Common constraint mistakes SU P P LY C H AIN AN ALYTIC S IN P YTH - PowerPoint PPT Presentation

Common constraint mistakes SU P P LY C H AIN AN ALYTIC S IN P YTH ON Aaren St u bber � eld S u ppl y Chain Anal y tics Mgr .

Dependent demand constraint Conte x t Prod u ction Plan Planning for 2 prod u cts ( A , and B ) Planning for prod u ction o v er 3 months ( Jan - Mar ) Prod u ct A is u sed as an inp u t for prod u ction of prod u ct B Constraint Problem For each u nit of B , w e m u st also ha v e at least 3 u nits of A SUPPLY CHAIN ANALYTICS IN PYTHON

Dependent demand constraint For each u nit of B , w e m u st also ha v e at least 3 u nits of A 3 B ≤ A 3(2) ≤ A 6 ≤ A Common Mistake : B ≤ 3 A 3 B = A SUPPLY CHAIN ANALYTICS IN PYTHON

Code e x ample from pulp import * # Define Objective demand = {'A':[0,0,0],'B':[8,7,6]} model += lpSum([costs[p][t] * X[(p, t)] costs = {'A':[20,17,18],'B':[15,16,15]} for p in prod for t in time]) # Initialize Model # Define Constraint So Production is >= Demand model = LpProblem("Aggregate Production Planning", for p in prod: LpMinimize) for t in time: model += X[(p, t)] >= demand[p][t] # Define Variables time = [0, 1, 2] prod = ['A', 'B'] X = LpVariable.dicts( "prod", [(p, t) for p in prod for t in time], lowBound=0, cat="Integer") SUPPLY CHAIN ANALYTICS IN PYTHON

Code e x ample contin u ed for t in time: model += 3*X[('B',t)] <= X[('A',t)] SUPPLY CHAIN ANALYTICS IN PYTHON

E x tended constraint For each u nit of B , w e m u st also ha v e at least 3 u nits of A and acco u nt for direct to c u stomer sells of A . A 3 B + Demand ≤ A SUPPLY CHAIN ANALYTICS IN PYTHON

Combination constraint Conte x t Wareho u se distrib u tion plan 2 w areho u ses ( WH 1, and WH 2) We ship 2 prod u cts ( A , and B ) from each w areho u se Wareho u se WH 1 is small and can either ship 12 A prod u cts per a w eek or 15 B prod u cts per a w eek Constraint Problem What combinations of A , or B can be shipped in 4 w eeks ? SUPPLY CHAIN ANALYTICS IN PYTHON

1 w eek onl y: (1/12) A + (1/15) B ≤ 1 Correct Form (1/12) A + (1/15) B ? ≤ (1/12)(32) + (1/15)(20) ≤ 4 (32/12) + (20/15) ≤ 4 4 ≤ 4 Common Mistakes 12 A + 15 B ≤ 4 (1/12) A + (1/15) B = 4 SUPPLY CHAIN ANALYTICS IN PYTHON

from pulp import * import pandas as pd demand = pd.read_csv("Warehouse_Constraint_Demand.csv", index_col=['Product']) costs = pd.read_csv("Warehouse_Constraint_Cost.csv", index_col=['WH','Product']) # Initialize Model model = LpProblem("Distribution Planning", LpMinimize) # Define Variables wh = ['W1','W2'] prod = ['A', 'B'] cust = ['C1', 'C2', 'C3', 'C4'] X = LpVariable.dicts("ship", [(w, p, c) for c in cust for p in prod for w in wh], lowBound=0, cat="Integer") SUPPLY CHAIN ANALYTICS IN PYTHON

Code e x ample contin u ed # Define Objective model += lpSum([X[(w, p, c)]*costs.loc[(w, p), c] for c in cust for p in prod for w in wh]) # Define Constraint So Demand Equals Total Shipments for c in cust: for p in prod: model += lpSum([X[(w, p, c)] for w in wh]) == demand.loc[p, c] SUPPLY CHAIN ANALYTICS IN PYTHON

Code e x ample contin u ed Constraint model += ((1/12) * lpSum([X['W1', 'A', c] for c in cust]) + (1/15) * lpSum([X['W1', 'B', c] for c in cust])) <= 4 SUPPLY CHAIN ANALYTICS IN PYTHON

E x tend constraint Wareho u se WH 1 is small and either ship 12 A prod u cts per a w eek , 15 B prod u cts per a w eek , or 5 C prod u cts per a w eek . What combinations of A , B , or C can be shipped in 4 w eeks ? (1/12) A + (1/15) B + (1/5) C ≤ 4 SUPPLY CHAIN ANALYTICS IN PYTHON

S u mmar y Common Mistakes Dependent constraint Combination selection constraint Ho w to e x tend constraints Check constraint b y pl u gging in a v al u e SUPPLY CHAIN ANALYTICS IN PYTHON

Let ' s practice SU P P LY C H AIN AN ALYTIC S IN P YTH ON

Capacitated plant location - case st u d y P 2 SU P P LY C H AIN AN ALYTIC S IN P YTH ON Aaren St u bber � eld S u S u ppl y Chain Anal y tics Mgr .

Capacitated plant location model Modeling Prod u ction at regional facilities T w o plant si z es ( lo w / high ) E x porting prod u ction to other regions Prod u ction facilities open / close SUPPLY CHAIN ANALYTICS IN PYTHON

Decision v ariables What w e can control : ij x = q u antit y prod u ced at location _ i _ and shipped to _ j _ is y = 1 if the plant at location _ i _ of capacit y _ s _ is open , 0 if closed s = lo w or high capacit y plant SUPPLY CHAIN ANALYTICS IN PYTHON

Constraints Total Prod u ction = Total Demand x = D for j = 1,..., m ∑ i =1 n j ij n = n u mber of prod u ction facilities m = n u mber of markets or regional demand points SUPPLY CHAIN ANALYTICS IN PYTHON

Constraints Total Prod u ction ? Total Prod u ction Capacit y ∑ j =1 m ∑ s =1 ij is is x ? K y is K = potential prod u ction capacit y of plant _ i _ of si z e _ s _ SUPPLY CHAIN ANALYTICS IN PYTHON

from pulp import * # Initialize Class model = LpProblem("Capacitated Plant Location Model", LpMinimize) # Define Decision Variables loc = ['A', 'B', 'C', 'D', 'E'] size = ['Low_Cap','High_Cap'] x = LpVariable.dicts("production_", [(i,j) for i in loc for j in loc], lowBound=0, upBound=None, cat='Continuous') y = LpVariable.dicts("plant_", [(i,s) for s in size for i in loc], cat='Binary') # Define Objective Function model += (lpSum([fix_cost.loc[i,s]*y[(i,s)] for s in size for i in loc]) + lpSum([var_cost.loc[i,j]*x[(i,j)] for i in loc for j in loc])) SUPPLY CHAIN ANALYTICS IN PYTHON

Code e x ample contin u ed # Define the Constraints for j in loc: model += lpSum([x[(i, j)] for i in loc]) == demand.loc[j,'Dmd'] for i in loc: model += lpSum([x[(i, j)] for j in loc]) <= lpSum([cap.loc[i,s]*y[(i,s)] for s in size]) SUPPLY CHAIN ANALYTICS IN PYTHON

S u mmar y Capacitated Plant Location Model : Constraints Total Prod u ction = Total Demand Total Prod u ction ≤ Total Prod u ction Capacit y SUPPLY CHAIN ANALYTICS IN PYTHON

Re v ie w time SU P P LY C H AIN AN ALYTIC S IN P YTH ON

Sol v e the P u LP model SU P P LY C H AIN AN ALYTIC S IN P YTH ON Aaren St u bber � eld S u ppl y Chain Anal y tics Mgr .

Common modeling process for P u LP 1. Initiali z e Model 2. De � ne Decision Variables 3. De � ne the Objecti v e F u nction 4. De � ne the Constraints 5. Sol v e Model call the solve() method check the stat u s of the sol u tion print optimi z ed decision v ariables print optimi z ed objecti v e f u nction SUPPLY CHAIN ANALYTICS IN PYTHON

Sol v e model - sol v e method .solve(solver=None) solver = Optional : the speci � c sol v er to be u sed , defa u lts to the defa u lt sol v er . SUPPLY CHAIN ANALYTICS IN PYTHON

# Initialize, Define Decision Vars., Objective Function, and Constraints from pulp import * import pandas as pd model = LpProblem("Minimize Transportation Costs", LpMinimize) cust = ['A','B','C'] warehouse = ['W1','W2'] demand = {'A': 1500, 'B': 900, 'C': 800} costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264, ('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250} ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust], lowBound=0, cat='Integer') model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust]) for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c] # Solve Model model.solve() SUPPLY CHAIN ANALYTICS IN PYTHON

Sol v e model - stat u s of the sol u tion LpStatus[model.status] Not Sol v ed : The stat u s prior to sol v ing the problem . Optimal : An optimal sol u tion has been fo u nd . Infeasible : There are no feasible sol u tions ( e . g . if y o u set the constraints x ≤ 1 and x ≥ 2). Unbo u nded : The object f u nction is not bo u nded , ma x imi z ing or minimi z ing the objecti v e w ill tend to w ards in � nit y ( e . g . if the onl y constraint w as x ≥ 3). Unde � ned : The optimal sol u tion ma y e x ist b u t ma y not ha v e been fo u nd . 1 2 Keen , Ben Ale x. “ Linear Programming w ith P y thon and P u LP Part 2.” _ Ben Ale x Keen _, 1 Apr . 2016, benale x keen . com / linear - programming -w ith - p y thon - and - p u lp - part -2/._{{5}} SUPPLY CHAIN ANALYTICS IN PYTHON

# Initialize, Define Decision Vars., Objective Function, and Constraints from pulp import * import pandas as pd model = LpProblem("Minimize Transportation Costs", LpMinimize) cust = ['A','B','C'] warehouse = ['W1','W2'] demand = {'A': 1500, 'B': 900, 'C': 800} costs = {('W1','A'): 232, ('W1','B'): 255, ('W1','C'): 264, ('W2','A'): 255, ('W2','B'): 233, ('W2','C'): 250} ship = LpVariable.dicts("s_", [(w,c) for w in warehouse for c in cust], lowBound=0, cat='Integer') model += lpSum([costs[(w, c)] * ship[(w, c)] for w in warehouse for c in cust]) for c in cust: model += lpSum([ship[(w, c)] for w in warehouse]) == demand[c] # Solve Model model.solve() print("Status:", LpStatus[model.status]) Status: Optimal SUPPLY CHAIN ANALYTICS IN PYTHON

Print v ariables to standard o u tp u t : for v in model.variables(): print(v.name, "=", v.varValue) Pandas data str u ct u re : o = [{A:ship[(w,'A')].varValue, B:ship[(w,'B')].varValue, C:ship[(w,'C')].varValue} for w in warehouse] print(pd.DataFrame(o, index=warehouse)) loop model v ariables store v al u es in a pandas DataFrame SUPPLY CHAIN ANALYTICS IN PYTHON