Engineering Economics

Dr. Karim KobeissiDr. Karim Kobeissi

Chapter 5: Breakeven & Payback Analysis

Introduction Breakeven analysis is performed to determine the value of a

variable or parameter of a project or alternative that makes two elements equal, for example, the sales volume (Q) that will equate revenues and costs.

A breakeven study is performed for two alternatives to determine when either alternative is equally acceptable.

The breakeven analysis approach is commonly used for make-or-buy decisions. This means the company contracts to buy the product from the outside , or makes it within the company. The alternative to buy usually has no fixed cost and a larger variable cost than the option to make. Where the two cost relations cross is the make-buy decision quantity. Amounts above this indicate that the item should be made, not purchased from outside.

Terminology and Symbol P = value or amount of money at a time designated as the present or time 0. Also P

is referred to as present worth (PW), present value (PV), net present value (NPV),

discounted cash flow (DCF), and capitalized cost (CC); monetary units, such as

dollars

F = value or amount of money at some future time. Also F is called future worth

(FW) and future value (FV); dollars

A = series of consecutive, equal, end-of-period amounts of money. Also A is called

the annual worth (AW) and equivalent uniform annual worth (EUAW); equivalent

uniform annual cost (EUAC) dollars per year, euros per month

n = number of interest periods; years, months, days

i = interest rate per time period; percent per year, percent per month

t = time, stated in periods; years, months, days

Breakeven Point

The parameter (or variable) can be an amount of revenue, cost, supply, demand, etc. for one project or between two alternatives

One project - Breakeven Quantity is identified as QBE. Determined using linear or non-linear math relations for revenue and cost.

Between two alternatives - Determine one of the parameters P, A, F, i, or n with others constant

Value of a parameter that makes two elements equal

Cost-Revenue Model ― One Project

Quantity, Q — An amount of the variable in question, e.g., units/year, hours/month Breakeven value is QBE

Fixed cost, FC — Costs not directly dependent on the variable, e.g., buildings, fixed overhead, insurance, minimum

workforce costVariable cost, VC — Costs that change with parameters such as

production level and workforce size. These are labor, material and marketing costs. Variable cost per unit is v

Total cost, TC — Sum of fixed and variable costs, TC = FC + VC

Revenue, R — Amount is dependent on quantity sold

Revenue per unit is r

Profit, P — Amount of revenue remaining after costs

P = R – TC = R – (FC+VC)

Breakeven for linear R and TC

Set R = TC and solve for Q = QBE

R = TC rQ = FC + vQ

FC r – v

When variable cost, v, is lowered, QBE decreases

(moves to left)

QBE =

Example: One Project Breakeven Point

Solution: Find QBE and compare to 15,000; calculate Profit

QBE = 75,000 / (8.00-2.50) = 13,636 units/month

Production level is above breakeven Profit

Profit = R – (FC + VC) = rQ – (FC + vQ) = (r-v)Q – FC

= (8.00 – 2.50)(15,000) – 75,000 = $ 7500/month

A plant produces 15,000 units/month. Find breakeven level if FC =

$75,000 /month, revenue is $8/unit and variable cost is $2.50/unit.

Determine expected monthly profit or loss.

Breakeven Between Two AlternativesOften breakeven analysis involves revenue or cost variables common to

both alternatives, such as price per unit, operating cost, cost of

materials, or labor cost. The next slide illustrates the breakeven concept

for two alternatives with linear cost relations. The fixed cost of

alternative 2 is greater than that of alternative 1. However, alternative

2 has a smaller variable cost, as indicated by its lower slope. The

intersection of the total cost lines locates the breakeven point, and the

variable cost establishes the slope. Thus, if the number of units of the

common variable is greater than the breakeven amount, alternative 2 is

selected, since the total cost will be lower. Conversely, an anticipated

level of operation below the breakeven point favors alternative 1.

Breakeven Between Two Alternatives With Linear Cost Relations

Selection between alternatives is based on this guideline:- If the anticipated level of the common variable is below the breakeven value, select the alternative with the higher variable cost (alternative 1). - I f the anticipated level of the common variable is above the breakeven point, select the alternative with the lower variable cost (alternative 2).

Selection of alternative is based on anticipated value of common variable:

Value BELOW breakeven; select higher variable cost

Value ABOVE breakeven; select lower variable cost

Breakeven Between Two Alternatives

Instead of plotting the total costs of each alternative and estimating the

breakeven point graphically, it may be easier to calculate the

breakeven point numerically using engineering economy expressions

for the PW (present worth ), or AW (annual worth).

The AW is preferred when the variable units are expressed on a yearly

basis, and AW calculations are simpler for alternatives with unequal

lives.

Example #1Investment (A) costs $10,000 today and pays back $11,500 two years from

now. Investment (B) costs $8,000 today and pays back $4500 each year for

two years. If the interest rate of 5% is used for comparison, which

investment is superior?

Solution:

Since the two alternatives do the same job and have the same

lifetimes, we will compare them by converting each to its cash

value today. The superior alternative will have the highest present worth.

P(A)= –$10,000 + ($11,500)(P/F,5%,2) = –$10,000 + ($11,500)(0.9070) = $431

P(B) = –$8,000 + ($4500)(P/A,5%,2) = –$8,000 + ($4,500)(1.8594) = $367

$431 > $367, so Alternative (A) is superior.

Example #2

Alternative A (Make):

First Cost = 18000

Salvage Value = 2000

Per Unit Cost = 0.4

Alternative B (Buy):

Per Unit Cost = 1.5

Minimum Attractive Rate of Return (MARR) = 15%

Life = 6 Years

Example # 2 (con)Perform a make/buy analysis where the common

variable is X, the number of units produced each year. AW (the annual worth is chosen because the two alternatives have unequal lives) relations are:

AWmake = -18,000(A/P,15%,6) +2,000(A/F,15%,6) – 0.4X

AWbuy = -1.5X

Solution: Equate AW relations, solve for X

-1.5X = -4528 - 0.4X X = 4116 per year

X, 1000 units per year

Breakeven value of X

1 2 3 4 5

AWbuy

AWmake

If anticipated production > 4116, select make alternative (lower variable cost)

AW, 1000 $/year

8

7

6

5

4

3

2

1

0

The following steps determine the breakeven point of the common

variable and the slope of a linear total cost relation.

1. Define the common variable and its dimensional units.

2. Develop the PW or AW or EUAC or EUAW relation for each

alternative as a function of the common variable.

3. Equate the two relations and solve for the breakeven value of the

variable.

Example #3How many kilometers must be driven per year for leasing and buying to cost

the same? Use 10% interest and year-end cost. Leasing: $0.15 per

kilometer Buying: $5000 purchase cost, 3-year life, salvage $1200, $0.04

per kilometer for gas and oil, $500 per year for insurance.

Let (x) be the common variable = number of kilometers driven

EUAC (leasing) = $0.15x

EUAC (buying) = $0.04x + $500 + ($5 k)(A/P,10%,3) – ($1.2 k)(A/F,10%,3)

= $0.04x + $500 + ($5 k)(0.4021) – ($1.2 k)(0.3021) = $0.04x + $2148

Setting EUAC (leasing) = EUAC (buying) and solving for x

$0.15x = $0.04x + $2148

x = 19,527 km that must be driven to break even

Payback AnalysisPayback analysis is used to determine the amount of

time (Payback Period - np) , usually expressed in

years, required to recover the first cost of a

project.

The investment decision criteria for this technique

suggests that if the calculated payback period is less

than some maximum value acceptable to the

company, the project is accepted.

Types of Payback Analysis There are two types of payback analysis as determined by the

required return:

1- No Return Payback; i= 0%: Also called simple payback, this is the

recovery of only the initial investment. So no return is expected for

the investment made.

2- Discounted Payback; i > 0%: The time value of money is considered

in that some return, for example, 10% per year, must be realized in

addition to recovering the initial investment.

Caution!!It is important to understand that payback analysis neglects all cash

flows after the payback period of (np) years. Consequently, it is

preferable to use payback as an initial screening method or

supplemental tool rather than as the primary means to select an

alternative. The reasons for this caution are:

• No-return payback neglects the time value of money, since no return

on an investment is required.

• Either type of payback (No Return Payback & Discounted Payback)

disregards all cash flows occurring after the payback period. These

cash flows may increase the return on the initial investment A

different alternative may be selected using payback.

Computation of Payback Period (np)

The equations used to determine the payback period (np) differ for each type of payback

analysis (No Return Payback & Discounted Payback). For both types, the terminology is

(P) for the initial investment in the asset, project, contract, etc., and NCF for the estimated

annual net cash flow.

NCF = cash inflows - cash outflows

To calculate the payback period for i = 0% or i > 0%, determine the pattern of the NCF

series. Note that n p is usually not an integer. For t _ 1, 2, . . . , (np) ,

Eqn. 1

Eqn. 2

Eqn. 3

Eqn. 4

Example: Payback Analysis System #1 System #2

First cost, $ 12,000 8,000NCF, $ per year 3,000 1,000 (year 1-5)

3,000 (year 6-14)Maximum life, years 7 14

Problem: Use (a) no-return payback, and (b) discounted payback at 15% to select a system.

(a) Solution: np1 = 12,000 / 3,000 = 4 years

0 = -8,000 + 5(1,000) + 1(3,000) np2 = 6 years (found by trial and error)

4 < 6 Select system # 1

Example: Payback Analysis (con) System #1 System #2

First cost, $ 12,000 8,000NCF, $ per year 3,000 1,000 (year 1-5)

3,000 (year 6-14)Maximum life, years 7 14

(b) Solution: System 1: 0 = -12,000 + 3,000(P/A,15%,np1)

np1 = 6.6 years

System 2:

0 = - 8,000 + [1,000(P/A,15%,5) + 3,000(P/A,15%,np2 - 5)(P/F,15%,5)] np2 = 9.5 years

6.6 < 9.5 Select system #1

Summary of Important PointsBreakeven amount is a point of indifference to accept or reject a project

One project breakeven: accept if quantity is > QBE

Two alternative breakeven: if level > breakeven, select lower variable cost alternative (smaller slope)

Payback estimates time to recover investment.Return can be i = 0% or i > 0%

Use payback as supplemental to PW or other analyses, because np neglects cash flows after payback, and if i = 0%, it neglects time value of money Payback is useful to sense the economic risk in a project