Analysis of

Production
Systems
Engineering
Economy and
Costing
 Introduction – Product & Process Decision and
Design

New products and services – vital to economic survival of

business organization.

Most companies continually redesign existing product - to

reflect market demands, changing technologies and
other inputs.

Created by new improved process.

Important functions in organization and greatly affect

operations system.
Product development

 Idea Generation
 Idea Screening
 Concept Development and Testing
 Business Analysis
 Prototype Development
 Market Testing
 Commercialization plan
 Launch
 Post-Launch Evaluation
 Product Maintenance and Improvement
New product decision and design
 Continued introduction of
new products or services –
assures maintenance of
sales volume over a
period of time.
Introduction
It relates the volume of
sales of a product – to
time that has elapsed
since its introduction into
market place.
Product Life Acceptance

Cycle Product life cycle involves

4 stages,

Maturity

Decline
Product life cycle
Stages of product life cycle
 Stage 1 – Introduction stage (Product A)
- new and unknown to public
- field troubles call for redesign or production changes
- high priced.

 Stage 2 – Acceptance stage

- product improved
- sales volume increases at accelerated pace
- price decreases
- production efficiencies and reduced costs.
 Stages of product life cycle
 Stage 3 – Maturity stage
- Product – dependable and trade name becomes accepted.
- Company should decide on, design and start marketing new and different
product – to sustain production capacity and sales volume.
- Product B – entering stage 2 while A entering stage 3

 Stage 4 – Decline stage

- product – faced with new competition
- benefited from new technological breakthroughs
- popularity and sales volume decline.
- company involved in R & D of new products.
- Product B entering stage 3 while A entering stage 4).
Economic Analysis
 It determines the economic viability of the business organization.

Break even analysis

 Minimum volume of sales must be produced and sold to break even after
paying all expenses.
Min. Volume – Break-even point (BEP)

 Producing and selling more than BEP – to make profit.

 Fixed cost: remain constant regardless of volume of products (rent, property

taxes, depreciation, insurance, salaries to staff)

 Variable cost: fluctuate directly with changes in the output volume of

products.
BEP = Fixed cost / (unit selling price – variable cost per unit)
Break-even analysis
Break even analysis
 In simple terms, the break-even point is the juncture where total cost
and total sales (revenue) are equal. This point is important for every
company to know because, from this point, the company starts to
become profitable. If total cost and total revenue are equal at this point,
that means the units produced would generate zero profit.
 That means at this point,
Revenue – Total Cost = 0

N = Fixed Cost / (Price per unit – Variable Cost)

 Total cost = Variable Cost * N + Fixed Cost

 Revenue = Price per unit * N
Contribution Margin
 Break-even analysis also deals with the contribution margin of
a product. The excess between the selling price and total
variable costs is known as contribution margin.
 It represents the revenue collected to cover the fixed costs.

Note: In the calculation of the contribution margin, fixed costs are

not considered.
Problem
General Electric is considering the production of new, energy saving
light bulb. The selling price is $10.00, and variable cost is about
$2.00 per light bulb. If total fixed costs are $20 million, the break-
even point, in units of output sold, or light bulbs is

BEP = $20,000,000 / ($10.00 - $2.00) = 2,500,000

That is, if GE produces 2,500,000 light bulbs, total costs equal total
revenue

Total costs: $20,000,000 + $2(2,500,000) = $25,000,000

Total revenue: $10 x 2,500,000 = $25,000,000
 In the same problem what will happen if selling price
of bulb is $12.00?

Obviously, BEP will be lower.

BEP = $20,000,000 / ($12.00 - $2.00) = 2,000,000 bulbs

Selling price is subject to market constraints – cannot be raised.

BEP – lowered by – reducing production costs (through reduced

scrap, more efficient use of machine and labour)
Break Even Point Example

Details of the two companies are given in the below table.

In $ Company X Company Y
Fixed Cost 30000 50000
Price per unit 100 90
Variable Cost per
40 30
unit

Find the Break-even point (units) of both Company X and

Company Y?

BEP = FC / (Price – VC)

 Putting the data in the formula, we get –

In $ Company X Company Y
Price per unit 100 90
Variable Cost per unit 40 30
Contribution
60 60
Margin/unit

 Now, the BEP would be –

In $ Company X Company Y
Fixed Cost (A) 30000 50000
Contrib. Margin/unit (B) 60 60
BEP (A / B) 500 833.33

That means beyond 500 units, Company X and beyond 833.33 units,
Company Y would be able to make profits.
Break Even Point Example
A new word-processing machine is contemplated by Short-Life
Underwriters, Inc., to accommodate insurance policy typing and
printing, the fixed costs of energy, depreciation, labor, printing paper,
and disc supply amount to $19,700, and the variable costs are S3
policy. The average revenue from an insurance policy drafted is
$200.
a) How- many policies should be drafted in order to break even?
b) What is each policy's contribution to fixed costs and profit?
Solution
a) Break Even point = $19700/($200-$3)
= 100 policies

b) Contribution = $200 - $3 = $197

Break Even Point Example

In order to facilitate a faster return of rented cars. Hertz Rent-A-

Car has installed a special express counter with a capability of
handling 60 car returns/hour. The fixed costs involved are $5,000.
The variable cost of direct labor, depreciation, and material
(including computer terminal charges) amounts to $1.30/car return.
The surcharge to the customer using this express counter is
$1.40/car return.
a) What is the break-even point, expressed in number of returns?
b) What is the break-even point in hours?
Solution

a) Break Even point = $5000/($1.4-$1.3)

= 50000 cars

b) BEP in hours = 50000/60

= 833.33 hours
Break Even Point Example

Details of the two companies are given in the table below.

Calculate the Fixed Cost ?

In $ Company X Company Y
Fixed Cost ? ?
Price per unit 120 140
Variable Cost per unit 60 70
BEP (units) 500 600
Solution

In $ Company X Company Y

Contrib. Margin/unit (A) 60 70

BEP (B) 500 600

Fixed Cost (A * B) 30000 42000

Limitations

BEP Analysis is based on many assumptions, thus most of the

companies don’t rely much on BEP alone. They compute
many other ratios to make a concrete decision. Here are a few
limitations of BEP Analysis –
 There are many assumptions we make while computing
Break-even point.
 we assume that the price of a product remains constant
all the time; but in reality, the price changes at the
different output level of the product.
 We also assume that fixed costs remain the same always;
but even fixed cost changes after a certain output.
Limitations

 One of the chief assumptions of BEP analysis to equate

sales as output. But there may be a difference between
sales and output. Not all output is not meant to get sold and
there is a chance as well for normal or abnormal loss where
few outputs get wasted.
 Most of the businesses have more than one product. Thus,
it’s become tougher to calculate BEP.
 Variable cost per unit doesn’t always remain constant. It
changes when the production is being done in a large chunk
and thus the firm gets the inputs at a much lower rate.
Depreciation

 Spreading the cost of a fixed asset over the life, or expected years
of use, of the asset.

 Annual charge reflecting the decline in value of an asset – due to

causes (wear and tear, obsolescence, inadequacy).

 Depreciation is a method of asset cost allocation that apportions

an asset's cost to expenses for each period expected to benefit
from using the asset.
Depreciation
Causes of depreciation
Depreciation

Due to Physical Conditions, Due to functional conditions

• Wear and tear • Inadequacy

• Physical decay • Obsolescence
• Accident
• Poor maintenance
  Based on years
 Straight Line Method or fixed installment
method
 Written Down Method or Reducing Balance
Method or Diminishing Balance method
Types of  Sum-of-the-Years-Digits method
Depreciation  Double Declining Depreciation method
methods  Based on units produced
 Production Units method
 Depletion method
 Based on hours
 Machine Hour method
 Depreciation
Straight Line Method (Fixed installment method)

 Amount of depreciation is distributed over the useful life of the machine in

equal periodic instalments.
(Cost – Salvage)
 Depreciation, D=
Estimated life of machine in years

 It doesn’t consider the maintenance and repair charges.

A drilling machine is purchased for Rs. 45,000 and the assumed life in 10 years.
The scrap value is taken as Rs. 5,000. Calculate the yearly depreciation by straight
line method.
C – Rs. 45,000, n – 10 years, S – Rs. 5,000
Yearly depreciation cost,
D = (45000 – 5000)/10 = Rs. 4,000
Straight Line Method
Problem
A second-hand bike was purchased for Rs. 60,000. Its useful life was estimated
as 12 months but after 3 months it was sold for Rs. 35,000. If the depreciation
followed here was Straight line method, find the where the deal was loss or
profit. Find how much Rupees it gained or loss by the deal.
 Sol:
(Cost – Salvage) 60,000−0
 Depreciation = = = 5000
Estimated life of machine in years 12

 Book value (end of 1st month) = 60,000-5000 = Rs. 55,000

 Book value (end of 2nd month) = 55,000-5000 = Rs. 50,000
 Book value (end of 3rd month) = 50,000-5000 = Rs. 45,000
Book value at the end of 3rd month is more than the deal amount. So, the
deal was considered to be a loss.
 Loss = 45000-35000 = Rs. 10,000
 Depreciation
Units of Production Method

 Units-of-production depreciation method depreciates assets

based on the total number of hours used or the total number of
units to be produced over its useful life.

The formula for the units-of-production method

 Depreciation Expense = (Number of units produced / Life in
number of units) x (Cost – Salvage value)
Units of Production Method Problem
A machine was bought for Rs. 20,00,000. Its residual value was Rs.
2,00,000. The machine can produce 1.5 lakhs unit in its lifetime. The
machine was used for 10 years, and the products were produced as
follows,

Years Units Produced

1-3 20,000 units per year
4-7 15,000 units per year
8-10 10,000 units per year

If the depreciation followed here is unit production method, find the book
value at the end of 10th year.
Units of Production Method Problem
Sol:
(𝐶𝑜𝑠𝑡−𝑆𝑎𝑙𝑣𝑎𝑔𝑒 𝑣𝑎𝑙𝑢𝑒) (20,00,000−2,00,000)
Depreciation per unit = = = 12 Rs/unit
𝑢𝑠𝑒𝑓𝑢𝑙 𝑙𝑖𝑓𝑒 𝑖𝑛 𝑢𝑛𝑖𝑡𝑠 1,50,000

Depreciation during 1st -3rd year

 Depreciation per unit * units produced during that year
= 12*20,000 = Rs. 2,40,000 per year
 Total depreciation during 1st -3rd year = 2,40,000*3 = Rs. 7,20,000
 Book value at the end of 3rd year
= 20,00,000 – 7,20,000
= Rs. 12,80,000
 Contd…
Depreciation during 4th -7th year
 Depreciation per unit * units produced during that year
= 12*15,000 = Rs. 1,80,000 per year
 Total depreciation during 4th -7th year = 1,80,000*4 = Rs. 7,20,000
 Book value at the end of 7th year
= 12,80,000 – 7,20,000
= Rs. 5,60,000
Depreciation during 8th -10th year
 Depreciation per unit * units produced during that year
= 12*10,000 = Rs. 1,20,000 per year
 Total depreciation during 4th -7th year = 1,20,000*3 = Rs. 3,60,000
 Book value at the end of 7th year
= 5,60,000 – 3,60,000
= Rs. 2,00,000
Depreciation - Units of Production Method
Example
 Depreciation

Double-declining balance method

 Annual rate of depreciation is derived by –

doubling the straight line rate and applying it, year by year to the portion of
asset value still not depreciated.

 With 10 year depreciation period – depreciation rate would be 2(1/10) = 0.20.

2(𝐵𝑜𝑜𝑘 𝑣𝑎𝑙𝑢𝑒)
 Double-declining balance depreciation =
𝐿𝐼𝑓𝑒𝑡𝑖𝑚𝑒
Book value – value of asset less the accumulated depreciation.
Calculation is done for each year over the lifetime of the asset.
 Depreciation
Double-declining balance method
A new machine costs $20,000 and has a depreciable (accounting) life of 5 years,
with a salvage value of $5,000. What are the values of the annual double-declining
depreciation?
Year Depreciation Beginning Depreciation Accumulated Ending
Ratio Book value Charge Depreciation Book Value
1 2/5 = 0.40` x 20,000 = 8000 8000 12000
2 2/5 = 0.40 x 12,000 = 4800 12800 7200
3 2/5 = 0.40 x 7,200 = 2200 15000 5000
4 0 x 5,000 = 0 15000 5000
5 0 x 5,000 = 0 15000 5000
Total = 15000
 In this method, one does not depreciate beneath the salvage value.
 Machine is fully depreciated by the third year; depreciation must be 0 in years 4
and 5
 Depreciation

Double-declining balance method

Problem:
The company is considering the purchase of a second-hand scanning microscope
at a cost of $10,500, with an estimated salvage value of $500 and a projected
useful life of four years. Determine the straight-line (SL) and double declining
balance (DDB) depreciation.
Depreciation

Solution:
Cost - $10,500 salvage - $500 estimated life – 4 years

Straight line method (SL)

(10500 – 500)
Depreciation, D= = 2,500 per year
4
 Depreciation

Double-declining balance method

Ending book value = initial cost – accumulated depreciation

Year Depreciation Beginning Depreciation Accumulated Ending

Ratio Book value Charge Depreciation Book Value

1 2/4 = 0.50 x 10,500 = 5,250 5,250 5,250

2 2/4 = 0.50 x 5,250 = 2,625 7,875 2,625
3 2/4 = 0.50 x 2,625 = 1,312.5 9,187.5 1,312.5
4 2/4 = 0.50 x 1,312.5 = 656.25 9843.75 656.25
Total = 9,843.75
 Depreciation

Double-declining balance method

Problem:
ABC Limited purchased a Machine costing $12500 with a useful life
of 5 years. The Machine is expected to have a salvage value of
$2500 at the end of its useful life. Let’s calculate the depreciation
using the Double Declining Balance method.
 Depreciation

Year Dep. Beginning Dep. Accumulated Ending Book Value

Ratio Book Charge Dep.
Value
1 2/5 = 0.4 x 12500 = 5000 5000 7500
2 2/5 = 0.4 x 7500 = 3000 8000 4500
3 2/5 = 0.4 x 4500 = 1800 9800 2700
4 2/5 = 0.4 x 2700 = 1080 10880 1620
Total = 10880
 Depreciation

Double-declining balance method

Problem:
The Farma company provides the following information: Cost of the
equipment: $500,000, Salvage value: $50,000, Useful life: 5 years.
Prepare a depreciation schedule using double declining balance
method.
Depreciation
 Capital Budgeting Techniques
Investment decisions are generally called capital budgeting decisions

 Net present value considers – “Time value of money”

 Money grows over time, when it earns interest.
 Money in hand today is worth more than a money in future.

Discounted Cash Flow (DCF)

Investment decisions made by – taking into account the interest that the
money in hand can earn if invested in a bank or somewhere else.

Non-Discounted Cash Flow (NDCF)

 Interest is not taken into consideration.
 Time value of money is not considered.
 Capital Budgeting Techniques

• NPV – Net Present Value

Discounted cash flow
• IRR – Internal Rate of Return

Non – discounted cash flow Payback Period

 Present value
Example
If Rs. 120 is to be received after 1 year, What is the PV of Rs. 120 today?
If Rs. 120 is to be received after 5 year, What is the PV of Rs. 120 today?
If Rs. 120 is to be received after 15 year, What is the PV of Rs. 120 today?
Note: Discounted rate is 6% per year
Solution:
FV = $ 120k = 6 % = 6/100 = 0.06
1 FV
PV= FV x (1 + k)n = (1 + k)n

120
PV (n=1): PV = = 113.2
(1 + 0.06)1
120
PV (n=5): PV = = 89.67
(1 + 0.06)5
120
PV (n=15): PV = = 50.07
(1 + 0.06)15
 Net Present Value Method (NPV)
Problem:
The initial cost of a project is Rs 4,000. The forecast of year end cash
inflows is Rs 1900, Rs 1600, Rs 1400, Rs 1200 and Rs 1100
respectively during the 5 years of its life. If the rate of interest is
10%, determine the net present value of the project.
Solution:
Initial cost = Rs. 4,000 Interest rate (k) = 10% = 0.1 Period (n) = 5 years
Cash inflow (1st year) = Rs. 1,900 Cash inflow (4th year) = Rs. 1,200
Cash inflow (2nd year) = Rs. 1,600 Cash inflow (5th year) = Rs. 1,100
Cash inflow (3rd year) = Rs. 1,400

cash inflows are different in each and every year.

FV
Present value of cash inflows, PV=
(1 + k)n
Net Present Value Method (NPV)
Total present value for 5 years,
1900 1600 1400 1200 1100
= + + + +
(1 + 0.1)1 (1 + 0.1)2 (1 + 0.1)3 (1 + 0.1)4 (1 + 0.1)5

= 1,727 + 1,322 + 1,052 + 820 + 682

= Rs. 5,603

NPV = (Total present value of cash inflows

– Total present value of cash outflows)

NPV = 5,603 – 4,000 = Rs. 1,603

(Note: If NPV is positive, project should be accepted otherwise it should be rejected)
 Net Present Value Method (NPV)
Problem:

Two project A & B, both have same initial cost Rs. 40,000.

The forecast of year end cash inflows for project 1 is Rs 19000, Rs

16000, Rs 14000, Rs 12000 and Rs 11000 respectively during the 5
years of its life.

The forecast of year end cash inflows for project 2 is Rs 18000, Rs

17000, Rs 16000, Rs 9000 and Rs 10000 respectively during the 5
years of its life.

If the rate of interest for both project is 10%, determine the best
project using NPV.
Net Present Value Method (NPV)
FOR PROJECT A
Total present value for 5 years,
19000 16000 14000 12000 11000
= + + + +
(1 + 0.1)1 (1 + 0.1)2 (1 + 0.1)3 (1 + 0.1)4 (1 + 0.1)5

= 17272.7 + 13,223.1 + 10,518.4 + 8196.1 + 6830.1

= Rs. 56040.4

NPV = (Total present value of cash inflows

– Total present value of cash outflows)

NPV = 56040.4 – 40,000 = Rs. 16,040.4

Net Present Value Method (NPV)
FOR PROJECT B
Total present value for 5 years,
18000 17000 16000 9000 10000
= + + + +
(1 + 0.1)1 (1 + 0.1)2 (1 + 0.1)3 (1 + 0.1)4 (1 + 0.1)5

= 16363.6 + 14049.5 + 12,021 + 6147.1 + 6209.2

= Rs. 54790.4

NPV = (Total present value of cash inflows

– Total present value of cash outflows)
NPV = 54790.4 – 40000 = Rs. 14,790.4

Therefore, project A is selected because of its high NPV value.

Internal Rate of Return Formula
𝐶𝐹𝑛
 0 = σ𝑁
𝑛=0 (1+𝐼𝑅𝑅)𝑛

Or
𝐶𝐹1 𝐶𝐹2 𝐶𝐹3 𝐶𝐹𝑛
 0 = 𝐶𝐹0 + 1 + 2 + 3 +…+
(1+𝐼𝑅𝑅) (1+𝐼𝑅𝑅) (1+𝐼𝑅𝑅) (1+𝐼𝑅𝑅)𝑛
Where:
𝐶𝐹0 = Initial Investment/outlay
𝐶𝐹1 𝐶𝐹2 𝐶𝐹3 … 𝐶𝐹𝑛 = Cash flows
n = Each period
N = Holding period
IRR = Internal Rate of Return
Rules to Calculate Internal Rate of Return
Formula

There are certain rules to keep in mind while calculating the internal rate
of return formula. They are:
 Set the NPV to zero and solve for the discount rate i.e. the internal rate of
return
 The initial investment is always negative because it represents an outflow
 Each subsequent cash flow could be positive or negative, depending on the
estimated cash flow determined by the project in the future.
An investor made an investment of $500 and got $570 next year. Calculate the internal
rate of return on the investment.

Solution:
Invested amount, 𝐶𝐹0 = -$500 (negative, because money went out)
Cash inflow after 1 year, 𝐶𝐹1 = $570
Using internal rate of return formula,
𝐶𝐹1
0 = 𝐶𝐹0 + (1+𝐼𝑅𝑅) 1

570
0 = -500+ (1+𝐼𝑅𝑅)1

500+500*IRR = 570
IRR = 70/500
IRR = 0.14 = 14%
Therefore, the internal rate of return on the investment = 14%.
X bought a house for $250,000. He plans on selling the house 1 year later for
$350,000, after deducting any realtor's fees and taxes. Calculate the internal rate of
return on the complete transaction.

SOLUTION:
Invested amount, 𝐶𝐹0 = -$250,000 (negative, because money went out)
Cash inflow after 1 year, 𝐶𝐹1 = $350,000
Using internal rate of return formula,
𝐶𝐹1
0 = 𝐶𝐹0 + (1+𝐼𝑅𝑅) 1

350000
0 = -250000+ (1+𝐼𝑅𝑅)1

250000+250000*IRR = 350000
IRR = 100000/250000
IRR = 0.4 = 40%
Therefore, the internal rate of return on the investment = 40%.
 Internal Rate of Return Method (IRR)

An engraving machine (project proposal) costs Rs. 64, 800. It has an

economic life of 3 years and is expected to generate cash inflows a detailed
in table.
Year 1 2 3
Cash inflow 32000 28000 24000
Find the internal rate of return (IRR) for the project proposal
Solution
Initially, let us set a rate of return as 20% and calculate NPV
1
Rate = 20%, PVF =
(1+k)𝑛
1
Present value for 1st year cash inflow = 32000* = Rs. 26,656
(1+0.2)1
1
Present value for 2nd year cash inflow = 28000* = Rs. 19,432
(1+0.2)2
1
Present value for 3rd year cash inflow = 24000* = Rs. 13,896
(1+0.2)3
Total cash inflow = Rs. 59,984
NPV = 59984 - 64800 = - 4816
 Internal Rate of Return Method (IRR)
 NPV is negative, rate assumed is higher. Therefore, try for lower rates (16%
and 14%)
 NPV for 16% = - 1028, NPV for 14% = 996
 NPV = 0 lies between 16% and 14%
 Rate 15% - NPV is zero.
 The internal rate of return of the project is 15%
Payback period
 An investment project is accepted or rejected on the basis of
payback period.
 Period of time that a project requires to recover the money
invested in it.
 Unlike NPV and IRR, this method does not take into account the
time value of money.
 Payback period of a project is shorter than or equal to the
management’s maximum desired payback period – the project is
accepted, otherwise rejected.
Payback period – even cash flow
Same cash flow for every period,

*Incremental cash flow if an old asset (e.g., machine or equipment)

is replaced by a new one.
Example:
The ABC company is planning to purchase a machine known as machine X.
Machine X would cost $24,000 and would have a useful life of 10 years with zero
salvage value. The expected annual cash inflow of the machine is $11,000.
Compute payback period of machine X and conclude whether or not the machine
would be purchased if the maximum desired payback period of company is 3 years.
Solution:
Annual cash inflow is even in this project,
Payback period = $24,000/$11,000 = 2.18 years
purchase of machine X is desirable because its payback period is 2.18 years which
is shorter than the maximum payback period of the company.
 Payback period – even cash flow
Example
Due to increased demand, the management of XYZ Company is considering to
purchase a new equipment to increase the production and revenues. The useful life of
the equipment is 10 years and the company’s maximum desired payback period is 4
years. The inflow and outflow of cash associated with the new equipment is given
below:
Initial cost of equipment: $36,500
Annual cash inflows:
Sales: $70,000
Annual cash Outflows:
 Cost of ingredients: $42,000
 Salaries expenses: $12,500
 Maintenance expenses: $1,000
Non cash expenses:
Depreciation expense: $6,000
Use payback method and decide whether the company should purchase the new
equipment or not.
Payback period – even cash flow
Solution
 Computation of net annual cash inflow (by deducting the total of
cash outflow from the total of cash inflow associated with the
equipment)
$70,000 – ($42,000 + $12,500 + $1,000) = $14,500
 Payback period of the equipment
= $36,500/$14,500 =2.52 years
 Depreciation is a non-cash expense and has therefore been
ignored while calculating the payback period of the project.
 The equipment should be purchased because the payback period
of the equipment is 2.52 years which is shorter than the maximum
desired payback period of 4 years.
 Payback period – Advantages and Disadvantages
Advantages
 Investment project (with short payback period) promises - quick inflow of
cash - useful capital budgeting method for cash poor firms.
 Project with short payback period - improve the liquidity position of the
business quickly.
 Investment with short payback period - makes the funds available soon to
invest in another project.
 Short payback period reduces - risk of loss caused by changing economic
conditions and other unavoidable reasons.
 Payback period is very easy to compute.
Disadvantages
 Time value of money – not take into account.
 useful life of the assets and cash inflow after payback period - not considered.