Ifs in PuLP

[priscila24]

I am running a PuLP with multiple If's, and it is optimizing in a way where the solution always meets the first condition.

Code:

import numpy as np
import pandas as pd
from pulp import *

idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]
d = {
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx),
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx),
'output':pd.Series([0,0,0,0.087,0.309,0.552,0.682,0.757,0.783,0.771,0.715,0.616,0.466,0.255,0.022,0,0,0,0,0,0,0,0,0], index=idx)}
cfPV = pd.DataFrame(d)


idx = [0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23]
d1 = {
'day': pd.Series(['01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14', '01/01/14'], index=idx),
'hour':pd.Series(['00:00:00', '01:00:00', '02:00:00', '03:00:00', '04:00:00', '05:00:00', '06:00:00', '07:00:00', '08:00:00', '09:00:00', '10:00:00', '11:00:00', '12:00:00', '13:00:00', '14:00:00', '15:00:00', '16:00:00', '17:00:00', '18:00:00', '19:00:00', '20:00:00', '21:00:00', '22:00:00', '23:00:00'], index=idx),
'output':pd.Series([0.528,0.512,0.51,0.448,0.62,0.649,0.601,0.564,0.541,0.515,0.502,0.522,0.57,0.638,0.66,0.629,0.589,0.544,0.506,0.471,0.448,0.438,0.443,0.451], index=idx)}
cfWT = pd.DataFrame(d1)


prob = LpProblem ("System", LpMinimize)

CPV = LpVariable ("PVCapacity",0) #PV Capacity in kW
CWT = LpVariable ("WTurCapacity",0) #WT Capacity in kW
CBA = LpVariable ("BatteryCapacity",0) #Battery Capacity kW

prob+= 63.128*CPV + 88.167*CWT + 200*CBA, "TotalCostSystem"

xEne = 0
xREin = 0
xBin = 0
xBout = 0
SOCB = 0
xPEMin = 0
xOvEn = 0
xSum = 0

CPEM = 230

for i in idx:

xEne = (CPV*cfPV['output'][i]+CWT*cfWT['output'][i])

#Low limit for Variables
prob += (CPV*cfPV['output'][i]+CWT*cfWT['output'][i]) >= 0
prob += xREin >= 0
prob += xBin >= 0
prob += xBout >= 0
prob += SOCB >= 0
prob += xPEMin >= 0
prob += xOvEn >= 0
prob += xSum >= 0
prob += CBA >= SOCB
prob += xBin <= (CBA - SOCB)
prob += xBout <= SOCB

#Cases

#Case 1 xEne > CPEM
if xEne >= CPEM:

xREin = CPEM
xBout = 0
xOvEn = xEne - CPEM

#Case 1.1 xOvEn < CBA - SOCB
if (value(xOvEn) <= (CBA - value(SOCB))):
xBin = xOvEn

#Case 1.2 xOvEn > CBA -SOCB
else:
xBin = CBA - SOCB

#Case 2 xEne < CPEM
else:
xREin = xEne
xBin = 0
xOvEn = 0

#Case 2.1 SOCB > CPEM - xREin
if (value(SOCB) >= (CPEM - value(xREin))):
xBout = (CPEM - xREin)

#Case 2.2 SOCB < CPEM - xREin
else:

xBout = SOCB

SOCB = SOCB + xBin - xBout
xPEMin = xREin + xBout

xSum += xPEMin

prob += xSum >= 5000


prob.writeLP("PVWTBattSyste.lp")

prob.solve()

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The solution is always meeting xEne>= CPEM, and even if I change this condition for something that it is impossible to meet (let's say xEne>= 5000000) it acts as the condition is true and follows that path.

Any recommendations or comments that could help me?

Thank you in advance.

Priscila Castillo Cazares
JEMES CiSu student
TUHH, Hamburg, Germany