TELBOY produces displays and batteries for smartphones across 3 floors. Floor 1 produces display glass and ITO, Floor 2 assembles batteries, and Floor 3 final assembly. The problem aims to maximize total weekly profits from 2 products subject to limited production hours. The optimal solution is to produce 2.25 batches of Product 1 and 2.8 batches of Product 2 for a maximum profit of $12,000 per week. Changing the available production hours could increase the maximum potential profit.

1. COMPANY PHONES
KAREN DANIELA CUERVO CELY
FABIAN ESTEBAN GONZALEZ GARAVITO

2. TELBOY
(TELEPHONY BOYACÁ)
TELBOY producing for high-end mobile Smartphone, including the display and
batteries which are assembled to the phone body. 3 floors, the first is where the
glass and ITO for mobile occurs, where the second and third stack which is
assembled is manufactured.
Product 1: Display glass and ITO.
Product 2: Battery for Smartphone.
The product 1 of the required capacity in the plants 1 and 3; 2 needs the product
of the work of levels 2 and 3.

3. Determine what production rates should have two products in order to
maximize total profits, subject to the restrictions imposed by limited
production capacity available in the 3 levels.
Facts:
● Number of production hours
available per week in each plant to
manufacture the products.
● Hours of manufacture are used to
produce each batch (50 units) of
each new article in each of the
plants.
● Earnings per batch of each new
product.

4. Formulation as a linear programming
problem
X1 = number 1 batch of product is manufactured by week.
X2 = Product Lot Number 2 being manufactured per week.
Z = total weekly earnings generated by these products.
PLANT
Batch production time,
hours
product
1 2
Available production
time per week, hours.
1
2
3
4 0
0 5
4 3
9
14
20
Earnings per Lot $6000 $4000

5. Maximize Z= 6X1 + 4X2
4X1<= 9
5X2<=14
4X1 + 3X2 <= 20
X1>= 0, X2>=0

6. 2, 25
2,8 4x1 <=9
5x2 <= 14
4x1 + 3x2 <=20

7. Z = 6X1 + 4X2
Z = 6(2)
Z= 12
X1 = 2
Z = 4X2
12 = 4X2
12/4 = X2
X1 = 3
When z=0 when Z=12

8. Max Z= 6X1 + 4X2

9. X1 = 2,25
X2 = 2,8

10. Maximize Z = 6X1 + 4X2
Sujeta a
4X1 <= 9
5X2 <=14
4X1 + 3X2 <= 20
X1>= 0, X2>=0
Minimize W = 9Y1 + 14Y2 + 20Y3
Sujeta a
4Y1 + 4Y3 >= 6
5Y2 + 3Y3 >= 4
Y1, Y2, Y3 >= 0

11. SIMPLEX DUAL

12. If the value of the number of hours of production batches of the product number
1 which is manufactured by week 9-11 changed the gain is increased.