the arrangements of tubular form is given by assignment 2011

8/6/2019 The Arrangements of Tubular Form is Given by Assignment 2011

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0903949 Mike Crosbie wc 4th

/April/2011

1) Let the number of UKSAT be x and let the number of EUROSAT be y

The arrangements of Tubular form is layout below:

Alpha Beta Chips Ass. AndPacking

Profit()

Monthlyoutputsupply

UKSAT (X) 12 15 30 6.50 9200EUROSAT(Y) 24 16 45 10.00 10000Total available figurefor the twocomponents

288000 240000 576000

2)

Linear programming problem: is a solution method to problems where an objective

has to be optimized subject to constraint the factors concerns are numerical and they

have relationships. LP can be formulated in standardised manner, example, it start

with objective or functions noting how many unknown or decision variable appears.

To find out the limit constraints that can be produced, the factors can put in

mathematically form as follows:

Let the number of UKSAT produced be x and

Let the number of EUROSAT produced be y

a) Maximise: X+Y (for part a)

b) Maximise profit= 6.50x + 10y, Profit (objectivity function)

S.T.: 1)12x +24y 288000 Alpha constraints

2) 15x + 16y


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3) Graph is a method of solving linear programming problems with unknown or

decision variables. Problem with unknowns need to be solved with care,

examples, limitation. Maximum and minimum problems can be dealt also with

graphically and the methods can also deal with limitations of the (>_) or less

(


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straight lines presenting different combinations of UKSAT and EUROSAT which yield

the contributions.

Follow that contribution line for shows the optimum production plan to provide the

maximum possible contribution would be point C where x= 4800, and y= 9600 which

yielding contributions of 127200 i.e. 6.50*4800 + 10*9600= 1272 00

4)

a) Optimum solution is the point where either UKSAT or EUROSAT gets maximise

profit, but before that equation would be:

a ) If y= mx+c

X= 0 and y= 10

y= 0-12/24= 1x/2

y= 1/2x+C

y= 10

10=1/2x +12

20=x+24

x= 24-20

x=4

If x= 4, y= 10

,(b) = (4, 10)

If y= 0, x=0

y=12.8, x=19.2

Y= mx+c, then, m=0-12.8/19.2

y= 2/3x+12

0=[-1/2-(-2/3]x +12-12.8


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-1x/2+2x/3-0.8=0

x/6-0.8=0+0.8

x/6*6=0.8*6

X=4.8

y= 9.6

(C) (4.8, 9.6)

[-2/3-[-15/16]x+12.8-15=0

-2x/3+15x/16-2.2=0

13x/48*48-2.2=0+2.2

13x/48*48=2.2*48

13x=105.6

X=8.1

y= -15/16*8.1+15=0

y= -121.5+240/16

y=7.4

(D) (8.1, 7.4)

(E) (9.2,6.4)

(a) (0, 10)


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Extremely points limiting the feasible region

X= Y= = Alpha Beta Packing =

A= (4,5) 9200 10000 350400 298000 726000 No spareB=(1,5) 4000 10000 288000 220000 576000 Spare

Alpha,Packing

C=(1,3) 4800 9600 288000 225600 576000 Spare A,P

D= (2,3) 8100 7400 274800 239900 576000 SparePacking

E= (3,4) 9200 6400 264000 240400 564000 N0 Spare

Optimum solution: 6.50x + 10y

Max profit

Extreme points 6.50x + 10y

A:(0,10000) 100,000

B:(4000,10000) 126000

C: (4800,9600) 127200

D: (8100, 7400) 126650

E: (9200, 6400) 123800

d) The table shows optimum solution is from point C where the output combination

extreme point is 4800 and 9600 which produced a profit of 127200.

5) Now, the UKSAT and EUROSAT firms can find Alpha shadows price in arithmetic

methods:

Alpha: 12x +24y 288001 (original 288000+1) ---equation (1)

Beta chip: 15x + 16y 240000 (unchanged) ---------equation (2)

So, 12*15 + 24* 15 = 288001*15

15* 12 + 16* 12 = 240000* 12

180x + 360y = 4320015


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180x+ 192 = 2880000

Now, do second equation minus the first equation as the x values cancels out

168y = 1440015

y= 8571.517857

Now, y value can be substituted into the second equation to find the new value of x.

15x+ 16*8571.517857= 240000

15x +137144.2857-137144.2857= 240000-137144.2857

15x= 1202855.7143

X= 6857.047619

Now, substituting 6857.047619 0f x and 8571.517857 of y into the objective function

gives a new contribution.

Therefore, 6.50(6857.047619) + 10(8571.517857) =130285.99

Original contribution= (127200)

Difference = 3085.99

Now, the firms find the Beta shadows:

12x+24y 288000 ----equation (1)

15x+16y 240001 -----equation (2)

180x+360y= 4320000

180x+192y =2880012

Do first equation minus from the second equation to find the new value of x

168y = 1439988

y= 8571.357143

Substitute the value of y into the first equation to find the new value of x.


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12x +24*8571.357143

12x +205712.5714-205712.5714 = 288000-205712.5714

12x= 82287.42857

X= 6857.285714

Substituting 6857.285714 of x and 8571.357143 of y into the objective function gives

a new contribution. 6.50(6857.285714) + 10(8571.357143) = 130285.93

Original contribution = (127200)

Difference =3085.93

So, the shadow prices increase for Alpha is greater than that for Beta chip. Anyway

both profits have increased.

Section B

2) The six existing products are shown in the table below:

a) Profit and Research Costs

Check the scattered diagrams on graphical papers .

To calculate the values of a and b is done in the general form of the equation fo r

a straight line where: y= a +bx

Using simultaneous:

an+ b=y----equation (1)

ax+bx=xy---equation (2)

Where n =the number of pairs of figure. The use of these equations will become:

6a+850b=396 ----------equation (1)

850a+126900=60570 equation (2)

Simultaneous equations are solved algebraically by eliminating one of the variables

and solving for the other. The value calculated is then substituted in one of the


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equations to find the one originally eliminated. So, equation one multiple 141.7 and

then deducted from equation two to eliminate a and therefore find the value of b:

850a+126900b=60570

850a+120445b=56113.2

Eliminate a and remains with 6455b=4456.8

b= 0.69x

This value is substituted in one of the original equation to find the value of a

therefore:

396-(850*0.69)/6a=

a= -31.75 Therefore, correlation coefficient is -31.75 which shows a strong

negative linear.

Therefore, the regression line is:

y= 0.69x-31.75

The three values of x chosen will be:

x -31.75+0.69x120 51.05141.7 66.02

180 92.45

3) Pearson product moment correlation coefficient r, and the coefficient of

determination r, for each module,

2b) Profit and Research solutions

y x y x xy

38 100 1444 10000 3800105 200 11025 40000 2100055 150 3025 22500 8250

60 120 3600 14400 720044 120 1936 14400 528094 160 8836 25600 15040396 850 29866 126900 60570

3a)The coefficient of correlation for this relation is,


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r= nxyxy (y) *y(y)

=6*60570 850*396 /6*126900 850 *6*29866 396

r= 363420 336600 / 761400 722500 *179196 156816

= 26820 / 38900

= 26820 / 29505.62658

r=0.909

r= 0.909 (3.d.p)

There is a strong positive correlation linear.

R= 0.90898

So, R= 0.826

r is called the Pearson product, moment correlation Coefficient and calculated as

follow:

y= a + bx, where a is intercept and b is the gra dient.

b= n (n()

6*60570- 850*396/6*126900-850*850= 363420-336600/761400-722500=

26820/38900= 0.6895

Therefore, b= 0.69 (2 d .p)

a= ybx

a= 39660.68946 8506=3966586.0416= 396 -586.0416= 190.0416= -31.6735

Therefore, a= -31.75

y= -31.75+0.69x


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Using the regression equation/lines to predict the likely profitability of a new product

launched after 250000 has been spent on research and development would be:

4a) 250,000 expenditure results in (0.69x250-31.75) = 140,750

2b) Profit and marketing Budget 000 for the equation of the least square regressionline:

y x y x xy

y x y x xy

38 20 1444 400 760

105 40 11025 1600 4200

55 20 3025 400 1100

60 30 3600 900 180044 20 1936 400 880

94 40 8836 1600 3760

396 170 29866 5300 12500

6a+170b=396---equation (1)

170a+5300b= 12500---equation (2)

So, 170a+ 5300b=12500---equation (2)

170a+4811=11206.8 Eq1 *28.3 (1)

489b=1293.2

b= 2.64x

Therefore, to find a becomes: 396-(170*2.64)/6a

a= -8.8

Therefore, the regression line is: y= -8.8+2.64x

The equation of the regression line of the best fit, the line which gets as near as

possible, on the average to the scattered point are chosen as:


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15, 28.3 and 35,

x 2.65x-8.815 30.828.3 6635 84

3b) the coefficient of correlations for this is:

R = nxy (xy)/nx (x) *n*y (y)

R =6*12500 170*396/6*5300 170 *6*29866 396

R= 75000 67320 31800 28900 *179196 156816

R= 7680/2900 *22380

R = 7680/8056.2

R = 95

Therefore, R=90.25%

In a similar way the regression y on x is also calculated below:

b=xy x2y/nxy (x)

b= 6*12500 170*396/6*5300 170

b= 75000 67320/31800 28900

b =2.64x

a = 396-(2.64*170)/6

a = -8.8

Therefore, y=

y= 2.65x-8.8


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4b) Using the regression equation/lines to predict the likely profitability of a new

product what the profit might result from a new product marketing spend of 27000

would be:

y= 2.65x- 8.8

y= 2.65*27-8.8

y=71.28-8.8

62,450 these profits result from a new product.

5)

In question one, movements of one another between variables is positive correlation

and therefore: the equation for y= a +bx

The calculation for drawing the mathematically correct line of best fit on graph will be

used by plotting three values of x. The lowest, highest and mean,

The three values of x chosen will be:

y= 0.69x-31.75and2.64x-8.8. Therefore, correlation coefficient is -31.75 and

-8.8 which shows a strong negative linear. Note: Normally automatically

produce the correct sign (+-) for the regression coefficient b; in this case,

minus. Thus, for profit and research x values are120, 141.7 and 180, and in

profit and marketing budget are 15, 28.3 and 35. Each of these values will be

substituted into the calculated regression l ine and the result values plotted on

the graph.

In 2nd question there are a and b which are constants and a represents the fixed

elements and b the slope of the line, that is to say, the ratio of the vertical increase in

y and to horizontal increase in x. Using simultaneous.

In 2nd that result also may be interpreted that in the problem r=83% , the variation in

actual faulty parts delivered may be predicted by change in the actual value s of x

amount spent on research and development, factor other than cha nges in the values

of x account for 17% of the variation in y.


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In third R=90.25% of the variation in actual faulty parts delivered may be predicted

by changes in the actual values of x amount spent on research and development,

factor than changes in the values of x account for 9.75% of the variation in y.

In fourth: When UKSAT products of x values chosen for: 120, 141.7 and 180

respectively, the value of all lie within the range of x values. It is called interpolation.

This is reliable estimate. if all values of 120, 141.7 and 180 lie outside the range of x

values, then that range would have been called extrapolation and they might not be

reliable estimated.

The same process in EROSAT products of x values chosen 15, 28.3 and 35 were

also being like UKSAT explained above.

the arrangements of tubular form is given by assignment 2011

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