Production Costing and Cost Control Lecture Notes

Bahir Dar University
Bahir Dar Institute of Technology
Faculty of Mechanical and Industrial
Engineering
MSc. In Industrial Engineering (Specialization in
Production Engineering & Management)
Lecture Note: Production Costing and Cost
Control (IEng 6031)
Tomas C.
2017/18

Outline
1. Introduction to Engineering Economics
 Engineering economics & decision making
2. Production Costs
 Cost elements & cost accounting
3. Production Cost Estimation
 Cost estimation methods
4. Cost and Budget Control
2

1.1 Basic Concepts of Engineering Economics
Economy
 Is the study of how limited resources are used
optimally to satisfy unlimited wants
 Resources are known as factors of production used
to produce goods and services
3
Types of
Resources
Land Labor Capital
Money,
Physical,
Human
Entrepreneur

Basic Concepts CNTD…
Engineering Economy
 It is an economic approach in solving problems in
engineering projects particularly for the
assessment of project's investment.
 It is a methodology that provides the necessary
tools to describe and evaluate engineering projects
according to their economic characteristics.
 EE involves the systematic evaluation of
economic merits of proposed solutions to
engineering problems.
4

Engineering Economy…
 To be economically acceptable (i.e) affordable,
solutions to engineering problems must
demonstrate a positive balance of long-term
benefits over long-term costs.
 Solutions to engineering problems must also;
– Promote the well being and survival of an organization
– Embody creative and innovative technology and ideas
– Permit identification and scrutiny of their estimated
outcomes
– Translate profitability to the bottom line through a valid
and acceptable measure of merit 5

Importance of studying Engineering Economy
 Engineer must take into account the monetary
value in their invention as they have to position a
firm to be the most profitable one
 The aim is to balance the trade off between cost
and performance in the most economical manner
 Engineers play a major role in capital investment
decisions based on their technical knowledge
 To prevent a conflict between engineers and
financial department
6

Engineering Economy Decisions and factors…
 Factors affected the decision making process are
from combination of economic and non economic
factor
– Economic factors: production cost, total
revenue, financial sources
– Non economic factor: social responsibilities,
environment, laws, politics
7

Engineering Economic project
8
Categories of
Investments made by
Engineers
Profit- enhancing
Programs
Cost-control
Programs
Public Improvement
Programs
A company may expand its
production, product line, or
services to increase sales
Examples
New-product
development
Production capacity
expansion
Improved customer
service
Engineers are often asked to
correct errors in systems that cost
money.
-they will have to provide a
solution that will save more
money in the long run.
Examples;
Improving efficiency
Streamlining operations
Eliminating waste
Reducing liabilities
Government entities often
make investments, the goal
is not to increase profits
rather to increase public
satisfaction
Examples;
Improve public
participation in stg…
Increased public safety
Improved infrastructure

Role of Engineers in Business
9
What role do engineers play within a firm?
What specific tasks are assigned to the engineering staff, and
what tools and techniques are available to it to improve a
firm's profits?
Engineers are called upon to participate in a variety of
decision-making processes, ranging from design-
manufacturing and marketing to financing decisions.
We will restrict our focus, however, to various economic
decisions related to engineering projects. We refer to
these decisions as engineering economic decisions.

10
Engineering Economic Decisions
Investment
Manufacturing

11
Create & Design
• Engineering Projects
Evaluate
• Expected
Profitability
• Timing of
Cash Flows
• Degree of
Financial Risk
Analyze
• Production Methods
• Engineering Safety
• Environmental Impacts
• Market Assessment
Decision
• Impact on
Financial Statements
• Firm’s Market Value
• Stock Price
Impact of Engineering Projects on Financial
Statements
How a successful engineering project affects a firm's
Market value

Decision-Making Process
12
1. Recognize a decision
problem
2. Define the goals or
objectives
3. Collect all the relevant
information
4. Identify a set of feasible
decision alternatives
5. Select the decision
criterion to use
6. Select the best alternative

1.2 Time value of money
The time value of money means that we
can’t compare amounts of money from two
different periods without adjusting for this
difference in value.
 The concept that a dollar received today is
worth more than a dollar received in the future
and, therefore, comparisons between sums in
different time periods cannot be made without
adjustments to their values.
13

TVM CNTD…
Time Value of money is related to:
 Labor
 Materials
 Machinery & Equipment
 External service
All these factors that have direct
relationship with our products or services
will be affected by time value of money.
14

TVM CNTD…
Time Value of money examples
If a given sum of money is deposited in a
savings account; it earns interest.
If it is used to start a business, it earns profit
If it is used to purchase a share in a
business, it earns dividends.
If it is used to purchase an office building or
apartment house, it earns rent.
15

TVM CNTD…
Definitions of terms wrt Time Value of money
 Interest: the money earned by the original sum of money,
regardless of whether the earned money is referred to as
“interested”, “profits”, “dividends”, or “rent” in ordinary
commercial parlance.
 Interest rate: the time rate at which a sum of money earns
interest (it is usually expressed in percentage form).
 Investment: the productive use of money to earn interest.
 Capital: the money that earns interest.
 The interest earned by the original capital can itself be
invested to earn interest, and this process can be continued
indefinitely. This capacity of money to enlarge itself with
the passage of time is referred to as the time value of
money.
16

TVM CNTD…
Why money loses value over time
 There are several reasons.
 Most obviously, there is inflation which reduces the
buying power of money.
 But quite often, the cost of receiving money in the future
rather than now will be greater than just the loss in its real
value on account of inflation.
 The opportunity cost of not having the money right now
also includes the loss of additional income that you could
have earned simply by having received the cash earlier.
17

TVM CNTD…
Why money loses value over time…
 Moreover, receiving money in the future rather than now
may involve some risk and uncertainty regarding its
recovery. For these reasons, future cash flows are worth
less than the present cash flows.
 Time Value of Money concept attempts to incorporate the
above considerations into financial decisions by facilitating
an objective evaluation of cash flows from different time
periods by converting them into present value or future
value equivalents. This ensures the comparison of 'like
with like'.
18

TVM CNTD…
Consider that you begin to invest on
establishing one manufacturing industry and
decided to pay for all building parts except
the generator house which costs $11,100.00
now.
So you decide to put some savings from this
year’s investment thinking that the
generator house can be built after three
years. 19

TVM CNTD…
With $11,100 in hand you go down to your
local bank, open up a savings account, and
deposit the money.
 You look forward to earning interest over the
next three years until you need the money.
 You are shocked and disappointed to learn
from the bank that interest rates on savings
accounts are rather low—0.3 percent.
20

TVM CNTD…
Note that we multiply by 1 + 0.003 because
the interest rate is .3 of a percent.
The periodic rate is 0.3 percent.
So at the end of the first year, you’ll have
$11,133.30 in your account.
NB: This amount will be the principal for
next year determination.
22

TVM CNTD…
How inflation affects savers
You’ve already put the money for the
generator house in the bank and, indeed,
you’re going to earn $100.20 profit from
your compound interest.
However, you pointed out that construction
and production cost has been rising rapidly
over the last few years- 8% each year.
27

TVM CNTD…
In this case, is the cost of construction 3
years from the beginning,
 P is the amount paid at the beginning of the first
year,
 r is the inflation amount per year and
 t is the number of time periods.
29

TVM CNTD…
Comparing the $11,200.20 you would have
in your savings account with the estimated
final year construction cost of $13,982.80,
you can see how high inflation hurts savers.
 This leads to two observations:
 First, savers want more interest than the prevailing rate
of inflation.
 Second, current interest rates are exceptionally low.
 Interest rate has been deliberately driven lower by
government policy…why?
 b/c in the hope that people will be discouraged from
saving and that they’ll go out and work with it so that
workers employed.
31

TVM CNTD…
Present value of a future sum of money
Now that you know that construction cost is
likely to be $13,982.80 three years from
now as you finish the other construction,
you need to know how much money to save
right now so that you’ll have the correct
amount when it’s needed.
We can calculate this just by rearranging the
future value formula to solve for the present
value, Px. 32

Next class….. Depreciation
Assignment-1
1. Write a one page report on
1. Opportunity Costs and Sunk Costs including
their relevance in engineering economics
2. Cash Flow and Cash-Flow Diagrams
2. Do the following exercise (see next slide)
3. Read more on how to calculate P, F, i, n...
36

Exercise 1
37
Given the following set of payments:
$ 800 at the beginning of year 2, and
$ 500 at the beginning of year 6
On the basis of an 8% interest rate, this set of
payments is to be transformed to an equivalent set
consisting of the following:
A Payment of X at the beginning of year 5, and
A payment of 2X at the beginning of year 9.
Find the payments under the second set, and
verify the values.

1.3 Depreciation
Introduction
Any equipment which is purchased today
will not work for ever.
This may be due to wear and tear of the
equipment or obsolescence of technology.
Hence, it is to be replaced at the proper time
for continuance of any business.
38

Dep. Int. CNTD…
The replacement of the equipment at the
end of its life involves money.
This must be internally generated from the
earnings of the equipment.
The recovery of money from the earnings of
an equipment for its replacement purpose is
called depreciation fund since we make an
assumption that the value of the equipment
decreases with the passage of time. 39

Dep. Int. CNTD…
Thus the word “depreciation” means
decrease in value of any physical asset with
the passage of time.
 Points:
 Book value
 Salvage value
 Market value
 Scrap value
Read more on their difference 40

Methods of Depreciation
A firm in its accounting records is
constrained to follow an officially approved
depreciation method.
However, in its informal records it may
prefer to follow some other method it
considers more realistic, there by obtaining
a more accurate appraisal of the value of its
assets and the profitability of its operations.
41

Methods of Depreciation….cont’d
1. Straight Line Method of Depreciation
In this method of depreciation, a fixed sum is
charged as the depreciation amount
throughout the lifetime of an asset such that
the accumulated sum at the end of the life of
the asset is exactly equal to the purchase value
of the asset.
42

Straight CNTD…
Here, we make an important assumption
that inflation is absent.
Let
P or Bo = first cost of the asset,
F or L = salvage value of the asset
n = life of the asset,
Bt = book value of the asset at the end of the period t,
Dt = depreciation amount for the period t.
43

Straight CNTD…
The formulae for depreciation and book value
are as follows:
Dt = (P – F)/n…constant
Book value at tth Period will be
Bt = Bt–1 – Dt = P –t[(P – F)/n]...show the
derivation?
44

Straight CNTD…
EXAMPLE
A company has purchased an equipment
whose first cost is Br. 1,00,000 with an
estimated life of eight years. The estimated
salvage value of the equipment at the end of
its lifetime is Br. 20,000. Determine the
depreciation charge and book value at the end
of various years using the straight line method
of depreciation. 45

Straight CNTD…
Solution
P = Br. 1,00,000
F = Br. 20,000
n = 8 years
Dt = (P – F)/n
= (1,00,000 – 20,000)/8
= Br. 10,000 46

Straight CNTD…
 In this method of depreciation, the value of Dt is
the same for all the years.
The calculations pertaining to Bt for different values
of t are summarized in Table below:
47
End of year
(t)
Depreciation
(Dt)
Book Value
(Bt = Bt–1 – Dt)
0 100,000
1 10,000 90,000
2 10,000 80,000
3 10,000 70,000
4 10,000 60,000
5 10,000 50,000
6 10,000 40,000
7 10,000 30,000
8 10,000 20,000

Straight CNTD…
If we are interested in computing Dt and Bt
for a specific period (t), the formulae can be
used.
In this approach, it should be noted that the
depreciation is the same for all the periods.
EXAMPLE
Consider Example and compute the depreciation and
the book value for period 5.
P = Rs. 1,00,000
F = Rs. 20,000
n = 8 years
48

Straight CNTD…
D5 = (P – F)/n
= (100,000 – 20,000)/8
= Br. 10,000 (This is independent of the time
period.)
Bt = P – t (P – F)/n
B5 = 1,00,000 – 5 (1,00,000 – 20,000)/8
= Br. 50,000
49

2. Sum-of-the-Years-Digits Method of
Depreciation
In this method of depreciation also, it is
assumed that the book value of the asset
decreases at a decreasing rate.
If the asset has a life of eight years, first the
sum of the years is computed as:
Sum of the years = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8
= 36
= n(n + 1)/2
50

Sum-of-the-Years CNTD…
The rate of depreciation charge for the first
year is assumed as the highest and then it
decreases.
The rates of depreciation for the years 1–8,
respectively are as follows:
8/36, 7/36, 6/36, 5/36, 4/36, 3/36, 2/36, and
1/36.
51

4
52
See book for
derivation

EXAMPLE
Consider the previous example and
demonstrate the calculations of the sum-of-
the-years-digits method of depreciation.
Solution
P=Bo = Br. 100,000
F=L = Rs. 20,000
n = 8 years
Sum = n (n + 1)/2
= 8* 9/2
= 36
53

The rates for years 1–8, are respectively
8/36, 7/36, 6/36, 5/36, 4/36, 3/36,2/36 and
1/36.
The calculations of Dt and Bt for different
values of t are summarized in table using
the following formulae:
Dt = Rate (P – F)
Bt = Bt–1 – Dt
54

Table: Dt and Bt under Sum-of-the-years-digits
Method of Depreciation
55
End of year
(t)
Depreciation
(Dt)
Book Value
(Bt)
0 1,00,000.00
1 17,777.77 82,222.23
2 15,555.55 66,666.68
3 13,333.33 53,333.35
4 11,111.11 42,222.24
5 8,888.88 33,333.36
6 6,666.66 26,666.70
7 4,444.44 22,222.26
8 2,222.22 20,000.04

If we are interested in calculating Dt and Bt
for a specific t, then the usage of the
formulae would be better.
EXAMPLE
Consider the previous example and find the
depreciation and book value for the 5th year using the
sum-of-the-years-digits method of depreciation.
Solution
P = Br. 1,00,000
F = Br. 20,000
n = 8 years 56

Other depreciation method …read
59
3. Units of production method
4. Decline balance method , double decline
balance method
5. Other…. Read more on given materials

Assignment -2
Prepare a brief note on:
(a) The relationship between inflation and
depreciation,
(b) The relationship between cost and
depreciation,
(c) The relationship between depreciation and
taxation.
(d) The difference b/n book value, salvage
value, market value and scrap value
60

1.4 Cost comparison of alternation
methods
Every need that arises in our industrial
society can be satisfied in multiple ways.
For example, there are alternative
manufacturing processes for producing a
commodity, and there are alternative
designs for constructing bridge.
Consequently, the engineering economist
has the task of identifying the most
desirable way of satisfying each need that
arises.
61

Cost comparison CNTD…
Generally, the disbursements associated
with each alternative method span a period
of several years, and therefore, it is
necessary to weave the time value of money
into any investigation.
There are several techniques of cost
comparison.
62

In the following elaborations, it is to be
understood that the alternative methods differ
only with respect to cost and alike with respect
to income, serviceability, general convenience,
etc.
For example, in determining whether a
commodity is to be produced manually or by
automated equipment, we assume that the
quality of the product is identical under the
two methods. 63

1. Selection of interest rate
The first problem in a cost comparison of
alternative methods is to select the interest
rate on which the calculations are to be
based.
The criterion that governs this selection will
emerge from the very simple example that
follows:
64

Example
A firm must purchase a new machine, and it has a
choice of two models, X and Y. The initial cost is
Br. 50,000 for X and Br. 62,000 for Y. The
estimated annual operating cost, which may be
treated as a lump-sum payment made at the end of
the year, is Br. 13,000 for X and Br. 10,000 for Y.
Both models have an anticipated service life of 5
years and zero salvage value, and they are alike in all
other respects. 65

If the firm purchases model X in preference to
model Y, the 12,000Br capital thus saved will
be placed in an investment that will last 5
years and will return invested capital with
interest at 10%in the form of five equal end-
of-year payments. This investment has
approximately the same degree of risk as that
inherent in purchase of the machine. Which
model is preferable? 66

Solution
The distinguishing consequences that stem
from the two mutually exclusive events are
as follows:
Buy model X:
Annual income from investment is:
= 12,000(A/Pu,5,10%)
= Br. 3166 67

Buy model Y:
Annual savings in operating cost is:
= 13,000 – 10,000
= Br. 3000
We conclude that model X is preferable
because the decrease in operating cost that
results from use of model Y does not justify
the higher initial cost.
68

2. Description of simplified model
It is a classical type, most of its assumptions
are ideal, but is the basis to begin the future
complex comparisons.
Our immediate objective is to formulate
standard techniques of cost comparison.
To avoid making our task prohibitively
arduous (difficult), we shall construct a
simplified model of the industrial world
with the following characteristics: 69

1. All economic and technological conditions remain
completely static, except where changes are
explicitly described.
As a result, interest rates and costs remain constant
as time elapses, and each asset is replaced with an
exact duplicate when it is retired.
2. The future can be seen with certainty.
Consequently, all forecasts and projections prove to
be accurate in every respect.
70

3. Interest is compounded annually.
4. All disbursements and receipts associated with an
asset occur at the beginning or end of a year.
The process of retiring an asset and replacing it
with exact duplicate may be viewed as a
renewal of the original asset, and each asset
may be considered to have an endless
succession of lives.
Therefore, the set of payments that occurs in
the first life will recur in all ensuring lives. 71

In our cost comparison, we shall disregard
the effects of taxation at present.
After we have formulated the techniques of
cost comparison, we shall alter our model to
make it conform more closely to reality.
The payments associated with a given
method of performing a task are classified
as ordinary payments and capital payments,
according to whether they have long term or
short term payments, respectively. 72

Thus, expenditures for the daily operation
of a machine are ordinary payments, and
an expenditure for major repairs that
extends the life of the machine is a capital
payment.
73

3. Present worth of costs
Where two alternative assets are to be
compared with respect to cost, we may
establish a basis of comparison in this
manner:
 Select a period of time that encompasses an
integral number of lives of each asset.
 Select the beginning of this time period as the
valuation date, and find the value at this date of
the entire set of payments associated with each
asset during this time period.
74

 This value is called the present worth of costs,
and the period of time selected is known as the
analysis period.
 The present worth of costs is abbreviated PW,
with a subscript to identify the method.
See the example below:
75

Example
Two types of equipment are available for
performing a manufacturing operation; the
cost data associated with each type are
recorded in the accompanying table. Applying
an interest rate of 8%, determine which type is
more economical.
76

Table: cost data
Solution
 We shall compute the present worth of costs.
 Select a 12-years analysis period; this
encompasses one life of type A and two lives of
type B. 77
Type A Type B
First cost, Br. 88,000 45,000
Salvage value, Br. 7,500 4,000
Annual maintenance, Br. 4,300 5,200
Life, years 12 6

The capital payments that occur during this
period are recorded in Figure below, where
expenditures are shown below the base line and
income above it.
78
Br. 88,000
0
Br. 7,500
12
For Type A
Br. 45,000
0
Br. 4,000
12
Br. 4,000
Br. 45,000
For Type B
6
Years
Years

 With respect to type A, the salvage value
pertaining to the first life falls within the
analysis period, but the first cost of the second
life falls beyond this period, therefore, we don’t
further analyze for the second cost.
 Similar comments apply with respect to type B.
 Payments for annual maintenance are treated as
lump-sum, end-of-year expenditures.
79

PWA= 88,000 + 4300(Pu/A,12) – 7500(P/F,12)
= 88000 + 4300(7.53608) - 7500(0.39711)
= Br. 117,430
PWB= 45000 + (45000-4000)(P/F,6) + 5200(Pu/A,12) – 4000(P/F,12)
= 45000 + (41000)(0.63017) + 5200(7.53608) – 4000(0.39711)
= Br. 108,440
 Conclusion
 Type B equipment is more economical.
80

When taxation on depreciation is considered
See example …word file
Next remaining two parts of the chapter + cost
comparison (other methods than Pw)
prepare note in group (2-3 students) for next
round class and present
81

1.5 Economy Analysis of industrial
operations
Every industrial operation must be carefully
scrutinized/analyzed to establish the most
efficient way in which it can be performed.
There are numerous techniques for
obtaining the maximum economic
efficiency of an operation.
82

Economy CNTD…
Definitions pertaining to costs
Let X denote a variable that strongly
influences the cost of an operation.
 For example, X may denote the number of
machine parts that are handled daily by a repair
shop or the number of passengers transported
daily by public bus.
Now assume that X undergoes a relatively
small change.
83

Economy CNTD…
A cost that remains constant during this
change is a fixed cost with respect to X, and
one that changes is a variable cost.
As an illustration:
Let X denote the number of units of a standard
commodity that a firm produces per month.
With respect to X, rent paid for the factory is a
fixed cost.
The cost of raw materials for this commodity is a
variable cost. 84

Economy CNTD…
Intermediate between these extremes are
semi-variable costs, which are composite
costs that contain both fixed and variable
elements.
For example:
 Let X denote the volume of production in an
industrial plant.
 The cost of the electric power in this plant is a
semi-variable cost because certain elements are
independent of X while others are function of X. 85

Economy CNTD…
Variable costs can often be divided into
three sub groups:
Those that are directly proportional to X,
Those that are inversely proportional to X, and
Those that vary in some more complex manner.
86

Economy CNTD…
The first two types are known as directly
varying costs and inversely varying costs,
respectively.
As an illustration (for directly varying cost):
Assume that a firm manufactures a standard
commodity, that it is required to pay royalties for
the use of a patent, and that the agreement
stipulates the inventor is to receive a fixed sum
of each unit manufactured.
If X denotes the number of units produced
annually, the royalty payment is a directly
varying cost.
87

Economy CNTD…
The unit cost of an operation is the result
obtained by dividing the total cost by X.
For example, the unit cost may be:
 the cost of manufacturing one unit of a
commodity, or
 the cost of constructing 1km of a highway.
88

Economy CNTD…
Standard costs are pre-established costs that
are designed to gauge the efficiency of an
operation.
As the operation proceeds and the true costs
become known, they are compared with the
standard costs.
Several methods of securing costs are
widely used.
89

Economy CNTD…
By one method, they are obtained
statistically on the basis of historical data.
This method has serious deficiencies:
First, if the operation was performed poorly in
the past, the use of these standard costs may
serve to perpetuate inefficiency.
Second, new technology may permit a
reduction in costs, thereby rendering the
historical data irrelevant.
90

Economy CNTD…
As a result of these deficiencies:
 Many analysts prefer to obtain standard costs by
a method that equates them to currently
attainable costs.
Standard costs derived from historical data
are generally objective/fair; those that are
based on present possibilities are
necessarily judgmental.
91

Economy CNTD…
Minimization of cost
Let C denote the total cost of a project or
operation, and assume that is the only
variable that influences C.
Also assume that C is composed exclusively
of directly varying costs, inversely varying
costs, and fixed costs. Then;
C = aX + b/X + c
= aX + bX-1 + c ……….. (1.1)
Where a, b and c are constants. 92

Economy CNTD…
The following figure is a graphic
demonstration of Kelvin’s law:
 Lines 1, 2 and 3 are a plot of aX, b/X, and
aX + b/X, respectively (The fixed costs are
excluded because they don’t affect Xo).
96

Economy CNTD…
As the above diagram reveals, line 3 has its
sag/fall point vertically above the point at
which lines 1 and 2 intersect one another.
Typical illustration:
In the design of a steel bridge, it is necessary to
decide how many piers are to be used. The
center-to-center distance between adjacent piers
is termed the span, and the span may be taken
as the independent variable X.
97

Economy CNTD…
 The cost of the bridge consists mainly of the
cost of the steel and the cost of the piers.
 The weight of the steel is minimized by using
a short span, and the number of piers is
minimized by using a long span.
 The problem is to find the particular span that
yields the minimum total cost under these
conflicting criteria.
 Lets consider an example:
98

Economy CNTD…
Example
A steel bridge is to have a length of 420m. The cost
of the steel and of an individual pier are estimated to
be as follows: Cs = 1,200,000 + 47,000X and Cp =
840,000 + 560X, where Cs is the cost in dollars of
the steel, Cp is the average cost in dollars of a pier,
and X is the span in meters. All other costs are
independent of the span. Determine the most
economical span for this bridge.
99

1.6 Decision making with probability
We shall now investigate the role played by
probability in managerial decision
making.
102

Decision making CNTD…
Calculation of expected value
Where there is uncertainty concerning the
cost of an industrial operation or the profit
that occurs from an investment, a decision
is often made on the basis of the expected
cost or profit.
The expected value of a random variable is
calculated by applying the following equation:
103

Example
A 1-day construction job can be performed by two
alternative methods: A and B. Both methods, the
cost of the job is dependent on the state of the
weather, and for estimating purposes we recognize
three states of weather: sunny, cloudy and rainy. The
costs corresponding to the two methods and the three
states of weather are recorded in the following table.
The contracting firm must decide 3 days in advance
which method it will apply. 105

It has obtained a weather forecast for the day of
construction, and this forecast gives the probabilities
in the last column of the table below. Which method
of construction should be applied?
106
Weather Construction cost, $ Probability
Method A Method B
Sunny 2200 1500 0.15
Cloudy 2500 2100 0.40
Rainy 3000 4000 0.45

Solution
Multiplying the possible costs by their respective
probabilities, we obtain the following results:
Method A:
Expected cost = 2200(0.15) + 2500(0.40) + 3000(0.45) = $2680
Method B:
Expected cost = 1500(0.15 + 2100(0.40) + 4000(0.45) = $2865
 On the basis of the forecast, the firm should apply method A.
 As these calculations disclose, the lower cost by method B
corresponding to sunny and cloudy weather is more than offset by its
much higher cost corresponding to rainy weather. 107

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