Hello guys, i'm back. Today i'd like to
share how to create mathematical modelling for optimation purpose. In
optimations, some factors need to be calculated to obtain the maximum or
minimum point.
INTRODUCTIONS
Sego Njamoer is a snack made from rice and mushroom served in
onigiri style. The word ‘Sego
Njamoer’ comes from Javanesse language means ‘Mushroom Rice’. Sego
Njamoer also being used as trademark of the company owned by Njamoer
Management, established in 2009 by Mahendra Ega Higuitta, Young Entrepreneur
and student of Engineering Physics Department Sepuluh Nopember Institute of
Technology. With many innovations, Sego Njamoer has collected many achievements
and award in business competitions, such as LA Business contest, PIMNAS 24,
Shell eco wire, and many others.
Production House of
Sego Njamoer addressed in Gebang Wetan, stored main ingerdients of the product
: Rice and Mushroom, storedhouse of the package, plastic spoon, and product
seasoning (such as flour, salt, pepper, etc). Not only for warehouse, PH also
used as outlet that sell Sego Njamoer Product.
Figure 1 Sego njamoer and its mascot (called Jembi)
Since June 2011, Sego Njamoer owned nine outlets in Surabaya. Four outlets located in ITS area (Ria Swalayan, BEM FTI, Center Canteen, and Gebang), and the others are located in Faculty of Medicine Airlangga University (Unair A), Faculty of Economy and Business Airlangga University (Unair B), Surabaya State University (Unesa), Stella Maris Senior High School (SMA Stella Maris), and Computer High School (Stikom). Based on managers data report, each outlet has different average amount demand in each day. Highest demand of the product is in BEM FTI outlet with 350 product per day. Followed by Unair A (260 portions), Outlet with the lowest demand is SMA Stella Maris which is 80 products. Complete report of each Sego Njamoer daily demand are listed in Table 1 below :
Figure 2 Map of sego njamoer outlet
Table 1 List of product average demand for
each outlet
No
|
Outlet
|
Average demand in one day
|
1
|
Ria Swalayan
|
150
|
2
|
BEM FTI
|
350
|
3
|
Center Canteen
|
200
|
4
|
Gebang
|
200
|
5
|
Unair A
|
260
|
6
|
Unair B
|
150
|
7
|
Unesa
|
100
|
8
|
SMA Stella Maris
|
80
|
9
|
Stikom
|
150
|
Main ingerdients (rice and mushroom) is not produced by the
company. Rice are collected from many food stall, and mushroom is providable
from market. Based on main ingerdients supply, maximum product that can be
served and selled are 1500 portions. Package of the prduct ordered from print
shop which is printed 100.000 packages per month. Fuel cost is one variable
that should be calculated because some outlet is far from production house.
Fuel cost for each oulet is shown in Table 2
Table 2 Daily fuel cost for each oulet
No
|
Outlet
|
Fuel cost per day (x1000 rupiahs)
|
1
|
Ria Swalayan
|
1
|
2
|
BEM FTI
|
1
|
3
|
Center Canteen
|
1
|
4
|
Gebang
|
0
|
5
|
Unair A
|
3
|
6
|
Unair B
|
3
|
7
|
Unesa
|
5
|
8
|
SMA Stella Maris
|
10
|
9
|
Stikom
|
4
|
Based
on the explanation above, how to obtain the maximum profit in one day?
SOLUTIONS :
Maximum supply in 1 day :
1500 products
Income in one day based
on maximum supply :
1500 x 3000 = 4.500.000 =
4.500 (x1000)
Package
ordered in 1 month are 100.000 packages, assume that there are 30 days in one
month
Objective : How to obtain
the maximum profit in 1 day?
Modeling :
Step 1 : Decision Variables :
Let xi = Amount
of product selled in each outlet (i)
x1 = Ria
Swalayan
x2 = BEM FTI
x3 = Center
Canteen
x4 = Gebang
x5 = Unair A
x6 = Unair B
x7 = Unesa
x8 = SMA
Stella Maris
x9 = Stikom
Step 2 : Objective Functions Maximize profit per each day :
z = x1 + x2
+ x3 + x4 + x5 + x6 + x7
+ x8 + x9
Step 3 : Define the Constrains :
Maximum demand in each
outlet
x1 <
150
x2 <
350
x3 <
200
x4 <
200
x5 <
260
x6 <
150
x7 <
100
x8 <
80
x9 < 150
Maximum
product that can be supplied :
x1 + x2
+ x3 + x4 + x5 + x6 + x7
+ x8 + x9 <
1500
Fuel
cost for each outlet (fuel cost greater than income is not allowed) :
x1 + x2
+ x3 + 3x5 + 3x6
+5x7 +10x8 +4x9 < 4200
Maximum package :
30 (x1 + x2
+ x3 + x4 + x5 + x6 + x7
+ x8 + x9) < 100.000
Each
variable should be positive
x1 >
0
x2 >
0
x3
>
0
x4 >
0
x5 >
0
x6 >
0
x7 >
0
x8 >
0
x9 > 0
Complete model :
z(xi) = x1
+ x2 + x3 + x4 + x5 + x6
+ x7 + x8 + x9
x1 < 150
x2 < 350
x3 < 200
x4 < 200
x5 < 260
x6 < 150
x7 < 100
x8 < 80
x9 < 150
x1 + x2
+ x3 + x4 + x5 + x6 + x7
+ x8 + x9 < 1500
x1 + x2
+ x3 + 3x5 + 3x6
+5x7 +10x8 +4x9 < 4200
3(x1 + x2
+ x3 + x4 + x5 + x6 + x7
+ x8 + x9) < 10.000
x1 >
0
x2 >
0
x3
>
0
x4 >
0
x5 >
0
x6 >
0
x7 >
0
x8 >
0
x9 > 0
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