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GEEN 1300Introduction to Engineering Computing

Class meeting #8Monday, Sept 21st

circular calculations and the iterative solver

solving nonlinear equations live solution using bisection spherical tank problem TABLE function application

Spreadsheet Problem Solving

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Homework #4, due WednesdaySection Test 1, next Monday, Sept 28th, 7–9pm,

Math 100 and ECCR 1B40Coverage: Class #1 through #8 today, Labs 1-5

Text, pages 1-274

pp

What is a circular calculation?

Calculation involvingone or many formulas

Answer“x”

Estimate“x”

for the solution to be correct,the estimate “x” going into thecalculation has to be equal tothe answer “x” coming out ofthe calculation

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the calculationA simple example:

1

xsin x

estimateanswer

2

This is also called

an problem x g x

or a substitution problem

or a fixed-point iteration problem

Note: by rearranging the above to x g x 0

we get a standard equation-solving problem

f 0

33

f x x g x 0

or an f x 0 problem

and we use different methods to solve it (later!)

For our simple example,

1

xsin x

estimateanswer

Circular Calculation of x=1/sin(x)

Estimate Answer

0.5 2.086

1.1492.086

1.096

1.124

1.149

1.096

Circular Calculation of x=1/sin(x)

1.000

1.500

2.000

2.500

x v

alu

e

44

1.109

1.117

1.113

1.115

1.124

1.109

1.117

1.113

0.000

0.500

1 2 3 4 5 6 7 8

try

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Excel-based solution using formula copying

Copy formulas down

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Excel-based solution using the iterative solver

Office Button Excel Options Formulas

Set up Excel for iterative solving first

Default settings T i l tti

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Default settings Typical settings

then click OK

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Setting up the iterative calculation on the spreadsheet

Initial estimatefor x

g(x) based ong( )what’s in B2

the result of thefirst try

enter a “pointer” formula to

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enter a “pointer” formula tocell B3 in place of the initialestimate

this “closes the loop” on the circular calculation

as soon as you press Enter,the calculation proceeds byone iteration

and “Calculate” is shown onthe Status bar to indicate thatthe Status bar to indicate thatyou are in an iterative situation

Note: the spreadsheet willnot update automaticallynow you have toforce it to re-calculate with

Pressing the F9 key afew times will causethe calculation toconverge to the correct

( i t l )

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the F9 [ Calculate ] keyanswer (approximately)

since “estimate” x agreeswith “answer” x, the equationis solved, numerically

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There are (at least) a couple issues here:

1) if you want to reset the calculation to the startingestimate (0 5) and run through it again you haveestimate (0.5) and run through it again, you haveto replace the pointer formula in B2 with 0.5, andthen put the pointer formula back in -- not convenient!

2) this type of calculation will not always convergein fact, it might just as likely blow up on you –

h t t d if thi h ?

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what to do if this happens?

Issue 1: How to reset the calculation

Enter labels “Reset” in C1and “Initial” in D1.

Transfer these labels toEnter “true” in C2

Transfer these labels tocells C2 and D2 as names[Formulas Create from Selection Top Row]

and 0.5 in D2

Instead of the pointer formula in B2, enter the formulashown below that uses Excel’s IF function.

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Excel’s IF function:

IF(logical expression, true alternative, false alternative)

a logical expression evaluates to a true/false result

examples: A1>2

AND(A3=4,A1<=2)

TRUE

FALSE

These are Excel’slogical constants

relational expression

logicalexpression

1111

FALSE

IF(Reset,Initial,B3)

If the value in Reset is TRUE, the value in cell Initial isthe result, otherwise, the value in cell B3 is the result

When Reset is TRUE,

and then when Reset is set to FALSE,

and then after pressing F9 a few times

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and then, after pressing F9 a few times,

we have the solution again.

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Automating the iterative solution

Experimentation with F9 shows that this calculationconverges in about 10 iterations.

Now, if you reset the calculation and enter FALSE again,the solution should appear without having to press F9.

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Showing the Developer tab

Office Button Excel Options Popular

14This will be necessary for VBA work in the next segment of the course.

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Automating the reset process

Instead of typing TRUE and FALSE repeatedly, we canautomate this with a checkbox directly on the spreadsheet.

click onceclick onceon thecheckboxthen dragout arectangleon the

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on theworksheet

Edit the title

Automating the reset process

Right-click the boundaryof the checkbox andclick on Format Control...

Type Reset into theCell Link field

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Cell Link field

and click OK.Click away from the checkboxfield, then

you can control the value in the Reset cell by clickingthe Initialize checkbox on and off.

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Issue 2: What to do if the calculation blows up

For example, with our original equation,

1

xsin x

solve for the x in the sin(x) term,

1 1 1x sin

x

This is an alternate form of the same equation. x g x

Try the formula copy technique on this form.

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Note: initial estimateof 0.5 won’t work, but1.1 (near the solution)will

Replace the initial estimate of 1.1 with the pointer formula,=B3,

P F9 th ti dPress F9 three times, and . . .

the calculation blows up!

So, when you are setting up a circular calculation, you

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always have to rearrange your equation in theform. You do this by solving for one of the x’s in theequation. If your formulation blows up, you can usuallyfind one that will converge by solving for one of the otherx’s in the equation. In other words, we were just luckythat our first formulation converged to the right answer!

x g x

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What’s behind this convergence and blow-up business?A typical graph of g x vs x will reveal the problem.

the initial estimate, x1,is plugged into g(x) toyield g(x1)

Note: the solution is where the g(x)curve intersects the 45-degree line,shown as a dashed line

g(x)

1g x 2g x g(x1) becomes x2,

the next estimate

the pattern repeats:x2 is plugged into g(x)to yield g(x2), whichbecomes x

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x1x 2x 3x becomes x3

evidently, the circular calculation is proceeding towardthe solution, indicated by the circle

This is a similar scenario, but with a different g(x) curve.

Notice that the processis diverging away fromthe solution – in otherwords, it’s blowing up!

g(x)

1x

1g x

2x3x

2g x

Now, observe what’sdifferent between thetwo examples.

The slope of the g(x)line in the first exampleis less than the slope

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x is less than the slopeof the 45-degree line,less than 1.The slope in this example is > 1

The condition required for the solution to converge isthat the absolute value of the slope (or derivative) ofg(x) at or near the solution is less than one.

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Now let’s apply this to a more complicated problem.

Steam in equilibrium with water

Equation of state – the ideal gas law

abs

mPV RT

MW

P pressure, Pa

V volume, m3

m mass, kg

MW molecular wt, kg/kgmolMW

Vapor-liquid equilibrium – the Antoine equation

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Blog P A

T C

Temperature scale conversion

R gas law constant

absT absolute temperature,Kelvins

T temperature, degC

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absT T 273.15

Problem to be solved: m, V known, find absP, T , T

known parameters: R, MW , A, B, C

A,B,C Antoine equationparameters

Strategy for solution

estimate T absT T 273.15

RTabsT

absRTmP

MW V

P

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BT C

A log P

answer

2222

T

circular arrangement

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Parameter values3Pa m

R 8314kgmol K

kgMW 18.02

kgmol

m 3 755 kg 3V 3 142 m

A 12.559 B 2354.7 C 280.709

m 3.755 kg V 3.142 m

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Initial estimate T 0 C

Solution using formula copying

convergesconvergesquickly!

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Using the iterative solver

Set up the calculation, starting with an estimate of 0 degC

Make sure that the Iterative Solver is set up properly

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Enter a pointer formula to Tans in the Test cell instead ofthe initial estimate

First iteration And, just like before, we canadd an initialization switch

Press F9 three times

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Converged!

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And what about another approach to the solution?

estimate P abs

P MW VT

m R

absT

absT T 273.15

abs

TB

AT CP 10

answer

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Panswer

Looks plausible

Try it!

ka-boom!This formulation is unstable andwill not converge to the answer.

That’s the story on solving iterative problems using Excel

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That’s the story on solving iterative problems using Excel.

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Solving single nonlinear equations in one unknown

precautions:

there may be multiple solutionsthere may be no solutionthere may be one solution that’s hard to findthere may be one solution that s hard to find

Simple example 1

xsin x

or x sin x 1 0

x g x form f x 0 form

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When possible, it’s useful to observe the behavior of f(x)graphically.

0 00

0.50

1.00

For our example, two solutions evidentfor x between 0 and 4

2 00

-1.50

-1.00

-0.50

0.00

f(x)

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-3.00

-2.50

-2.00

0 0.5 1 1.5 2 2.5 3 3.5 4

xNote: a solution is wheref(x) = 0

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For most engineering problems, multiple solutions are justa nuisance, since only one of the solutions will makephysical sense. That may not be the case in abstractmath problems.

So, we introduce a real problem:

R

V h

Vh R h

2 3

3

b g

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This diagram describes liquid stored in a spherical tank.Spherical tanks are commonly used. The volume ofliquid in the tank is related to the inside radius of the tankand the liquid depth by the equation shown above. If weknow values for R and h, computing V is straight-forward.

On the diagram, you will notice a level gauge on the sideof the tank. This “sight gauge” allows one to measure h.Then, the procedure would be to use the formula tocompute V. This is tedious, and may be difficult for

ti l th t d ’t h t t h i loperating personnel that don’t have strong technicaltraining. One could then have a graph of V versus hthat could be read to determine V. This is commonlydone. A better solution would be to put a scale directlyon the sight gauge that is calibrated in volume.

To make such a scale, we would decide on a set of

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volume “tick marks” for the scale and need to knowwhere to place them. In other words, we would knowa value of volume, V, and need to find h to mark thetick.

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Developing a calibration scale for a spherical tank

2h 3R hV

3

The fundamental equation can be rearranged tothe following f(x)=0 form:

3 2 3Vh 3Rh 0

or f h 0

This is a cubic equation, which will have three roots.Let’s consider an example to investigate the naturef th t

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of these roots.

We will try a radius of 10 feet (3.05m) and a liquidvolume of 500 gal (1.9m3).

50

100

There are three roots:one at about –0.5mone at about 0.5manother at about 9m

-50

0

f(h

)

another at about 9m

onlyfeasiblesolution

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-150

-100

-2 0 2 4 6 8 10 12

h

feasible rangeof h for tank

Multiple roots, but only one makes sense!

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So, if we confine ourselves to the feasible range for h[from tank empty (h=0) to tank full (h=2R)]there will be only one root.

Let’s say we start by making two estimates for h thatare certain to enclose or bracket the true solutionare certain to enclose or bracket the true solution.

These would be the ones above: h=0 and h=2R

Then we could evaluate f(h) for each of these estimates.

Editorial note: Since we have the graph of f(h) vs h,

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we can do a much better job of estimating the solutionthan these two estimates. But we usually don’t havea graph for every solution, so the graph here is justused as a way to illustrate the method.

-20.000

0.000

20.000

1f h

Notice that f(h1) and f(h2) areof opposite sign. That’s important.It means that there is a zerocross-over in f(h) between h1 andh2 – a solution to the equation.

-80.000

-60.000

-40.000

f(h

)

h2 a solution to the equation.

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-120.000

-100.000

0 1 2 3 4 5 6 7

h

1h 2h

2f h

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Now, we would like a better estimate for the solution, basedon our first two. A simple method (perhaps the simplest) isto use the midpoint between h1 and h2.

1 2mid

h hh

2

mid 2

We can then evaluate f(hmid) and see where it’s locatedon the graph.

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-20.000

0.000

20.000

1f h

-80.000

-60.000

-40.000

f(h

)

midf h

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-120.000

-100.000

0 1 2 3 4 5 6 7

h

1h2h

2f h

midh

20

-20.000

0.000

20.000

1f h

Because f(hmid) has the same sign (negative) as f(h2),hmid can become h2 the method can be repeated.

midf h

As the method is repeated overand over again, hmid will approachthe true solution.

-80.000

-60.000

-40.000

f(h

)

2f h

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-120.000

-100.000

0 1 2 3 4 5 6 7

h

1h2h

2f h

2hmidh

This is the basis for the bisection method for finding thesolution of a nonlinear equation. It is a simple, yet effectivemethod. Now, let’s implement it with an Excel spreadsheet.

First, we create column headings and put in the initialestimates for h1 and h2:1 2

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Next, we need to construct a formula for f(h) that makesreference to the cell to the left for the value of h. We dothat for f(h1) first. 3 2 3V

f h h 3Rh 0

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Since this formula makes reference to the cellto the left for h, it can be copied directly for f(h2).

Notice that the signs of f(h1) and f(h2) are opposite.That’s good. It means that there’s a solution in between.

Now we enter a formula for the midpoint and copy the f(h)Now, we enter a formula for the midpoint and copy the f(h)formula to evaluate f(hmid).

and

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Note that the sign of f(hmid) is negative, as is the sign of f(h2).We could have predicted that from our graph. This meansthat hmin will replace h2 for the next round (or iteration).

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But, in general, hmid will not always replace h2. Considerthe following scenarios:

0

5

10(x

)

f(x1)<0f(x2)>0

-10

-5

0

-1 1 3 5

x

f(

x2x1 xmid

( 2)f(xmid)<0xmid replaces x1

5

10

f(x1)<0

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-10

-5

0

5

-1 1 3 5

x

f(x

)

x2x1 xmid

f(x1)<0f(x2)>0f(xmid)>0xmid replaces x2

and these too:

0

5

10

f(x

)

f(x1)>0f(x2)<0f(x )>0

-10

-5

-1 1 3 5

x

x2x1 xmid

f(xmid)>0xmid replaces x1

5

10

f(x1)>0

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-10

-5

0

-1 1 3 5

x

f(x

)

x2x1 xmid

f(x1) 0f(x2)<0f(xmid)<0xmid replaces x2

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The strategy that emerges from these examples is:

hmid should replace h1 if f(hmid) is the same sign as f(h1)otherwisehmid should replace h2 if f(hmid) is the same sign as f(h2)

To implement this on the spreadsheet, we have to developa way to check whether the signs of two quantities, x and y,are the same, or different.

The easiest way to do this is to check the conditionx y 0

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If this condition is TRUE, x and y have to have the same sign.

This strategy is now implemented in the second row of thespreadsheet table using Excel’s IF function.

Note: a simple counter formula, =A11+1, has been enteredinto cell A12.

1 midf h f h 0 Select hmid if TRUE and h1 if FALSE

This implements the strategy whether to continue with thecurrent h1 or replace it with hmid.

4646And this implements the strategy whether to continue with thecurrent h2 or replace it with hmid.

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Notice that, this time, hmid replaced h2 for round 2 ofthe methodthe method.

The rest of the formulas are “operational” and canjust be copied down to complete the 2nd round.

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just be copied down to complete the 2 round.

What should happen in round 3?

We call the first row of the method the initialization row.It gets things off the ground.

We call the second row of the method the operational row.It can be copied downward to extend the method many rounds.

4848The three current estimates, h1, h2 and hmid arequite close now. We’re approaching the solution.

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After 20 rounds, things look like this:

The three estimates differ in the 5th significant figure,meaning that we have found the solution to better than1 part in 10 000 That’s close enough for engineering work!1 part in 10,000. That s close enough for engineering work!

This technique is called a “live solution” method becauseit presents the solution to the equation without having torun anything in Excel. To illustrate this, if we change theliquid volume from 500 gal (1.89 m3) to 1000 gal (3.79 m3),the new solution appears instantly as 0.6525 m:

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To make things more convenient to see, we put a pointerformula in cell B7 to bring up the value from cell F30 intoview.

pointer formulafor solution toequation

We’ve created a live solution of the equation, f(h)=0,and that solution is dependent on the value of theliquid volume that we specify in cells B5 and B6.

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We can now view this as a computing “engine” thatproduces a value of h, given a value of V:

V hBisection solutionof f(h) = 0

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We can now use this computing engine to solve the f(h)equation for a range of liquid volumes, employing Excel’sTABLE function. First, we create column headings

and a column of liquid volume

The TABLE function will put thesevolume values, one at a time, in

ll B5 th i t ll f l

and a column of liquid volumevalues from 100 to 31300 (justbefore the tank is full). UseHome Fill Series.

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cell B5, the input cell for volumein gallons in the upper left-handcorner of the spreadsheet.

Then, we need a pointer formula in cell K2 to get thedepth in feet. That is =B8.

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Then, we select the range of cells req’d for the TABLEfunction:

and, from the menu,

Data Data TableWhat-If Analysis

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Note: the column input cell field is usedbecause the volume values are in acolumn

Click OK!

and we now have the solution to the f(h) equation –more than 300 times!

We can make a calibration plot ofthese values:

20.00

Spherical Tank Calibration

6.00

8.00

10.00

12.00

14.00

16.00

18.00

Liquid Depth ‐feet

Series1

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and we can add a couple columns to the rightto display feet and inches, for ease of measurement.

0.00

2.00

4.00

0 5000 10000 15000 20000 25000 30000 35000

Volume ‐ gallons

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Formula for integral feet:

Formula for remainderin inches:

5555

and (finally!) we’re readyto mark our calibrationtape in even incrementsof liquid volume.

Some final comments on the bisection live solution method:

1) By replacing the initial estimates of h1 and h2 with someother x1 and x2, and by replacing all the f(h) formulaswith some other f(x) formula, we can use the samespreadsheet to solve another equation.

To get the appropriate accuracy in the solution, we mighthave to adjust the number of iterations (how far the tableis copied down) for another equation. [Later, we’ll showhow this can be automated using VBA.]

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[Obviously, we would have to change the basic datain the upper left-hand corner of the spreadsheet toreflect the new equation being solved.]

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2) The bisection method is one of the simplest methodsfor solving nonlinear equations. The scheme we havedeveloped can be adapted to more sophisticatednumerical methods, such as false position (regula falsi ), and Newton’s method (which uses derivatives)

Excel also has built-in tools that allow for morestreamlined solution of equations.

( )Although bisection is simple, and converges slowly, itis stable and reliable and suitable for mostengineering problems.

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These are Goal Seek and the Solver, and wewill introduce them in the next class.

They have advantages, and disadvantages, whencompared to the live solution method introduced here.

It will be advantageous to know all the approaches.