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Mechanical Engineering Design Project MECH 390 Lecture 7

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Page 1: Mechanical Engineering Design Project - Concordia …users.encs.concordia.ca/~nrskumar/Index_files/Mech390/Lecture... · Mechanical Engineering Design Project ... 2/27/2017 Chapter

Mechanical Engineering Design Project

MECH 390

Lecture 7

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3/2/2018 Chapter 7-Project Planning 2

Activity Duration Preceded by

A 4 —

B 5 —

C 4 A

D 5 A

E 6 A

F 4 D,C

G 7 F,B

H 4 G,E

Another Example of Constructing a Network Diagram*

Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

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3/2/2018 Chapter 7-Project Planning 3

Example Problem of Finding the Critical Path*Path Duration

A-E-H 14

A-C-F-G-H 23

A-D-dummy-F-G-H 24

B-G-H 16

Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

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3/2/2018 Chapter 7-Project Planning 4

Alternate Method for Determining Critical Path• This approach is more efficient for larger networks

• Use forward sweep to find earliest start (ES) for each activity

• Use backward sweep to find latest start (LS) for each activity

• Calculate total float (TF) for each activity

TF = LS - ES

• Critical Path consists of Activities with TF = 0

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3/2/2018 Chapter 7-Project Planning 5

Earliest Start*

• Earliest Start (ES) is the earliest time an activity can start. It is found by tracing forward (from tail to head of each activity arrow) from the project Start event to the tail of the selected activity. When several paths are possible, use the longest path as determined by the sum of the activity durations on that path.

• For activity F, ES = 9

Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

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3/2/2018 Chapter 7-Project Planning 6

Project Duration*

• Continue until we have ES for all activities that terminate in the project Finish event. When duration of each of those activities are added to their respective EStimes, the largest of the resulting sums is defined as the Project Duration.

• For activity H, ES=20

• Project Duration =20+4=24Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

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3/2/2018 Chapter 7-Project Planning 7

Latest Start*

• Latest Start (LS) is the latest time an activity can start and still have the project completed within the Project Duration time.

• LS is found by tracing backwards (from head to tail of each activity) from the project Finish event to the tail of selected activity. Make sure you reach the tail of the selected activity via the head of that activity. When several paths are possible, use the longest path as determined by the sum of the activity durations on that path. The Project Duration minus the length of this longest path is the LS for the selected activity.

• For activity A, LS=24-24=0

Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

Page 8: Mechanical Engineering Design Project - Concordia …users.encs.concordia.ca/~nrskumar/Index_files/Mech390/Lecture... · Mechanical Engineering Design Project ... 2/27/2017 Chapter

3/2/2018 Chapter 7-Project Planning 8

Total Float

• Total Float for each activity is the difference between the latest start and the earliest start.

TF = LS - ES

• The activities for which TF=0 define the critical path.

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3/2/2018 Chapter 7-Project Planning 9

Summary of Total Float Calculations*

• Critical Path consists of A-D-F-G-H

Activity Duration ES LS TF

A 4 0 0 0

B 5 0 8 8

C 4 4 5 1

D 5 4 4 0

E 6 4 14 10

F 4 9 9 0

G 7 13 13 0

H 4 20 20 0

Project Duration 24 24

Start

Finish

A

4

C 4

F 4

H

4

G

7

D 5

E 6

B

5

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3/2/2018 Chapter 7-Project Planning 10

Total Float Calculations

Activity Duration Earliest Start Latest Start Total Float

A 3 0 0 0

B 3 3 3 0

C 4 0 1 1

D 1 4 5 1

E 3 6 13 7

F 2 6 6 0

G 2 8 8 0

H 4 10 10 0

I 1 4 13 9

J 3 10 16 6

K 5 14 14 0

Project Duration 19

For project shown in Fig. 7.12

Start

A3

Finish

B

3

E

3

H 4

G

2

I

1

J3

K5

F2

D1

C4

Thus, A-B-F-G-H-K is the critical pathStart

A3

Finish

B

3

E

3

H 4

G

2

I

1

J3

K5

F2

D1

C4

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3/2/2018 Chapter 7-Project Planning 11

Example of Critical Path Determination

A

1

B2

D 4H

5

E6

G

1

L

3

M

4

O

7F7

J 2

K 5I

10 N6

C3

• Identify critical path for the diagram

• Construct the table for ES and LS for each activity

• The ones with TF as 0 will form the critical path

• A,B,D,E,F,G,L,M

Activity ES LS TF

A 0 0 0

B 1 1 0

C 1 4 3

D 3 3 0

E 7 7 0

F 7 7 0

G 13 13 0

H 3 8 5

I 3 12 9

J 14 15 1

K 16 17 1

L 14 14 0

M 17 17 0

N 21 22 1

O 21 21 0

Project

Duration

28

A

1

B2

D 4

H5

E6

G

1

L

3

M

4

O

7

F

7

J 2

K 5I

10 N6

C3

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3/2/2018 Chapter 7-Project Planning 12

• Based on Critical Path Method

• Replaces single estimate of activity duration by

a probability distribution

• Allows estimate of probability of completing

project by a specified time

Program Evaluation and Review Technique

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3/2/2018 Chapter 7-Project Planning 13

Typical Beta Distributions for Activity Durations• to-optimistic estimate; the shortest time within which this activity can be completed

assuming everything goes right. This is the left terminus of the pdf.

• tm-the most likely time required to complete the activity. This is the mode of the pdf.

• tp- pessimistic estimate; the longest time it will take this activity to be completed assuming everything goes wrong. This is the right terminus of the pdf.

to

tm t

p

(a) Skewed-left Beta distribution

to

tm t

p

(b) Skewed-right Beta distribution

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3/2/2018 Chapter 7-Project Planning 14

PERT Procedure

• Calculate the expected duration te (mean) of each activity

• Calculate the variance s2 of each activity

• Use te to determine the expected project duration Te and identify the critical path

6

4 pmo

e

tttt

2

2

6

op tts

Start

A1-3

-3

Finish

B

2-3-4

E

2-3-6

G

1-2-4

I

0-1-3

J1-3-6

K3-

5-12

F

1-2

-7

D1-

1-5

C3-4-10

H

3-4

-6

• To, tm, tp, for each activity are shown in the diagram

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3/2/2018 Chapter 7-Project Planning 15

Determination of Critical Path for PERT Example Calculation

Activity Expected

Time

Variance Earliest

Start

Latest

Start

Total

Float

A 3.00 0.44 0.00 0.50 0.50

B 3.00 0.11 3.00 3.50 0.50

C 4.83 1.36 0.00 0.00 0.00

D 1.67 0.44 4.83 4.83 0.00

E 3.33 0.44 6.50 14.84 8.34

F 2.67 1.00 6.50 6.50 0.00

G 2.17 0.25 9.17 9.17 0.00

H 4.17 0.25 11.34 11.34 0.00

I 1.17 0.25 4.83 14.34 9.51

J 3.17 0.69 11.34 18.17 6.83

K 5.83 2.25 15.51 15.51 0.00

Project

Duration

21.34

Table 7.5

Start

A1-3

-3

Finish

B

2-3-4

E

2-3-6

G

1-2-4

I

0-1-3

J1-3-6

K3-

5-12

F

1-2

-7

D1 -

1-5

C3-4-10

H

3-4

-6

• Critical path is C,D,F,G,H,K

• This is different from the one we got with CPM

• Te = 21.34

• T for project and t for activity

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3/2/2018 Chapter 7-Project Planning 16

Probability of Completing Project by Specified time Ts

• Calculate the variance of the project duration as the sum of the variances of the activities on the critical path

• Use the standard normal variable z to find the probability of completing the project in a specified time Ts

• For Ts=20 From Table 5.1 Pr (z < -0.57) = 0.285 (28.5% chance)

T

esS

TTz

s

362

001250250001440361

222222

.

......

)pathcritical(

T

KHGFDCT

s

sssssss

570362

342120.

.

.TTz

T

esS

s

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3/2/2018 Chapter 7-Project Planning 17

Activities for Space Station PERT Problem

Activity Description

A Construct shell of module

B Order life support system and scientific

experimentation package from same supplier

C Order components of control and navigational system

D Wire module

E Assemble control and navigational system

F Preliminary test of life support system

G Install life support in module

H Install scientific experimentation package in module

I Preliminary test of control and navigational system in

module

J Install control and navigational system in module

K Final testing and debugging

Start

A25

-30-

45

F

1-1-1

G4-5-7

E

5-7-12

C20-25-35

B

10-15-20

D

3-3-5

H2-2

-3

J8-

10-

14

I

4-4-6

K

6-8-15Finish

• Given table and pert diagram for space station, find the probability of completing the space station in over 60 days

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3/2/2018 Chapter 7-Project Planning 18

Summary of PERT Calculations for Space Station Project

Activity te s2 ES LS TF

A 31.67 11.11 0.00 0.50 0.50

B 15.00 2.78 0.00 19.50 19.50

C 25.83 6.25 0.00 0.00 0.00

D 3.33 0.11 31.67 32.17 0.50

E 7.50 1.36 25.83 25.83 0.00

F 1.00 0.00 15.00 34.50 19.50

G 5.17 0.25 35.00 42.83 7.83

H 2.17 0.03 35.00 35.50 0.50

I 4.33 0.11 33.33 33.33 0.00

J 10.33 1.00 37.67 37.67 0.00

K 8.83 2.25 48.00 48.00 0.00

Project Duration 56.83

• Calculate the variance of the project duration as the sum of the variances of the activities on the critical path

σt = σC2 + σE

2 + σI2 + σJ

2 + σK2

= (6.5+1.36+.11+1.0+2.25)1/2 = 3.31

• Use the standard normal variable z to find the probability of completing the project in a specified time Ts

• For Ts=60

(60-56.83)/3.31

= 0.96

• From Table 5.1

Pr (TS >60) = 1 - Pr (z < 0.96) = 1 - 0.830 = 0.170 or 17%

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3/2/2018 Chapter 7-Project Planning 19

Another PERT Example Problem*

Activity te s2 ES LS TF

A 4.32 1.00 0 0 0

B 5.40 1.56 0 8.64 8.64

C 4.32 1.00 4.32 5.40 1.08

D 5.40 1.56 4.32 4.32 0

E 6.48 2.25 4.32 15.12 10.80

F 4.32 1.00 9.72 9.72 0

G 7.56 3.06 14.04 14.04 0

H 4.32 1.00 21.60 21.60 0

Project Duration 25.92

Start

Finish

A

2-4-8

C

2-4

-8

F2-

4-8

H

2-4-8G

3.5-7-14

D2.

5-5-

10

E 3-6-12

B

2.5-5-10

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3/2/2018 Chapter 7-Project Planning 20

Probability of Completing Project by Specified time Ts*

• Calculate the variance of the project duration as the sum of the variances of the activities on the critical path

• Use the standard normal variable z to find the probability of completing the project in a specified time Ts

• For Ts=30

• From Table 5.1

Pr (z < 1.48) = 1 - Pr (z < -1.48) = 1 - 0.067 = 0.932

T

es TTz

s

762001063001561001

22222

......

HGFDAT

ssssss

481762

922530.

.

.TTz

T

es

s

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3/2/2018 Chapter 7-Project Planning 21

Calendarized Version

3

C4

2

A3

Start 1 D1

I

1

B

34

E

3

5G

2

8

6

J3

7

K5

9 FinishH4

F2

• CPM is good for identifying the critical path

• It does not display time based relationship like Gantt chart

• Calendarizing is done by chronoligcally numbering the nodes

• Length of activity proportional to time

• All activities (except dummy) are horizontal

• Horizontal vertical dashed line to enable connection

• Between J and 9 or D and 4

• Dummy activity vertical

• (between 6 and 8)

• Nodes from which many activity emerge will be split

• Nodes 1, 3, and 4 are split

• Time omitted on activity but complimented by an master duration scale

3

C

2A

Start 1

D

I

B

4E

5G

8

6

J

7K

9 FinishHF

dummy

0 62 4 8 10 12 14 16 18 20

Start 1 3

4

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3/2/2018 Chapter 7-Project Planning 22

Closure

• Purpose of planning

• Gantt chart

• PERT and CPM

• Calenderized chart combines the benefit of CPM and Gantt

• Commercial software packages available for these

• A plan is just a plan

• Gives an overall tentative estimate of what we think might happen

• So it is always good to review once in a while and update

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3/2/2018 Chapter 8-Engineering Economics 23

Definition of an Engineer

An engineer is someone who can do for $1 what any fool can do for $2.

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3/2/2018 Chapter 8-Engineering Economics 24

Fundamental Problem of Engineering Economics• Trade offs between initial costs and downstream costs

• You are to decide between 2 motors, Ajax and Baylock

• There are difference in initial cost

• Annual maintenance cost

• At the end of 3 years, retrofitting required for baylockmotors

• Ajax is energy efficient at 3000/year while baylockrequires 3500/year

• Electricity charge increases at 5%/year

• Ajax motors are more reliable and conveyors will not have problems so production is 500/year higher than baylock motors

• Additionaly at the end of five years, ajax motors have a salvage value of 4000, while baylock none

• Which motors you will use

• Four Decision making methods

1. Low initial Cost approach (straight away choose baylock motors because it has the lowest initial cost

2. Low Life cycle cost approach

3. Average annual rate of return

4. Payback period

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3/2/2018 Chapter 8-Engineering Economics 25

Decision Rules Ajax Blaylock

($1000) ($1000)

Initial Cost 30.00 20.00

Rebuilding at End of 3rd Year - 3.00

Salvage Value -4.00 -

Maintenance (5 years) 5.00 10.00

Productivity Benefit (5 years) -2.50 -

Electricity

year 1 3.00 3.50

year 2 3.15 3.68

year 3 3.31 3.86

year 4 3.47 4.05

year 5 3.65 4.25

Total Costs 45.08 52.34

• Lowest initial cost rule

• Ajax = 30.00; Blaylock = 20.00;

• Choose Blaylock since it has the least first cost

• Total Life-Cycle Cost Decision Rule

• Ajax = 45.08; Blaylock = 52.34

• Choose Ajax since it has the least total life-cycle cost

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3/2/2018 Chapter 8-Engineering Economics 26

Annual Rate of Return Decision Rule• Incremental Investment = Initial Cost of More Expensive Option - Initial Cost of Less Expensive

Option = 30.00 - 20.00 = 10.00

• Downstream Benefits = Avoidance of Rebuilding Cost + Income from Salvage + 5 Years Savings on Maintenance Costs + Productivity Enhancement + 5 Years Savings on Electricity Costs =

3.00 + 4.00 + (10.00 - 5.00) + 2.50 + (3.50 - 3.00) + (3.68 - 3.15) + (3.86 - 3.31) + (4.05 - 3.47) + (4.25 -3.65)= 17.26

• Average Annual Benefit = Downstream Benefits/Project Duration = 17.26/5 = 3.45

• Annual Rate of Return = [Annual Benefits / Incremental Investment] 100 = 34.5 %.

• Choose Ajax unless another opportunity for investing $10,000 provides a greater rate of return.

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3/2/2018 Chapter 8-Engineering Economics 27

PaybackPeriod (time to recover incremental investment)

• Benefits in 1st Year = 1.00 + 0.50 + 0.50 = 2.00

(maint) + (prod) + (elec)

• Total Benefits in 2 Years

= 2.00 + 1.00 +0.50 +0.53 = 4.03

(yr 1) + (maint) + (prod) + (elec)

• Total Benefits in 3 Years

= 4.03 + 1.00 +0.50 +0.55 + 3.0 = 9.05

r 2) + (maint) + (prod) + (elec)+ (rebuild)

• Total Benefits in 4 Years

= 9.05 + 1.00 +0.50 +0.58 = 11.13

(yr 3) + (maint) + (prod) + (elec)

• The payback period is 3.5 years. Select Ajax unless another opportunity to invest the $10,000 provides a quicker payback.

Ajax Blaylock

($1000) ($1000)

Initial Cost 30.00 20.00

Rebuilding at End of 3rd Year - 3.00

Salvage Value -4.00 -

Maintenance (5 years) 5.00 10.00

Productivity Benefit (5 years) -2.50 -

Electricity

year 1 3.00 3.50

year 2 3.15 3.68

year 3 3.31 3.86

year 4 3.47 4.05

year 5 3.65 4.25

Total Costs 45.08 52.34

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3/2/2018 Chapter 8-Engineering Economics 28

Time Value of Money• How much would I have to give you one year from now to make that payment equally

attractive as a $100 payment today?

• That premium represents the interest rate (i) you could earn on the best investment available to you.

• Compound Interest Formula

• P=present value of a transaction

• F=future value of the same transaction after n years

• i=interest rate

• Present Value Formula

• P=present value of a transaction

• F=future value of the same transaction

• i=discount rate

• T=Present Worth Factor

F P in

1

P F in

1 T iFP i nn

, ,

1

P FTFP i n , ,

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3/2/2018 Chapter 8-Engineering Economics 29

Uniform Series Present Worth Factor

Ti

i iUP i n

n

n, ,

1 1

1

TUP, . ,

.

. .

.

0 20 5

5

5

1 0 20 1

0 20 1 0 20

2 99

• Maintenance and productivity improvement occur at each year and rebuilding and salvage costs at once in five years

• We can either add these cost year by year (maint and prod) or simplify it using a formula to calculate the overall

Ajax Blaylock

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Initial Cost 30.00 20.00

Maintenance

TUP, . , .0 20 5 2 99

1.00 2.99 2.00 5.98

Total Present Value Of Costs 32.99 30.46

Select Blaylock since the Net Present Value is lower than for Ajax.

Note that a single calculation using TUP gives the same result as five

calculations using TFP.

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3/2/2018 Chapter 8-Engineering Economics 30

Present Value Calculation ($1,000)*

• Use discount rate i=0.20

Select Blaylock since the Net Present Value is lower than for Ajax.

Ajax Blaylock

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Initial Cost 30.00 20.00

Maintenance in 1st year

TFP, . , .0 20 1 0833

1.00 0.83

Maintenance in 2nd year

TFP, . , .0 20 2 0 694

1.00 0.69

Maintenance in 3rd year

TFP, . , .0 20 3 0577

1.00 0.58

Maintenance in 4th year

TFP, . , .0 20 4 0 482

1.00 0.48

Maintenance in 5th year

TFP, . , .0 20 5 0 403

1.00 0.40

Total Present Value Of Costs 32.98

Ti

i iUP i n

n

n, ,

1 1

1

T iFP i nn

, ,

1

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3/2/2018 Chapter 8-Engineering Economics 31

Example 8.3.1• A company is faced with the decision of selecting either machine A or B, both of which are capable of doing

the same job. Machine A costs $1,250 with annual maintenance and operating charges of $150 for the first ten years and $180 for the following ten years. Machine B costs $1,500 and annual operating and maintenance expenses are $100 for the 20-year period. Both machines have zero salvage value at the end of their economic lives. By how much should the manufacturer of machine A increase or decrease the purchase price of the machine so that it is competitive with machine B at a discount rate of 15 percent?

Machine A Machine B

Future/Annual

Value

Present

Value

Future/Annual

Value

Present

Value

Capital Cost 1250 1500

Annual Operating Cost

(years 1-10)

TUP,0.15,10 = 5.02 150 753

Annual Operating Cost

(years 11-20)

TUP,0.15,10 = 5.02

TFP,0.15,10 = 0.247

180 223

TUP,0.15,10TFP,0.15,10 = 1.24

Annual Operating Cost

(years 1-20)

TUP,0.15,20 = 6.26 100 626

Total present value of costs 2226 2126

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Geometric Series Present Worth Factor• If annual costs are increasing at a constant annual rate e over the period n, the present value of

the entire stream is found from

• If costs are decreasing at a constant annual rate, e will be negative

• TGP is multiplied by the cost G during the first year to obtain the present value of the entire stream.

Ti e

i e iGP i e n

n n

n, , ,

1 1

1

TGP, . , . ,

. .

. . ..0 20 0 05 5

5 5

5

1 0 20 1 0 05

0 20 0 05 1 0 203 25

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3/2/2018 Chapter 8-Engineering Economics 33

Present Value Calculation Including TGP ($1,000)• Use discount rate i=0.20

Select Blaylock since the Net Present Value is lower than for Ajax.

Ajax Blaylock

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Initial Cost 30.00 20.00

Rebuilding at End of 3rd year

TFP, . , .0 20 3 0577

- - 3.00 1.73

Salvage Value

TFP, . , .0 20 5 0 403

-4.00 -1.61 - -

Maintenance

TUP, . , .0 20 5 2 99

1.00 2.99 2.00 5.98

Productivity Benefit

TUP, . , .0 20 5 2 99

-0.50 -1.50 - -

Electricity

TGP, . , . , .0 20 0 05 5 325

3.00 9.75 3.50 11.37

Total Present Value Of Costs 39.63 39.08

Ti e

i e iGP i e n

n n

n, , ,

1 1

1

Ti

i iUP i n

n

n, ,

1 1

1

T iFP i nn

, ,

1

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3/2/2018 Chapter 8-Engineering Economics 34

Annualized Cost Analysis• Instead of converting all costs to

their equivalent present value, we could instead convert all costs to their equivalent annual value.

• Although the numerical results of the two methods will be different, the least cost alternative using the present value method will also be the least cost alternative using the annualized cost method.

• Capital Recovery Factor

• Sinking Fund Factor

• Generalized Sinking Fund Factor

• Geometric Series Sinking Fund Factor

T

T

i i

iPU i n

UP i n

n

n, ,, ,

1 1

1 1

T

i

iFU i n n, ,

1 1

Ti i e

i i eGU i n

n n

n, ,

1 1

1 1

T

i i

iFU i n

n m

n, ,

1

1 1

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Annualized Cost Analysis Ajax Blaylock

Future/

Present

Value

Annual

Value

Future/

Present

Value

Annual

Value

Initial Cost

TPU , . , .0 2 5 0 334

30.00 10.02 20.00 6.68

Rebuilding at End of 3rd year

TFU , . , , .0 2 5 3 0194

- - 3.00 0.58

Salvage Value

TFU , . , .0 2 5 0134

-4.00 -0.54 - -

Maintenance 1.00 2.00

Productivity Benefit -0.50 - -

Electricity

TGU , . , . , .0 2 0 05 5 1086

3.00 3.26 3.50 3.80

Total Equivalent Annual Costs 13.24 13.06

• Capital Recovery Factor

Occur initially (capital)

• Sinking Fund Factor

Occurs downstream (recovery)

• Generalized Sinking Fund Factor

Occurs downstream (rebuilding where m=3)

• Geometric Series Sinking Fund Factor

Geometrically increasing transaction (electricity)

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Variations to Basic Engineering Economics Problem

• Unequal lifetimes

• Effect of taxes

• Non-uniform downstream transactions

• Accounting for inflation

• Downstream transactions occurring other than at end of year

• Long project lifetimes

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Example of Unequal Lifetimes Problem

Ajax (n=6) Blaylock (n=4)

($1000) ($1000)

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Initial cost - 30.00 - 20.00

Salvage value

33506200 .T ,.,FP

-4.00 -1.34 - -

Maintenance

32636200 .T ,.,UP

1.00 3.33 - -

58924200 .T ,.,UP - - 2.00 5.18

Electricity

67536050200 .T ,.,.,GP

3.00 11.02 - -

75924050200 .T ,.,.,GP - - 3.50 9.66

Total present value of costs 43.01 34.84

• Modified Ajax/Blaylock problem• n(Ajax)=6 n(Blaylock)=4• no rebuilding• no productivity benefit

• Table shows the regular calculation of the decision making process• However if you look at ajax electricity Vs

blaylock electricity, Ajax is more expensive eventhough it consumes less power

• Because it has a longer life• This is unfair

• Alternatively we can do the process for 4 years• But this will be against Ajax as well

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Three Approaches to Dealing with Unequal Lifetimes

• Multiple Lifetimes

• Annualized Costs

• Conversion of Present Values

T

T

i i

iPU i n

UP i n

n

n, ,, ,

1 1

1 1

T

i

iFU i n n, ,

1 1

Ti i e

i i eGU i n

n n

n, ,

1 1

1 1

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Present Value Calculation for Common Multiple of Lifetimes (n=12)

• Modified Ajax/Blaylock problem• n(Ajax)=6 n(Blaylock)=4

• no rebuilding

• no productivity benefit

• Multiple Lifetimes• Select a analysis period common to lifetime

of both.

• In this case, 4 and 6 years, common term will be 12

• Buy Blayblock at end of 4 and 8 years

• Buy Ajax and end of 6 years

• At 12 year period, Ajax is the better choice

Ajax ($1000) Blaylock ($1000)

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Capital Cost (at n=0) 30.00 20.00

Capita Cost (at n1=4)

( TFP, . , .0 2 4 0482 )

20.00 9.64

Capital Cost (at n2=6)

( TFP, . , .0 2 6 0335 )

30.00

10.05

Capital Cost (at n3=8)

( TFP, . , .0 2 8 0233 )

20.00

4.65

Salvage Value (at n2=6)

( TFP, . , .0 2 6 0335 )

-4.00 -1.34 -

Salvage Value (at n4=12)

( TFP, . , .0 2 12 0112 )

-4.00 -0.45

Maintenance

( TUP, . , .0 2 12 4 439 )

1.00 4.44 2.00 8.88

Electricity

( TGP, . , . , .0 2 0 0512 5324 )

3.00 15.97 3.50 18.63

Total Present Value Of Costs 58.67 61.80

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Annualized Value Calculation for Unequal Lifetimes

Ajax (n=6) Blaylock (n=4)

($1000) ($1000)

Future/

Present

Value

Annual

Value

Future/

Present

Value

Annual

Value

InitialCost

( 3010620 .T ,.,PU )

( 3860420 .T ,.,PU )

30.00

9.02

20.00

7.73

Salvage Value

( 1010620 .T ,.,FU )

-4.00 -0.40 -

Maintenance 1.00 2.00

Electricity

( 1051605020 .T ,.,.,GU )

( 0661405020 .T ,.,.,GU )

3.00

3.32

3.50

3.73

Total Present Value Of Costs 12.94 13.46

• Modified Ajax/Blaylock problem• n(Ajax)=6 n(Blaylock)=4

• no rebuilding

• no productivity benefit

• Annualized Costs• All costs expressed as equivalent annual

present value costs

• Inverse factors of TUP, TFP, TGP calculated

• No special treatment needed for difference in lifetime

• Buy Ajax and end of 6 years

• At 12 year period, Ajax is the better choice

T

T

i i

iPU i n

UP i n

n

n, ,, ,

1 1

1 1

Ti

iFU i n n, ,

1 1

Ti i e

i i eGU i n

n n

n, ,

1 1

1 1

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Conversion of Present Values*

• Third approach is to Convert any Present Value for period n1 into an equivalent Present Value based on another period n2 using

• Convert P(Ajax) from n1=6 to n2=4 and compare to P(Blaylock)

• Since P4 (Ajax) < P4(Blaylock), choose Ajax

• We can also do the same for Blayblock for 6 years and compare as well

2211 n,i,PUnn,i,PUn TPTP

48.33589.293.12)(

93.12301.001.43)(

4,2.0,4

6,2.0,6

XTAjaxP

XTAjaxP

UP

PU

Ajax (n=6) Blaylock (n=4)

($1000) ($1000)

Future/

Annual

Value

Present

Value

Future/

Annual

Value

Present

Value

Initial cost - 30.00 - 20.00

Salvage value

33506200 .T ,.,FP

-4.00 -1.34 - -

Maintenance

32636200 .T ,.,UP

1.00 3.33 - -

58924200 .T ,.,UP - - 2.00 5.18

Electricity

67536050200 .T ,.,.,GP

3.00 11.02 - -

75924050200 .T ,.,.,GP - - 3.50 9.66

Total present value of costs 43.01 34.84

T

T

i i

iPU i n

UP i n

n

n, ,, ,

1 1

1 1

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The Effect of Taxes

TotalIncome

OperatingExpenses

Depreciation

Taxable Income

Taxes

• Corporation subject to income taxes and type of expense have an effect on the tax rate (mortgage Vs.rental)

• So economic attraction of design options may be influenced

• Figure shows total income on the big block and companies can take deductions on the said income• Operating expenses (payroll, electricity, utility bills etc

• Depreciation (industrial equipment loosing value over time)

• There are different methods of calculating depreciation

• Taxable income is total income – the detections (operating + depreciation)

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MACRS Depreciation Schedule Depreciation Period (years)

Year 3 5 7 10 15

1 0.333 0.200 0.143 0.100 0.050

2 0.445 0.320 0.245 0.180 0.095

3 0.148 0.192 0.175 0.144 0.095

4 0.074 0.115 0.125 0.115 0.077

5 0.115 0.089 0.092 0.069

6 0.058 0.089 0.074 0.062

7 0.089 0.066 0.059

8 0.045 0.066 0.059

9 0.065 0.059

10 0.065 0.059

11 0.033 0.059

12-15 0.059

16 0.030

• Depreciation has two components

• Telephone (15 years) computer (5 years)

• Modified Accelerated Cost Recovery System

• % of initial investment that can be depreciated using MARCS shown in table

• Assumed as investments made in the middle of the year, so half year at first and half year after the depreciation period is over

• The total for each adds up to one (means, entire investment is depreciated at the end of the period, even though there might be some salvage value remaining

• When you consider the difference between initial investment and salvage value and divide it over the depreciation period, this is called the Straight line approach

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Marginal Tax Rates

Taxable Income Marginal Tax Rates

< 50,000 0.15

50,000 - 75,000 0.25

75,000 - 100,000 0.34

100,000 - 335,000 0.39

335,000 - 10,000,000 0.34

10,000,000 - 15,000,000 0.35

15,000,000 - 18,333,333 0.38

>18,333,333 0.35

• Once we know the depreciation that we can get for that year, we have to apply the tax reate to the allowable depreciation

• This will determine the tax implications of the capital cost of various design alternatives

• Marginal tax rates shown in table

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Including Income Tax Effects in Economic Analysis(assume 34% tax bracket and MACRS depreciation schedule with n=5)

Ajax Blaylock

Fut./Ann.Val

.

Pres.

Val.

Fut,/Ann.Val

.

Pres.

Val.

Initial Cost 30.00 30.00 20.00 20.00

Rebuilding at End of 3rd year (TFP,0.2,3

= 0.577)

- - 3.00 1.73

Salvage Value (TFP,0.2,5 = 0.403) -4.00 -1.61 - -

Maintenance (TUP,0.2,5 = 2.99) 1.00 2.99 2.00 5.98

Productivity Benefit (TUP,0.2,5 = 2.99) -0.50 -1.50 - -

Electricity (TGP,0.2,0.05,5 = 3.25) 3.00 9.75 3.50 11.37

Depreciation

1st year @ 0.200 -6.00 -4.00

2nd year @ 0.320 -9.60 -6.40

3rd year @ 0.192 -5.76 -3.84

4th year @ 0.115 -3.45 -2.30

5th year @ 0.115 -3.45 -2.30

6th year @ 0.058 -1.74 -1.16

Tax Benefits from Deprec. @ 0.34

1st year(TFP,0.2,1=0.833) -2.04 -1.70 -1.36 -1.13

2nd year (TFP,0.2,2=0.694) -3.26 -2.26 -2.18 -1.51

3rd year (TFP,0.2,3=0.579) -1.96 -1.13 -1.31 -0.76

4th year (TFP,0.2,4=0.482) -1.17 -0.56 -0.78 -0.38

5th year (TFP,0.2,5=0.402) -1.17 -0.47 -0.78 -0.31

6th year (TFP,0.2,6=0.335) -0.59 -0.20 -0.39 -0.13

Tax Benefits from Maint.+Rebuild.+Elec.

@ 0.34

-4.33 -6.49

Tax Burdens from Product.+ Salv. @ 0.34 1.06

Total Present Value Of Costs 30.04 28.37

Depreciation for the Ajax motors is based on neglecting salvage value. Assume no depreciation allowed for

rebuilding Blaylock motors.

• Top half of the table is exactly the same as before where we calculate present cost

• % calculated using MACRS at 5 years

• assumed company income is 500000 usd (so 34% tax rate)

• Tax benefit of Depreciation is negative

• Maintenance, rebuild and electricity are operating expenses so direct apply 34% negative again (tax benefit)

• Production and salvage cost incur payment of taxes so hence they are kept positive

• Considering the taxes, Blayblock is the better alternative

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Example 8.7.1

• You recommend spending $10,000 on equipment that will increase sales of your product by $1,000 a year and reduce annual operating costs by $800. The equipment has a 10 year lifetime but can be depreciated in three years according to MACRS. Your company uses a 10% discount rate and is in the 25% tax bracket.

• What is the pretax present value of this investment?

• What is the after-tax present value of this investment?

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Future/Annual Present

Initial Cost -10,000

Additional annual income TUP,0.1,10 = 6.145 1,000 6,145

Annual savings in operating costs TUP,0.1,10 = 6.145 800 4,916

Pre-tax present value 1,061

Additional tax on additional income rate = 0.25 -1,536

Lost tax deduction on operating costs rate = 0.25 -1,229

Depreciation in year 1 on initial cost 0.333 3,330

Depreciation in year 2 on initial cost 0.445 4,450

Depreciation in year 3 on initial cost 0.148 1,480

Depreciation in year 4 on initial cost 0.074 740

Tax benefits of depreciation in year 1 rate = 0.25 833

Tax benefits of depreciation in year 2 rate = 0.25 1,113

Tax benefits of depreciation in year 3 rate = 0.25 370

Tax benefits of depreciation in year 4 rate = 0.25 185

Discounted tax benefits of depr. in year 1 TFP,0.1,1 = 0.909 757

Discounted tax benefits of depr. in year 2 TFP,0.1,2 = 0.826 919

Discounted tax benefits of depr. in year 3 TFP,0.1,3 = 0.751 278

Discounted tax benefits of depr. in year 4 TFP,0.1,1 = 0.683 126

After-Tax Present Value 376

• Pretax calculations show 10 years of additional income and reductions in operating costs when discounted at 10%

• Pretax present value is 1061

• Tax on additional income (6145*.25)

• Tax loss due to reduction in operating cost (4916*0.25)

• Since happen each year, applied directly to present cost

• Depreciation from MACRS table and tax benefit from depreciation

• These values converted to present value using TFP rule

• This gives the after tax present value as 376

Example 8.7.1

Ti

i iUP i n

n

n, ,

1 1

1 T iFP i n

n, ,

1

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Accounting for Inflation

Entity Units 1990 amount 2000 amount

Textbook Nominal (current) dollars $30 $60

Hourly Wage Nominal (current) dollars $5 $10

Textbook Real (constant) 1990 dollars $30 $30

Hourly Wage Real (constant) 1990 dollars $5 $5

Textbook Real (constant) 2000 dollars $60 $60

Hourly Wage Real (constant) 2000 dollars $10 $10

Consider a product whose real cost has just kept pace with inflation. There are

several ways to describe its cost.

Regardless of the way that is looked at, a person has to work 6 hours to get the

book, both in 1990 and 2000

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Determining the Real Discount Rate• The present value of a future amount F must be the same, regardless

of whether F is expressed in terms of nominal dollars or real dollars.

• But if the annual inflation rate is f

• If the Ajax motor salvage value is 4000 at the end of 5 years, and inflation at 3%

• Freal is 4000/(1+.03)5 is 3450

• 20% appreciation over 5 years got us P for ajax motors as 4000/(1+.2)5

is 1610

• From this we can get the ireal as (inom – f)/(1+f)

• For inom as 20% and f as 3% ireal will be 16.5%

• And the present value taking into account the inflation will be

F

F

freal

nomn

1

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• Investing $20,000 in new equipment will provide benefits of $6,000, $9,000 and $12,000 at the end of the first, second, and third years respectively. If the real discount rate is 8% and the inflation rate is 6%, is this a good investment?

• Calculate future real value Freal

• Calculate present value with ireal as .08

• Alternatively, calculate inom from the equation ireal = (inom – f)/(1+f) which ives inom

as .145

• And use the future nominal value

• Both methods give the same answer for the present value as 107

• Since this is positive, it is a good investment

Transformation Factors Future Real Value Present Value

Investment -20.000

Benefits in Year 1 TFP,ireal,1 = 0.926 5.660 5.241

Benefits in Year 2 TFP,ireal,2 = 0.857 8.010 6.867

Benefits in Year 3 TFP,ireal,3 = 0.794 10.075 7.998

Present Value 0.107

Transformation Factors Future Nominal Value Present Value

Investment -20.000

Benefits in Year 1 TFP,inom,1 = 0.874 6.000 5.241

Benefits in Year 2 TFP,inom,2 = 0.763 9.000 6.867

Benefits in Year 3 TFP,inom,3 = 0.667 12.000 7.998

Present Value 0.107

Example 8.8.1

F

F

freal

nomn

1

ireal = is .08

T iFP i nn

, ,

1

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Summary of Transformation Factors

Factor Name Comments

n

n,i,FP iT

1 present worth factor

n

n

n,i,UPii

iT

1

11

uniform series present

worth factor

The inverse of this is known as the capital

recovery factor

nn

n,i,APii

niiT

1

112

arithmetic series present

worth factor

n

nn

n,e,i,GPiei

eiT

1

11

geometric series

ei

in

T n,e,i,GP

1

present worth factor

ei

11

nn,i,FUi

iT

sinking fund factor

11

1

n

mn

m,n,i,FUi

iiT

generalized

sinking fund factor

m is the year in which the transaction F

occurs; n is the period over which the annual

payments U are made.

eii

eiiT

n

nn

n,e,i,GU

11

11

geometric series sinking

fund factor

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• We have not seen Arithmetic series present worth factor

• Suppose, the cost of mainteneance of blayblock is 2000 for first year, and increases by 200 each year, then

• The first part 5980, we have already calculated

Summary of Transformation Factors

nn

n,i,APii

niiT

1

112

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Summary of Transformation Factors

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Summary of Transformation Factors

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Preliminary Cost Estimates

• Table 9-50 of Perry’s Chemical Engineering Handbook, 7th Ed. provides preliminary capital cost estimates for many types of mechanical equipment.

• Capital cost C of a system of size Q is

where Cref, Qref, and n are given in Perry’s.

• Be careful to distinguish between FOB, delivered, and installed costs. Also, adjustments may have to be made to account for inflation and regional differences.

C CQ

Qref

ref

n

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Preliminary Cost Estimates - Example Problem

• What is the capital cost of a 500 psi, 1000 hp reciprocating compressor?

• From Perry’s Table 9-50,

Qref = 300 hp

Cref = $133,000

n = 0.84000367

hp 300

hp 1000000133

840

,$,$C

.

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Cost Estimate for Machining Operation

Category Cost ($)

Feed 1.48

Rapid Traverse 0.11

Load and unload 0.92

Setup 0.43

Tool change 0.49

Tool depreciation 0.13

Tool resharpening 1.48

Rebrazing 0.16

Tip cost 0.10

Grind wheel 0.02

Total $5.33

• For design projects, we need calculation of fabrication costs

• They depend on material being machined, equipment used, tool, operation and rate at which it is performed, desired tolerances, surface finish, and labor

• As example, if we want to consider turning a 3.5” dia 19” long shaft on a lathe using a brazed carbide tool. Shaft is 430 steel

• Cost is summarized in the table. And each cost in itself a detailed calculation

• Tool change cost can be calculated from

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Costs and profit

• After calculating the direct and indirect costs, when you provide the customer, you also need to take into account other factors like • Time spent on design, selecting hardware etc

• Man hour * hourly wage * benefits

• Then indirect cost like overheads, utility, telephone, power, etc

• In addition, the company needs to make profit. It could be a % of total costs

• Moreover, an economic safety factor to handle unexpected fluctuations