Python's pulp linear programming introduction and examples

Pulp http://pythonhosted.org/PuLP/main/basic_python_coding.html Water Supply problem

1 Questions

The water supply company has three reservoirs for A,b,c to four residential quarters ,a and B water supply to all communities,and C only water supply to a ethyl C, Maximum water supply in the reservoir (thousand tonnes )

Reservoir	A	B	C
Maximum quantity of water supply (thousand tonnes)	50	60	50

the water use situation of the community is

Community	A	B	C	Ding
Basic water consumption (KT)	30	70	10	10
Additional water consumption (KT)	50	70	20	40

The reservoir water supply income is RMB /thousand tons , the expense cost is: other expenses 450/ thousand tons, the diversion management fee:

Cost of diversion from reservoir to community: Yuan/thousand tons	A	B	C	Ding
A	160	130	220	170
B	140	130	190	150
C	190	200	230

asked How to allocate water supply to the reservoir, the company can make the most profit.

2 solving problems Ideas

to minimize the cost of diversion management, the profit is the highest .

Set : Reservoir i to the cell J Water Supply , the total Diversion management fee is Z

The target function is: z = 160.0*x11 + 130.0*x12 + 220.0*x13 + 170.0*x14 + 140.0*x21 + 130.0*x22 + 190.0*x23 + 150.0*x24 + 190.0*x31 + 200.0*x32 + 230.0*x33

to make Z the smallest value .

the constraints are:

1 #Coding=utf-82 #Coding=utf-83  fromPulpImport*4 5 6 defget_re ():7     Pass8 9 Ten defGetResult (C, con): One  A #Set Object -Prob = Lpproblem ('Mypro', Lpminimize) - #set the variable and set the variable minimum value the  -x11 = lpvariable ("X11", lowbound=0) -x12 = lpvariable ("X12", lowbound=0) -X13 = Lpvariable ("x13", lowbound=0) +x14 = Lpvariable ("x14", lowbound=0) -x21 = Lpvariable ("x21", lowbound=0) +x22 = Lpvariable ("x22", lowbound=0) Ax23 = Lpvariable ("x23", lowbound=0) atx24 = Lpvariable ("x24", lowbound=0) -X31 = Lpvariable ("x31", lowbound=0) -x32 = Lpvariable ("x32", lowbound=0) -x33 = Lpvariable ("x33", lowbound=0) -  -X =[X11, X12, X13, x14, x21, x22, x23, x24, x31 , x32, x33] in  -     #C = [the ",", ",", " ,", ",", "," to  +  -  the  *  $ Panax Notoginseng #objective Function -z =0 the      forIinchRange (len (X)): +z + = x[i]*C[i] A     #print (z) theProb + =Z +  - #load constraint variable $Prob + = x11+x12+x13+x14 = = Con[0]#constraint conditions 1 $Prob + = x21+x22+x23+x24 = = Con[1] -Prob + = X31+x32+x33 = = Con[2] -  theProb + = x11+x21+x31 <= con[3] -Prob + = x11+x21+x31 >= con[4]Wuyi  theProb + = x12 + x22 + x32 <= con[5] -Prob + = x12 + x22 + x32 >= con[6] Wu  -Prob + = x14 + x24 <= con[7] AboutProb + = x14 + x24 >= con[8] $  - #solving -  -Status =prob.solve () A  +     #print (Lpstatus[status]) the     Print(Value (prob.objective))#Calculation Results -  $  the #Show Results the For I in Prob.variables (): the Print (i.name + "=" + str (i.varvalue)) the      forIinchprob.variables (): -         in  the  the if __name__=='__main__': Aboutc = [160, 130, 220, 170, 140, 130, 190, 150, 190, 200, 230] thecon = [50, 60, 50, 80, 30, 140, 70, 50, 10] theGetResult (C, con)

Output Result:

Optimal24000.0x11=0.0x12=50.0x13=0.0x14=0.0x21 =0.0 x22=50.0x23=0.0x24=10.0x31=50.0x32=0.0x33=0.0

Python's pulp linear programming introduction and examples