Question: I'm trying to simultaneously enforce sequential duration and spacing constraints to vector solution output in Gekko. Normally, this would be fairly straightforward using window logic, but my time array (in weeks) has gaps per the "week" array below (e.g., it goes [13, 14, 17...]). I was able to get the spacing requirement (s) in weeks to work by looking up the index of the next sequential week using the code below (seems to work for all values of "s"), but I'm unsure how to factor in "d" (the number of consecutive weeks that must be run) into the existing solution (complete reproducible example below). The general solution output I'm looking for would look like this for s=1 and d=2: [13, 14, 17, 18, 33, 34, 50, 51...] import numpy as np import pandas as pd from gekko import GEKKO m = GEKKO(remote=False) m.options.NODES = 3 m.options.IMODE = 3 m.options.MAX_ITER = 1000 lnuc_weeks = [0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0] min_promo_price = [3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,3] max_promo_price = [3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5, 3.5,3.5, 3.5, 3.5, 3.5, 3.5, 3.5] base_srp = [3.48, 3.48, 3.48, 3.48, 3.0799, 3.0799, 3.0799, 3.0799,3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799, 3.0799] lnuc_min_promo_price = 1.99 lnuc_max_promo_price = 1.99 coeff_fedi = [0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589,0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589, 0.022589] coeff_feao = [0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995, 0.02929995] coeff_diso = [0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338, 0.05292338] sumproduct_base = [0.20560305, 0.24735297, 0.24957423, 0.23155435, 0.23424058,0.2368096 , 0.27567109, 0.27820648, 0.2826393 , 0.28660598, 0.28583971, 0.30238505, 0.31726649, 0.31428312, 0.31073792, 0.29036779, 0.32679041, 0.32156337, 0.24633734] neg_ln = [[0.14842000515],[0.14842000512],[0.14842000515],[0.14842000512],[-0.10407483058],[0.43676249024],[0.43676249019],[0.43676249024],[0.43676249019],[0.43676249024],[0.43676249019], [0.026284840258],[0.026284840291],[0.026284840258],[0.026284840291], [0.026185109811],[0.026284840258],[0.026284840291],[0.026284840258]] neg_ln_ppi_coeff = [1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879,1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879, 1.22293879,1.22293879, 1.22293879, 1.22293879, 1.22293879] base_volume = [124.38, 193.2, 578.72, 183.88, 197.42, 559.01, 67.68, 110.01,60.38, 177.11, 102.65, 66.02, 209.83, 81.22, 250.44, 206.44, 87.99, 298.95, 71.07] week = pd.Series([13, 14, 17, 18, 19, 26, 28, 33, 34, 35, 39, 42, 45, 46, 47, 48, 50, 51, 52]) n = 19 x1 = m.Array(m.Var,(n), integer=True) #LNUC weeks i = 0 for xi in x1: xi.value = lnuc_weeks[i] xi.lower = 0 xi.upper = lnuc_weeks[i] i += 1 x2 = m.Array(m.Var,(n)) #Blended SRP i = 0 for xi in x2: xi.value = 5 m.Equation(xi >= m.if3((x1[i]) - 0.5, min_promo_price[i], lnuc_min_promo_price)) m.Equation(xi <= m.if3((x1[i]) - 0.5, max_promo_price[i], lnuc_max_promo_price)) i += 1 x3 = m.Array(m.Var,(n), integer=True) #F&D x4 = m.Array(m.Var,(n), integer=True) #FO x5 = m.Array(m.Var,(n), integer=True) #DO x6 = m.Array(m.Var,(n), integer=True) #TPR #Default to F&D i = 0 for xi in x3: xi.value = 1 xi.lower = 0 xi.upper = 1 i += 1 i = 0 for xi in x4: xi.value = 0 xi.lower = 0 xi.upper = 1 i += 1 i = 0 for xi in x5: xi.value = 0 xi.lower = 0 xi.upper = 1 i += 1 i = 0 for xi in x6: xi.value = 0 xi.lower = 0 xi.upper = 1 i += 1 x7 = m.Array(m.Var,(n), integer=True) #Max promos i = 0 for xi in x7: xi.value = 1 xi.lower = 0 xi.upper = 1 i += 1 x = [x1,x2,x3,x4,x5,x6,x7] neg_ln=[m.Intermediate(-m.log(x[1][i]/base_srp[i])) for i in range(n)] total_vol_fedi =[m.Intermediate(coeff_fedi[0]+ sumproduct_base[i] + (neg_ln[i]*neg_ln_ppi_coeff[0])) for i in range(n)] total_vol_feao =[m.Intermediate(coeff_feao[0]+ sumproduct_base[i] + (neg_ln[i]*neg_ln_ppi_coeff[0])) for i in range(n)] total_vol_diso =[m.Intermediate(coeff_diso[0]+ sumproduct_base[i] + (neg_ln[i]*neg_ln_ppi_coeff[0])) for i in range(n)] total_vol_tpro =[m.Intermediate(sumproduct_base[i] + (neg_ln[i]*neg_ln_ppi_coeff[0])) for i in range(n)] simu_total_volume = [m.Intermediate(( (m.max2(0,base_volume[i]*(m.exp(total_vol_fedi[i])-1)) * x[2][i] + m.max2(0,base_volume[i]*(m.exp(total_vol_feao[i])-1)) * x[3][i] + m.max2(0,base_volume[i]*(m.exp(total_vol_diso[i])-1)) * x[4][i] + m.max2(0,base_volume[i]*(m.exp(total_vol_tpro[i])-1)) * x[5][i]) + base_volume[i]) * x[6][i]) for i in range(n)] [m.Equation(x3[i] + x4[i] + x5[i] + x6[i] == 1) for i in range(i)] #Limit max promos m.Equation(sum(x7)<=10) #Enforce spacing and duration d=2 s=1 for s2 in range(1, s+1): for i in range(0, n-s2): f = week[week == week[i] + s2].index if len(f) > 0: m.Equation(x7[i] + x7[f[0]]<=1) m.Maximize(m.sum(simu_total_volume)) m.options.SOLVER=1 m.solve(disp = True) df = pd.concat([pd.Series(week), pd.Series([i[0] for i in x7]), pd.Series([i[0] for i in simu_total_volume])], axis=1) df.columns = ['week', 'x7', 'total_volume'] df[df['x7']>0] 

Answer: Below is a minimal example of duration and spacing constraints that may help. The decision variable is when to start the promo. The promo selection is post-processed after st is optimized. from gekko import GEKKO import numpy as np # Initialize the model m = GEKKO(remote=False) # Define weeks and parameters weeks = np.arange(1,11) # Week numbers [1,2,...,9,10] n_weeks = len(weeks) # Number of weeks d = 3 # Duration constraint: number of consecutive weeks s = 2 # Spacing constraint: number of weeks between events # Define variables # start promo location st = m.Array(m.Var,n_weeks,integer=True,lb=0,ub=1) # Objective Function m.Maximize(sum(st)) # Always start on first week available m.fix(st[0],1) # Don't start at the end if duration constraint doesn't allow it for i in range(n_weeks-d+1,n_weeks): m.Equation(st[i]==0) # Spacing Constraint for i in range(n_weeks-d-s): m.Equation(sum(st[i:i+d+s]) <= 1) # Solve the problem m.options.SOLVER=1 m.solve(disp=True) # Output the solution print("Weeks:", weeks) print("Start Promo:", [int(si.VALUE[0]) for si in st]) x = np.zeros(n_weeks) for i in range(0,n_weeks-d+1): if (int(st[i].VALUE[0])==1): x[i:i+d]=1 print("Promo occurrence:", [int(xi) for xi in x])  The solution respects the duration (d=3) and spacing (s=2) constraints. The start of a promo is not scheduled unless there is sufficient duration to meet the minimum duration constraint. Weeks: [ 1 2 3 4 5 6 7 8 9 10] Start Promo: [1, 0, 0, 0, 0, 1, 0, 0, 0, 0] Promo occurrence: [1, 1, 1, 0, 0, 1, 1, 1, 0, 0]