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电商平台建站,seo关键词优化软件官网,wordpress修改页面地址,金华企业网站建设公司Spatial Data Analysis（六）：空间优化问题使用pulp库解决空间优化问题： pulp是一个用于优化问题的Python库。它包含了多种优化算法和工具，可以用于线性规划、混合整数线性规划、非线性规划等问题。Pulp提供了一个简单…

Spatial Data Analysis（六）：空间优化问题

使用pulp库解决空间优化问题：

pulp是一个用于优化问题的Python库。它包含了多种优化算法和工具，可以用于线性规划、混合整数线性规划、非线性规划等问题。Pulp提供了一个简单的方式来定义优化问题，包括变量、约束和目标函数，并且可以使用多种求解器进行求解。Pulp也提供了可视化工具来展示优化问题的结果。Pulp是一个开源项目，可以在GitHub上获取它的源代码。

空间优化（一）：p-中值问题

这个问题需要p设施的位置，同时最小化服务所有需求的总加权距离。
每个节点都有一个关联的权重，表示该节点的需求量。

目标函数： 最小化所有设施和需求节点的需求加权总和。

决策变量： 将设施放置在何处以及哪个设施位置为哪些需求节点提供服务

限制：

每个节点由 1 个设施提供服务
仅当某个位置存在设施时，节点才可以由该设施提供服务。
我们必须放置p设施
每个节点要么是一个设施，要么不是。

pip install -q pulp

from pulp import *
import numpy as np
import geopandas as gp
from scipy.spatial.distance import cdist
import matplotlib.pyplot as plt

#read a sample shapefile
georgia_shp = gp.read_file("https://raw.githubusercontent.com/Ziqi-Li/GEO4162C/main/data/georgia/G_utm.shp")

georgia_shp.shape

(172, 18)

创建一个需求和一个设施变量，表示每个需求和设施的索引。
需求节点：所有县
facility：设施将建在一些选定的县之上

#create a demand and a facilities variable, indicating the indices of each demand and facility.
#demand node: all counties
#facility: Facilities will be built on top of some chosen countiesdemand = np.arange(0,172,1)
facilities = np.arange(0,172,1)

计算距离矩阵d_ij(n×n)

#Calculate a distance matrix d_ij (n by n)
coords = list(zip(georgia_shp.centroid.x,georgia_shp.centroid.y))d = cdist(coords,coords)

每个县(hi)的需求是总人口

#the demand for each county (h_i) is the total populatoion
h = georgia_shp.TotPop90.values

声明设施变量；生成的变量名称为：X_1,X_2,…

# declare facilities variables;the resulting variable names are: X_1,X_2,...
X = LpVariable.dicts('X_%s',(facilities),cat='Binary')# declare demand-facility pair variables; the resulting variable names are Y_0_1, Y_0_2,...
Y = LpVariable.dicts('Y_%s_%s', (demand,facilities),cat='Binary')

要放置的设施数量

#Number of facilities to place
p = 3 #change this and re-run the code.#Create a new problem
prob = LpProblem('P_Median', LpMinimize)

目标函数：最小化所有设施和需求节点的加权需求距离总和
(h_i: i 处的需求；d_ij: i 和 j 之间的距离)
“for”循环用于迭代序列

# Objective function: Minimizing weighted demand-distance  summed over all facilities and demand nodes
# (h_i: demand at i; d_ij: distance between i and j)
# A "for" loop is used for iterating over a sequenceprob += sum(sum(h[i] * d[i][j] * Y[i][j] for j in facilities) for i in demand)

这个约束表明我们必须精确放置 p 个设施

# This constraint indicates we must place exactly p facilitiesprob += sum([X[j] for j in facilities]) == p

这一约束意味着需求节点 i 只能由一个设施提供服务

# This constraint implies that a demand node i can only be serviced by one facilityfor i in demand:prob += sum(Y[i][j] for j in facilities) == 1

这个约束意味着需求节点 i
仅当 j 处有设施时才能由 j 处的设施提供服务
它隐式地消除了 X[j] = 0 但 Y[i][j] = 1 时的情况
（节点 i 由 j 提供服务，但 j 处没有设施）

# This constraint implies that that demand node i
# can be serviced by a facility at j only if there is a facility at j
# It implicitly removes situation when X[j] = 0 but Y[i][j] = 1
# (node i is served by j but there is no facility at j)for i in demand:for j in facilities:prob +=  Y[i][j] <= X[j]

%%time# Solve the above problem
prob.solve()print("Status:", LpStatus[prob.status])

Status: Optimal
CPU times: user 1.35 s, sys: 64 ms, total: 1.42 s
Wall time: 11.5 s

# The minimized total demand-distance. The unit is person * meter (total distance travelled)
print("Objective: ",value(prob.objective))

Objective:  469538765110.4489

# Print the facility node.
rslt=[]
for v in prob.variables():subV = v.name.split('_')if subV[0] == "X" and v.varValue == 1:rslt.append(int(subV[1]))print('Facility Node: ', subV[1])

Facility Node:  126
Facility Node:  30
Facility Node:  82

# Get the geomerty of the facility nodes.
fac_loc = georgia_shp.iloc[rslt,:]

fac_loc

	AREA	PERIMETER	G_UTM_	G_UTM_ID	AREANAME	Latitude	Longitud	TotPop90	PctRural	PctBach	PctEld	PctFB	PctPov	PctBlack	X	Y	AreaKey	geometry
126	7.315030e+08	117190.0	130	128	GA, Crisp County	31.92540	-83.77159	20011	48.4	10.0	12.47	0.30	29.0	40.66	805648.4	3537103	13081	POLYGON ((787012.250 3547615.750, 820243.312 3...
30	1.385270e+09	274218.0	32	31	GA, Fulton County	33.78940	-84.46716	648951	4.2	31.6	9.63	4.13	18.4	49.92	733728.4	3733248	13121	POLYGON ((752606.688 3785970.500, 752835.062 3...
82	9.179670e+08	121744.0	84	84	GA, Jenkins County	32.78866	-81.96042	8247	53.8	7.7	13.10	0.21	27.8	41.51	970465.7	3640263	13165	POLYGON ((989566.750 3653155.750, 981378.062 3...

#Plot the faclities (stars) on top of the demand map.
fig, ax = plt.subplots(figsize=(5,5))georgia_shp.centroid.plot(ax=ax,markersize=georgia_shp.TotPop90/1000)#markersize is proportional to the population
fac_loc.centroid.plot(ax=ax,color="red",markersize=300,marker="*")

在这里插入图片描述

空间优化（二）：集合覆盖问题

在此模型中，设施可以为距设施给定覆盖距离 Dc 内的所有需求节点提供服务。问题在于放置最少数量的设施，以确保所有需求节点都能得到服务。我们假设设施没有容量限制。

pip install -q pulp

from pulp import *
import numpy as np
import geopandas as gp
from scipy.spatial.distance import cdistimport matplotlib.pyplot as plt

#read a sample shapefile
georgia_shp = gp.read_file("https://raw.githubusercontent.com/Ziqi-Li/GEO4162C/main/data/georgia/G_utm.shp")

georgia_shp.shape

(172, 18)

创建一个需求和一个设施变量，表示每个需求和设施的索引。
需求节点：所有县
facility：我们可以在一些县建造设施

#create a demand and a facilities variable, indicating the indices of each demand and facility.
#demand node: all counties
#facility: we could build facilities in some countiesdemand = np.arange(0,172,1)
facilities = np.arange(0,172,1)

计算距离矩阵d_ij(n×n)

#Calculate a distance matrix d_ij (n by n)
coords = list(zip(georgia_shp.centroid.x,georgia_shp.centroid.y))
d = cdist(coords,coords)

阈值覆盖距离

# Threshold coverage distance
Dc = 100000 #100km coverage, change this and re run the code.

创建一个变量，指示节点 i 是否可以被设施 j 覆盖。

#Creata a variable (alpha in the lecture slide pg.28), indicating  whether a node i can be covered by facility j.
a = np.zeros(d.shape)
a[d <= Dc] = 1
a[d > Dc] = 0

声明设施变量 Xj

# declare facilities variables Xj
X = LpVariable.dicts('X_%s',(facilities),cat='Binary')

创建一个最小化问题

#Create an minimization problem
prob = LpProblem('Set_Covering', LpMinimize)

目标函数：我们要最小化放置设施的数量

# Objective function: we want to minimize the number of placed facilities
prob += sum([X[j] for j in facilities])

该约束意味着每个需求节点 i 需要至少由设施服务

# This constraint implies every demand node i needs to be served by at least facility
for i in demand:prob += sum(a[i][j]*X[j] for j in facilities) >= 1

%%time
# Solve the above problem
prob.solve()print("Status:", LpStatus[prob.status])

Status: Optimal
CPU times: user 22.5 ms, sys: 1.05 ms, total: 23.6 ms
Wall time: 66.4 ms

# The minimal number of facilities with the defiened coverage.
print("Objective: ",value(prob.objective))

Objective:  8.0

# Print the facility nodes.
rslt = []
for v in prob.variables():subV = v.name.split('_')if subV[0] == "X" and v.varValue == 1:rslt.append(int(subV[1]))print('Facility Node: ', subV[1])

Facility Node:  102
Facility Node:  120
Facility Node:  145
Facility Node:  150
Facility Node:  30
Facility Node:  38
Facility Node:  9
Facility Node:  97

# Get the geomerty of the facility nodes.
fac_loc = georgia_shp.iloc[rslt,:]

#Plot the faclities (stars) on top of the demand map.
fig, ax = plt.subplots(figsize=(5,5))georgia_shp.centroid.plot(ax=ax)
fac_loc.centroid.plot(ax=ax,color="red",markersize=300,marker="*")