Simple computer simulation (without graphics)
- Status Closed
- Budget $30 - $250 USD
- Total Bids 13
Project Description
Simple simulation project
The description may look quite complicated, but at the end of this description I put some simplified algorithm showing that the tast is not that complicated)
Assume we have a conveyor which is formed in a kind of a circle. This circle is divided to some small sectors and in each sector some product can be put (at stand A).
The Conveyor length is L
Conveyor sector length is X and there are L/X sectors on conveyor
Conveyor runs with speed V.
As was told earlier on stand A, operator can put product (one by one) on a conveyor.
Let assume that he can fill every sector with 1 product (I mean into sector which is not occupied by another product (some product may circulate on a conveyor more than one turn)
Now on the way there are some spurs (slopes). These slopes are used to collect products from conveyor.
Let us assume there is some way, which can force product to go from a conveyor sector into a slope.
Let us also assume that there are Y slopes located around conveyor. Number of slopes shall be the same as number of sectors (see bellow explanation).
If operator put some product on conveyor then this product is dedicated to arrive in one of a slope (randomly selected).
A slope requirement is that it can receive a next product after the first one comes to the end of a slope. Let then assume that a time product runs on a slope is “Z”.
So in case a product shall be diverted into a slope the machine is checking if a slope is not occupied (time Z). In case it is occupied then the product is not diverted and it shall make one more circle on conveyor before it can again try to go into a slope.
Other assumptions.
1. The product at the moment when it is put on conveyor is randomly assigned to some slope.
2. Machine has to know “P” time before product is opposite a slope if it can be diverted in to a slope. So when a product is diverted into a slope then the next one can be put minimum Z+P time later.
3. The operator A may be able to fill the conveyor with product every sector or percentage of sectors (eg 80% of sectors will be filled) – value B
The calculation result should show how large part of products have to make one or more circles before it go into a slope.
The calculation should be done by a computer simple program (standalone). On the display should be shown:
1. Fields to put all values mentioned above (with field description)
2. Indicator saying (in percent I in pieces actual number of product which have to rotate more than one full turn.
3. Start/Stop button
4. Possibility to place other values to above mentioned fields
5. List of slopes with number of products arriving into slopes.
5. For simplification we can assume that conveyor is forming ideal circle, slopes are placed around entire conveyor homogeneously and that the number of slopes = number of sectors – 1 (because there is also stand A which occupies one sector).
Just for your information:
V- can be [url removed, login to view]
Number of slopes – 10 to 200
X value - [url removed, login to view]
In other words the task should be described as follows (simplified description):
0. Put counter to 0 value
1. The user puts a product on a sector opposite stand A (if this sector is empty).
2. Program randomly is assigning some spur for it.
3. It is calculated the time when the product arrives opposite slope
4. When slope is not occupied then product is simply disappearing. In case slope is occupied then the product makes one or more turns unlit a slope is free
5. If product could not go into a slope then increase counter
6. If a product, which is making additional circle will finally go into a slope then decrease counter.
7. Display:
a. Counter
b. Number of products which are actually on conveyor
c. Ratio between Counter and number of product on conveyor.
8. Goto 1
Before purchasing program it should be possible to demonstrate that program work properly (maybe by using remote desktop)
See drwaig included
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