1 chapter 4 planning models operations analysis using ms excel

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Chapter 4Chapter 4

Planning ModelsPlanning Models

Operations Analysis Using MS Excel

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Planning modeling

Chapter Outline:

1. The basic planning problem

2. The basic pricing problem

3. Nonlinear cost and demand functions

• XP Function

• Mathematical model of an XY Function

• Spreadsheet Model of XY Function

• Approximating the cost with a Cubic Function

4. Preparing a Five year Plan

5. The Impact of Pricing

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Growth RatesUnits sold 50%Fixed costs 10%

Year 1 2 3 4 5Unit selling price $60 $60 $60 $60 $60Fixed cost $1,500 $1,650 $1,815 $1,997 $2,196Variable cost $45 $45 $45 $45 $45Number of units 80 120 180 270 405Total cost $5,100 $7,050 $9,915 $14,147 $20,421Revenue $4,800 $7,200 $10,800 $16,200 $24,300Profit -$300 $150 $885 $2,054 $3,879

Discount rate 5% 10% 15% 20%Present value $5,343 $4,327 $3,537 $2,915

Assumptions

Present Value of Profit

The Basic Planning ProblemDown is a skeleton model for a five year projection of profit for a corporation.

Assumption

•Selling price is $60 (not change over next five years)

•Fixed cost $1500 (grow at constant rate)

•Number of units – 80 (grow at a constant rate)

•Variable cost $45 per unit (not change over next five years)

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The NPV function calculates the net present value based on a series of cash flows. The syntax of this function is=NPV(rate,value1,[value2],[...])

The annual cash flows are the (profit = revenues minus costs) generated from the investment during its lifetime.

NPV compares the value of a dollar today to the value of that same dollar in the future, taking inflation and returns into account. These cash flows are discounted or adjusted by incorporating the uncertainty and time value of money.

NPV is one of the most robust financial evaluation tools to estimate the value of an investment.NPV > 0 the investment would add value to the firm, so the project may be accepted NPV < 0 the investment would subtract value from the firm, so the project should be rejected NPV = 0 the investment would neither gain nor lose value for the firm, so we should be indifferent in the decision whether to accept or reject the project.

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Present Value of Profits

$0

$500

$1,000

$1,500

$2,000

$2,500

5% 10% 15% 20%

Profit

-$1,000

$0

$1,000

$2,000

$3,000

$4,000

1 2 3 4 5

The Basic Planning Problem

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The Basic Planning Problem

Data table is created to view What-if Analysis on the growth rates for fixed costs and units sold

$1,919 10% 20% 30% 40% 50% 60% 70% 80% 90% 100%5% -$730 -$98 $611 $1,402 $2,279 $3,249 $4,315 $5,483 $6,758 $8,144

10% -$1,091 -$459 $250 $1,041 $1,919 $2,888 $3,955 $5,123 $6,397 $7,78315% -$1,474 -$842 -$133 $658 $1,535 $2,505 $3,571 $4,739 $6,014 $7,40020% -$1,881 -$1,249 -$540 $251 $1,129 $2,098 $3,165 $4,333 $5,607 $6,99325% -$2,311 -$1,680 -$971 -$180 $698 $1,668 $2,734 $3,902 $5,177 $6,56230% -$2,767 -$2,135 -$1,426 -$635 $243 $1,212 $2,279 $3,446 $4,721 $6,10735% -$3,248 -$2,616 -$1,907 -$1,116 -$239 $731 $1,797 $2,965 $4,240 $5,62640% -$3,756 -$3,124 -$2,415 -$1,624 -$746 $224 $1,290 $2,458 $3,732 $5,11845% -$4,290 -$3,658 -$2,949 -$2,159 -$1,281 -$311 $755 $1,923 $3,198 $4,58450% -$4,853 -$4,221 -$3,512 -$2,721 -$1,843 -$874 $193 $1,361 $2,635 $4,021Gr

owth

Rat

es fo

r Fixe

d Co

st

Growth Rates For Units Sold

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The Basic Pricing ProblemThe central issue in pricing is determining how quantity sold depends on the price.

Demand in most cases is elastic, that is when the price increases, the demand decreases.

The simplest assumption is that demand is a linearly decreasing function of price.

Organization assumptions or judgments:

1. At price $70, sales will be 2,400

2. One dollar increase in priceincrease in price = 37 units decrease in the saledecrease in the sale

3. Using the above information to express the number of units sold as a function of price.

PRICE UNIT

(PRICE– 70) = 0 70 2400

(PRICE – 70) > 0 INCREASE DECREASE

(PRICE – 70) < 0 DECREASE INCREASE

4. Change in number of units = - 37 * (PRICE - 70)

5. Using the second assumption, then

6. Number of unit = 2400 + Change in number of units

7. Number of units = 2400 - 37 * (PRICE - 70) = 4990 – ( 37 * PRICE )

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Price 70Quantity 2400Cost 134000Revenue 168000Profit 34000

Basic Pricing Model

The Basic Pricing Problem

=4990 – 37 * Price

=50000 + 35 * Quantity

=Price * Quantity

=Revenue – Total Cost

Marketers are interested in a price range of $60 to $100. They expect fixed cost to be $50000, with a unit cost of $35.

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Price vs Profits

-$10,000

$0

$10,000

$20,000

$30,000

$40,000

$50,000

50 55 60 65 70 75 80 85 90

Price vs Quantity

0

1,000

2,000

3,000

4,000

50 55 60 65 70 75 80 85 90

34000 2,40050 -$2,900 50 3,14055 $9,100 55 2,95560 $19,250 60 2,77065 $27,550 65 2,58570 $34,000 70 2,40075 $38,600 75 2,21580 $41,350 80 2,03085 $42,250 85 1,84590 $41,300 90 1,660

Price vs Profits Price vs Quantity

Maximum profit occurs around $42,750 at the price $85. More accurate value can be generated by entering more numbers in the column providing the input to the table. However exactness might be misleading because of the uncertainty in the demand function.

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Given Variable Cost Function

0

1000

2000

3000

4000

5000

6000

0 50 100 150

Most applications in real life have nonlinear relationships. They follow a curved, nonlinear XY functions.

EX.EX.• Torrington Corporation, deciding whether they should

introduce a new product.• Production prepares a cost estimate for making up to

150 units. (variable cost will not be a linear function of quantity)

• Graph is prepared to show the curve representing variable cost

• The problem is to develop formulas to give the cost values for any value of the quantity.

Nonlinear Cost and Demand Function

Quantity Variable Cost

0 0

50 3000

100 4500

150 4900

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Mathematical Models of an XY Functions

Suppose Torrington wishes to determine the cost associated with the Suppose Torrington wishes to determine the cost associated with the quantity 70quantity 70.

Find the slope of any line segment. Considering the cost at 50 with 3,000; then slope is calculated as;

SlopeSlope = (Y2-Y1)/(X2-X1)

= (4,500-3,000)/(100-50) = 30

CostCost = Y1 + Slope ×( X-X1)

Quantity (X) = 70, X1=50, and Cost Y1= 3,000;

Then Cost for XCost for X = 3,000 + 30 × (70 – 50) = 3,600

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Nonlinear cost for Torrington

Number of unitsVariable

cost Slope0 0 60

50 3000 30100 4500 8150 4900 0

Unit selling price $50Number of units 112Fixed cost $1,000Left 100Up 4500Slope 8Variable cost $4,596Total cost $5,596Revenue $5,600Profit $4

Com

e fr

om ta

ble

abov

e


Spreadsheet Models of an XY Functions

• Cells A3 to C7 (Number of units, variable cost, slope) contain a lookup table that Excel uses to find the necessary parameters for a given number of units.

• Cells B9 to B11 are user-entered data.(Input factors)• Cells B12 to B14 use the lookup table to find

required items of data.• Cell B15 computes the variable cost for the number

of units entered in cell B10,Variable cost = UP + ( NUMBER OF UNIT – LEFT ) * SLOPE using the formula =B13 + ( B10-B12)*B14

• Cell B16 computes the total cost by adding the fixed and variable costs using =B11 + B15

• Cell B17 compute revenue using =B9*B10• Cell B18 profits using =B17-B16

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Units vs ProfitUnits Profits

$40 -$1,000

10 -$1,10020 -$1,20030 -$1,30040 -$1,40050 -$1,50060 -$1,30070 -$1,10080 -$90090 -$700

100 -$500110 -$80120 $340130 $760140 $1,180150 $1,600


Spreadsheet Models of an XY Functions

The table in the left shows the data table comparing number of units with profits

Training ExerciseTraining ExerciseCalculate profits against number of units being sold, where number of units start from 0 up to 150 with increment of 10 units.

Find the break-even point using Goal seeker?

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Approximating the Cost with a Cubic Function

Curves are a good facility for representing nonlinear Curves are a good facility for representing nonlinear functions. Polynomial is a class of functions that are often functions. Polynomial is a class of functions that are often satisfactory. satisfactory.

• The linear (first-order) function has the following form:2 + (10 * X)

• Quadratic function assumes the form:-24 + (56 × X2)

• Cubic function assumes the form:+(5.3 × X2) + ( 21.6 × X3)

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The general approach

1- Try polynomials, quadratic, cubic, and so on, on the spreadsheet representing the situation.

2- Calculate the values for the given curve

3- Calculate the square of the deviations (differences), add them

4- Minimize the sum with Excel Solver by allowing the coefficient of function to be changed.

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Approximating the Cost with a Cubic FunctionNumber of units

Variable cost Cubic

Squared Deviations Coefficients of Cubic Function

0 0 -36 1,272 0 1 2 325 1,500 1,611 12,338 -35.6607 73.8081 -0.3263 0.000350 3,000 2,883 13,75575 3,750 3,812 3,844

100 4,500 4,432 4,670125 4,700 4,774 5,542150 4,900 4,873 724

Total Deviations 42,145

1. Cell A3 uses the formula “=AVERAGE(A2, A4)”. 2. Column C calculates the cubic function based on the current coefficients in

cells F3 to I3. For example Cell C2 uses the formula “=$F$3 + ($G$3*A2)+($H$3*A2^2) +($I$3*A2^3)”.

1. Column D calculates the squared difference between the variable cost and the cube function. For example, cell D2 uses the formula “=(B2-C2)^2”.

2. Cell D10 shows the original sum of deviations.

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-1,000

0

1,000

2,000

3,000

4,000

5,000

6,000

0 25 50 75 100 125 150

Number of units

Variable cost Cubic

Squared Deviations

0 0 -36 1,27225 1,500 1,611 12,33850 3,000 2,883 13,75575 3,750 3,812 3,844

100 4,500 4,432 4,670125 4,700 4,774 5,542150 4,900 4,873 724

Total Deviations 42,145

• The target cell D10 needs to be minimized. There are no constraints. The cells to vary are the cubic coefficient in cells F3 to I3.

• The better way to judge how good the approximation is to compare the given curve with the calculated curve.

• If the management feels that the approximation is not good enough, a fourth, fifth order polynomial or other type of function can be tried.

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Preparing a Five-Year plan

Nonlinear cost for Torrington Using Cubic FunctionUnit selling price $50 Coefficients of Cubic FunctionNumber of units 112 0 1 2 3Fixed cost $1,000 -35.6607 73.8081 -0.3263 0.0003Variable cost $4,629Total cost $5,629Revenue $5,600Profit -$29

• The lookup table is no longer required as cubic equation replaces it

• Cell B7 in the cubic model now computes the variable cost using the cubic function = $D$6+($E$6*B5)+($F$6*B5^2)+($G$6*B5^3)

• This Modified Model can be used to perform scenario analysis

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• The management is particularly interested in three scenarios

• They are also interested in growth over the next five years

Growth in Sales

Growth in Fixed Cost

Best, optimistic 30% 10%

Realistic 20% 20%

Worst, pessimistic 10% 30%

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Scenario SummaryCurrent Values: Optimistic Realistic Pessimistic

Changing Cells:Sales 30% 30% 20% 10%Fixed_Cost_Growth 10% 10% 20% 30%

Result Cells:Present_value $20,111 $20,111 $9,066 $831

Notes: Current Values column represents values of changing cells attime Scenario Summary Report was created. Changing cells for eachscenario are highlighted in gray.

Nonlinear cost for Torrington Using Cubic Function

Growth Rates Coefficients of Cubic FunctionSales 30% 0 1 2 3Fixed Cost 10% -35.6607 73.8081 -0.3263 0.0003

1 2 3 4 5Unit selling price $50 $50 $50 $50 $50Number of units 112 145 188 244 317Fixed cost $1,000 $1,100 $1,210 $1,331 $1,464Variable cost $4,629 $4,871 $4,629 $3,622 $1,700Total cost $5,629 $5,971 $5,839 $4,953 $3,164Revenue $5,600 $7,250 $9,400 $12,200 $15,850Profit -$29 $1,279 $3,561 $7,247 $12,686

Present value of profit


Year


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The Impact of Pricing

• Apply the cubic function approach to the analysis of pricing

• The Marketing suggests the following pegs to approximate the price versus quantity

• The first step is to develop a cubic function for quantity based on price.

Price Quantity

20 250

40 150

60 100

80 60

Price Quantity CubicSquared

Deviations Coefficients of Quantity Cubic Function20 250 251 0.36 0 1 2 340 150 150 0.16 444.6295 -12.7211 0.1683 -0.000960 100 100 0.0080 60 60 0.01

Total Deviations 0.53

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The Impact of PricingTorrington Using Price Function Coefficients of Variable Cost Function

0 1 2 3

Growth Rates -35.6607 73.8081 -0.3263 0.0003Sales 30%Fixed Cost 10% Coefficients of Quantity Cubic Function

0 1 2 3444.6295 -12.7211 0.1683 -0.0009

1 2 3 4 5Unit selling price $50 $50 $50 $50 $50Number of units 121 157 204 265 344Fixed cost $1,000 $1,100 $1,210 $1,331 $1,464Variable cost $4,737 $4,861 $4,408 $3,110 $961Total cost $5,737 $5,961 $5,618 $4,441 $2,425Revenue $6,049 $7,850 $10,200 $13,250 $17,200Profit $312 $1,889 $4,582 $8,809 $14,775

Present value of profit


Year

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