Full schedule

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There are times when you do not get a stock whole solution, ie a number with decimals is not appropriate because it is inconsistent with the situation that is posing the problem. Consider, for example, a company that makes pants and distributes to retail (shops that sell directly to end consumers and not to other companies so that these in the stock market turn sold to other companies). This company must deal with whole pieces (pants stock prices completed), so it would not be valid for a response that indicates that you must sell to gain 1’75 stock quote pants.
Therefore, call scheduling problem entirely to one in which demand that all the variables are. Thus, the stock market approach of a whole programming problem is as follows:
Maximize F (x) x C
subject to: x ‘b
x ‘0
xi whole with i 1 ,…, n
where “C” attached to the weighting variable (ie earnings per pantalon), “A” quantities of the factors to be used (ie quantity of cloth, yarn), “b” the maximum amount available for each resource and “x” variable scope (ie number of trousers).
Example: Company Costuritas S.L. manufactures and markets jeans and shirts, this can be purchased at most 50 mA price stock denim fabric a day. He knows that a jacket is required for 1.5 mA cloth pants and a 2 m .
The requirements for personnel responding to the following table:
Personal Cutters ironing Time Time Time Sewing
1 1/3h Pants 2 1/3h 1 1/6h
Shirts 3 1/3h 3 1/2h 1 1/4h
The maximum number of cutters of cloth can hire is 12, due to the availability of machines, the sewing is 20 and of the press 15.
It has been estimated demand, and in the best possible scenario has been obtained the following sales forecast: 18 pants, 16 shirts a day.
The benefit for each pantalon sold is 15 stock quotes ‘, per shirt, earn 12’5’. Want to know that production is to maximize the annual profit of this company.
From the above data are the problem, taking into account that:
x1 trousers, shirts x2
Function to maximize: 15 x1 12’5 x2
Number of cutters and time spent on each unit: 1 / 3 buy stock x1 3 / 4 x2’12
Number of sewing and time spent on each unit: 2 / 3 x1 3 / 2 x2’20
Number of press and time spent on each unit: 1 / 6 x1 1 / 4 x2’15
Fabric Requirements: 2 x1 1.5 x2’50
Maximum production of trousers a day: x1’18
Maximum production of shirts per day: x2’16
Non-negativity constraints: x1, x2 ‘0
15 maximize x1 12’5 x2
subject to: 1 / 3 exchange x1 3 / 4 x2’12
2 / 3 x1 3 / 2 x2’20
1 / 6 x1 1 / 4 x2’15
2 x1 1.5 x2 ’50
x1 ’18
x2 ’16
x1, x2 ‘0
The order for this system gives the following solution: The maximum daily benefit is 336’67 ‘, the production that maximizes the benefit is 18 pants / day and 5 shirts a day. As can be seen both as pants shirts are a whole number.

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