Download LECTURE 2

BREAKEVEN ANALYSIS

Introduction

What is Break-even Analysis?


Break-even in comparing alternative propositions

Break-even in single project analysis

Break-even in decision making
Optimisation
1
INTRODUCTION

Break-even analysis – a powerful management tool

A tool for cost comparison
 Example: How can we choose between two different options for
a required piece of equipment?
A tool for single project analysis
 Example: How many units are required to be sold before the
project yields a positive profit?
A tool for decision making
 Example: is an investment in a marketing initiative that is
believed to have a certain benefit worth undertaking?


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COMPARING ALTERNATIVES

In situations where the alternatives are affected in some way by a
common variable

Total cost of Option 1 = TC1

Total cost of Option 2 = TC2

There exists a common, independent decision variable affecting
both Options – ‘x’
TC1  f1(x)
TC2  f 2 (x)
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EQUIPMENT SELECTION EXAMPLE

2 pump options
 Electric: Capital cost + Annual maintenance + Energy cost
 Diesel: Capital cost + Hourly maintenance + Hourly operator cost
+ Energy cost
 4 year project life
 12% interest rate

Which is the lowest cost option?
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PROBLEM SOLVING PROCESS

Identify the common, independent decision variable

Translate the cost information for each option into cost function form

Do the number crunching

Solve analytically or graph both cost functions

Locate the break-even value (the intersection of the two cost
functions)
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SOLUTION 1

Common, independent decision variable
 ‘h’, pump operational hours per year

Cost function for Pump 1
 Initial cost
 Annual maintenance cost
 Energy cost
 Annual Equivalent
 Annual amount
 Hourly rate
Cost function for Pump 2
 Initial cost
 Maintenance cost
 Energy cost
 Operator cost
 Annual Equivalent
 Hourly rate
 Hourly rate
 Hourly rate

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SOLUTION 2

Common cost function:
Total Annual Equivalent Cost = Annual Cost + Hourly Rate * h

Equation of a straight line
y (TAEC) = m (Hourly rate). x (h) + c (Annual cost)

Result is two straight lines, one for each option
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SOLUTION 3 – NUMBER CRUNCHING

Pump 1
Initial Capital cost
Annual Equivalent
Annual maintenance cost
Total Annual cost
= 1,800 m.u.
= Initial Cost * A/P(12,4)
= 1,800 * 0.3292
= 592.56 m.u.
= 360.00 m.u.
= 952.56 m.u.
Hourly rate
= 1.10 m.u. / hour
Total Annual Equivalent cost
= 952.56 + 1.10*h ………(1)
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SOLUTION 4 – NUMBER CRUNCHING

Pump 2
Initial Capital cost
Annual Equivalent
Total Annual cost
= 550 m.u.
= Initial Cost * A/P(12,4)
= 550 * 0.3292
= 181 m.u.
Hourly rate
= 0.60 + 1.40 + 0.35 m.u. / hour
= 2.35 m.u. / hour
Total Annual Equivalent cost
= 181.00 + 2.35*h ………(2)
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SOLUTION 5 – SOLVE

Analytical
 Total Annual Equivalent cost
= 952.56 + 1.10*h ………(1)
 Total Annual Equivalent cost
= 181.00 + 2.35*h ………(2)
 Break-even is when these are equal, i.e.
3000
2500
2000
1500
1000
500
1000
900
800
700
600
500
400
300
200
0
100
Total Annual Equivalent
Cost
Alternative Analysis
952.56 + 1.10*h = 181.00 + 2.35*h
771.56 = 1.25*h
h = 617.25
Annual operational hours
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MULTIPLE – ALTERNATIVE PROBLEMS


The same solution approach applies
Reduce all problems to common cost function
Graphical solution is best way of visualising the solution
M ultiple Alternatives
2500
2000
1500
V < 50
Blue
50 < V < 150 Green
150 < V
Red
1000
500
0
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BREAK-EVEN IN A SINGLE PROJECT

Definition of Costs

Fixed: “A cost is said to be fixed if it does not change in response
to changes in the level of activity”

Variable: “The cost that is directly associated with the production
of one unit”
Total Cost
Total
Cost
(Ct)
Ct  Cf  v *Cv
Cv
Cf

Volume (v)
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COST – VOLUME – PROFIT EXAMPLE



Telephone:
Annual line rental charge
Cost per call
Cost for 100 calls Line rental + call cost
Cost for 500 calls Line rental + call cost
Total
Cost
(Ct)
75
25.00 m.u.
0.10 m.u.
35.00 m.u. [0.35]
75.00 m.u. [0.25]
Average Cost
“Average cost is the total
cost of providing a product
or service, divided by the
number that are provided.”
35
25
100
500
Volume (v)
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LINEARITY OF VARIABLE COSTS

Variable costs = f (volume), but the relationship is not linear

Limitations on linearity

Bulk purchase price break point

Demand fluctuations

Economic climate

Production capability

Efficiency & Productivity changes

Technology changes
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REALISTIC COST FUNCTIONS
Fixed
Cost
Variable
Cost
+
Volume
Volume
Total
Cost
=
Relevant Range
Volume
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CVP ANALYSIS

Profit (P) = Sales Revenue (SR) – Total Costs (Ct)
SR = Selling Price (Sp) * Volume (V)
Ct = Fixed Costs (Cf) + Variable Costs (CV)

Marginal cost: “The cost of providing one additional unit/item

Cv = Marginal Cost (Cv) * Volume


P  S pV  (C f  CvV )
P  (S p  Cv )V  C f

Break-even when P=0

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BREAK-EVEN ANALYSIS
Profit
P  S pV  (C f  CvV )
Gradient
= (Sp - Cv)
P  ( S p  Cv )V  C f
At Breakeven P = 0
Cf
Vbe 
Break-Even
Volume
Cf
( S p  Cv )
Volume
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SINGLE PRODUCT DECISIONS

You buy and sell a product which sells for 15.00 m.u. each. The
cost for you to purchase the product is 3.00 m.u. In order for you to
trade you require premises and equipment which, in total, represent
a fixed cost to you of 25,000 m.u. Your total planned volume for the
year of the product is 4,000 units.
1)
2)
3)
4)
How many units do you need to sell to break-even?
How many units do you need to sell to make 1,000 m.u. profit?
Would it be worth the introduction of advertising at a cost of
6,000 m.u. to increase sales to 4,450?
What impact would a 10% drop in selling price have on the
break-even volume?
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SOLUTION - 1
Problem 1) How many units to breakeven?
P = (Sp - Cv) * V - Cf
Definition of Break-Even : P = 0
Vbe 
Cf
( S p  Cv )
Profit
Break-Even
Volume = 2084
£25k
Break-Even
Volume = 2083.3
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SOLUTION - 2
Problem 2) How many units to make £1000 profit?
P = (Sp - Cv) * V - Cf
Volume for Profit = £1000
Vbe 
Cf
( S p  Cv )
Profit
Volume = 2167
£26k
Volume = 2166.6
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SOLUTION - 3
Problem 3) Would it be worth the introduction of
advertising at a cost of £6,000 to increase
sales to 4450?
P = (Sp - Cv) * V - Cf
Profit for V = 4000 is £23,000
Profit for V = 4450 is £28,400
Gain in Profit = £5,400
Cost to Achieve Gain = £6,000
Hence Not Worth Pursuing!
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SOLUTION - 4
Problem 4) What impact would a 10% drop in selling
price have on the break even volume. ?
P = (Sp - Cv) * V - Cf
V
Cf
(0.9 * S p  Cv )
Profit
Increase = 298 units
or 14.3%
£25k
Break-Even
Volume = 2381
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CONTRIBUTION
Problem 3) Would it be worth the introduction of
advertising at a cost of £6,000 to increase
sales to 4450?
P = (Sp - Cv) * V - Cf
Profit for V = 4000 is £23,000
Profit for V = 4450 is £28,400
Gain in Profit = £5,400
Cost to Achieve Gain = £6,000
There is an alternative way of solving this.
23
CONTRIBUTION
Problem 3) Would it be worth the introduction of
advertising at a cost of £6,000 to increase
sales to 4450?
P = (Sp - Cv) * V - Cf
Per unit Profit = (Sp - Cv) = £12
Increase in Volume with Advertising : 450 units
Increase in Profit = £12 * 450 = £5,400
Cost to Achieve Gain = £6,000
“£12 is the contribution or profit margin per unit”
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CONTRIBUTION
Marginal Contribution = Selling Price - Variable Cost
Revenue /
Cost
Sales Revenue
Contribution
Cf
Variable Costs
Volume
Total Contribution = (Selling Price - Variable Cost) * Volume
25
OPTIMISATION ANALYSIS



Some cost components vary directly with a common decision
variable while others vary inversely with the decision variable
In such cases an optimum (lowest cost) exists
The general form of such a cost function is:
TC  A  B.x 

C
x
Where:
 x = common
 decision variable
 TC = Total cost
 A, B, C = constants
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OPTIMISATION ANALYSIS

The general form can be solved analytically and/or graphically
Optimisation Analysis
9000
8000
7000
6000
5000
4000
3000
2000
1000
0
dTC
C
 A 2 0
dx
x
200
x
C
A

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ALTERNATIVE OPTIONS - 1
Alternative Options Optimisation
1950
1850
1750
1650
1550
1450
1350
1250
1150
1050
950
850
750
650
550
450
350
250
150
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
50

Single cross-over
Lowest cost option changes once
Total cost

De cis ion variable
28
ALTERNATIVE OPTIONS - 2
Alternative Options Optimisation
1000
950
900
850
800
750
700
650
600
550
500
450
400
350
300
250
200
150
100
35000
30000
25000
20000
15000
10000
5000
0
50

Double cross-over
Lowest cost option changes twice
Total cost

De cis ion variable
29
ALTERNATIVE OPTIONS - 3
Alternative Options Optimisation
1000
950
900
850
800
750
700
650
600
550
500
450
400
350
300
250
200
150
100
45000
40000
35000
30000
25000
20000
15000
10000
5000
0
50

No cross-overs
Lowest cost option never changes
Total cost

De cis ion variable
30
OPTIMISATION CASE STUDY
Sometown Compressors
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