pbl report-1

8/7/2019 PBL Report-1

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INSTITUTO TECNOLGICO Y DE ESTUDIOS SUPERIORES DE MONTERREY

CAMPUS SANTA FE

PBL- Report

Matemticas para ingeniera II Prof.Francisco Prez

Biaani Cervantes Luna A01015566

Tania Gonzlez Mata A01018297

Diana Lugo Pin A01015694

31 de enero de 2011

PBL Reporte: Ttulo

Abstract

The purpose of this PBL is to evaluate the funcionality of different methods, and to

evaluate which one of them is the best or has the major productivity in an industrial process. For

this, we need to evaluate the function for each graphic and therefore obtain the integral to see if a

process has more productivity than other, in this case, machine A and machine B have different

cycles and conditions. Another aspect to consider is the way in which the graphics need to be

ordered and maximize productivity and the method to obtain functions for both graphics.

Introduction

Parabola

A parabola is the set of all points in the

plane equidistant from a given line (the conic

section directrix) and a given point not on the line

(the focus). The focal parameter (i.e., the distance

between the directrix and focus) is therefore given

by , where is the distance from the vertex to

the directrix or focus.


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Integral

An integral is a mathematical object that can be interpreted as an area or a generalization

of area. Integrals, together with derivatives, are the fundamental objects of calculus.

Problem definition

The problem to be solved in this workshop consists on finding, first, the formula of a

parabola given the plot, and second, the areas under the two curves (parabola, line) between

specific intervals making use of integrals. That way, we will be able to find production rates and

determine which one of the proposed methods is more effective, or in its case, find another

method that results in a better production.

Input data

This graphic represents the behavior of

the production vs. time, in the machine A. The

machines production has a complete cycle of

eight hours, in which production increases during

the first four hours, the major production happens

in the fourth hour, and during the last four hours

it decreases. This machine can work eight hours

but must rest two hours.

This second graphic represents the

behavior of production vs. time, in the

machine B. The production is constant and

it can continue working indefinitely.

Also we know that in five days

(120 hrs) the machine can work with three

cycles:

y Machine A (8hrs) and machine B

(2hrs) with twelve repetitions in order to complete the five days.

y Machine A (4hrs) and machine B (2hrs) with twenty repetitions in order to complete the

five days.


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y Machine A (6hrs) and machine B (2hrs) with fifteen repetitions in order to complete the

five days.

Assumptions

We assumed that machine An in x=0 the production was 900 units, in x=4 the maxim

production was 1900 units and in x=8 the production had turned back to 900 units. Also, we

assumed that the function in the graphic corresponds with the parabola and it can be determined

following the equation: .

We assumed that the function of machine B corresponds to y=800

Data Analysis

Byusing the points in the graphic we can determine the equation ofthe parabola in the following way:

As result we obtained the following equation:

By plotting the equation, we realized that the equations are likely accurate.

x y

0 900

4 1900

8 900


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In order to find the area under the curve for

obtaining the production units, we have to integrate

the equations.

Then, we have to evaluate the functions according with the three cycles. Also, we can

integrate them determining the range in the graphic.

y Machine A (8hrs) and machine B (2hrs) with twelve repetitions in order to complete the

five days.

y Machine A (4hrs) and machine B (2hrs) with twenty repetitions in order to complete the

five days.

y Machine A (6hrs) and machine B (2hrs) with fifteen repetitions in order to complete the

five days.

After these procedures we searched others methods in order to achieve a major

production. We found two different cycles which results in major productions:


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y The first: we tried to found the cycle by random calculations taking into account that we

needed to eliminate the decrease in the machine A and we have to obtain a great number

of repetitions. So, we found this cycle:

y The second: we realized that the functions intersect themselves in y=1600, in this way

we can obtain a great production if we start the machine B when machine A decreases to

y=1600 for avoiding the following decrease of it. By the plotting and by equation

balancing, we can obtain the x value when y=1600.

Results

As we can observe, of the established cycles, the major production was reached by the

last one, with 172500 units. Because the machine A was stopped before to decrease until

y=1600. And the greatest cycle that we found was: machine A (7hrs) machine B (2hrs) which

results in 173388.88 units.

Conclussions


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According to the results, we could determine the functions for the graphics assuming it

was a polinomial funtcion of order 2, and see if the methods applied were lucrative and which

one was the best, for this, using integrals was the best option to determine the area under the

curve, to see it graphically and analitically, and comparing the results for each methos in units

sqaured, at last, we used the hours as intervals for the integral and evaluate how much production

was created after 120 hours in the cycle.

Reference

MathWorld. (2011). Mth. Disponible en: http://mathworld.wolfram.com/Integral.html

MathWorld. (2011). Mth. Disponible en: http://mathworld.wolfram.com/Parabola.html

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