Download Linear Programming Transportation

Linear Programming
By: Jordan Pickett and Diana
Vlassenko
Transportation Scenario
The officers of a highschool senior class are
planning to rent buses and vans for a class trip.
Each bus can transport 40 students, requires 3
chaperones, and costs $1200 to rent. Each van
can transport 8 students, requires 1 chaperone,
and costs $100 to rent. The officers must plan to
accommodate at least 400 students. Since only
36 parents have volunteered to serve as
chaperones, the officers must plan to use at
most 36 chaperones. How many vehicles of
each type should the officers rent in order to
minimize the transportation costs? What are the
minimal transportation costs?
Variables and Information
Buses=X
40 students, 3 chaperones
$1200 to rent
400 students
36 parents
Vans=Y
8 students, 1 chaperone
$100
Constraints
40x + 8y ≥ 400
3x + 1y ≤ 36
Objective function: We are trying to minimize so we can
get the lowest transportation cost possible. So we would
want to get the least amount of buses and vans that
would make it the lowest price to rent. Using the
previous constraints listed, and remembering x= buses
and y= vans, the below objective function will get us that
answer.
1200x + 100y= cost
POINTS: (10,0) (12,0) (8,10)
Linear Programming Transportation Graph
Calculations
POINTS: (10,0) (12,0) (8,10)
OBJECTIVE FUNCTION: 1,200x+100y=cost
SOLUTIONS:
1,200(10)+100(0)=12,000
1,200(12)+100(0)=14,000
1,200(8)+100(10)=10,600
ANSWER: 1,200(8)+100(10)=10,600
This would be the answer meaning we need 8
buses and 10 vans to seat the 400 students and
36 parents using the least amount of money
possible (minimizing). The other solutions were:
1,200(10)+100(0)=12,000
1,200(12)+100(0)=14,000
Those solutions would be more expensive, so they
wouldn’t be the answer we’re looking for
considering we want the cheapest one. We got
the solutions by taking the coordinate points that
surrounded the feasible region and plugging
them into the objective function. Then we took
the smallest cost number and made that our
answer.
In Conclusion…
In order to minimize transportation costs,
the high school would need to rent 8
buses and 10 vans, which would equal
$10,600.