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I understand how to solve transportation problem with only members in the chain, but how can I solve the transport problem with multiple members in the chain? Thank you.

KingInYellow
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  • I do not understand what do you mean by "to solve transportation problem with only members in the chain" and "with multiple members in the chain". For what I get, this is a classical transportation problem where you have finite capacity at each site. You may want to define the flow from each origin to each destination on each level, keeping the capacities and the demands into account (Hint: $x_{ij}$ is the quantity from $i \in {1,2,3}$ to $j \in {A,B,C}$). – Libra Jan 28 '14 at 12:32

2 Answers2

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The question is tagged homework, so I'll help you but I won't give a complete solution.

The first thing you need to consider is the objective - what are you minimising/maximising, which, according to what constraints?

Getting the answers to these questions down on paper, expressed with algebra, can make the method to gaining a solution easier to see than a few tables and a crazy diagram!

OJFord
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  • Sorry for the misinformation. The arrow should point from all origin point so all route is possible. This diagram comes from my teacher. He didn't put all arrow in this because he think it may make it hard to read. – KingInYellow Jan 28 '14 at 10:42
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Since this question is now old enough to no longer be a homework problem, here's my answer. The transportation problem is typically formulated as a single-level depot-customer problem. This multi-level transportation problem is better known as a "minimum-cost flow problem" (see https://en.wikipedia.org/wiki/Minimum-cost_flow_problem).

Here is a solution coded in R for your problem:

library(igraph)
g <- graph_from_literal(Origin -- One:Two:Three, One:Two:Three -- A:B:C:D, A:B:C:D -- I:II:III, I:II:III -- W:X:Y:Z, W:X:Y:Z -- Destination)
E(g)$weight <- c(0,0,0,4,3,6,5,5,5,7,5,2,5,3,4,6,4,6,5,3,7,3,5,3,4,5,8,4,3,6,5,5,5,7,5,2,5,3,4,0,0,0,0)
E(g)$capacity <- c(200,250,300,rep(c(120,200,200,150),3),rep(c(250,200,150),4),rep(c(50,190,120,200),3),50,190,120,200)

max_flow(g,"Origin","Destination")

The max flow is 560 (you are market demand limited). Here are the flow values:

[1] 200 210 150   0   0  50 150   0   0 100 110   0   0 150   0   0   0   0   0
[20]   0   0   0 150 150 100  60 100  50   0   0  50   0   0 120  90   0 190   0
[39]  60  50 190 120 200

Thus, Supplier 1 ships 200 units, Supplier 2 ships 210 units, Supplier 3 ships 150 units. Plant C gets 50 units from 1, 100 units from 2, and 150 units from 3. Plant D gets 150 units from 1 and 110 units from 2. You can match the remaining values to the remaining edges in the graph to see the complete solution.

Michael Tiemann
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