If you have a business, the obvious reason is to make money, but in order to do that you need to do your homework. It is a good idea to organize your production such that you maximize your profit. With that in mind, consider the following:
A furniture company has a factory that produces tables and chairs. This factory is equipped with 10 saws, 6 lathes, and 18 sanding machines. To produce a chair requires 10 minutes of saw time, 5 minutes on a lathe, and 5 minutes of sanding. Tables take 5 minutes on a saw, 5 minutes on a lathe, and 20 minutes of sanding. A chair sells for $10 and a table sells for $20. How many tables and chairs should the factory produce per hour to yield the highest revenue, and what is that revenue?
Assuming that people were available to operate equipment.
Each hour the factory producing all chairs could turn out 60 chairs (limited by saws and saw time 10 saws at ten minutes) 1 chair x 6 per hour x 10 saws = 60 chairs (maximum) at $10 = $600
Each hour the factory producing all tables - limited by lathe time (six lathes at five minutes) = 12 x 6 = 72 tables (maximum). Saw time does not factor.
Lathe and sanding time for chairs does not factor. The time for a single sander per table however is twenty minutes (3 per hour) but with 18 sanders sanding time could be reduced by a factor of four so the number of tables increases to 12 tables per hour at $20 = $240
Iļæ½d find a table sanding solution but until then stick to all chairs.
To yield the highest revenue first lay off the workers and sell the equipment. Then buy tables and chairs from a factory in China that are branded with your logo. warehouse them in your former workshop and hire at minimum wage some kids to ship for you.
The factory capacity of chairs exclusively would be 60 each hour or $600.
If the factory spent full time with tables they could make 72 tables each hour (six lathes turning out a table every five minutes = 6 x 12 = 72 tables per hour.
With six lathes but only six of ten saws fully occupied plus all 18 sanding machines fully occupied for each hour the built capacity is three tables per hour per sanding machine. That totals 54 tables each hour = $1080
Since the lathes and sanding tools are fully occupied for the full hour there is a choke point there. Chairs amount to fewer dollars so this time the better choice is tables. I get 54 tables, 0 chairs $1080/hour
If you want to maximize your yield, sell all of the sanding stuff and the tablesaws and buy a few more lathes. Then, teach people how to turn stuff at 200/day. Your constraint is only how many lathe's you can buy with used sanders and tablesaws.
So far the only constraint is me. If I sacrifice tables to make chairs to put all the tools at full capacity that somewhere gives me a requirement of at least three chairs to one table to make more than making all tables. I just get in my own way trying to make the ratio work. There ought to be a reasonably simple formula I can make, but I haven't been able to set it up. I'm still making tables to get my "nut" for the mortgage. The rest I'll work out by the time, as has been pointed out, the Pacific East returns me their quote. With my workers at middle age or above if I outsource the health insurance savings alone (about 34%) should guarantee a profit.
My apologies to those that don't like this type of problem, but it does kinda represent something that might actually occur in a production shop. ;-)
Here is my solution:
There are several ways to do it. You can plot it, but you don't get a unique solution because the total available time is not evenly distributed. That is where the constraint comes in. What I did is the following:
First, I wrote an equation for each machine;
Saw: 600=10C +5T, which solved for t yields T=120-2C
Lathe: 360=5C +5T solved for T T=72-C
sanders: 1080=5C+20T or T=(1080-5C)/20
I then took each tool and plotted the number of tables vs the number of chairs for each one in columns. The easiest way from that point is brute force since it is a computer doing all of the work. Going across the columns, I took the number of chairs times $10 and the minumum value of tables across the machines for this number of chairs (this takes into account the constraints) times $20 and added them to get the total profit for that combination. The combination that gives the largest income is 48 tables and 24 chairs for a total income of $1200. There is excess capacity on the saws in this scenario.
