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Linear Programming Problem with Example

Essay by   •  November 17, 2017  •  Essay  •  1,189 Words (5 Pages)  •  1,268 Views

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  1. Defining The Problem Of Beyond Era (Pvt) Ltd

The beyond era pvt ltd is company they produce and selling water bottle to market. Last month they imported new 2 water bottle product machine from china.

  1. Water bottles filing machine
  2. Water bottles packing machine

Those days very hot and have high demand for water bottle. They produce two size wetter bottles.

  1. 500 ML bottle
  2. 1000 ML bottle

The marketing team has concluded that the company could sell as much of either product per day as could be produced by these machine. However, those tow machine hours are limiting factors. According to machine manufacturing company those tow machine only can work per day as follows

  1. Filing Machine  12 H
  2. Packing machine 10 H

So beyond era pvt ltd management thinking about how many no of water bottle pack both tow size should be produced to maximize their profit .this is product mix type linear programming problem. so we have to select best product mix using linear programmer model .

  1. Identified Some Date To Solved This Problem
  1. How many bottles in one pack? 100 PCS
  2. Number of time of produce time use in each machine to produce each 2 bottle pack? it’s as follows  

MACHINE

TIME REQUIRED TO PRODUCE  100 PCS BOTTLE PACK

AVAILABLE HOURS PER DAY

500 ML BOTTLE PACK

1000 ML BOTTLE PACK

FILING

2M

4M

12*60 = 720M

PACKING

3M

2M

10*60 = 600M

  1. How much each bottle pack profit?
  • 500 ml 100 Pcs pack Rs 3000/=
  • 1000 ml 100 Pcs pack Rs 5000/=

According to above data beyond era pvt ltd management want to know that what is the best product mix to produce to reach the maximum profit .

  1. Mathematical Formulation

Beyond era management is looking best product mix of both products to maximize their profit. So, we identified this is product mix linear programming problem . using graphical and simplex method we will advice to management to select their best production mix to maximize their profit .  

Data we have collect as follows

MACHINE

TIME REQUIRED TO PRODUCE  100PCS BOTTLE PACK

AVAILABLE HOURS

500 ML BOTTLES PACK(X)

1000 ML BOTTLES PACK(Y)

PER DAY

FILING

2M

4M

12*60 = 720M

PACKING

3M

2M

10*60 = 600M

PROFIT PER UNIT

RS 3000

RS 5000

 

  • The objective is to maximize profit.
  • The constraints are
  • The hours of filing machine time use cannot exceed 12 hours per day.
  • The hours of packing machine time use cannot exceed 10 hours per day
  • The decision variables representing the actual decision we will make are
  • X = number of 500 ml bottles pack to be produced per day
  • Y = number of 1000 ml bottles pack to be produced per day.

So we can develop mathematical function as follows

  • Profit maximization = Rs3000X  +  Rs5000Y
  • Filing machine time using =  2X + 4Y  <  720
  • Packing machine time using = 3X + 2Y < 600
  • Both of these constraints restrict production capacity and affect to profit
  •  The number of production X and Y greater than or equal to 0 so X,Y > 0

  1. Develop A Solution

3.1) Graphical solution

To use above data we solved linear programmed using graphical method as follows.

We can short collect data like this

X

Y

Z

3000

5000

 

 

FILING

2

4

<

720

PACKING

3

2

<

600

  • The fist step in solving the problem is to identify a set of region of feasible solution
  • To do this we plot each constraint equation on a graph.
  • When 1000ml bottles production is 0 500ml bottles production by fully using filling machine as follows

2x + 4y    <= 720

2x + 4(0) <= 720

         2x   <= 720

            X   = 720/2

            X   = 360

  • Similarly no 500ml bottle

2x + 4y    <= 720

2(0) + 4Y <= 720

         4Y  <= 720

            Y   = 720/4

            Y  =  180

  • After solving above tow equations we can make flowing table for filing machine

FILING

X

Y

360

0

0

180

  • Same as we can make table for packing machine as follows

3x + 2y    <= 600

3x + 2(0) <= 600

         3x   <= 600

...

...

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