Shortest Path Problem and/ Dummy nodes? The 2019 Stack Overflow Developer Survey Results Are InDynamic programming problem and shortest path problemBlack and Yellow edges and Shortest pathShortest acyclical path between two nodes, negative weights allowedModified Shortest Path ProblemShortest path between three nodes in a graphHow to find shortest path assigning weights to intermediate nodesShortest path in a graph with weighted edges and verticesWhich algorithm could I use for shortest path problem with distinct sets of must-pass nodes?Shortest Hamilton Path Planar ProblemModeling problems using graphs
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Shortest Path Problem and/ Dummy nodes?
The 2019 Stack Overflow Developer Survey Results Are InDynamic programming problem and shortest path problemBlack and Yellow edges and Shortest pathShortest acyclical path between two nodes, negative weights allowedModified Shortest Path ProblemShortest path between three nodes in a graphHow to find shortest path assigning weights to intermediate nodesShortest path in a graph with weighted edges and verticesWhich algorithm could I use for shortest path problem with distinct sets of must-pass nodes?Shortest Hamilton Path Planar ProblemModeling problems using graphs
$begingroup$
I am currently studying shortest path problems and for one of the assignment problems, I just can't figure out. The problem is below:
A guy wants to rent out his yacht from May 1 to September 1. He places an ad in the newspaper and gets 10 offers with different time periods (ex: July 1 to Aug 1) and different Dollar amounts.
My assignment is "simple": graph it as a "shortest path problem". What I don't understand is if the guy wants to maximize his revenue, shouldn't this be a widest path problem instead? Why shortest path?
Also, is there any sense in using dummy nodes to signify that during time period xxx-yyy, there is no offer to choose from? To illustrate with an example, let's say he has an offer from July 1st to July 15th; no offers from the 15th to the 30th, and then an offer from the 30th to August 15. Should I put a "dummy" node from the 15 to 30?
Thanks so much for your help.
graph-theory operations-research
asked Mar 23 at 5:39
MozzaMozza
82
$endgroup$
add a comment |
$begingroup$
I am currently studying shortest path problems and for one of the assignment problems, I just can't figure out. The problem is below:
A guy wants to rent out his yacht from May 1 to September 1. He places an ad in the newspaper and gets 10 offers with different time periods (ex: July 1 to Aug 1) and different Dollar amounts.
My assignment is "simple": graph it as a "shortest path problem". What I don't understand is if the guy wants to maximize his revenue, shouldn't this be a widest path problem instead? Why shortest path?
Also, is there any sense in using dummy nodes to signify that during time period xxx-yyy, there is no offer to choose from? To illustrate with an example, let's say he has an offer from July 1st to July 15th; no offers from the 15th to the 30th, and then an offer from the 30th to August 15. Should I put a "dummy" node from the 15 to 30?
Thanks so much for your help.
graph-theory operations-research
asked Mar 23 at 5:39
MozzaMozza
82
$endgroup$
add a comment |
$begingroup$
I am currently studying shortest path problems and for one of the assignment problems, I just can't figure out. The problem is below:
A guy wants to rent out his yacht from May 1 to September 1. He places an ad in the newspaper and gets 10 offers with different time periods (ex: July 1 to Aug 1) and different Dollar amounts.
My assignment is "simple": graph it as a "shortest path problem". What I don't understand is if the guy wants to maximize his revenue, shouldn't this be a widest path problem instead? Why shortest path?
Also, is there any sense in using dummy nodes to signify that during time period xxx-yyy, there is no offer to choose from? To illustrate with an example, let's say he has an offer from July 1st to July 15th; no offers from the 15th to the 30th, and then an offer from the 30th to August 15. Should I put a "dummy" node from the 15 to 30?
Thanks so much for your help.
graph-theory operations-research
asked Mar 23 at 5:39
MozzaMozza
82
$endgroup$
I am currently studying shortest path problems and for one of the assignment problems, I just can't figure out. The problem is below:
A guy wants to rent out his yacht from May 1 to September 1. He places an ad in the newspaper and gets 10 offers with different time periods (ex: July 1 to Aug 1) and different Dollar amounts.
My assignment is "simple": graph it as a "shortest path problem". What I don't understand is if the guy wants to maximize his revenue, shouldn't this be a widest path problem instead? Why shortest path?
Also, is there any sense in using dummy nodes to signify that during time period xxx-yyy, there is no offer to choose from? To illustrate with an example, let's say he has an offer from July 1st to July 15th; no offers from the 15th to the 30th, and then an offer from the 30th to August 15. Should I put a "dummy" node from the 15 to 30?
Thanks so much for your help.
graph-theory operations-research
graph-theory operations-research
asked Mar 23 at 5:39
MozzaMozza
82
asked Mar 23 at 5:39
MozzaMozza
82
edited Mar 23 at 16:49
Mozza
asked Mar 23 at 5:39
MozzaMozza
82
asked Mar 23 at 5:39
MozzaMozza
82
asked Mar 23 at 5:39
MozzaMozza
82
82
add a comment |
add a comment |
1 Answer
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$begingroup$
Make a graph with 12 nodes, 1 for each lease offer, a start node and an end node. Draw a directed edge from a given node to another node if you can take the lease at the new node after you finish the one at the given node. The edges should be weighted with the negative of the monetary value of the lease. Then the path across the graph from the start to end node with minimum cost(shortest path) is a feasible set of leases that generates the most revenue.
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
$endgroup$
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
Make a graph with 12 nodes, 1 for each lease offer, a start node and an end node. Draw a directed edge from a given node to another node if you can take the lease at the new node after you finish the one at the given node. The edges should be weighted with the negative of the monetary value of the lease. Then the path across the graph from the start to end node with minimum cost(shortest path) is a feasible set of leases that generates the most revenue.
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
$endgroup$
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
add a comment |
$begingroup$
Make a graph with 12 nodes, 1 for each lease offer, a start node and an end node. Draw a directed edge from a given node to another node if you can take the lease at the new node after you finish the one at the given node. The edges should be weighted with the negative of the monetary value of the lease. Then the path across the graph from the start to end node with minimum cost(shortest path) is a feasible set of leases that generates the most revenue.
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
$endgroup$
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
add a comment |
$begingroup$
Make a graph with 12 nodes, 1 for each lease offer, a start node and an end node. Draw a directed edge from a given node to another node if you can take the lease at the new node after you finish the one at the given node. The edges should be weighted with the negative of the monetary value of the lease. Then the path across the graph from the start to end node with minimum cost(shortest path) is a feasible set of leases that generates the most revenue.
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
$endgroup$
Make a graph with 12 nodes, 1 for each lease offer, a start node and an end node. Draw a directed edge from a given node to another node if you can take the lease at the new node after you finish the one at the given node. The edges should be weighted with the negative of the monetary value of the lease. Then the path across the graph from the start to end node with minimum cost(shortest path) is a feasible set of leases that generates the most revenue.
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
answered Mar 23 at 17:25
MeowBlingBlingMeowBlingBling
1226
1226
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
add a comment |
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
1
1
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
$begingroup$
Thanks so much. finally makes sense cause the objective is to maximize, so if we minimize the negative, we get the same. YOU'RE A GENIUS!
$endgroup$
– Mozza
Mar 23 at 18:03
add a comment |
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