2019-05-16

● Today:

● Pulse meeting 4.

● Short lecutre about investment analysis.

● Seminar 4 with Viktorija Badasjane.

● Next lectures, 2019-05-20:

● Change Management with Viktorija Badasjane and Staffan Andersson.

● Last lecture, 2019-05-23:

● Automation with Mikael Hedelind.

● Final presentation: On Monday 2019-05-27.

● Technical report deadline: Monday 2019-05-20, not later than 23:59 to san.aziz@mdh.se.

Technical report deadline:

Monday 2019-05-20, not later

than 23:59 to san.aziz@mdh.se.

Investment analysisSan Aziz

2019-05-16

Agenda

● Fundamental economy and investment analysis.

● Methods for investment analysis.

● LCC-calculations.

Fundamental economy

● Profit

● Revenues

● Costs

Profit = Revenues - Costs

Fundamental economy

Machine X

Capacity: 16 000 details/year

Selling price: 30:-/detail

Operator salary: 100 000:-/year

Operating costs: 150 000:-/year

Revenue = 16 000 x 30 = 480 000:-/year

Cost = 250 000:-/year

Yearly profit = 480 – 250 = 230 000:-/year

Fundamental production economy

Profit

Reduce cost Increase Revenue

Non-productive time Production cost Sales volume Sales price

• Minimize downtime

• Minimize rework

• Minimize scrap

• Minimize inventory

• Minimize overtime

• Minimize cycle times

• Maximize OEE

• Secure Supply chain

Methods for investment

analysis.

Methods for investment analysis

Annuity method

Net Present Value

Pay-back method

Annuity method

Annuity method

This method indicates how profitable an

investment is per year.

Annuity method

c

c = CA- xix(1+i)n

(1+i)n-1

S

(1+i)n[ ]

C = Total cost per year

CA= Acquisition cost

r = Yearly profit

i = Interest rate

s = Rest value

n = Economic life time

ccc

S

Y1 Y2 Y3 Yn

r1 r2 r3 rn

CA

Year

Kr

Acquisition cost

(Investment cost)

Cost per year

Profit per yearRest value

Annuity method

Example

Machine cost: 500 000:-

Capacity: 16 000 details/year

Selling price: 30:-/detail

Operator salary: 100 000:-/year

Operating costs: 150 000:-/year

Economic life: 5 years

Rest value: 50 000:-

Interest rate: 10%

Acquisition cost, CA = 500 000:-

Rest value, s = 50 000:-

Revenue = 16 000 x 30 = 480 000:-/year

Cost = 250 000:-/year

Yearly profit, r = 480 – 250 = 230 000:-/year

Interest rate, i = 10%

Economic life, n = 5 years

Annuity method

c = 500 - x = 124000: -/year0,1x1,15

1,15-1

50

(1+0,1)5[ ]

230000-124000 = 106 000:-/year (Profitable investment)

c = CA- xix(1+i)n

(1+i)n-1

S

(1+i)n[ ]

Yearly profit, r = 480 – 250 = 230 000:-/year

Net Present Value

Net Present Value

This method indicates how profitable an

investment is at present.

NPV = r +(1+i)n-1

ix(1+i)n

S

(1+i)n- CA

Net Present Value

Example

Machine cost: 500 000:-

Capacity: 16 000 details/year

Selling price: 30:-/detail

Operator salary: 100 000:-/year

Operating costs: 150 000:-/year

Economic life: 5 years

Rest value: 50 000:-

Interest rate: 10%

Acquisition cost, CA = 500 000:-

Rest value, s = 50 000:-

Revenue = 16 000 x 30 = 480 000:-/year

Cost = 250 000:-/year

Yearly profit, r = 480 – 250 = 230 000:-/year

Interest rate, i = 10%

Economic life, n = 5 years

NPV = 230 x + - 500 = 403 > 0, profitable investment at present.(1+0,1)5-1

0,1x(1+0,1)5

50

(1+0,1)5

NPV = r +(1+i)n-1

ix(1+i)n

S

(1+i)n- CA

Pay-back

Pay-back

After how long time an investment will pay

back invested money?

Pay-back

Y1 Y2 Y3 Yn

r1 r2 r3 rnr1

r2

r3

T = CA

r

CA

CA= Acquisition cost

r = yearly profit

T= Pay-back time

Acquisition cost

(Investment cost)

Profit per year

rn

Pay-back

Example

Machine cost: 500 000:-

Capacity: 16 000 details/year

Selling price: 30:-/detail

Operator salary: 100 000:-/year

Operating costs: 150 000:-/year

Economic life: 5 years

Rest value: 50 000:-

Interest rate: 10%

CA = 500 000:-

Rev = 16 000 x 30 = 480 000:-/year

Cost = 250 000:-/year

r = 480 – 250 = 230 000:-/year

T = = 2.2 years500

230

T = CA

r

Example from industry

Pay-back

Pay-back

Machine X

Theoretical output: 650 components/hour.

Actual output: 500 components/hour.

Loss is in production: 1:- /component.

Loss in production: 650 – 500 = 150 components/hour that is equal to 150: -.

Pay-back

Småstopp KB w603

0

100

200

300

400

500

600

700

800

900

7 8 4 6 14 5 2 1 21 13 9 17

Stegba

lk 18 12 16 22 3 10 11 15 19 20

Station

Ant

al

Short stops

Qu

an

tity

Station 7 causes about 38% of the speed loss in

the machine X.

0.38 x 150 components (from the slide before) =

57 components/hour

The cost of loss from station 7 is 57:-/h, because

the loss in production is 1:- /component (from the

slide before).

Pay-back

Station 7

Stops or rejects

Station 4

Long spring not in place

Short spring falls out

Short spring not in place

Moving contact not mounted

Counter cylinder to fast

Assembly method

Motor not running

Wrong measure on spring

Wrong measure on spring

Contact moves the spring

Contact get stuck in feeder

Station 3

These two causes 80% of the problems

in station 7.

0.80 x 57 (from the slide before) = 45:-/h

The loss is 45:-/h because of these two

causes.

A modification would cost about 50 000:-

Pay-backWeekly working hours: 120 h

Planning factor: 0.8

Availability: 0.85

Actual production time: 120 x 0.8 x 0.85 = 82 h

Improvement potential: 45:-/h (from ishikawa

diagram)

Investment cost: 50 000:- (from ishikawa diagram)

Pay-back: = 13.5 weeks50 000

3700

Potential savings: 45 x 82 = 3700: -/ week

LCC - calculations

It is about the total cost of the life of

an equipment, for example

maintenance cost.

LCC - calculations

LCC-calculations

Machine X Machine Y

Aquisition cost 98500 € 66000 €

Maintenance cost 12500 €/y ?

Life length 25 years 30 years

MTTF 420 h 300 h

MTTR 3 h 3,5 h

Cost of downtime 500 €/h 500 €/h

Operations cost 12500 €/y 14500 €/y

Operating hours 4300h/y 4300h/y

LCC ? 905640€

Time span for comparison: 20 years

How to calculate LCC for Machine X?

LCC = CA + tc(CO + CM + CDT)

CDT = Downtime costCA = Aquisition cost

tc = time of comparison

CO = Operations cost

CM = Maintenance cost

LCC-calculations

LCC-calculationsMachine X Machine Y

Aquisition cost, CA 98500 € 66000 €

Maintenance cost, CM 12500 €/y ?

Life length 25 years 30 years

MTTF 420 h 300 h

MTTR 3 h 3,5 h

Cost of downtime, CDT 500 €/h 500 €/h

Operations cost, CO 12500 €/y 14500 €/y

Operating hours 4300h/y 4300h/y

LCC ? 905640€

LCCX = 98500 + 20(12500 + 12500 + 15357) = 905640€

LCC = CA + tc(CO + CM + CDT)

CDT(X) =MTTR x Cost of Downtime x Operating hours

MTTF=3𝑥500𝑥4300

420= 15357€/y

Time span for comparison, tc: 20 years

LCC-calculations

Machine X Machine Y

Aquisition cost, CA 98500 € 66000 €

Maintenance cost , CM 12500 €/y ?

Life length 25 years 30 years

MTTF 420 h 300 h

MTTR 3 h 3.5 h

Cost of downtime , CDT 500 €/h 500 €/h

Operations cost , CO 12500 €/y 14500 €/y

Operating hours 4300h/y 4300h/y

LCC ? 905640€

Time span for comparison, tc : 20 years

How to calculate maintenance cost, CM for Machine Y ?

LCC-calculations

Machine X Machine Y

Aquisition cost, CA 98500 € 66000 €

Maintenance cost , CM 12500 €/y ?

Life length 25 years 30 years

MTTF 420 h 300 h

MTTR 3 h 3.5 h

Cost of downtime , CDT 500 €/h 500 €/h

Operations cost , CO 12500 €/y 14500 €/y

Operating hours 4300h/y 4300h/y

LCC ? 905640€

Time span for comparison, tc : 20 years

LCCY = 66000 + 20(14500 + CM + 25083) = 905640€

CDT(Y)= 3.5x500x4300/300 = 25083 €/y

CM = ((905640-66000)/20) – 14500 – 25083 = 2399 €/y

LCC = CA + tc(CO + CM + CDT)

Remember…

LCC - calculations

MTTF = MTBF