1 Solver Finding maximum, minimum, or value by changing other cells Can add constraints Don’t need...

1

Solver

• Finding maximum, minimum, or value by changing other cells

• Can add constraints

• Don’t need to “guess and check”

2

Solver

3

Using Solver, Excel’s Solver

1. EXCEL’S SOLVER

The utility Solver is one of Excel’s most useful tools for business analysis. This allows us to maximize, minimize, or find a predetermined value for the contents of a given cell by changing the values in other cells. Moreover, this can be done in such a way that it satisfies extra constraints that we might wish to impose.

Example 1. The size limitations on boxes shipped by your plant are as follows. (i) Their circumference is at most 100 inches. (ii) The sum of their dimensions is at most 120 inches. You would like to know the dimensions of such a box that has the largest possible volume. Let H, W, and L be the height, width, and length of a box; respectively; measured in inches. We wish to maximize the volume of the box, V = HWL, subject to the limitations that the circumference C = 2H + 2W 100 and the sum S = H + W + L 120.

This problem is set up in the Excel file Shipping.xls. We will outline its solution with screen captures and directions. First, enter any reasonable values for the dimensions of the box in Cells B7:D7.

Shipping.xls

Using Solver. Using Solver. Excel’s SolverExcel’s SolverUsing Solver. Using Solver. Excel’s SolverExcel’s Solver

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To use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver.

Using Solver, Solver

W

H

L

FRAGILECrush slowly

Shipping.xls

Using Solver. Using Solver. Excel’s Solver: page 2Excel’s Solver: page 2Using Solver. Using Solver. Excel’s Solver: page 2Excel’s Solver: page 2

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Computer Problem?

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To use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver.

Enter cell that computes volume.

Select Max.

Enter cells that contain dimensions

Click on Add.

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Using Solver, Solver

Shipping.xls


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Enter cell that computes circumference.

Select <=.

Enter the limiting number.

Click on OK.

The requirement that the circumference be at most 100 inches is called a constraint. We want to have the contents of Cell E7 be at most 100.

Repeat the above process to add the constraint F7 <= 120, then click on Solve.

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Click on Keep Solver Solution.

Click on OK.

Click on Solve.

Using Solver, SolverUsing Solver. Using Solver. Excel’s Solver: page 4Excel’s Solver: page 4Using Solver. Using Solver. Excel’s Solver: page 4Excel’s Solver: page 4

Shipping.xls (material continues) ICT

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Using Solver, SolverThe dimensions that maximize

volume are now shown in Cells B8:D8. The maximum volume, the value of the circumference and the sum of the dimensions are now displayed. For a maximum volume of 43,750 cubic inches, the box should be 25 inches high, 25 inches wide, and 70 inches long.

Shipping.xls


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In rare cases; such as very large or small initial values of H, W, or L; you may need to add the constraints B7 >= 0, C7 >= 0 and D7 >= 0.

CT

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Using Solver, SolverUsing Solver. Using Solver. Excel’s Solver: page 6Excel’s Solver: page 6Using Solver. Using Solver. Excel’s Solver: page 6Excel’s Solver: page 6


Rush! shipping company limits the size of the boxes that it accepts by limiting their volume to at most 16 cubic feet (27,648 cubic inches). For it to ship a box, each dimension must be between 3 and 54 inches. (i) Modify Shipping.xls and use Solver to find the dimensions of a Rush! box which will accept the longest possible item. Hint: Use different initial values for each dimension. (ii) What is the maximum length of such an item? Note that the longest item which can be shipped in a box has a length of

.222 LWH

Shipping.xls ICT

Show ex3-sep14-shipping.xls

9

Solver

• Sensitive to initial value

• Use graphical approximation to help solve project

• Use to verify/solve Questions 1 - 3

• Use to solve Questions 6 - 8

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Integration

• Revenue as an area under Demand function

• qDqqR

-1.2

-10 q

D(q)

Demand Function

Revenue

qD(q)

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Integration

• Total possible revenue- The total possible revenue is the money that

the producer would receive if everyone who wanted the good, bought it at the maximum price that he or she was willing to pay. This is the greatest possible revenue that a seller or producer could obtain when operating with a given demand function

-1.2

-8

Demand Function

Total PossibleRevenue

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Integration• Consumer surplus – revenue lost by charging less/ Some buyers would have

been willing pay a higher price for the good than we charged. The total extra amount of money that people who bought the good would have paid is called the consumer surplus

• Producer surplus – revenue lost by charging more/ some potential customers do

not buy the good, because they feel that the price is too high. The total amount of this lost income, which we will call not sold, is represented by the area of the region under the graph of the demand function to the right of the revenue rectangle.

-1.2

-8q

D(q)

Revenue

ConsumerSurplus

NotSold

Demand Function

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Integration

• Approximating area under graph

- estimating areas of rectangles (by hand)

- Using Midpoint Sums.xls (using Excel)

- Using Integrating.xls (using Excel)

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Integration

• Approximating area (Midpoint Sums)

- Notation

- Meaning

bafSn ,,

given interval :,

givenfunction :

rectangles with sum :

ba

f

nSn

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Integration


- Process

Find endpoints of each subinterval

Find midpoint of each subinterval

nxxxxx ...,,,,, 3210

nmmmm ...,,,, 321

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Integration


- Process (continued)

Find function value at each midpoint

Multiply each by and add them all

This sum is equal to

nmfmfmfmf ...,,,, 321

imf x xmfxmfxmfxmf n ...321

bafSn ,,

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Integration


Ex1. Determine where .

5.1,0,3 fS 246 xxxf

5.1

0.1

5.0

0

3

2

1

0

x

x

x

x

25.1

75.0

25.0

3

2

1

m

m

m

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Integration


Ex1. (Continued)

375.2

5.025.15.025.25.025.1

5.025.15.075.05.025.0

5.1,0, 3213

fff

xmfxmfxmffS

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IntegrationApproximating area (Midpoint Sums.xls)

Ex1. (Continued)

=6*x-4*x^2

• Show• ex2-n-100Area

Example.xlsm

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EXAMPLE 2 - Modify sheet n = 20 in Area Example.xls, so that it computes the sum S100(f, [0, 4]), with 100 subintervals, for f(x) = 2x x2/2.

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Integration

• Approximating area (Integrating.xls)

- File is similar to Midpoint Sums.xls

- Notation: or or …

b

adxxf

b

adttf

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Integrationshow ex3-Integrating.xlsm


Ex3. Use Integrating.xls to compute 4

1

6/61 dxe x

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IntegrationApproximating area (Integrating.xls)

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Integration


Ex3. (Continued)

So . Note that is the p.d.f. of an exponential random variable with parameter . This area could be calculate using the c.d.f. function .

3331.04

1

6/61 dxe x 6/

61 xe

6 aFbF XX

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Integration


Ex3. (Continued)

3331.0

1535182751.0486582881.0

11

146/16/4

ee

FFaFbF XxXX

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Integration

• Approximating area

- Values from Midpoint Sums.xls can be positive, negative, or zero

- Values from Integrating.xls can be positive, negative, or zero

dqqDqDqdqqDmaxq

soldq

soldsold

soldq

)(SoldNot)()(Surplus

Consumer

0

.)(Revenue

PossibleTotal

0

dqqDmaxq

Integration, Applications


Integration. Integration. Applications: page 6Applications: page 6Integration. Integration. Applications: page 6Applications: page 6

Revenue computations for an arbitrary demand function work in the same way as those for the buffalo steak dinners.

Let D(q) give the price per unit for a good,that would result in the sale of q units, and let qmax be the maximum number of units that could be sold at any price. That is, D(qmax) = 0. The total possible revenue is given by

If qsold units are sold, then the revenue will be qsoldD(qsold). The following formulas give consumer surplus and lost revenue from units not sold.

It is clear that

revenue + consumer surplus + not sold = total possible revenue.

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Integration

• Ex4. Suppose a demand function was found to be

. Determine the consumer surplus at a quantity of 400 units produced and sold.

196.321225.00001392. 2 qqqD

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revenue + consumer surplus at 400 units

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Integration

• Ex. (Continued)

Calculate Revenue at 400 units

60.83569$

924.208400

400400

)(

D

qDqqR

qDqqR soldsoldsold

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Integration

• Ex. (Continued)

$107,508.80 – $83,569.60 = $23,939.20

So, the consumer surplus is $23,939.20

)(})({Surplus

Consumer

0

soldsold

q

qDqdqqDsold

Integration, Evaluation


Integration. Integration. Evaluation: page 6Evaluation: page 6Integration. Integration. Evaluation: page 6Evaluation: page 6

First Fundamental Theorem of Calculus

Let f and F be well behaved functions (continuous in the sense defined in the Help Me Understand link on page 82) that are defined on the closed interval [a, b]. Assume that f is the derivative of F on the open interval (a, b). In this case,

).()()( aFbFdxxfb

a

The expression F(b) F(a) is given a standard notation.

This is read as “F(x) evaluated from a to b.”

.)()()(ba

xFaFbF

The study of differentiation and integration is called calculus. It is evident that a relationship between these two branches of calculus is a major accomplishment.

IT C

Differentiation Antidifferentiation

1 uu xuxdx

d

C

u

xdxx

uu

1

1

xx eedx

d Cedxe xx

xx

dx

d 1ln Cxdx

xln

1


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FORMULAS

If the First Fundamental Theorem of Calculus is to be of any use in business problems, we must be able to find antiderivatives. The only available tools come from what we know about differentiation. Every differentiation formula translates into a formula for antidifferentiation. We will start with our four rules for the differentiation of specific types of functions.


C

Differentiation Antidifferentiation

)()( xfdx

daxfa

dx

d dxxfadxxfa )()(

)()()()( xgdx

dxf

dx

dxgxf

dx

d dxxgdxxfdxxgxf )()()()(

Products of functions do not work well with differentiation or antidifferentiation.

.)()()()(so),()()()( dxxgdxxfdxxgxfxfdx

dxf

dx

dxgxf

dx

d


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Differentiation formulas that allowed us to split up functions into smaller parts yield antidifferentiation formulas that can be used to split up indefinite integrals. Suppose that a is a constant and that both f and g are differentiable functions with antiderivatives.


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Integral Additivity. If a function f is integrable on some closed interval containing the numbers a, b, and c (in any order), then all three of the following integrals exists and

.)()()( c

b

b

a

c

a

dxxfdxxfdxxf




0

1a b c

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