EMP5102B - Week 6 Presentation

8/13/2019 EMP5102B - Week 6 Presentation

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EMP5102B: Introduction to EngineeringManagement

WEEK 6


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Agenda

Midterm # 1 ReviewEconomic Decision Making Models

2EMP5100 - WEEK 5 - RIDHIMA SHARMA


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EconomicDecision Making

EMP5100 - WEEK 5 - RIDHIMA SHARMA 3


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Time value of Money

$1 today is worth more than $1 tomorrow



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Interest and Interest Formulas

Interest Rate: The ratio of borrowed money to the feecharged for its use over a period. The ratio is usually

expressed as a percentage



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Money Flow Over Time


P F

A A A A A

1 2 3. . .

n

i = nominal annual interest rate

n = number of interest periods, usually annual

P = principal amount at a time assumed to be the Present

A = single amount in a series of nequal amounts at the end of each interest

periodF = amount, ninterest period hence, equal to the compound amount of P, or

the sum of the compound amounts of A, at the interest rate i


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Single Payment Compound AmountFormula


When interest is permitted to compound, the interest earnedduring each interest period is added to the principal amount atthe beginning of the next interest period.

We can show how this applied using the following table:


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Single Payment Compound AmountFormula


Single Payment Compount-Amount Formula

Yr

Amount at

Beginning of

Year

Interest Earned

During the YearCompound Amount at end of Year

1 P Pi P + Pi = P (1+i)

2 P (1+i) P(1+i)i P(1+i) + P(1+i)i = P(1+i)2

3 P(1+i)2 P(1+i)2i P(1+i)2+ P(1+i)2i = P(1+i)3

4 P(1+i)n-1 P(1+i)n-1i P(1+i)n-1+ P(1+i)n-1i = P(1+i)n= F


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Single Payment Compound-AmountFormula

Referred to as (F/P, i, n)

Used to express the equivalence between a present amount, P, and a future amount, F,at an interest rate I for n years




F = amount, ninterest period hence, equal to the compound amount of P, or the sumof the compound amounts of A, at the interest rate i


(1 )Single Payment Compound-Amount Factor

( , ,)

Eqn 1

Eqn 2


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Single Payment Present-AmountFormula

Also referred to as (P/F,i,n)

We can express Eqn 1 in terms of the PV




F = amount, n interest period hence, equal to the compound amount of P, or the sumof the compound amounts of A, at the interest rate i


1( 1 )

Single Payment Present-Amount Factor

( , ,)

Eqn 3

Eqn 4


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Ordinary Annuity

In some situations, a series of receipts or payments occurringuniformly at the end of each periodmay be encountered. Theseare called ordinary annuities

The sum of the compound amounts can be determined as

follows:


1 1

For derivation: http://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuity
http://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuityhttp://www.mathalino.com/reviewer/derivation-formulas/derivation-formula-future-amount-ordinary-annuity


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Ordinary Annuity

We can also express the relationship as:

F = A(F/A,i,n)

We can also express the relationship in terms of A as:

or A=F(A/F,i,n)


1

1


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Ordinary Annuity

We can also express the relationship in terms of the PresentValue, P

We can also express the relationship in terms of A as:

or A=P(A/P,i,n)


(1 ) + = (+)+

(1 )


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Ordinary Annuity

Rearranging and solving for P, we get

or P=P(P/A,i,n)


1 1

(1)


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Equivalence Calculations

Allows us to compare 2 or more situations In terms of money, two monetary amounts are

equivalent when they have the same value in exchange.

There are three factors involved in equivalence of sums

of money: (1) the amount of the sums

(2) the time of occurrence of the sums

(3) the interest rate



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Example:

At an interest rate of 10% with n = 8 years, a P of $1 isequivalent to an F of what amount?



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F = P(F/P,i,n) = $1(F/P,10,8) = $2.144

Practically, this can mean that $1 today, is equivalent to

$2.144 8 years from now given an interest rate of 10%compounded annually.



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Example:

At an interest rate of 12% with n = 10 years, a F of $1 isequivalent to a P of what amount?



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P = F(P/F,i,n) = $1(P/F,12,10) = $0.322

Practically, this can mean that $1 10 years from now, is

equivalent to $0.322 today given an interest rate of 12%compounded annually.

So if we dont want to exceed a total cost of $1, 10 yearsfrom now, we should not spend more than $0.32 today



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Example:

At an interest rate of 8% with n = 20 years, and an A of $1is equivalent to a F of what amount?



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F = A(F/A,i,n) = $1(F/A,8,20) = $45.762

Practically, this means that $1 spent each year for 20

years will result in a total cost of $45.76 20 years fromnow given an interest rate of 8%



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Example:

At an interest rate of 12% with n = 6 years, and an F of $1is equivalent to an A of what amount?



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A = F(A/F,i,n) = $1(A/F,12,6) = $0.1232

This means that $0.1232 must be received each year for

6 years to be equivalent to the receipt of $1, 6 yearshence.



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Example:

At an interest rate of 9% with n = 10 years, and an A of $1is equivalent to a P of what amount?



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P = A(P/A,i,n) = $1(P/A,9,10) = $6.4177

This means that an investment of $6.4177 today must

yield an annual benefit of $1 each year for 10 years if theinterest rate is 9%.



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Example:

At an interest rate of 14% with n = 7 years, and an P of $1is equivalent to a A of what amount?



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A = P(A/P,i,n) = $1(A/P,14,7) = $0.233

This means that $1 can be spent today to capture an

annual savings of $0.233 per year over 7 years if theinterest rate is 14%



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Evaluating a Single Alternative

Sometimes we have situations where we need to acceptor reject a single alternative (i.e. do we proceed withthis design). In such a case, the decision will be basedon the relative merit of the alternative

We will assume that all costs/savings will occur at theend of the year



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Present Equivalent Evaluation


Disbursements and Savings for a Single Alternative

Item Date Disbursements Savings

Initial Cost 1-1-2005 $28,000 -

Savings, First Year 1-1-2005 - $9,500

Savings, Second Year 1-1-2007 - $9,500

Overhaul Cost 1-1-2007 $2,500

Savings, Third Year1-1-2008

- $9,500

Savings, Fourth Year1-1-2009

- $9,500

Salvage Value1-1-2009

- $8,000


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$28,000 $25,000

$9,500 $9,500 $9,500 $9,500

$9,500

$8,000


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Present Value = PV of Initial Cost + PV of Savings (Yr 1) +PV of Savings (Yr 2) + PV of Overhaul Cost+ PV of Savings (Yr 3) + PV of Savings (Yr4) + PV of Salvage Value

Present Value = -28,000(P/F,12,0) + $9,500(P/F,12,1) +$9,500(P/F,12,2) - $2,500(P/F,12,2) +$9,500(P/F,12,3) + $9,500(P/F,12,4) +

$8,000(P/F,12,4)



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Present Value = -28,000(P/F,12,0) + $9,500(P/F,12,1) +$7,000(P/F,12,2) + $9,500(P/F,12,3) +$17,500(P/F,12,4)

Present Value = -28,000(1.00) + $9,500(0.8929) + $7,000(0.7972) +$9,500(0.7118) + $17,500(0.6355)

Present Value = $3,946

Because the PV is greater than $0, this is a desirable undertakingat an interest rate of 12%



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In general, the present value can be given by:

=

( 1 )



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Annual Equivalent Evaluation

We know that the relationship between the present and the annual value is

given by:

A = P(A/P,i,n)

Or

(1)1 1

So using the present equivalent value that we calculated, we can form an annual

equivalent amount

= ( 1 ) (+)

+



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Annual Equivalent Evaluation

=( 1 ) (1)1 1

Annual Equiv= ($3,946)(A/P,12,4)Annual Equiv= $1,299

What this tells us is that if $28,000 is invested on Jan 1,

2005, a 12% return will be received plus an equivalent of$1,299 on Jan 1,2006, 2007, 2008, and 2009.



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Future Equivalent Evaluation

We know that the relationship between the present and the future value is

given by:

Or

So using the present equivalent value that we calculated, we can form a futureequivalent amount

= ( 1 )( 1 ) = ( 1 )


( , ,)

(1 )


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Future Equivalent Evaluation

=( 1 )

Future Equiv= ($3,946)(F/P,12,4)Future Equiv= $6,211

What this tells us is that there will be a difference of

$6,211 between future equivalent savings and futureequivalent costs. Because this amount is great than zero,it is a desirable venture at 12%


A l E i l C f A


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Annual Equivalent Cost of an Asset

A useful application of the annual equivalent costpertains to the cost of an asset. The cost of an asset ismade up of two components:

(1) The cost of depreciation

(2) The cost of interest on the undepreciated balance

We will assume straight line depreciationthis meansthat the asset will depreciate by a fixed amount every

year


A l E i l t C t f A t


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Example:An asset has a first cost of $5,000, a salvage value of$1,000, and a service life of 5 years, and the interest rateis 10%. What is the present cost?




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Example:An asset has a first cost of $5,000, a salvage value of$1,000, and a service life of 5 years, and the interest rateis 10%. What is the annual equivalent cost?




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Calculate the Depreciation Cost:

$5,000 $1,000

5 $800

We know that:

cost of any asset = cost of depreciation + cost of interest on undepreciated balance

At the beginning of year 1:

Cost of Depreciation = $800

Undepreciated Balance = $5000

Cost of Asset = $800 + $5,000(0.1) = $800 + $500 = $1,300




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Cost of Depreciation = $800

Undepreciated Balance = $4,200

Cost of Asset = $800 + $4,200(0.1) = $800 + $420 = $1,220


Cost of Asset = $800 + $3,400(0.1) = $800 + $340 = $1,140


Cost of Asset = $800 + $2,600(0.1) = $800 + $260 = $1,060


Cost of Asset = $800 + $1,800(0.1) = $800 + $180 = $980



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Alternatively, Annual Equivalent Cost can be expressed as such:

( )( , , ) ()


B k E E i E l ti


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Break-Even Economic Evaluation

Break-Even Analysis may be graphical or mathematicalin nature

Useful in relating FC and VC to the number of hours ofoperation, the number of units produced, or other

measures of operational activity. It identifies the range of the decision variable within

which the most desirable economic outcome may occur

Especially useful when the cost of two or more

alternatives is a function of the same variable.


M k B E l ti


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Make or Buy Evaluation

Often, a manufacturing firm has the choice of making orbuying a certain component for use in the productbeing produced.

In this case, the firm faces a make-or-buy decision


M k B E l ti


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Example:Suppose that a firm finds that it can buy from a vendorthe electric power supply for the system that it producesfor $8/unit. Alternatively, suppose that it can

manufacture an equivalent unit for a variable cost of$4/unit. It is estimated that the additional fixed cost inthe plant would be $12,000 per year if the unit ismanufactured.

Finding the number of units/year for which the cost ofthe two alternatives breaks would help in making thedecision.


M k B E l ti


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Make Alternative:

Given:

Fixed Costs: $12,000 in plant

Variable Costs: $4/unit

Let N be the number of units produced




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Total Cost = Fixed Costs + Variable Costs

$12,000$4




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Buy Alternative:

Given:

Fixed Costs: $0 (there are no fixed costs for the Buy option)

Variable Costs: $8/unit





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$ 0 $ 8

$8




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Break EvenMathematical Analysis

Break Even occurs when TCbuy= TCmake

$12,000 + $4N = $8N

$12000$4

3,000units




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Which Option is Better?

$12,000 $4 $12,000 $4 3,001 $24,004

$8$8(3,001)= $24,008

From the above, we can see that in excess of the break evenquantity, the Make Option is better


Lease or Buy Evaluation


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Lease-or-Buy Evaluation

Often, a company has to consider the decision to leaseor buy a piece of equipment.

In this case, the firm faces a lease-or-buy decision




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Lease-or-Buy Evaluation

Example:

Assume that a small electronic computer is needed for dataprocessing in an engineering office. Suppose that the computercan be leased for $50/day, which includes the cost ofmaintenance. Alternatively, the computer can be purchased for

$25,000. The computer is estimated to have a useful life of 15years with a salvage value of $4,000 at the end of that time. It isestimated that the annual maintenance costs will be $2,800. Ifthe interest rate is 9% and it costs $50/day to operate thecomputer, how many days of user per year are required for the

two alternatives to break even?




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Lease Alternative:

Given:

Fixed Costs: $0 (no fixed costs)

Variable Costs:

$50/unit to lease computer

$50/unit to operate computer

Let N be the number of computer units required




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$0$50$50

$100




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Buy Alternative:

Given:

Initial Cost of Computer = $25,000

Salvage Value = $4,000

Useful life = 15 years

Interest Rate = 9%

Variable Costs: $50/day to operate





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( )( , , ) ()

25,000 4,000 0.1241 4,000 0.09 2,800 $50 5,766$50




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Break Even occurs when TCbuy= TClease

$100N = $5,766 + $50N

$5,766$50

115

So, for levels of use exceeding 115 days/year, it would be more

economical to purchase the computer. IF the level of use isanticipated to be below 115 days per year, the computer should

be leased


Equipment Selection Evaluation


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Example:

Suppose that a fully automatic controller for a machine centercan be fabricated for $140,000 and that it will have an estimatedsalvage value of $20,000 at the end of four years. Maintenancecosts will be $12,000/year, and the cost of operation will be

$85/hour As an alternative, a semiautomatic controller can be fabricated

for $55,00. This device will have no salvage value at the end of a4-year service life. The cost of operation and maintenance isestimated to be $140/hour.

Assume an interest rate of 10% - which machine should weselect?




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Machine A

Given:

Fabrication Cost: $140,000

Salvage Value: $20,000

Maintenance Cost: $12,000

$80/unit to operate machine

Let N be the number of operating hours




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Machine B

Given:

Fabrication Cost: $55,000

Salvage Value: $0

Maintenance Cost: $12,000

$80/unit to operate machine

Let N be the number of operating hours


Equipment Selection Evaluation - Machine


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Equipment Selection Evaluation MachineA


( )( , , ) ()

140,000 20,000 0.3155 20,000 0.10 12,000 $85 51,800$85



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EMP5102B - Week 6 Presentation