revision linear optimisation. a miller can buy wheat from three suppliers: airey farm, berry farm...
TRANSCRIPT
Revision
Linear Optimisation
A miller can buy wheat from three suppliers: Airey Farm,
Berry Farm and Cherry Farm. In each case the wheat is
contaminated with two things – bran and husks. When combined, the wheat must contain no more than 5% bran and no more than 4% husks. The miller wishes to
make 50 tonnes of wheat in total, by purchasing from each
farm. Each farm’s wheat contains the following amounts
of bran and husks.
Identify the decision variables
• Let x = tonnes of wheat purchased from Airey Farm
• Let y = tonnes of wheat purchased from Berry Farm
• Let z = tonnes of wheat purchased from Cherry Farm
• Write z in terms of x and y
Define the constraint due to bran
• 3% of x + 5% of y + 7% of z is less than or equal to 5% of 50 tonnes.
€
0.03x + 0.05y + 0.07z ≤ 0.05 × 50
€
3x + 5y + 7(50 − x − y) ≤ 5(50)
€
4x + 2y ≥100
Define the constraint due to husks
• 5% of x + 2% of y + 6% of z is less than or equal to 4% of 50 tonnes.
€
0.05x + 0.02y + 0.06z ≤ 0.04 × 50
€
5x + 2y + 6(50 − x − y) ≤ 4(50)
€
x + 4y ≥100
Non-negativity
€
x ≥ 0
€
y ≥ 0
Supply of wheat from Cherry Farm must be non-negative too.
€
50 − x − y ≥ 0
€
x + y ≤ 50
Graph the solution region
100
50
25
50
Find the vertices of the solution region
100
50
25
50
(0, 50)
(14.29, 21.43)
(33.33, 16.67)
Define the objective function
• Cost = $70x + $60y + $40z• Cost = $70x + $60y + $40(50 - x - y)
• Cost = 30x + 20y + 2000
Substitute each vertex’s coordinates into the objective
function• Cost = 30x + 20y + 2000• A(0, 50) Cost = $3000• B(14.29, 21.43) Cost = $2857.30• C(33.33, 16.67) Cost = $3333.30
• B gives the minimum costs so purchase
14.29 from Airey, 21.43 from Berry and 14.28 from Cherry
Question 2Barmy Bicycle Builders makes bicycles at two factories and ships them to two separatedistributors.Factory A produces 100 bicycles; Factory B produces 150.Distributor X wants 70 bicycles;Distributor Y wants 90 bicycles.Furthermore the shipping costs between factories (in dollars per bicycle) and distributorsare shown in the table below.
Factory A Factory BDistributor X $6 $7Distributor Y $3 $5
Define the decision variables
• Let x = number of bicycles shipped to distributor X from factory A
• Therefore 70 - x is the number shipped to X from factory B
• Let y = number of bicycles shipped to distributor Y from factory A
• Therefore 90 - y is the number shipped to y from factory B
Summarise the decision variables, supply and demand
in a tableFactory A Factory B Total
demand from each distribut
or
Distributor X
Distributor Y
Total supply
from each factory
Summarise the decision variables, supply and demand
in a tableFactory A Factory B Total
demand from each distribut
or
Distributor X
x 70 - x 70
Distributor Y
y 90 - y 90
Total supply
from each factory
100 150
Define constraints based on the fact that all shipments must be non-negative amounts
• Factory A:
€
x + y ≤100
Define constraints based on the fact that all shipments must be non-negative amounts
• Factory B:
€
70 − x + 90 − y ≤150
€
x + y ≥10
Define constraints based on the fact that all shipments must be non-negative amounts
• Non-negativity:
x ≥0y≥0
70 −x≥0 ⇒ x≤7090 −y≥0 ⇒ y≤90
Graph the solution region
100
100
Determine the vertices
100
100 (0, 10)
(0, 90)
(10, 90)
(70, 30)
(70, 0)
(10, 0)
Determine the objective function
• Factory A to distributor X: x bicycles at $6
• Factory A to distributor Y: y bicycles at $3
• Factory B to distributor X: 70 - x bicycles at $7
• Factory B to distributor Y: 90 - y bicycles at $5
• Total cost =
€
6x + 3y + 7(70 − x) + 5(90 − y)
€
Cost = − x − 2y + 940
Determine the cost for each vertex
• A(0, 10) Cost = 920• B(0, 90) Cost = 760 • C(10, 90) Cost = 750• D(70, 30) Cost = 810• E(70, 0) Cost = 870• F(10, 0) Cost = 930
• Minimum costs occur when x = 10 and y = 90
Amount to ship from each factory
• Ship 10 from factory A to distributor X
• Ship 90 from factory A to distributor Y
• Ship 60 from factory B to distributor X
• Ship 0 from factory B to distributor Y
