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A multi plant problem. A company consists of two factories A and B . Each factory makes two products: standard and deluxe. Each unit of product yields the following profit. Each factory use two processes, grinding and polishing for producing its product. A multi plant problem. - PowerPoint PPT Presentation
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A multi plant problem
A company consists of two factories A and B. Each factory makes two products: standard and deluxe
Each factory use two processes, grinding and polishing for producing its product
standard deluxeunit profit 10 15
Each unit of product yields the following profit
A multi plant problem
factory a factory Bstandard deluxe standard deluxe
grinding 4 2 5 3polishing 2 5 5 6
The grinding and polishing times in hours for a unit of each type of product in each factory are
Factory A has a grinding capacities of 80 hours per week and polishing capacity of 60 hours per week
Factory B has a grinding capacities of 60 hours per week and polishing capacity of 75 hours per week
A multi plant problem
Availability of raw material
Each product (standard or deluxe) requires 4 kg of a raw material
The company has 120 kg of raw material per week
120 kg.
Factory A is allocated 75 Kg
Factory B is allocated 45 Kg
A possible scenario
Mathematical model for factory A
The two type of products are the decision variables for FACTORY A
Objective function is the profit to be maximize
standard = x1, deluxe = x2
max 10 x1 + 15 x2
x1 , x2 >= 0
Constraints: Availability of raw material
4 x1 + 4 x2 <= 75
Kg of raw material for unit of standard product
Kg of raw material for unit of deluxe product
Unit profit of standard product
Unit profit of standard deluxe
Mathematical model for factory A (2)
Constraints:
4 x1 + 2 x2 <= 80
Technological constraints
Grinding process
2 x1 + 5 x2 <= 60Polishing process
max 10 x1 + 15 x2
4 x1 + 4 x2 <= 75
4 x1 + 2 x2 <= 80
2 x1 + 5 x2 <= 60
x1 , x2 >= 0
Overall model for factory A
4 x1 +
4 x2 =
75
Geometric representation of F
Let draw the set F of the feasible solutions for factory A
In the plane (x1, x2 ), draw the equations of the constraints
5
10
15
20
25
30
35
40
40
45
5 10 15 20 25 30 35 40 40 45x1
x2
The constraint 4 x1 + 2 x2 = 80
does not play any role in defining the feasible region: removing it does not change F
2 x1 + 5 x
2 = 60
4 x1 +
2 x2 =
80
Bad use of resources !
Feasible region
All non negative points constitutes the
4 x1 +
4 x2 =
75
Geometric representation of the profit
In the plane (x1, x2 ), draw the equation of the profit PTOT for increasing values
5 10 15 20 25 30 35 40 40 45x1
5
10
15
20
25
30
35
40
40
45x2
2 x1 + 5 x
2 = 60
PTOT = 10 x1 + 15 x2
=0
=150
=300
PTOT =0
PTOT = 150
PTOT = 300
They are parallel lines
Find the value of PTOT such that
the corresponding line “touch” the points
PTOT =300 does not touch any point in F
4 x1 +
4 x2 =
75
Geometric solution
In the plane (x1, x2 ), draw the parallel lines to the equation PTOT = 10 x1 + 15 x2 =0 until the last point is found that “touches” the feasible region
5 10 15 20 25 30 35 40 40 45x1
5
10
15
20
25
30
35
40
40
45x2
2 x1 + 5 x
2 = 60
PTOT =0
Raw material
hours2 x1 + 5 x2 = 60
4 x1 + 4 x2 = 75Optimal solution
PTOT = 10 x1 + 15 x2 = 112.5 + 112.5 = 225
5.7
25.11
A B C D E2 factory A3 standard deluxe4 unit profit 10 155 raw material 4 46 grinding 4 27 polishing 2 589 production 10 10
10 PROFIT 25011 raw constraint 80 75 raw availability12 grinding constraint 60 80 max grinding13 polishing constraint 70 60 max polishing
Excel table for factory A
datax1=C9, x2 =D9
Decision variables = level of production
Profit = C4*C9+D4*D9
Raw constraint = C5*C9+D5*D9
Grinding constraint = C6*C9+D6*D9
Polishing constraint = C7*C9+D7*D9
Using the Solver
constraints
Objective function = profit
Decision variables
Mathematical model for factory B
The two type of products are the decision variables for FACTORY B
Objective function is the profit to be maximize
standard = x3, deluxe = x4
max 10 x3 + 15 x4
x3 , x4 >= 0
Constraints:
Availability of raw material
4 x3 + 4 x4 <= 45
Mathematical model for factory B (2)
Constraints: Technological constraints
5 x3 + 3 x4 <= 60Grinding process
5 x3 + 6 x4 <= 75Polishing process
max 10 x3 + 15 x3
4 x3 + 4 x3 <= 45
5 x3 + 3 x4 <= 60
5 x3 + 6 x4 <= 75
x3 , x3 >= 0
Overall model for factory B
Geometric representation of F
Let draw the set F of the feasible solutions for factory B
In the plane (x3, x4 ), draw the equations of the constraints
5 x3 +
3 x4 =
60
5 10 15 20 30 40 50x3
5
10
15
20
30
40
50x4
4 x3 +
4 x4 =
45
5 x3 + 6 x
4 = 75
Feasible region
All non negative points constitutes the
Two constraints 5 x3 + 6 x4 =
75 and 5 x3 + 3 x4 = 60 do not play any role in defining the feasible region: removing them does not change F
Bad use of resources !
Geometric solution
5 10 20 30 40 50x3
5
10
15
20
30
40
50x4
4 x3 +
4 x4 =
45
In the plane (x3, x4 ), draw the parallel equations of the profit PTOT for increasing values
PTOT = 10 x3 + 15 x4
=0
=100
Find the value of PTOT such that
the corresponding line “touch” the points
PTOT =0 P
TOT = 100
Raw material
x3 = 0
4 x3 + 4 x4 = 45
PTOT = 112.5
Optimal solution =
0
25.11
A B C D E2 factory B3 standard deluxe4 unit profit 10 155 raw material 4 46 grinding 5 37 polishing 5 689 production 0 11,25
10 PROFIT 168,7511 raw constraint 45 45 raw availability12 grinding constraint 33,75 60 max grinding13 polishing constraint 67,5 75 max polishing
Excel table for factory B
datax3=C9, x4 =D9
Decision variables = level of production
Profit = C4*C9+D4*D9
Raw constraint = C5*C9+D5*D9
Grinding constraint = C6*C9+D6*D9
Polishing constraint = C7*C9+D7*D9
Note: the excel formulae are the same for factory A and B. The model is independent from data
Look at the company in this scenario
COMPANYstandard deluxe
production 11,25 18,75PROFIT 393,75
Overall production = sum of the production of factory A and factory B
Profit of the company = sum of the profits of factory A and factory B
This solution has been obtained with arbitrary allocation of resources
Changing the scenario
Factory A is allocated 90 Kg
Factory B is allocated 30 Kg
The solution has been obtained with arbitrary allocation of raw material, we can see what happens when allocation change
120 kg.
Total raw material
Changing the scenario: geometric view
5 10 1520 30 40 50 x3
5
1520
30
40
50x4
5 10 1520 30 40 50 x1
4 x1 +
4 x2 =
90
5101520
30
404550x2
2 x1 + 5 x
2 = 60
4 x1 +
2 x2 =
80
Factory A Factory B
5 x3 +
3 x4 =
60
4 x3 +
4 x4 =
30 5 x3 + 6 x
4 = 75
5 10 1520
x1
5101520
x2
5 10 1520
x3
5101520
x4
new optimum for A new optimum for B
5
5.17
5.7
0
PTOT = 250 PTOT = 112.5
Changing the scenario: excel view
Factory A
factory Astandard deluxe
unit profit 10 15raw material 4 4grinding 4 2polishing 2 5
production 17,5 5PROFIT 250raw constraint 90 90 raw availabilitygrinding constraint 80 80 max grindingpolishing constraint 60 60 max polishing
Profit is higher than the preceding scenario
factory Bstandard deluxe
unit profit 10 15raw material 4 4grinding 5 3polishing 5 6
production 0 7,5PROFIT 112,5raw constraint 30 30 raw availabilitygrinding constraint 22,5 60 max grindingpolishing constraint 45 75 max polishing
Factory B
Profit is lower than the preceding scenario
COMPANYstandard deluxe
production 17,5 12,5PROFIT 362,5
Look at the company in the new scenario
Overall production = sum of the production of factory A and factory B
Profit of the company = sum of the profits of factory A and factory B
This solution is worst than the preceding one
Mathematical model for the company
The two type of products produced in FACTORY A and B are the decision variables
Objective function is the overall profit to be maximize
max 10 x1 + 15 x2 + 10 x3 + 15 x4
standard in factory A= x1, deluxe in factory A = x2
standard in factory B= x3, deluxe in factory B= x4
x1 , x2 , x3 , x4 >= 0
Mathematical model for the company (2)
Constraints:
4 x1 + 2 x2 <= 80
5 x3 + 3 x4 <= 60
Technological constraints
Grinding process
2 x1 + 5 x2 <= 60
5 x3 + 6 x4 <= 75
Polishing process
Constraints: Availability of raw material
Factory A
Factory B
Factory B
Factory A
4 x1 + 4 x2 + 4 x3 + 4 x4 <= 120 Common constraint
Mathematical model for the company
max 10 x1 + 15 x2 + 10 x3 + 15 x4
4 x1 + 2 x2 <= 80
5 x3 + 3 x4 <= 60
2 x1 + 5 x2 <= 60
5 x3 + 6 x4 <= 75
4 x1 + 4 x2 + 4 x3 + 4 x4 <= 120
x1 , x2 , x3 , x4 >= 0
More than two variables: we can solve it with the Solver
Excel table for the company
x1=C10, x2 =D10, x3=E10, x4 =F10Decision variables = level of production
A B C D E F2 COMPANY3 standard deluxe4 unit profit 10 15 data for the company5 raw material 4 46 factory B factory A7 grinding 5 3 4 2 data for the factories8 polishing 5 6 2 59 standard deluxe standard deluxe
10 company production 0 11,25 10 1011 COMPANY PROFIT 418,7512 factory B factory A13 grinding constraint 33,75 max grinding 60 grinding constraint 60 max grinding 8014 polishing constraint 67,5 max polishing 75 polishing constraint 70 max polishing 601516 raw constraint 125 120 raw availability
Profit = C4*(C10+E10)+D4*(D10+F10)
Raw constraint = C5*(C10+ E10 )+D5*(D10+F10)
Setting the solver
Optimal solution for the company
A B C D E F2 COMPANY3 standard deluxe4 unit profit 10 15 data for the company5 raw material 4 46 factory B factory A7 grinding 5 3 4 2 data for the factories8 polishing 5 6 2 59 standard deluxe standard deluxe
10 company production 0 12,5 9,166667 8,33333333311 COMPANY PROFIT 404,1666712 factory B factory A13 grinding constraint 37,5 max grinding 60 grinding constraint 53 max grinding 8014 polishing constraint 75 max polishing 75 polishing constraint 60 max polishing 601516 raw constraint 120 120 raw availability
Optimal production: deluxe = 20.8, standard = 9.17
Profit = 404.16 Better than 393.75 obtained with the arbitrary allocation
