Un problema multi impianto
Un’azienda dispone di due fabbriche A e B. Ciascuna
fabbrica produce due prodotti: standard e deluxe
Ogni unità di prodotto da luogo ad un profitto unitario
riportato in tabella
standard deluxe
profitto unitario
10
15
Ogni fabbrica, A e B, gestisce due processi produttivi:
smerigliatura (grinding) e lucidatura (polishing)
Un problema multi impianto
I tempi di smerigliatura e lucidatura (espressi in ore per
unitá di ogni tipo di prodotto) nelle due fabbriche sono
diversi e riportati in tabella
smerigliatura
lucidatura
factory A
standard deluxe
4
2
2
5
factory B
standard deluxe
5
5
3
6
La fabbrica A ha macchinari per la smerigliatura con capacità di
80 ore settimanali e per la lucidatura con capacità di 60 ore
settimanali
La fabbrica B ha macchinari per la smerigliatura con capacità di
60 ore settimanali e per la lucidatura con capacità di 75 ore
settimanali
Un problema multi impianto
Disponibilitá di materiale grezzo
Ogni prodotto (standard o deluxe) richiede 4 kg di materiale
grezzo
L’azienda dispone di 120 kg di materiale grezzo a settimana
Determinare il livello di produzione ottimo (ovvero che
massimizza il profitto)
Un problema multi impianto
Disponibilitá di materiale grezzo
A possible scenario
75 Kg sono assegnati alla Fabbrica A
120 kg.
45 Kg sono assegnati alla Fabbrica B
Fabbrica A
Dunque abbiamo due
modelli matematici
Fabbrica B
Modello Matematico per la fabbrica A
Le variabili di decisione per la fabbrica A sono le quantità
di ciascun tipo di prodotti
standard = x1, deluxe = x2
x1 , x2 >= 0
La funzione obiettivo è il profitto che deve essere
massimizzato (max)
max 10 x1 + 15 x2
Profitto di un’unità
di prodotto standard
Vincoli:
Profitto di un’unità di
prodotto deluxe
Disponibilità di materiale grezzo
4 x1 + 4 x2 <= 75
Kg of raw material for
unit of standard product
Kg of raw material for
unit of deluxe product
Modello Matematico per la fabbrica A (2)
Ulteriori vincoli:
Vincoli di processo
Grinding process
4 x1 + 2 x2 <= 80
Polishing process
2 x1 + 5 x2 <= 60
max 10 x1 + 15 x2
4 x1 + 4 x2 <= 75
4 x1 + 2 x2 <= 80
2 x1 + 5 x2 <= 60
x1 , x2 >= 0
Modello
completo per la
fabbrica A
Soluzione geometrica: insieme amissibile
Grafico l’insieme F delle possibile soluzioni ammissibili
x2
45
40
I punti non negativi indicati con
costituiscono la
regione ammissibile
40
35
The constraint 4 x1 + 2 x2 = 80
does not play any role in
defining the feasible
region: removing it does
not change F
30
25
20
15
10
5
5
10 15 20 25 30 35 40 40 45
x1
Bad use of resources !
Soluzione geometrica: profitto Fabbrica A
In the plane (x1, x2 ), draw the equation of the profit PTOT for
increasing values
=0
x2
45
40
PTOT = 10 x1 + 15 x2
=150
40
35
They are parallel lines
=300
30
Find the value of PTOT such that
the corresponding line “touch”
the points
25
20
15
10
PTOT =300 does not touch any
point in F
5
5
10 15 20 25 30 35 40 40 45
x1
Soluzione geometrica:
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
x2
45
40
40
35
Optimal solution
2 x1 + 5 x2 = 60
Ore lavoro
4 x1 + 4 x2 = 75
materiale
30
25
11.25 


 7.5 
20
15
10
5
5
10 15 20 25 30 35 40 40 45
x1
PTOT = 10 x1 + 15 x2 =
112.5 + 112.5 = 225
Foglio Excel per analisi di scenario fabbrica A
A
2
3
4
5
6
7
8
9
10
11
12
13
B
unit profit
raw material
grinding
polishing
production
PROFIT
raw constraint
grinding constraint
polishing constraint
C
D
E
factory A
standard deluxe
10
15
4
4
4
2
2
5
10
250
80
60
70
data
Decision variables =
level of production
x1=C9, x2 =D9
10
Profit = C4*C9+D4*D9
75 raw availability
80 max grinding
60 max polishing
Raw constraint = C5*C9+D5*D9
Grinding constraint = C6*C9+D6*D9
Polishing constraint = C7*C9+D7*D9
Using the Solver
Objective function
= profit
Decision variables
constraints
Modello Matematico per la fabbrica B
The two type of products are the decision variables for
FACTORY B
standard = x3, deluxe = x4
x3 , x4 >= 0
Objective function is the profit to be maximize
max 10 x3 + 15 x4
Constraints:
Availability of raw material
4 x3 + 4 x4 <= 45
Modello Matematico per la fabbrica B (2)
Constraints:
Technological constraints
Grinding process
5 x3 + 3 x4 <= 60
Polishing process
5 x3 + 6 x4 <= 75
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
Soluzione geometrica: insieme amissibile
Let draw the set F of the feasible solutions for factory B
In the plane (x3, x4 ), draw the equations of the constraints
x4
All non negative points
constitutes the
50
Feasible region
40
30
20
15
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
10
5
Bad use of resources !
5
10 15 20
30
40
50
x3
Soluzione geometrica
In the plane (x3, x4 ), draw the parallel equations of the profit PTOT for
increasing values
=0
x4
PTOT = 10 x3 + 15 x4
50
=100
40
Find the value of PTOT such that
the corresponding line “touch”
the points
30
11.25 

Optimal solution = 
 0 
20
15
10
x3 = 0
Raw
4 x3 + 4 x4 = 45 material
5
5
10
20
30
40
50
x3
PTOT = 112.5
Foglio Excel per analisi di scenario fabbrica B
A
2
3
4
5
6
7
8
9
10
11
12
13
B
unit profit
raw material
grinding
polishing
production
PROFIT
raw constraint
grinding constraint
polishing constraint
C
D
E
factory B
standard deluxe
10
15
4
4
5
3
5
6
0
168,75
45
33,75
67,5
11,25
Raw constraint = C5*C9+D5*D9
Decision variables =
level of production
data
x3=C9, x4 =D9
Profit = C4*C9+D4*D9
45 raw availability
60 max grinding
75 max polishing
Note: the excel formulae
are the same for factory A
Grinding constraint = C6*C9+D6*D9 and B. The model is
Polishing constraint = C7*C9+D7*D9 independent from data
Uno sguardo “globale” sull’azienda
in questo scenario
production
PROFIT
COMPANY
standard
deluxe
11,25
18,75
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
The solution has been obtained with arbitrary allocation of
raw material, we can see what happens when allocation
change
Total raw material
Factory A is allocated 90 Kg
120 kg.
Factory B is allocated 30 Kg
Changing the scenario: geometric view
x2
50
45
40
x4
50
Factory A
40
30
30
20
15
10
5
20
15
Factory B
5
5 10 1520
30
40
50 x1
x2
20
15
10
5
5 10 1520
30
40
50 x3
x4
new optimum for A
17.5 


5


5 10 1520
x1 PTOT = 250
20
15
10
5
new optimum for B
 0 
 
 7 .5 
5 10 1520
x3 P = 112.5
TOT
Changing the scenario: excel view
unit profit
raw material
grinding
polishing
factory A
standard deluxe
10
15
4
4
4
2
2
5
production
PROFIT
raw constraint
grinding constraint
polishing constraint
17,5
250
90
80
60
Factory B
Profit is lower than
the preceding
scenario
Factory A
Profit is higher than
the preceding
scenario
5
90 raw availability
80 max grinding
60 max polishing
unit profit
raw material
grinding
polishing
production
PROFIT
raw constraint
grinding constraint
polishing constraint
factory B
standard deluxe
10
15
4
4
5
3
5
6
0
112,5
30
22,5
45
7,5
30 raw availability
60 max grinding
75 max polishing
Look at the company in the new scenario
production
PROFIT
COMPANY
standard
deluxe
17,5
12,5
362,5
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
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
Objective function is the overall profit to be maximize
max 10 x1 + 15 x2 + 10 x3 + 15 x4
Mathematical model for the company (2)
Constraints:
Technological constraints
4 x1 + 2 x2 <= 80
Factory A
5 x3 + 3 x4 <= 60
Factory B
Polishing process 2 x1 + 5 x2 <= 60
Factory A
5 x3 + 6 x4 <= 75
Factory B
Grinding process
Constraints: Availability of raw material
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 x 4
2 x1 + 5 x2
<= 60
<= 60
5 x3 + 6 x 4
4 x1 + 4 x2 + 4 x3 + 4 x4
<= 75
<= 120
x1 , x2 , x3 , x4 >= 0
More than two variables: we can solve it with the Solver
Excel table for the company
A
B
C
2
COMPANY
3
standard
4 unit profit
10
5 raw material
4
6
factory B
7 grinding
5
8 polishing
5
9
standard
10 company production
0
11 COMPANY PROFIT
418,75
12
factory B
13 grinding constraint
33,75
14 polishing constraint
67,5
15
16 raw constraint
125
D
E
F
deluxe
15
4
data for the company
factory A
3
4
6
2
deluxe
standard deluxe
11,25
10
max grinding
max polishing
2 data for the factories
5
10
factory A
60 grinding constraint
75 polishing constraint
60 max grinding 80
70 max polishing 60
120 raw availability
Decision variables = level of production x1=C10, x2 =D10, x3=E10, x4 =F10
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
2
3
4 unit profit
5 raw material
6
7 grinding
8 polishing
9
10 company production
11 COMPANY PROFIT
12
13 grinding constraint
14 polishing constraint
15
16 raw constraint
C
COMPANY
standard
10
4
factory B
5
5
standard
0
404,16667
factory B
37,5
75
120
D
E
F
deluxe
15
4
deluxe
data for the company
factory A
3
4
2 data for the factories
6
2
5
standard deluxe
12,5 9,166667
8,333333333
max grinding
max polishing
factory A
60 grinding constraint 53 max grinding 80
75 polishing constraint 60 max polishing 60
120 raw availability
Optimal production: deluxe = 20.8, standard = 9.17
Profit = 404.16
Better than 393.75 obtained with the arbitrary
allocation