多级制造和销售模型

Question

下面是R的公司制造和销售的流动模型图形的简化重复的例子

library(igraph) 

# Create graph 
graph= graph.formula(
    R --+ P1, 
    P1 --+ M1, 
    R --+ P2, 
    P2 --+ M2, 
    P1 --+ P3, 
    P2 --+ P3, 
    P3 --+ M2, 
    R --+ P4, 
    P3 --+ P5, 
    P4 --+ P5, 
    P5 --+ M3, 
    P5 --+ M4 
) 

# Change colors for pretty plot 
V(graph)$color= "gray" 
V(graph)[name== "R"]$color= "cyan" 
V(graph)[grepl(x= name, pattern= "M")]$color= "green" 
V(graph)[name %in% c("P1", "P2", "P4")]$color= "red" 

# Add sales volume as attribute and add to edge label in plot 
E(graph)[4]$sales= 100 
E(graph)[4]$label= paste("Sales:\n", E(graph)[4]$sales, "tons") 
E(graph)[6]$sales= 200 
E(graph)[6]$label= paste("Sales:\n", E(graph)[6]$sales, "tons") 
E(graph)[8]$sales= 500 
E(graph)[8]$label= paste("Sales:\n", E(graph)[8]$sales, "tons") 
E(graph)[11]$sales= 1000 
E(graph)[11]$label= paste("Sales:\n", E(graph)[11]$sales, "tons") 
E(graph)[12]$sales= 2000 
E(graph)[12]$label= paste("Sales:\n", E(graph)[12]$sales, "tons") 

# Add bill of material share as attribute and add to edge label in plot 
E(graph)[1:3]$share= 1.0 
E(graph)[1:3]$label= paste("Share:\n", E(graph)[1:3]$share*100, "%") 
E(graph)[7]$share= 0.8 
E(graph)[7]$label= paste("Share:\n", E(graph)[7]$share*100, "%") 
E(graph)[5]$share= 1 - 0.8 
E(graph)[5]$label= paste("Share:\n", E(graph)[5]$share*100, "%") 
E(graph)[9]$share= 0.4 
E(graph)[9]$label= paste("Share:\n", E(graph)[9]$share*100, "%") 
E(graph)[10]$share= 1 - 0.4 
E(graph)[10]$label= paste("Share:\n", E(graph)[10]$share*100, "%") 

# Add preliminary NA vol attribute to nodes and add label in plot 
V(graph)$vol= NA 
V(graph)$label= paste(V(graph)$name, "\nVolume:\n", V(graph)$vol, "tons") 

# Plot 
E(graph)$label.cex=0.8 
V(graph)$label.cex=0.8 
V(graph)$size=20 
layout= layout.reingold.tilford(graph, root=1) 
layout[3,2]=0 
layout[5,2]=0 
plot(graph, layout= layout) 

模型网络包括以下组件：

• R：制成产品所需的青色原料P1, P2, P4。
• P1...P5：红色产品直接从R（P1, P2, P4）或多级制造产品（P3, P5）中一步完成。每种产品都有一份配方材料清单。
• M...：产品P1...5销往的绿色市场。唯一的例外是P4这是没有出售，它只是作为P5（组合其中P3）的前身。

已知属性在此网络的边缘：

• Sales：销量吨P1, P2, P3, P5市场M1, M2, M3, M4。
• 配方Share：制造特定产品所需的前体量（％）。例如：要制造1吨的P1, P2, P4需要1吨原材料（因此需要100％）。要制造10吨P3 8吨（= 80％）P2和2吨（= 20％）需要P1。

我要寻找一个解决方案来计算体积为节点产品P1 ... P5和原材料R属性vol。目前它们设置为NA。 vol应与销售量和产品配方的份额保持一致。

评论：我正在寻找一个适用于这个简化示例的广义解决方案，但也适用于更复杂（更多节点和更多制造阶段）真实世界模型。我正在考虑一种多级传播算法。首先sales数据将被汇总，以产生出售给市场的产品的产品顶点vol属性。然后，另一个步骤将根据share计算上游产品的产量以达到上一步产量的要求等等。一般化的解决方案超出了我的知识范围R。任何想法如何完成任务？

Answer 1

经过大量的实验，看起来像我有第一个解决方案。可重现的例子可以用它来解决。到目前为止还没有在现实世界复杂的案例中进行测试。

1。重新编码任务了一下：

library(igraph) 

# Create graph 
graph= graph.formula(
    R --+ P1, 
    P1 --+ M1, 
    R --+ P2, 
    P2 --+ M2, 
    P1 --+ P3, 
    P2 --+ P3, 
    P3 --+ M2, 
    R --+ P4, 
    P3 --+ P5, 
    P4 --+ P5, 
    P5 --+ M3, 
    P5 --+ M4 
) 

# Change colors for pretty plot 
V(graph)$color= "gray" 
V(graph)[name== "R"]$color= "cyan" 
V(graph)[grepl(x= name, pattern= "M")]$color= "green" 
V(graph)[name %in% c("P1", "P2", "P4")]$color= "red" 

# Add sales volume as attribute and add to edge label in plot 
E(graph)$vol= NA # prefill 
E(graph)[4]$vol= 100 
E(graph)[6]$vol= 200 
E(graph)[8]$vol= 500 
E(graph)[11]$vol= 1000 
E(graph)[12]$vol= 2000 

# Add bill of material share as attribute and add to edge label in plot 
E(graph)[1:3]$share= 1.0 
E(graph)[7]$share= 0.8 
E(graph)[5]$share= 1 - 0.8 
E(graph)[9]$share= 0.4 
E(graph)[10]$share= 1 - 0.4 

# Add preliminary NA vol attribute to nodes and add label in plot 
V(graph)$vol= 0 

# Plot 
update.labels= function(graph){ 
    E(graph)[4]$label= paste("Vol:\n", E(graph)[4]$vol, "tons") 
    E(graph)[6]$label= paste("Vol:\n", E(graph)[6]$vol, "tons") 
    E(graph)[8]$label= paste("Vol:\n", E(graph)[8]$vol, "tons") 
    E(graph)[11]$label= paste("Vol:\n", E(graph)[11]$vol, "tons") 
    E(graph)[12]$label= paste("Vol:\n", E(graph)[12]$vol, "tons") 
    E(graph)[1:3]$label= paste("Share:\n", E(graph)[1:3]$share*100, "%", "\nVol:\n", E(graph)[1:3]$vol, "tons") 
    E(graph)[7]$label= paste("Share:\n", E(graph)[7]$share*100, "%", "\nVol:\n", E(graph)[7]$vol, "tons") 
    E(graph)[5]$label= paste("Share:\n", E(graph)[5]$share*100, "%", "\nVol:\n", E(graph)[5]$vol, "tons") 
    E(graph)[9]$label= paste("Share:\n", E(graph)[9]$share*100, "%", "\nVol:\n", E(graph)[9]$vol, "tons") 
    E(graph)[10]$label= paste("Share:\n", E(graph)[10]$share*100, "%", "\nVol:\n", E(graph)[10]$vol, "tons") 
    V(graph)$label= paste(V(graph)$name, "\nVolume:\n", V(graph)$vol, "tons") 
    graph 
} 
graph= update.labels(graph) 
E(graph)$label.cex=0.8 
V(graph)$label.cex=0.8 
V(graph)$size=20 
layout= layout.reingold.tilford(graph, root=1) 
layout[3,2]=0 
layout[5,2]=0 
plot(graph, layout= layout) 

解决方案：

# Aggregate sales volumes to markets 
this.vertices.names= V(graph)[grepl(x=name, pattern = "M")]$name 
for (i in this.vertices.names) { 
    V(graph)[name== i]$vol= sum(E(graph)[to(i)]$vol, na.rm=T) 
} 

# Calculate volumes along the network 
# do stepwise from farthest nodes to nearest (origin is "R") 
# 1.step: aggregate "from" edge vol attributes to product node vol attribute 
# 2.step: distribute node vol attribute to "to" edge vol attribute by vol of node * share of edge 

# Function to sort nodes even if they have same distance from root 
max.out.edges= function(graph, from.node.name) { 
    max(shortest.paths(graph= graph, v= V(graph)[name==from.node.name], mode="out") 
     [is.finite(shortest.paths(graph= graph, v= V(graph)[name==from.node.name], mode="out"))]) 
} 

# Function to create a list of nodes lists sorted with decending distance from root 
list.farthest.nodes= function(graph, from.node.name) { 
    ans= list() 
    sp= shortest.paths(graph= graph, v= V(graph)[name==from.node.name], mode="out") 
    max.distance= max(sp[is.finite(sp)]) 
    for (i in 0:max.distance-1) { 
    nodes= sapply(dimnames(sp)[[2]][which(sp==max.distance-i)], function(x) max.out.edges(graph, x), simplify= T) 
    ans= c(ans, list (names(nodes[order(nodes)]))) 
    } 
    ans[[1]]= NULL 
    ans[[max.distance+1]]= "R" 
    ans 
} 


farthest.nodes= list.farthest.nodes(graph, "R") 

for (i in farthest.nodes) { 
    print(paste("Levels:", i)) 
    for (j in i) { 
    print(paste("Single Level",j)) 
    if (!grepl(x=j, pattern="M")) { 
     V(graph)[name== j]$vol= V(graph)[name== j]$vol + sum(E(graph)[from(j)]$vol, na.rm=T) 
     if (!grepl(x=j, pattern="R")) { 
     E(graph)[to(j)]$vol= E(graph)[to(j)]$share * V(graph)[name== j]$vol   
     } 
    } 
    } 
} 

graph= update.labels(graph); plot(graph, layout=layout)