Ex1_py

Elevator Problem

Understanding and planning of the problem:

In this task we received 2 folders that include a json elevator call list and json building file that include the building details and the elevators details.
We need to find an elevator for any call in the optimal way , before we begin to write the algorithm , we want to present and explain the elevator problem .

There are some problems:

1. The calls list includes calls of people that want to go down , and calls of people that want to go up.
2. The building may have many floors which can affect the waiting time of each call.
3. Useage and supervision of many elevators in some cases.
4. every elevator has a different speed and "delay time".
5. We need to fill the calls in the optimal way , but we need to understand what is it the optimal way according to the problems above.

we will write some example and suggestions that can improve and take us to the optimal way do this calls list:

Our main targets in this algorithm:
1. Every person waiting time will be minimal .
2. Maximum handling of calls every moment .
3. All calls waiting time will be equal with no relation to the call type.
4. Maximum usage of all the elevators for minimal number of uncompleted calls.
In offine algoritem we start to perform the calls after we get all of them.

Our algorithm:

• some define - f=floors in the building. ll=list of calls (for example). el=number of elevators.

if we have one elevators so:

• We check if there is calls to go "up" and we assign the closest up call to the elevator.
• We take all the calls "up" and if there people in our stops that want to go down we take them.
• After we finish the calls "up" we check if there is down call above us .
• If there is a call down we go up and take the call.
• We do this until we don't have a calls to go down above us .
• Now we take all the poeple that want to go down to their destination.

else :if we have two or more elevators :

• First of all we count the number of calls down and calls up separately.
• Each elevator update her status according to the direction she's heading.(1 going up, -1 going down , 0 idle)
• If both of then equal approximately , we send el1/2 (half of elevators) to highest down calls ,and half of them to lowest up calls .
• After that , we will do same actions like the one's elevator algoritem.
• Each call have a status 1 for up -1 for down.
• We send the closest elevators in the same path as the call until we finish all the call in the list .