sprit Coding

euristic approach:

My goal is to minimize the routing time and to reduce the number of agents used in

routing.

The idea of the code is to start the iteration with number of vehicles equals to number of nodes

in the network. Let say 9. I need to have iteration loop for vehicle schedules, nodes are to be

scheduled to vehicles. In each iteration will have multiple variables that will be updated each

time the iteration happen. Let me explain what happen in the iteration in simple way. At the

first iteration, because the vehicles number is the same as number of nodes, each vehicle will

route one node only. And this will be done by Insertion based solution construction algorithm

(1.1 look below). And in the next iteration, vehicle numbers will decrease by one and another

vehicle will route two nodes. And so on and so forth, so in each iteration the following will

happen:

1. vehicle numbers will decrease by one. 2. Each vehicle will have a schedule (data structure: array , let say) 3. A node will remove from the removed vehicle to another schedule ( local search in

1.2).

4. Calculate routing time for each vehicle starting from source and end at destination 5. Remove a vehicle depend on the largest routing time. 6. Update the time spent in the routing for each vehicle ( apply A* search algorithm for

each vehicle)

The road network is like a graph, creating the network is already done as a function or

method in the jsprit. I want to have nodes as the same number as the vehicles. And the

links between them (roads) has time as weight (in minutes). I should edit the weight(time

in minutes) of links/edges between the nodes/ vertices. And the roads that go out from the

source have the same weight (larger than the largest weight of all roads. For instance, if a

road has weight of 10 minutes, these roads will have 15 minutes as weight, called upfront

cost).

The vehicles will start from a point (source) and end at different point (destination).

Each vehicle will have at least one node to route in roads. All vehicle located in the

source as a starting. We have to use A* search algorithm to find the shortest time path for

each vehicle to route from source to destination.

The variables are:

1- number of agents/vehicle.

2- Routing times.

3- Total routing time of all vahicles

4- Temp routing time

5- Varience (updated in each iteration)

6- Schedule for each vehicle, so, 15 schedules in this example.

7- Var. to save largest routing time

8- Counter starts from 0.

OUTPUT NEEDED:

Graph and tables shows:

Number of agents, total shortest routing time for all agents, solution table , detailed

solution table, problem table, graphical presentation plotter,

I need you to add weights on edges. I don’t think Jsprit offer adding weight on

edges. So, you have to make a new class to do that job, to add time on edges and let

the vehicle take the short time according to that time we add by ourselves, not

according the locations or time windows. In other words, the vehicles take the job/

services depending on the weight of the edge ( time) and the shortest time path is

taken from the time on edges also. I don’t need distance calculating. All the mater is

time and only time. that will be used in local search algorithm.

You have to comment the code very well and locate where is the algorithm used.

Important Algorithm will use:

1.1. Insertion based solution construction heuristic: Its main feature is the generation of the general solution of scheduling all basic nodes to all

agents we start the solution of n nodes and n agents. The solution will start generating a

schedule for each agent and assign a node to each agent. Thus, each agent will have a node in

its schedule. When all nodes are scheduled, then the construction ends. Local search algorithm

will start with schedules of one node at each and gradually decrease number of agent tours by

moving nodes at each step from a schedule to another according to local search algorithm

until all customers are served with the minimal number of agents and minimal routing time.

1.2.Local Search algorithm

Local Search LS algorithm is used to minimize the number of agents’ routes and decrease

the total travelled time. In our heuristic solution, moving nodes depends on number of

operators: such as, 𝜏𝛼#which is the number of nodes in an agent’s schedule, 𝑄%&is the

capacity of an agent 𝛼#, and i which is a counter. The basic idea of local search is that a

node, which has the largest routing time, will be transferred into the next lexicographic

neighbor or next lexicographic un-neighbor if the capacity of that neighbor’s agent is

enough to have it in its schedule.

Because the goals of the project are to minimize number of agents and minimize routing

time, in all scenarios, we will have only one solution to achieve both goals. Is that one

agent will pass by all nodes in the minimum routing time in the road network. That only

happens when the capacity of the agent is infinity and the number of nodes are infinity

also. But that will never happen. Reasons:

1. Because there is upfront cost that will effect on the agents’ tour time, the less use

the upfront cost, the less time spent in routing as total time for all trips. Upfront

cost is the reason that makes the total time increase high when we increase the

number of agents used in routing. So in all cases, the less number of agents used,

the less use of upfront cost.

2. The more agents used, the more upfront cost will be spent, the more total time for

all agents. ( the minimal time spent in routing per agent is more than or equal to

upfront cost).

3. The more agents used, the faster each node will be helped, the less time spent in

roads per agent,

HOW TO USE BOTH POINTS TO SOLVE THIS ISSUE? Is to take the average

number of agents so all we be average used.

The heuristic is considering the following:

1. The capacity of the vehicles is limited and all the same size let say Q.

2. The service time in each node is 0.

3. Agents are available 24 hours, means the time window of agents are open.

4. Time window of nodes are 24 hours available to be served.

A TS-based heuristic approach:

1. Let N be number of basic nodes and agents, 𝑅𝑇 array is the routing times

array for all agents of size N. Each agent will have an array, TT integer

variable for total routing time, TempTT integer variable for temporary total

time, variance integer array to calculate the difference of time between routing

time, N number of AgentSchedule array is agents’ schedule. The arrays size will

be equal to the capacity Q of the agents. TempLargeTime variable be the largest

routing time from the array RT, counter i,

2. Set the edges weights and upfront cost for every edge connected to source b0 and

set source and destination.

3. Do Constructive Heuristic algorithm

4. For each agent, iteration start:

4.1. Set i=1. This is a counter for number of agents’ arrays that has i nodes,

the i increment when TempLargeTime hold the last large routing time of

agent that has i nodes.

4.2.Calculate each agent shortest routing time by using A* search algorithm and save it in RT, save the largest value of RT in TempLargeTime.

4.3. Calculate the summation of all tour times of array 2 and save it in TT (and

TempTT for the first iteration).

4.4. Use Local Search Algorithm.

4.4.1. Move node/s that has the value in TempLargeTime, name it S, to the agent’s schedule that has the next lexicographic neighbor node, or the next lexicographic un-neighbor node and name it 𝑆* , and the capacity of that agent is enough to move the node/s into it’s schedule.

4.4.2. Update the 𝑆* schedule. 4.4.3. Delete S the old agent schedule. 4.4.4. Decrease the number of agents N.

4.5. Compare TT and TempTT and add the difference into variance array. then add TT into TempTT if this is the second iteration.

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