Post your CTRL+V!

#31
DSCF0763.jpg
 
#32
PROBLEM #1

DECISION VARIABLES:
x1=# of trainees hired in January
x2=# of trainees hired in February
x3=# of trainees hires in March

CONSTRAINTS
s.t.

.1(x1) < 30
.7(x1) - .1(x2) > 20
.7(x2) + .7(x1) - .1(x3) > 70
.7(x1) + .7(x2) + .7(x3) = 120

The greater than or less than signs are all (greater than or equal to)
...or (less than or equal to).

.1 comes from the ratio of 1/10 for the teachers.

.7 comes from the ratio for the success in trainees 7/10.

30 comes from the 130 available trained machinists at the beginning of
the year minus the 100 needed machinists for January.

you get 20 by saying that there are 150 needed machinists for the month
of February and we have 130 for January. Hence we need 20 more
machinists/teachers for February.

70 is derived from 200 total needed for March minus 130.

120 is derived from 250 total needed for April minus 130.

OBJECTIVE FUNCTION
January: 400(x1) + .7(x1)(700) + .7(x1)(700) + .1(x1)(700)
Breaks down to 1450(x1)

February: 400(x2) + .7(x2)(700) + .1(x2)(700)
Breaks down to 960(x2)

March: 400(x3) + .1(x3)(700)
Breaks down to 470(x3)



Yikes, I was working on my linear programming homework!
 
#44
hehehehe, I had copied part of a deposition to show a friend what this silly witness said:


Q. So the pain is really your low back?
A. It's between my butt cheeks.
Q. All right.
A. I have a problem on the toilet.

:nuts:
 
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