can be consumed/produced by multiple tasks i I s- /i I s+. Each task can consume/produce multiple materials; conversion coefficient ρ ιs Materials may have an initial inventory ζ s Customers have demands for final products ϕ st State s is stored in a dedicated tank with capacity γ s Find a schedule showing: When to begin each task Which processing unit to use for each task How much material to process for each task Challenge: Computational performance of MIP scheduling models Previous research has focused on the development of better models Goal: Develop tightening constraints based on specialized propagation algorithms Include recycle streams Consider minimum and maximum unit capacities P ROBLEM F ORMULATION Discrete time: Time points are fixed, and tasks begin/end at time points Continuous time: Time points can vary, so tasks can have any length Decision Variables: Integer Variable: Continuous Variables: B ijt = Batch size of task i in unit j starting at time t S st =Inventory level of material s in storage at time t Objectives:maximize profit, minimize cost, minimize makespan, … Constraints: Material balance and inventory capacity Unit capacities (minimum and maximum) Process at most one task at a time in a unit (This constraint is different for continuous time formulations) A TTAINABLE P RODUCTION If units have min and max capacities, some values of μ i are not feasible μ i is feasible if for some k, where α j k is the number of batches in unit j for range k. Otherwise, increase i to the nearest attainable amount i * ≥ i When a task can take place in multiple units, check all combinations of batches in units V ALID I NEQUALITIES R ESULTS Testing Test 36 discrete-time and 12 continuous-time problems Stop optimization after 30 minutes Algorithm was implemented and the MIPs were solved using GAMS 23.7/CPLEX 12.3 Results Adding the tightening constraints decreases the computational time for problems solved to optimality by a factor of >100 > twice as many problems are solved with tightening Algorithm runs in less than 10 seconds Similar results for discrete and continuous models R ECYCLE L OOPS 1. Identify loops and break using tear streams 2.Guess the tear stream doesn’t produce any material; backward propagate demand until reaching the tear stream 3.Check tear stream: T3 produces 50kg of S5, but T4 needs 52kg 4.Start over with updated cumulative production for S5 If a tear stream still does not produce enough material to meet demand after being updated, the problem is infeasible with the given initial inventories Heater Reactor 1 Reactor 2 Separator Heat/48Heat/52Heat/20 React1/50React2/50 React1/50React2/50 React1/80React2/80React3/80React3/55React2/70 Separate/80Separate/55 When? Where? How much? Solved to Optimality (avg. CPU time) Not Solved to Optimality (avg. optimality gap) D EMAND P ROPAGATION M ETHODS A LGORITHM Heat Separate Feed A Hot A Product 2 React2 Feed B Int AB React1 Int BC React3 Impure E 40% 60% 40% Product 1 Feed C 10% 90% 50% 80% 20% Heater Separator Reactors 1&2 T1: 50 T2: 30 50% ω = % 20% 50 μ = 100 D EMAND P ROPAGATION E XAMPLE 1.Customer demand: 15kg S4, 20kg S5, and 50kg S6 2a.T3 only produces S6; 2b. Check attainable production T2 produces both S4 and S5 for T2 and T3 to find μ* 3.T2 and T3 both need S3 4b. Check attainable production for T1 to find μ* 4a.T1 produces S5 T1 T2 T S1 S3 S5 S6 {U3} {U1,U2} 15 S4 S2 50% 75% 25% T1 T2: T3: S1 S3 S5 S6 {U3} {U1,U2} 15 S4 S2 50% 75% 25% U1U # of Batches Capacity: U kg, U kg T1 = 55 → T1 * = 60 Amount Produced T2 = 40 is attainable Capacity: U kg # of Batches T3 =50→ T3 * =60 Amount Produced T1 T2: 40 T3: S1 S3 S5 S6 {U3} {U1,U2} 15 S4 S2 50% 75% 25% T3 =50 T1: 60 T2: T3: S1 S3 S5 S6 {U3} {U1,U2} 15 S4 S2 50% 75% 25% T1 = 55 S1 T1T2 S2 S3 T3 S4 50% T4 S5 20% 80% S ζ S1 = 45 φ S4 = 50 ζ S1 = 45 S1 T1: 110T2: S2 S3 T3: S4 50% T4: S5 20% 80% S φ S4 = 50 ζ S1 = 45 S1 T1: 110T2: S2 S3 T3: S4 50% T4: S5 20% 80% S φ S4 = 50 U1U # of Batches Capacity: U kg, U kg * = 60 → ̂ = 75 Amount Produced Tighter Formulation Original Formulation I k : Tasks for which μ i is known S k : States for which ω s is known S T : set of tear states n: number of times tear stream has been updated P ARALLEL P RODUCTION P ATHS When a material can be produced by multiple tasks, there is no way to know how much of that state each task produces, so the backward propagation does not work Instead, find i solving an LP: min the amount task i produces s.t. the amount of each material s produced must be at least: (1) ω s (when ω s is known), and (2) the amount consumed T1 S1 S2 T2 S4 S3 90% 10% T3 T1 S1S2 T2 S4S3 90% 10% T3 Set ω s = 0 for tear states, calculate ω s for products, and set n = 0 Set I k = ∅ and S k = {products & tear states} Calculate μ i and then μ i * ∀ i ∉ I k s.t. S i + \ S k = ∅. Add these tasks to I k Calculate ω s ∀ s ∉ (S k \ S T ) s.t. I s - \ I k = ∅. Add these states to S k Does I k =I & S k =S? Is n≥ |S T |? StopSet n = n+1 Yes No Step 0Step 1Step 2 Step 3 Backward Propagation Update Tear Streams Continuous Time Model (Sundaramoorthy and Karimi, 2005) 149 s2.2% 0.45 s41.2 s With Tightening No Tightening Problems s1.3% 2.1% 1.4 s2.4 s 1.5% With Tightening No Tightening Problems Discrete Time Model (Shah et al., 1993) The valid inequalities are effective for long time horizons Varying the Time Horizon Testing Different Objectives The valid inequalities are more effective for cost minimization than for makespan minimization

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