Cash income comes from three crops (soybeans, corn and oats) and two kinds of poultry (cows and hens). Crops don't need investment, but each cow needs an initial investment of 400 dollars and each hen needs an initial investment of 3 dollars.
Each cow needs 1.5 mu of land, working hours in winter 100 hours, 50 hours in summer, and annual net income of 450 dollars; Accordingly, each hen does not occupy land, which is 0.6 hours in winter and 0.3 hours in summer, and the annual net income is 3.5 dollars. The henhouse can accommodate up to 3000 hens and the fence can accommodate up to 32 cows.
Planting an acre of soybeans, corn and oats requires 20, 35 and 10 hours of labor in winter, 30, 75 and 40 hours of labor in summer, and the annual income is 175, 300 and 120 dollars respectively.
Establish a mathematical model to help farmers determine the breeding plan and maximize the annual net income.
Grow soybeans, corn, oats, hens and cows to work.
Summer X 1 X2 X3 X4 X5 Y 1 (winter) /Y2 (summer)
Annual income C 1 C2 C3 C4 C5 D 1 (winter) /D2 (summer)
Annual net income: w
Summer consumption time: somh(i)
Winter consumption time: win (1)
Initial Investment: Expenditure (I)
Building area: area (i) (i= 1, 2,3,4,5)
Obviously, this is a linear programming problem.
Using the variables defined above, it is easy to obtain:
Objective function: max(w)= ∑X(i)*C(i)+∑Y(i)*D(i)
Constraint: 3500-∑iX(i)*winh(i)>=0.
4000-∑iX(i)*somh(i)>=0
5000 & gt=∑iX(I)* Expenditure (I)
100 >=∑iX(I)* area (i)
x( 14)& lt； = 3000 X(24)& lt； = 3000 X( 15)& lt； = 32 X(25)& lt； =32
X( 14), X(24), X( 15) and X(25) are all integers.
The way to maximize the annual income is: don't plant crops, don't raise livestock, and go to work on the farm all the time in a year, and you can get a maximum income of 37,200.
It can also be seen from the reduced cost that even if soybeans, corn and oats are planted, the annual income of raising hens and cows will reach 249,550.5,252,735 and 5.94 respectively, and it is more profitable to go out to work.
The argot procedure is as follows:
Model:
Set:
Person/1... 5/:x, value, area, winh, somh, spendh
spearh/ 1..2/:h，evalue
End set
Data:
Value =175,300,120,450,3.5; (Annual income)
winh=20，35， 10， 100，0.6； (Consumption time in winter)
somh=30，75，40，50，0.6； (Summer consumption time)
spendh=0，0，0，400，3； (initial investment)
Area = 1, 1, 1,1.5,0; (Floor space)
evalue=4.8，5. 1； (Income from winter and summer jobs)
End data
max = @ sum(people(I):x(I)* value(I))+@ sum(spearh(I):h(I)* evalue(I))；
h( 1)= 3500-@ sum(people(I):x(I)* winh(I)； (Remaining time in winter)
H(2)=4000-@sum (person (I): x (I) * some (I)); (Remaining time in summer)
h( 1)>=0;
h(2)>=0;
@ sum(people(I):x(I)* spendh(I))& lt； =5000;
@ sum(people(I):x(I)* area(I))& lt； = 100;
x(4)& lt； =3000;
x(5)& lt； =32;
@ gin(x(4))； @ gin(x(5))；
end