QNt130notes

QNT 130 Lecture Notes

Review of Basic Mathematics:

In this chapter, we present a review of basic mathematical concepts that the students have studied in their high school mathematics subjects or courses given in the first year of college. A sound review in mathematics is extremely useful in understanding applications in business. We cover topics ranging from order of arithmetic operations to functions and graphs.

2.1 Order of Arithmetic Operations:

In simplifying arithmetic expression it is conventional to follow the order given below:

a) Parenthesis in the sequence ( ), , ,b) Exponentsc) Multiplication or divisiond) Addition or subtraction

Example 1:

(12-4)¿(5−3 )+3¿6+√25+72+9−2×[ 4

√64 ]= {8÷2 }+3 ¿6

+√25+72+9−2×[ 4

√64 ]= [4+3 ]×6+5+49+9−2×

48

=7×6+5+49+9−2×12

=42+5+49+9−1=104______________________________________________

Example 2:

Solve [ (42 )×2 ]×6+14

Steps:

1

42=1616×2=3232×6=192192+14=206

Therefore ,[ ( 42 )×2 ]×6+14=206Example 3:

Simplify (9÷6 )×42−44

2.2 Rules of Exponents

Exponent is the power to which a number or variable is to be raised.

Example: 24=2×2×2×2=16

In general ap=a×a×a×a .. .. . .. .. . p times

Some Basic Rules of Exponents

1) a p×aq=ap+q

Example:

43×42=42+3=45=4×4×4×4×4=1024

2) (ap )q=a p×q

Example:

(23 )2=23×2=26=2×2×2×2×2×2=64

3) (a×b )p=ap×b p

Example:

(4×3 )3=43×33=64×27=1728

4)( ab )

p

=ap

b p

2

Example:

( 43 )

3

=43

33=64

27=2 .3704

5)

ap

aq=a p−q

Example:

54

52=54−2=52=25

6) a1p=p√a

Example:

8114=4√81=3

7) apq =

q√ap

Example:

8134=

4√813=27

8)a−p= 1

ap

Example:

6−3= 1

63= 1

216=0 .0046

9) a1=a

Example:

41=4

3

10) a0=1

Example:

100=1

11) 1p=1

Example:

16=1

Other Examples:

1)

(a2)3×(b3

4 )2a3×b3

=a6×b

32

a3×b3

¿a6−3×b32−3

=a3×b−32 =

a3

√b3

2)

(x4 )16 × y

23

(xy )23

=x46 × y

23

x23 × y

−23

=x23−2

3 × y23+2

3

¿ x0× y43 = y

43

2.4 Fractions:

A fraction is a ratio of the form

xy where y is different from zero and x and y are both whole

numbers. X is known as numerator and y is the denominator. Ratios

123

,3

11,−611 are examples of

such fractions.

4

Addition or subtraction of factors:

i. If the denominator is common for the fractions, add the numerators and divide by the common denominator to obtain the results.

Examples:

1621

+ 721

=16+721

=2321

1621

− 721

=16−721

= 921

ii. If the denominator is not common, then a procedure is followed to obtain common denominator. First, note that multiplying the numerator and denominator by the same non-zero number does not change the value of the fraction. Second, usual procedure of obtaining the common denominator is to multiply the numerator and the denominator by the ratio of the least common multiple (l.c.m) and the denominator of each fraction.

Example:

56+11

4

In this case, 12 is the l.c.m.

12÷6=2 , then

56=5×2

6×2=10

12

12÷4=3 , then

114=11×3

4×3=33

12

Therefore,

56+11

4=10

12+33

12=10+33

12=43

12

Example:

86+21

18−11

12

5

L.C.M =36

36¿6=6 then,

86=8×6

6×6=48

36

36¿18=2 then,

2118

=21×218×2

=4236

36¿12=3 then,

1112

=11×312×3

=3336

Therefore,

86+21

18−11

12=48

36+42

36−33

36=48+42−33

36=57

36

Multiplication and Division involving fractions:

The product of any two fractions is obtained by the ratio of the product of the two numerators and the product of two denominators.

Example:

73×

116

=7×113×6

=7718

−611

×32=

(−6 )×311×2

=−1822

When dividing a fraction by another fraction, take the reciprocal of the second fraction and multiply it by the first fraction, following the previous procedure.

Example:

27÷

95=

27×

59=

2×57×9

=1063

47÷

3(−5 )

=47×

(−5 )3

=4×(−5 )7×3

=−2021

10÷32=

101

×23=

10×21×3

=203

Mixed Fraction:

6

The sum of whole number and fraction less than one.

Example:

423=4+ 2

3=4

1+ 2

3=4×3

1×3+ 2×1

3×1=12

3+ 2

3=12+2

3=14

3

Problem: Express 2

37 as ratio.

Decimals: A number that involves three parts is usually known as a decimal number. The three parts are

i. Whole number ii. Decimal point and

iii. Numbers whose denominators are 10, or some powers of 10.

Example:

13.127

Here 13 is the whole number.

. is the decimal point.

The number after decimal are respectively 1 in “tenths” ( 110 )

digit, 2 in “hundredths” ( 1100 )

digit and 7 in “thousands” ( 11000 )

digit.

Thus,

13.127 =13

+110

+2100

+71000

=1×(10 )+3×(1 )+1×110

+2×1100

+7×11000

=10+3+0 .1+0.02+0 .007

Operations with decimals:

Addition or subtraction: These operations are carried out after aligning the numbers in their appropriate positions.

Example: 164.21+121.3719=164.21

7

+121. 3719285 .5819

Problem: Simplify 90.03-107.4271

Multiplication:

First, obtain the product of two numbers ignoring the decimal points. Count the “total” number of positions after the decimal points in both the numbers. Place the decimal point in the product number after the “total number of positions counting from right to left.

Example:

2.12¿0 .006

Step 1. 212×6=1272

Step 2. Total Number of positions =2+3=5

Step 3. 0.01272

Problem: Simplify 6.37 ¿4 .0712

Division: To divide a decimal by another, such as 62.744¿1 .24 , or 1 .24|62. 744 ,

First move the decimal point in the divisor to the right until the divisor becomes an integer, then move the decimal point in the dividend the same number of places;

124|6274 . 4This procedure determines the correct position of the decimal point in the quotient (as shown). The division can then proceed as follows:

12450 . 6

|6274 . 4

620744

744

8

Conversion from a given decimal to an equivalent fraction is straightforward. Since each place value is a power of ten, every decimal can be converted easily to an divided by a power of ten. For example,

84 .1=84110

9 .17=917100

0 .612=6121000

The last example can be reduced to lowest terms by dividing the numerator and denominator by 4, which is their greatest common factor. Thus,

0 .612=6121000

=612÷41000÷4

=153250 (In lowest terms)

Any fraction can be converted to an equivalent decimal. Since the fraction

ab means a÷b , we

can divide the numerator of a fraction by its denominator to convert the fraction to a decimal. For

example, to convert

38 to a decimal, divide 3 by 8 as follows.

80 .375|3.000

24 60

56 40

40

Changing Fractions to Decimals: To change a fraction to a decimal, simply do what the

operation says. In other words,

1320 means 13 divided by 20. So do just that (insert decimal points

and zeros accordingly):

200 .65

|13 . 00=0 .65

58=8

0 . 625|5 . 000

=0 . 625

Changing Decimals to Fractions: To change a decimal to a fraction,

9

1. Move the decimal point two places to the right2. Put that number over 1003. Reduce if necessary.

0 .65=65100

=1320

0 .05=5100

=120

0 .75=75100

=34

Read it: 0 .8

Write it:

810

Reduce it:

45

Finding Percent of a Number: To determine percent of a number, change the percent to a fraction or decimal (whichever is easier for you) and multiply. Remember, the word of means multiply.

What is 20% of 80?

20100

×80=1600100

or

0 .20×80=16 .00=16

What is 12% of 50?

12100

×50=600100

=6

or

0 .12×50=6 .00=6

10

What is

12 % of 18?

1/2100

×18= 1200

×18=18200

= 9100

or

0 .005×18=0 .09

Changing from Decimals to Percents: To change decimals to percents,

1. Move the decimal point two places to the right.2. Insert a percent sign.

0.75=75% 0.05=5%

Changing from Fractions to Percents: To change a fraction to a percent,

1. Multiply by 1002. Insert a percent sign.

12

:12×100=

1002

=50 %

25

:25×100=

2005

=40 %

Changing from Percents to Fractions: To change percents top fractions,

1. Divide the percent by 1002. Eliminate the percent sign3. Reduce if necessary

60%=

60100

=35

13 %=13100

Percentages less than 1:

0 .1%=0 . 1

100=0 . 001

0 .2 %= 0. 2100

=0 . 002

Other Applications of Percent: Turn the question word -for – word into an equation. For what substitute the letter x; for is substitute an equal sign; for of substitute a multiplication

11

sign. Change percents to decimals or fractions, whichever you find easier. Then solve the equation.

18 is what percent of 90?

18=x( 90)1890

=x

15=x

20 %=x

10 is 50% of what number?

10=0 .50( x )100 .50

=x

20=x

What is 15% of 60?

x=15100

×60=9010

=9

or

0.15(60)=9

Finding Percentage Increase or Percentage Decrease: To find the percentage change (increase or decrease), use this formula:

What is the percentage decrease of a $500 item on sale for $400?

12

Change

¿100 = Percentage Change

Starting point

Change: 500-400=100

What is the percentage increase of Jon’s salary if it went from $150 a week to $200 a week?

Change: 200-150=50

References to Handout 1

1. Mathematics for Business and Economics by Robert H. Nicholson, McGraw-Hill, 19862. Introductory Algebra by Alan Wise, Harcourt Brace Jovanovich, 1986

13

Change

¿100 =

100500

×100

Starting point

=

15×100=20 %

decrease

Change

¿100 =

50150

×100

Starting point

=13×100=33

13

% increase

Week 1. Review of Basic Mathematics

Problem Set 1

1) Evaluate the following:

(a)779+1 .64

(1, 060 )√36

(b) 0. 64−2.575(√ (0 .64 )( 0. 36 )

400)

(c) (182÷43 )(6+4 )

2) Simplify the following:

(a)( 2

3x−1

2)( 34

x1

2)

(b) [ (7 )0 (8 )1

3 ]−5

3) Factor the following.

a) x2−13 x+40

b) 16 x2−49

4) Simplify the following:

a)42

23+6

14+5

78

b) An investor bought 125 shares of stock at a price of $79

516 a share.

i) What was the cost of the stock?

ii) If this represents

35 of the total amount of investment, find the total

amount of investment.c) Complete the following blanks with correct answers

14

Decimal Fraction Percent

---- 13

----

0.1875 ---- -------- ---- 72

d) If a technology stock price has increased from $146 per share to $267 during 1998, find the percent increase in share price.

15

QNT 130 Lecture NotesWeek 2

y= f (x) is defined as a function of x if for one or more values of x, there corresponds one and only one value of y. The expression “f(x)” does not mean ‘ f times x ’. Instead, f (x) is another symbol for the variable y. Here f represents a particular rule by which x is converted into y.

In y= f (x), the set of values of x is called the domain of the function and the set corresponding to the y values is known as the range of the function. x, y are respectively known as independent and dependent variables in the context of the relation y =f (x).

y= f(x)

Example 1: y= x2 is a function.

y= x2

Example 2: y2=x is not a function.

If we solve for y, y=±√ x

Therefore, for a given value x, there are two values for y.

y2=x

Range y 1

Domain x 1

x 2

Range

16

Domain4

-4

Range y1=+x

y2=-x

Domainx

16

y=x

y2=16

y=16=4

Example 3. x2+y2=k2 is not a function.(circle)

y2 =k2-x2 , y=±√k2−x2

For a given x, we have two values of y.

Graphically, find if the following are functions are not.

Y Y

Y 1 Y = X2 Y = mX + c

X X Yes Yes

Y Y

Y=X3

X2 + Y2 = 16 X (0,0) X

NO NO

Range

-4 4

Domain

16

(0,0)

17

Y

Y= f(X)

X

YES

Note: Vertical Line Test: For a given value of x, find y by drawing a vertical line at x. If you find more than one value of y, then it is not a function.

WORKING WITH THE FUNCTION NOTATION

Evaluating functions given by a formula can involve algebraic simplification, as the following example shows. Similarly, solving for the input, or independent variable, involves solving an equation algebraically.

EVALUATING A FUNCTION

A formula like f(x)=

x2+15+x is a rule that tells us what the function f does with its input

value. In the formula, the letter x is placeholder for the input value. Thus, to evaluate f(x), we replace each occurrence of x in the formula with the value of the input.

Example1: Let g(x)=

x2+15+x . Evaluate the following expressions. Some of your answers

will contain a, a constant.

(a) g(3) (b) g(-1) (c) g(a)

(d) g(a-2) (e) g(a)-2 (f) g(a)-g(2)

Solution (a) To evaluate g(3), replace every x in the formula with 3.

18

g(3)=

32+15+3

=108=1.25

(b) To evaluate g(-1), replace every x in the formula with (-1).

g(-1)=

(−1)2+15+(−1 )

= 24=0 . 50

(c) To evaluate g(a), replace every x in the formula with a.

g(a)=

a2+15+a

(d) To evaluate g(a-2), replace every x in the formula with (a-2).

g(a-2)=

(a−2)2+15+(a−2)

=a2−4 a+4+15+a−2

=a2−4 a+53+a

(e) To evaluate g(a)-2, first evaluate g(a) (as we did in part (c)), then subtract 2:

g(a)-2=

a2+15+a

−2=a2+15+a

−21∗5+a

5+a=a2+1

5+a−10+2a

5+a=a2+1−10−2a

5+a=a2−2 a−9

5+a

(f) To evaluate g(a)-g(2), subtract g(2) from g(a):

g(2)=

22+15+2

=57

From part c, g(a)=

a2+15+a . Thus,

g(a)-g(2)=

a2+15+a

−57=

7( a2+1 )−5(5+a)7 (5+a )

=7 a2−5 a−187 (5+a )

19

Graphing Functions

1. a. Point method: Graph 2x+3y=6

Solution: Solve for y.

3 y=6−2 x ory=6−2 x

3

Therefore, y=2−2

3x

xy=2−2

3x

3 02 2

31 4

30 2-1 8

3-2 10

3-3 4

-4 -3 -2 -1 0 1 2 3 4

0

0.5

1

1.5

2

2.5

3

3.5

4

4.5

X

Y

20

b. Intercept method

In the function, 2x+3y=6,

If x=0, 3y=6, and so, y=

63 =2 (y-intercept)

If y=0, 2x=6, and so, y=

62 =3 (x-intercept)

The straight line is obtained by joining the two points (0,2) and (3,0).

If x=1, y=4/3 (one extra point)

0 0.5 1 1.5 2 2.5 3 3.5

00.51

1.52

2.5

X

Y

21

Example: Graph the function 4x+5y-20=0

2. Graph y=|x| Absolute value of x.

x y3 32 21 10 0-1 1-2 2-3 3

-4 -3 -2 -1 0 1 2 3 4

0

1

2

3

4

Series2

X

Y

22

3.Graph y=x2

x y3 92 41 10 0-1 1-2 4-3 9

-4 -3 -2 -1 0 1 2 3 40

2

4

6

8

10

X

Y

23

4. Graph the function y=2x ( the exponential function)x y3 82 41 20 1-1 1

2-2 1

4-3 1

8

-4 -3 -2 -1 0 1 2 3 40

2

4

6

8

10

X

Y

24

5. Graph the following exponential growth function P=Po eRt

where R= exponential growth rate t=time periodP= the value of the function at time t.Po = Value of P at t=0.

You are given

Po=8, e2.71828, R=0.02

t P

0 83 8.56 9.08 9.39

Example:

In 1985 the population of the United States was 234 million and the exponential growth rate was 0.8% per year. What will be the population in the year 2000?

Solution:

At t=0 (1985), the population was Po=234 million. Here R=0.008, therefore, the exponential equation is 234 e0.008t.

In the year 2000, we have t=15P2000=234 e0 . 008(15 )=234(1 .1275 )=263 . 8 million .

0 1 2 3 4 5 6 7 8 9

7

8

9

10

X

Y

25

Example: The cost of a first class postage stamp was 3 ¢ in 1932 and the exponential growth rate was 3.8% per year. What will be the cost of a first class stamp in year 2000.

Linear Functions and Applications in Business and Economics

Definition: A function with a constant rate of change, is known as a linear function.

In y=b0+ b1 x , b1 is called the slope of the linear function and it measures the change in the value of the dependent variable y as a result of one unit change in the value of the independent variable x. bo is known as the y-intercept and measures the value of y when x=0.

Note the following:

A horizontal line has zero slope.A vertical line has no slope or its slope is undefined. A line rising from left to right has positive slope. A line falling from left to right has a negative slope.

Examples: Q=50-2P+0.6Y is called a demand function where Q=quantity demanded of a product, P is the price in $ of the product and Y is the family income in thousands of dollars.

i) Draw the demand curve(graphical relationship between Q and P for fixed Y) for Y=50 at price levels of P=0,5,10,15 and 20.

ii) Repeat the above for Y=100 and for the same price levels on the same graph.iii) Comment on the demand curve in ii) in relation to the demand curve in i)

26

Solution

i) P Y Q0 50 805 50 7010 50 6015 50 5020 50 40

p

(II) 20 (I)

15

10

5

0 20 40 60 80 100 110 q

ii)P Y Q0 100 1105 100 10010 100 9015 100 8020 100 70

iii) It is evident from the graph that the demand curve in ii) is to the right of demand curve in i). This is due to increase in income. This is known as shift in demand curve.

Example: Determine the demand function for the price quantity data in the following table Q=f(P).

27

P in $’s 0 5 10 15 20 25 30 35 40Q in 000’s 80 70 60 50 40 30 20 10 0

Linear Consumption Function:

For the consumption function C=10+ 3

4Y

, where C stands for consumption expenditure in thousands of dollars, Y stands for disposable income in thousand of dollars, the slope 34 is the marginal propensity to consume (MPC) and the intercept $10,000 is known as subsistence level of consumption expenditure. Find i) the consumption expenditure at Y=20, Y=60 and ii) find the break even level of income.

Solution:

C

10

0 20 60 Y

C=10+ 3

4Y

at Y=20, C=10+ 3

4(20 )=10+15=25 .

(A dissaving of $5,000)

at Y=60, C==10+ 3

4(60 )=10+45=55.

(A positive saving of $5,000)

ii) For a breakeven income, i.e. C=Y, we have =10+ 3

4Y=Y

or 10=1

4Y

or Y=40.Linear Cost and Revenue Functions

28

Suppose that the fixed cost of production for a commodity is 10 dollars. The variable cost is 2 dollars per unit and the commodity sells for $4 per unit. Find the breakeven quantity. Also find the loss or gain for the company at production levels of 3 and 7 units pf output.

Q R C2 8 144 16 186 24 228 32 26

RC&R 25 C 20

15 10 5

0 2 4 6 8 Q

Note C=10+2Q R=4QTo find the break-even level,

4Q=10+2Q (i.e. set R=C) Solving, Q=5.

At Q=3, R=12 and C=16 loss of $4

At Q=7, R=28 and C=24 gain of $4

29

WEEK 2PROBLEM SET II

1. Determine which one of the following is a function? Explain why or why not?

a)

Domain Range3 146 168 2010 24

b)

Domain Range3 74 114 154 12

c) y=6x-14d) yx2+4

2. Given the function f(x)=2x+1 Find:i.

a. f(2)b. f(-1)c. f(0)

ii. Given the function f(x)=x/(1-x) find

a. f(0)b. f(1)c. f(1-x)

iii. If f(x1,x2)= 5 x1+ 6x2+15, find

f(4,3)f(0,1)f(0,0)

3.Graph the following functions.

i) y= x2+2xii) y=3x-1

30

iii) A freshman student has invested $10,000.00 on August 1,1998 at 10% rate of interest compounded continuously. Determine the total value (Principal +Interest) of this investment by August 1,2002.

4. Q=-10+20P-2R is an example of supply function, where Q is the quantity supplied of a product, P is the price in $ of the product and R is the interest rate in percentage terms.

i) Draw the supply curve ( graphical relationship between Q and P for a fixed R) for R=5, at price levels of P=2,4,6 and 8.

ii) Repeat the above for R=7 and for the same price levels n the same graph.iii) Comment on the supply curve in ii) in relation to the supply curve in i).

5. Suppose the fixed cost of production for a commodity is $45,000. The variable cost is 60% of the selling price of $15.00 per unit. Find the break-even quantity.

6. Using the following functions, complete the values in the table.

P=600-2QC(Q)=8000+200Q

Q Revenue Total Cost MR MC050100150200250300

MR: Marginal Revenue MC: Marginal Cost

31

LECTURE NOTESWEEK 3

Two equation models in Business:

i) Market Equilibrium: Consider the following demand and supply functions:

QD=f(PD) and QS=g(PS)

Here, the subscript D stands for the demand for the product by the consumers and the subscript S stands for the supply of the product by the producers. Q is quantity and P is the price.

Definition: Market equilibrium is a point where quantity demanded and quantity supplied are equal. Obviously at this equilibrium, the demand and supply price is equal. Therefore we can denote the equilibrium point by (P,Q) where PD=PS=P and QD=QS=Q. By solving the demand and supply functions simultaneously, we obtain the market equilibrium.

P

P*

0 Q* Q

Graphical Representation of Market Equilibrium

Example: The demand for automobiles is given as a linear function, P=32000-25Q and the linear supply function as P=11000+45Q.

Note: At equilibrium, PD=PS=P, QD=QS=Q and the price P, and the quantity Q are determined simultaneously.

Solution: Solving the demand and supply equations simultaneously,

P=32000-25QP=11000+45Q

32

Subtracting 0=21000-70Q or 21000=70Q or Q=21000

70=300 .

Substituting Q=300 in any one of demand or supply equations, we have P=32000-25(300)=32000-7500=24500.

Example:

a. If the demand for a brand of shoes is linear and of the form PD=84-0.14QD and the linear supply function has the form PS=18+0.02QS, find the equilibrium price and quantity for the firm’s shoes PD and PS are measured in dollars and QD and QS

designate number of pairs of shoes. b. If the demand function shifts to PD=96-0.14QD find the new equilibrium price and

quantity. Describe the change in demand by comparing the two demand functions.

ii) Further Consideration of Break-Even Analysis:

Another application of solving two linear equations in two variables is Break-Even Analysis. Here we determine the quantity that corresponds to the equality of total cost and total revenue. Once quantity is determined, the total cost which will be equal to total revenue can also be determined. Note that in case of non-linear revenue and cost functions, there may be more than one breakeven point.

Example:

A firms plans to sell the standard family size box for $2.40. Production estimates have shown that the variable cost of producing one unit of product is $2.16. Fixed costs of production are $3600. What is the breakeven volume of sales?

The total revenue function TR is linear and of the form

TR=(P)(Q)= 2.40Q

The total cost function TC is

TC=F+VQ= 3600+2.16Q, where V stands for the unit variable cost.

Breakeven quantity QE is established by equating total revenue and total cost functions and solving for QE . Thus,

TR=TC

2.40Q=3600+2.16Q

2.40Q-2.16Q=3600

33

QE=36002. 40−2 .16

=36000 . 24

=15000

All the intermediate steps are shown to demonstrate the relation between fixed cost, product price, and variable cost per unit of output. Revenue and cost corresponding to this quantity can be found by substituting QE into either the total revenue or total cost function as the dollar figures are equal. This is computed below using the total revenue function.

Total Cost= total revenue=(Price) (Quantity)

TC=TR=2.40 QE=(2.40)(15000)=$36,000.

This analysis has determined that the firm will earn a profit after the sale of the first 15,000 boxes. Total revenue and total cost at 15,000 boxes will be each equal to $36,000 and losses will be incurred at quantities below 15,000 boxes. These results apply unless changes occur in the price and cost data.

The following exercises will give additional practice in computing both demand and supply equilibrium and break-even points.

iii) Inventory Cost Functions and Economic Order of Quantity(EOQ):

Some Useful Definitions:

Inventory: This refers to idle goods or materials that are held by a company for future use.

Examples: Finished goods, raw materials and parts, work in progress and supplies.

The reason why organizations maintain inventory is that it is not exactly possible to predict uncertain demand, production times and sales levels. Maintaining inventory is convenient to do business, but it involves substantial costs. There are two important costs associated with maintaining inventory. One is holding cost and the other is ordering cost. Holding cost refers to the costs associated with maintaining an inventory investment, such as insurance, taxes and cost of capital locked-up in this inventory. This cost can e expressed as a percentage of the inventory investment or as a cost per unit. It is also referred to as carrying cost.

Ordering Cost: This is a fixed cost associated with preparing and placing an order. To derive the EOQ model, we first start with total cost of inventory.

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Inventory Models

Total Inventory Cost= Holding Cost(CARRYING COST)+ORDERING COST

Assumptions

1) Constant Demand2) No shortages were allowed.

Notation

1)Order Quantity Q for a time period of length T2)Average inventory level is Q/2 during this cycle period

q

q/2

0 T 2T 3T

T: Length of time required to deplete Q items

3) I= Annual holding cost rate (percentage)4)C= Unit cost of the inventory item5) Ch=Annual cost of holding one unit in inventory

Ch= I*C ( sometimes Ch is given directly)

Example:

C=$8 I=25% Ch=8*(0.25)=$2

Annual Holding Cost: (Q/2)*Ch .

6) D= Annual demand for the product 104,000

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7) C0=Fixed cost of placing an order ($32)8) N=Number of orders per year N=D/Q

Annual Ordering Cost=(D/Q)* C0

Total Annual Cost= T C (Q)=(1/2)Q Ch+(D/Q) C0

To find the economic order quantity, EOQ:

Using the above example, we find

TC=

Q2(2 )+104 , 000

Q(32 )=Q+ 3 ,328 , 000

QAnnual Holding, ordering and total costs for various values of order quantities, Q

Annual CostOrder Quantity Holding Ordering Total

5000 $5000 $666 $56664000 4000 832 48323000 3000 1109 41092000 2000 1664 36641824 1824 1824.56 3648.561000 1000 3328 4328

Note: From the above table, it is obvious that at Q=1824 the total cost is a minimum and at this level of Q, the holding ad ordering costs are equal to each other. Therefore this is the desired level of order quantity and known as Economic Order Quantity (EOQ).

This finding is also illustrated by the graph below.

Cost ($)

Annual Total Cost 6000

4000 Annual Inventory Holding

2000 Annual Ordering Cost

0 Q* 1000 2000 3000 4000 5000 6000 Order Quantity (Q)

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Both from the table and the graph, it is evident that at the desired level of order quantity, the annual holding cost equals annual ordering cost.

Therefore

Q2

Ch=DQ

C0

Multiplying by 2Q on both sides, we obtain

Q2 Ch=2 DC 0

Q2=2DC 0

Ch

and

Q =+√2 DC0

Ch

This is the Economic Order Quantity and written as EOQ=Q*=+√2DC 0

Ch . Using this formula, we obtain

1. Q*=√2(104 , 000 )322

=1824

Knowing Q*, we find the following aspects related to inventory management.

2. Average inventory level: Q*/2=1824/2=912.

3. Number of orders per year N= D

Q∗¿=104 , 000

1824=57 ¿

4. Cycle time in days:

360N

=36057

=6 .3 days

i.e. One order every 6.3 days. Here, 360 is given as the number of working days in a year.

Note: Cycle time is the length of time between placing of two consecutive orders.

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5. Reorder point: The inventory position at which a new order should be placed. By inventory position we mean the inventory on hand plus the inventory on order.

R=L*daily demand, where L stands for the time between placing an order and its receipt in the inventory system, known as the lead-time. If L=2 days in the above

example, the reorder point R=2×104 , 000

360≃578

Example: Suppose R&B beverage company has a soft-drink product that has a constant annual demand rate, 3600 cases. A case of the soft drink costs R&B $3. Ordering costs are $20 per order and holding costs are 25% of the value of the inventory. If there are 250 working days per year and 5 days of lead-time, find the following.

a) EOQb) Number of orders per yearc) Cycle time in daysd) Reorder pointe) Annual holding costf) Annual ordering costg) Total Annual cost

Example:

(Economic Order Quantity) Let the quantity Q, when purchased in each order minimize the total cost T incurred in obtaining and storing material for a certain time period to fulfill a given rate of demand for the material during the time period. The material demanded is 10,000 units per year; the cost price of material is $1 per unit; the cost of replenishing the stock of the material per order, regardless of the size Q of the order, is $25; and the cost of storing the material is 12.5% per year of the value of the average inventory (Q/2) on hand.

a) Show that T=10 , 000+250 , 000

Q+ Q

16 b) Find the economic order quantity and the total cost T corresponding to that value

of Q. c) Find the total cost when each order is placed for 2500 units.

ELASTICITY

Price Elasticity of demand is calculated in the following table, using the definition

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EP=ΔQQ

÷ ΔPP

= PQ∗ΔQ

ΔP

Where Q=400-4P. First we use the demand equation to produce the demand schedule:

P in $ Q in Units Point Elasticity =-4*(P/Q)90 40 -9.0070 120 -2.3350 200 -1.030 280 -0.4310 360 -0.11

Interpretation:

At P=90, a 1% increase in price will result in a 9% decrease in quantity demanded.

At P=70, a 1% increase in price will result in a 2.33% decrease in quantity demanded.

At P=50, a 1% increase in price will result in a 1% decrease in quantity demanded.

To show

ΔQΔP

=−4, consider Q=400-4P,

Then

Q1=400−4 P1

Q2=400−4 P2

Q1−Q2=−4 (P1−P2)Q1−Q2

P1−P2

=ΔQΔP

=−4

Note: -4 can also be seen directly as the slope of the linear demand function, Q=400-4P.

Example: Calculate the price elasticity of demand if Q=500(10-P) for each of the following values of P: i)P=2 ii)P=5 iii)P=6.

Note:

i)The sign of elasticity indicates the type( negative or positive) of relationship between the dependent and independent variables.

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ii) Elasticity can also be computed with respect to a variety of independent variables such as income, interest rate and prices of other goods.

iii) If the numerical value of elasticity is greater than unity, the demand is said to be elastic. That means one percent increase in price will result in more than one percent decrease in quantity demanded.

iv) If the numerical value of elasticity is equal to unity, the demand is said to be unit elastic. That means, a one percent increase in price will result in one percent decrease in the quantity demanded.

v) And if the numerical value of elasticity is less than unity, the demand is said to be inelastic. That means, a one percent increase in price will result in less than one percent decrease in quantity demanded.

Example: Q=5000-500P

EP=PQ(−500 )

P Q EP

2 4000 24000

(−500)=-0.25

Inelastic

5 2500 52500

(−500 )=-1

Unit elastic

6 2000 62000

(−500 )=-1.5

Elastic

REFERENCES

1.“Mathematics for Business and Economics” by Robert H. Nicholson, McGraw Hill, 1986.

2.“ An introduction to Management Sciences by Anderson, Sweeney, Williams, Eighth edition, West Publishing, 1994.

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Week # 4 QNT 130 Lecture Notes

Optimization:

This is to obtain the most desirable point. For example the output level where the profit is maximum, a given production levels at which cost is minimum.

I. Models of cost, revenue and profit maximization. Many business decisions are based on marginal analysis. For example, should an additional machine be purchased for maximization of profit or an additional dollar can be spent in order to maximize sales?

In order to understand the role of marginal analysis in optimization, we revisit the concepts of marginal cost and marginal revenue introduced on page 15 of week # 2 lecture notes.

Example 1 Demand Function: P =50 – 2.50 Q

Revenue Function: R (Q )=Q .P =50 Q -2.5Q2

Total Cost Function: C (Q)=25+25Q

Using these functions, the problem is to find the output level Q at which profit is maximum.

a) Algebraic Method: We follow the profit maximization rule which, states that a firm’s

profit is maximum at an output level where MC=MR . This is because, the firm

continues to produce as long as MC<MR . When MC>MR, its profit starts declining.

Therefore, the firms profit is maximum when MC=MR . We find below the MC and the MR of the above cost and revenue functions.

C (Q)=25+25QC (Q+ΔQ )=25+25 (Q+ΔQ )Now ,C (Q+ΔQ )−C(Q )=25+25(Q+ΔQ )−25−25Q=25 ΔQ

Dividing by ΔQ ,

C(Q+ΔQ)−C(Q )ΔQ

=25

For a small ΔQ , the left hand side is MC and therefore MC =25

Similarly, R(Q )=50 Q−2 .5 Q2

R(Q+ΔQ)=50(Q+ΔQ )−2 . 5(Q+ΔQ )2

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R(Q+ΔQ)−R(Q)=50(Q+ΔQ )−2 .5(Q+ΔQ )2−(50 Q−2 . 5Q2)

50 Q+50 ΔQ−2.5Q2+2QΔQ+(ΔQ )2 -50 Q+2 .5Q2

=50 ΔQ−2 . 5Q2−5 QΔQ−2 .5(ΔQ )2+2 . 5Q2

Now dividing throughout by ΔQ

R (Q+ΔQ )−R (Q )ΔQ

=50 ΔQ−5 QΔQ−2 .5( ΔQ )2

ΔQ

=50 Q−5Q−2 .5 ΔQ

For small ΔQ approaching zero, the left-hand side is MR and the right hand side becomes 50−5QThus, MR=50−5QNow applying the profit – maximization rule, we obtain

MC=MR25=50−5 Q , solving for Q , we obtain Q=5

The profit function Π (Q )=R(Q )−C(Q )

=50 Q−2. 5 Q2−(25+25 Q )

=50 Q−2. 5 Q2−25−25 Q

=25 Q−2. 5 Q2−25

At,Q=5

Π (5 )=25(5 )−2 .5 (52)−25=125−62 .5−25=37 . 5 is the value of maximum profit.

b) Trial and Error Method: referring to the completed table on page 15 of week # 2 lecture

notes, we find that the profit is maximum at an output level of Q=5 . We illustrate the profit maximization principle graphically below.

Note: The profit maximizing price is obtained by substituting the optimal Q into P=50−2 .5 Q .Thus P=50−2 .5 (5 )=50−12. 5=37 .5It is just a coincidence that price and profit are equal to 37.5

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R C C

25 R

0 5 Q

MR MC 50

25 MC

MR 0 5 Q

Example: A product has a demand function P=2000−40Q and a cost function C (Q)=3000+400 Q . Find the profit maximizing price and quantity. Find the maximum profit.

To solve by trial and error method use Q=15 , 20 and 25 .

Below we solve it by algebraic method:

C (Q)=3000+400QC (Q+ΔQ )=3000+400(Q+ΔQ )

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∴C (Q+ΔQ )−C (Q )=3000+400 (Q+ΔQ )−3000−400 Q

=400 Q+400 ΔQ−400 Q

Now dividing both the sides by ΔQ we get,

C(Q+ΔQ)−C(Q )ΔQ

=400= Marginal Cost

Similarly, R(Q )=PQ=(2000−40 Q )Q=2000 Q−400Q2

R(Q+ΔQ)=2000(Q+ΔQ )−40(Q+ΔQ )2

Now,

R(Q+ΔQ)−R(Q)=2000 Q+2000 ΔQ−40 Q2+2QΔQ+(ΔQ )2-

(2000 Q40Q2)

=2000 ΔQ−80QΔQ−40 (ΔQ )2−(2000Q−40 Q2 )¿2000 ΔQ−80 QΔQ−40(ΔQ )2

Now dividing by ΔQ on both sides,

R (Q+ΔQ )−R (Q )ΔQ

=2000−80Q−40 ΔQ

For small ΔQ , MR=2000−80 Q

Setting MR=MC for profit maximization,

2000−80 Q=4002000−400=80 Q⇒Q=1600 /80=20

Profit maximizing price P=2000−40Q=2000−40(20 )=1200

Profit Function Π (Q ), Π (Q )=R(Q ) -C(Q)=(2000Q-40Q2 )−(3000+400 Q)

=2000 Q−40 Q2−3000−400 Q

=1600 Q−40 Q2−3000

At Q=20 ,

Π (20 )=1600(20 )−40(202 )−3000

=32000−16000−3000=13000

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Constrained Optimization:

In this process of decision making in business, optimization generally involves constraints. For example firms maximize profit subject to resource constraints. Similarly, firms maximize return on portfolio of investments subject to cash and other asset constraints.

A commonly used model of constrained optimization deals with linear objective function, linear constraints involving non-negative variables. This is known as linear programming model. Although there are many methods of solutions, we consider below graphical method involving two variables.

Some useful definitions:

Constraint: An equation or inequality that rules out certain combinations of decision variables as feasible solutions.

Objective Function: All linear programs have a linear objective function that is to be either maximized or minimized. In many linear programming problems, the objective function will be used to measure the profit or cost of a particular solution.

Solution: Any set of values for the variables.

Optimal Solution: A feasible solution that maximizes or minimizes the value of the objective function.

Non-negativity Constraints: A set of constraints that requires all variables to be nonnegative.

Mathematical Model: A representation of a problem where the objective and all constraint conditions are described by mathematical expressions.

Linear Program: A mathematical model with a linear objective function, a set of linear constraints, and nonnegative variables.

Feasible Solution: A solution that satisfies all the constraints.

Feasible Region: The set of all feasible solutions

Slack variable: A variable added to the left-hand side of a less than or equal to constraint to convert the constraint into an equality. The value of this variable can usually be interpreted as the amount of unused resource.

Standard form: A linear program in which all the constraints are written as equalities. The optimal solution of the standard form of a linear program is the same as the optimal solution of the original formulation of the linear program.

Redundant Constraint: A constraint that does not affect the feasible region. If a constraint is redundant, it can be removed from the problem without affecting the feasible region.

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Extreme Point: Graphically speaking, extreme points are the feasible solution points occurring at the vertices or “corners” of the feasible region. With two-variable problems, extreme points are determined by the intersection of the constraint lines.

Surplus Variable: A variable subtracted from the left-hand side of a greater than or equal to constraint to convert the constraint into an equality. The value of this variable can usually be interpreted as the amount over and above some required minimum level.

Alternative Optimal solutions: The case in which, more than one solution provides the maximum / minimum value for the objective function.

Infeasibility: The solution in which there is no solution to the linear programming problem that satisfies all the constraints.

Unbounded: If the value of the objective function can be made indefinitely large in a maximization linear programming problem or indefinitely small in a minimization problem without violating any of the constraints, the problem is said to be unbounded.

Linear Programming:

Solution by graphical method:

Step 1: Show the constraint equations graphically and identify the feasible region.

Step 2: Find the corner points (extreme points) of the feasible region.

Step 3: Find the value of the objective function at each one of the corner points and identify the optimum solution.

A simple Maximization Problem:

Par, Inc., is a small manufacturer of golf equipment and supplies whose management has decided to move into the market for medium and high-price golf bags. Par’s distributor is enthusiastic about the new product line and has agreed to buy all the golf bags Par produces over the next 3 months.

After a thorough investigation of the steps involved in manufacturing a golf bag, management has determined that each golf bag produced will require the following operations:

1. Cutting and dyeing the material2. Sewing3. Finishing (inserting umbrella holder, club separators, etc.)4. Inspection and packaging

The following are the production requirements per golf bag.

Production Time(Hours)

Product Cutting & Dyeing Sewing Finishing Inspection and Packaging

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Standard bag 7/10 1/2 1 1/10Deluxe bag 1 5/6 2/3 1/4Available Resources 630 600 708 135

The accounting department estimated a profit contribution of $10 per every standard bag and $ 9 for every deluxe bag produced.

Solution:

Using the information given, we form below a mathematical model for the Par Inc. problem. (Linear programming model)

Max 10 x1+9 x2

Subject to (s.t.)

7

10x1+1 x2≤630

Cutting and dyeing (C & D)

1

2x1+

56

x2≤600 Sewing (S)

1 x1+

23

x2≤708 Finishing (F)

1

10x1+

14

x2≤135 Inspection and packaging (I & P)

x1, x2≥0

47

Step 1:

X2

1200

1000

F 800

S 600

5 400

4 3

200 Feasible Region 1 C & D 1 & P

0 200 400 600 800 1000 1200 1400

X1

Number of Standard Bags

Step 2:

To find the extreme points (corner points) of the feasible region 1, 2, 3, 4, and 5, we illustrate below the procedure of finding the corner point 3, which is the intersection of the constraint equation C & D and F.

C & D 7

10x1+1 x2=630

F 1 x1+

23

x2=708

To solve, consider,

48

−101−7 (C & D) :

−x1−107

x2=−107×630=−900

F : x1+

23

x2=708

Adding −16

21x2=−192

x2=192×21

16=252

Substituting x2=252 in F, We get

x1+23×252=708

x1=708−168=540

∴ (540, 252) is corner point 3. Similarly other corner points of the feasible region can be obtained and they are shown in step 3.

Step 3:

Extreme Point (Corner Point)Profit =10 x1+9 x2

1. (0,0) 10 (0) + 9 (0)=02. (708, 0) 10 (708) + 9 (0) =70803. (540, 252) 10 (540) + 9(252) =76684. (300, 420) 10 (300) + 9 (420) = 67805. (0, 540) 10 (0) + 9 (540) = 4860

It is clear from the above table that Par. Inc. maximizes its profit by producing 540 standard golf bags and 252 deluxe bags. The maximum profit value is $7668.00. The profit (540, 252) at which maximum profit is realized, known as optimum point.

Note 1: The sewing constraint:

12

x1+56

x2≤600 is a redundant constraint.

Note 2: At the optimum point, the following table presents the values of the slack variables.

Constraint Hours required

For x1=540 and x2=252Hours Available

Unused Hours (slack)

Cutting & dyeing 7/10 (540) + 1 (252) =630 630 0Sewing 1/2 (540) + 5/6 (252) = 480 600 120Finishing 1 (540) + 2/3 (252) = 708 708 0Inspection & packaging 1/10 (540) + ¼ (252) =117 135 18

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A Simple Minimization Problem:

M & D chemicals produces two products that are sold as raw material to companies manufacturing bath soaps and laundry detergents. Based on an analysis of current inventory levels and potential demand for the coming month, M & D’s management has specified that the combined production for product 1 and 2 must total at least 350 gallons. Separately, a major customer’s order for 125 gallons of product 1 must also be satisfied. Product 1 requires 2 hours of processing time per gallon while product 2 requires 1 hour of processing time per gallon, and for the coming month, 600 hours of processing time are available. M & D’s objective is to satisfy the above requirements at a minimum total production cost. Production costs are $2 per gallon for product 1 and $ 3 per gallon for product 2.

To find the minimum cost production schedule, we will formulate M & D chemicals problem as a linear program. Following a procedure similar to the one used for Par Inc., we first define the decision variables and the objective function for the problem.

Let x1= number of gallons of product 1 producedx2 = Number of gallons of product 2 produced

Since the production costs are $2 per gallon for product 1 and $3 per gallon for product 2, the objective function which corresponds to the minimization of the total production costs can be written as

Min 2 x1+3 x2

Next consideration the constraints placed on the M & D Chemicals problem. To satisfy the major

customer’s demand for 125 gallons of product 1, we know x1must be at least 125. Thus, we write the constraint as

1 x1≥125

Since the combined production for both products must total at least 350 gallons, we can write the constraint as

1 x1+1 x2≥350

Finally, since the limitation on available processing time is 600 hours, we include the constraint

2 x1+2 x2≤600

After adding the non-negativity constraints (x1 , x2 ¿0 ), we have the following linear program for the M & D Chemicals problem:

Min 2 x2+3 x2

s.t.

50

1 x1 ¿125 Demand for product 1 (In gallons)1 x1+1 x2≥350 Total Production (In gallons)

` 2 x1+1 x2≤600 Processing time (In hours)x1 ,x2¿0 Non-negative Constrained

Step 1:X 2

Gallons of x2

Product 2 Processing time

600 x1=125

500

400 1 x1+1 x2 =350 B

300 2 x1+1 x2=600 Production

200 C

100 A

0 100 200 300 400 500 600 X1 Gallons of Product 1

Step 2:

A (250, 100), B (125, 300), and C (125, 225) are the corner points of the feasible region ( shaded area) and they are obtained by solving the intersecting lines (constraint equations).

51

Step 3:

Extreme point Total costA (250, 100) 2 (250) + 3 (100) = 800B (125, 300) 2 (125) + 3 (300) = 1150C (125,225) 2 (125) + 3 (225) = 925

It is obvious that point A is optimum and the minimum cost is 800.

Note: At the optimum point (x1=250, x2 =100) the values of the surplus variables/slack variables are shown below.

Constraint Value of Surplus or Slack VariablesDemand for product 1 s1=125Total production s2=0Processing time s3=0

Problems:

1. Consider the following linear programming problem:

Max 3 x1+3 x2

s.t.

2 x1+4 x2¿12

6 x1+4 x2¿24

x1 , x2≥0

a. Find the optimal solution

b. If the objective function is changed to 2 x1+6 x2, what will the optimal solution be?

c. How many extreme points are there? What are the values of x1 and x2 at each extreme point?

2. Consider the following linear program:

Min 2 x1+2 x2

s.t.

52

1 x1+3 x2≤123 x1+1 x2≥131 x1−1 x2=3

x1, x2≥0

a. show the feasible regionb. What are the extreme points of the feasible region?c. Find the optimal solution using the graphical solution procedure.

Special Cases:

1. Consider the following linear program:

Max 1 x1−2 x2

s.t.

−4 x1+3 x2≤31 x1−1 x≤3x1, x2≥0

a. Graph the feasible region for the problem.b. Is the feasible region unbounded? Explain.c. Find the optimal solution.d. Does an unbounded feasible region imply that the optimal solution to the linear program

will be unbounded?

2. Does the following linear program involves infeasibility, unbounded, and / or alternative optimal solution? Explain.

Max 4 x1+8 x2

s.t.

2 x1+2 x2≤10−1 x1+1 x2≥8x1, x2≥0

Reference: An Introduction to Management Science, by Anderson, Sweeney, and Williams, eighth edition, West Publishing Co., New York

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