replacement problems

REPLACEMENT PROBLEMS

Introduction

In Industries all equipment are put to continuous use

which reduces the efficiency of the equipment.

A replacement is needed for the equipment if the

cost incurred in operating and maintaining the

equipment exceeds the benefit derived out of it.

The objective is to determine the optimal time at

which the equipment is to be replaced with new one.

MODELS OF REPLACEMENT

Replacement Models

Replacement of items that deteriorate with time,

Ignoring time value of money

Considering time value of

money

Items that fails suddenly

Individual replacement

policy

Group replacement

policy

REPLACEMENT OF ITEMS THAT DETERIORATE

WITH TIME, IGNORING TIME VALUE OF MONEY

Time value of money is not considered (Interest rate = 0)

C = Initial cost of equipment,

S(t) = Selling / Scrap value of equipment after ‘t’ years

f(t) = Operating and maintenance cost at time ‘t’

n = replacement period of the item

Annual Cost incurred at time ‘t’ = C – S(t) + f(t)

Total operating and maintenance cost incurred during n years =

Total cost (T) incurred on equipment during n years =

Average annual cost (TA) incurred in n years =

n

t

tf0

)(

n

t

tftSC0

)()(

n

T

•Replace the equipment at the end of the year having minimum TA value

Problems Small battery operated trucks are used in an industry for material handling purpose.

The operating cost, maintenance cost and resale value for six years are given in table.

Determine when the truck is to be replaced by a new one if initial cost of the truck is

Rs 75,000.

Year of

service

Annual Operating

cost (Rs)

Annual maintenance

cost (Rs)

Resale value

1 10,000 6,000 45,000

2 12,000 7,500 40,000

3 15,000 12,500 20,000

4 19,000 17,000 10,000

5 27,000 20,000 10,000

6 33,000 21,000 5,000

Solution

Year

end

(n)

C S(t) f(t)

f(t) T = C – S(t) + f(t)

TA = T/n

1 75000 45,00

0

16000 16000 46000 46000

2 75000 40,00

0

19500 35500 70500 35250

3 75000 20,00

0

27500 63000 118000 39333

4 75000 10,00

0

36000 99000 164000 41000

5 75000 10,00

0

47000 14600

0

211000 42200

6 75000 5,000 54000 20000

0

270000 45000

Replaced

Problem

A hand grinding machine (Type A) costs Rs 9500. Annual operating costs are Rs 220 for the first year and it increases by Rs 1900 every year. Find the optimal age at which the hand grinding machine of type A is to replaced along with corresponding average yearly cost of owning and operating it. Assume that the machine has no resale value and future costs are not discounted. Another hand grinding machine (Type B) costs Rs 10500. Annual operating costs Rs 410 for the first year and it increases by Rs 820 every year. You have one hand grinding machine of Type A which is one year old. Should you replace it with hand grinding machine of Type B and if so when ?

Type – A machine

Year

end

(n)

C S(t) f(t)

f(t) T = C – S(t) + f(t)

TA = T/n

1 9500 0 220 220 9720 9720

2 9500 0 2120 2340 11840 5920

3 9500 0 4020 6360 15860 5286

4 9500 0 5920 12280 21780 5445

Replaced

Type A machine as to be replaced at the end on third year

Type – B machine

Year

end

(n)

C S(t) f(t)

f(t) T = C – S(t) + f(t)

TA = T/n

1 10500 0 410 410 10910 10910

2 10500 0 1230 1640 12140 6070

3 10500 0 2050 3690 14190 4730

4 10500 0 2870 6560 17060 4265

5 10500 0 3690 10250 20750 4150

6 10500 0 4510 14760 25260 4210

Replaced

Avg Annual cost of Type B (Rs 4150) < Avg Annual cost of Type A (Rs 5286)

Hence, Type A can be replaced by Type B

To determine when type A as to be replace with type B:

The year at which Running Cost of Type A > Avg annual cost of type B (Rs 4150)

YEAR 1 2 3 4

Running cost 9720 11870 – 9720

= 2120

15860 – 1840

= 4020

21780-15860

= 5920

5920 > 4150

Replaced at the

end third year

REPLACEMENT OF ITEMS THAT DETERIORATE

WITH TIME, CONSIDERING TIME VALUE OF MONEY

The value of money changes at a period of time.

Done by Present value (PV) or Present worth (PW) analysis.

Interest rate is considered

C = Purchase price of the item (Capital Cost)

Rn = Running Cost in year ‘ n’

r = Rate of interest

Discount rate, r

V

1

1Present worth expenditure,

1

1)(n

n

n RVCnP

Fixed Annual

installment, )1(

)1().(

nV

VnPX

Items is to be replaced for minimum X value

Standard table to solve problems

Year

(n)

Vn-1

(PWF)

Rn Vn-1.Rn Vn-1.Rn C P(n) X

PWF – Present Worth factor

)1(

)1().(

nV

VnPX

1

1)(n

n

n RVCnP

Problem

A small workshop involved in sheet metal work has a

plan to purchase a small sheet bending machine. A

manufacturer offers two choices namely machine –A

and machine-B. The following details are supplied by

the manufacturer.

DETAILS MACHINE –A MACHINE - B

Initial Cost Rs 12000 Rs 6000

Running Cost Rs 1920 for first 5

years, increasing by

Rs 480 /yr thereafter

Rs 2880 for first 6

years, increasing by

Rs 480/yr thereafter

Worth of capital 10 percent per year

As a consultant to the workshop, you are required to suggest which

machine is to be purchased and why ?

Machine- A r = 10 %

V = {1/(1+0.1)} = 0.9091

Year

(n)

Vn-1

(PWF)


1 1 1920 1920 1920 12000 13920 13920

2 0.9091 1920 1745 3665 12000 15665 8205

3 0.8264 1920 1586 5251 12000 17251 6306

4 0.7513 1920 1442 6693 12000 18693 5361

5 0.6830 1920 1311 8004 12000 20004 4797

6 0.6209 2400 1490 9494 12000 21494 4486

7 0.5645 2880 1625 11119 12000 23119 4317

8 0.5132 3360 1724 12843 12000 24843 4233

9 0.4665 3840 1791 14634 12000 26634 4204

10 0.4241 4320 1832 16366 12000 28366 4212

Optimum

Machine - B

Year

(n)

Vn-1

(PWF)


1 1 2880 2880 2880 6000 8880 8880

2 0.9091 2880 2618 5498 6000 11498 6023

3 0.8264 2880 2380 7878 6000 13878 5073

4 0.7513 2880 2164 10042 6000 16042 4601

5 0.6830 2880 1967 12009 6000 18009 4319

6 0.6209 2880 1788 13798 6000 19798 4132

7 0.5645 3360 1897 15695 6000 21695 4051

8 0.5132 3840 1971 17665 6000 23665 4033

9 0.4665 4320 2015 19681 6000 25681 4054

Optimum

Machine B has least annual cost (4033) than Machine A (4204)

Hence Machine B is suggested

REPLACEMENT OF ITEMS THAT FAILS

SUDDENLY

Individual replacement policy:

An item is replaced as soon as it fails.

Group replacement policy:

A decision is taken to replace all item irrespective of its

failure.

An optimal group replacement period is decided.

Items which fails before optimal group replacement period

will be replaced individually.

Formulae used Pi = Probability that an item newly installed fails during ith

week

Ni = No. of replacements made at the end of ith week

N0 = Items which are newly installed

Expected life of each item = in days/weeks/months

Avg. no. of failure = Total no. of items / Expected life of

each item

133231404

1221303

11202

101

PNPNPNPNN

PNPNPNN

PNPNN

PNN

n

i

iiP1

Individual

replacement

Policy

Group

replacement

Policy

Problems

The following failure rates are for a resistor in an

electrical system. The number of resistor in the electrical

system is 1000 at the beginning.

End of week 1 2 3 4 5 6 7 8

Cumulative

Probability of failure 0.03 0.15 0.24 0.44 0.67 0.85 0.95 1

Cost of replacing an individually failed resistor is Rs 1.30

If all the resistors are replaced in a group, the cost per

resistor is 32 paise.

Determine the optimum group replacement period and

cost occurring during individual replacement and group

replacement.

Individual replacement policy

Pi = Probability of resistor that fails in ith week

Expected life of resistor = = 4.67 weeks

Avg, no of failures = 1000 / 4.67 = 214 (approx)

Weekly cost of individual replacement = Avg. no of failure x cost of

individual resistor = 214 x 1.30 = Rs 278.20

8

1i

iiP

P0=0 P1=0.03 P2=0.15- 0.03 =

0.12

P3 = 0.24 – 0.15 =

0.09

P4=0.44-0.24 =

0.20

P5=0.67-0.44 =0.23

P6=0.85-0.67 = 0.18 P7=0.95-0.85 =

0.10

P8= 1-0.95=0.05

Group replacement policy

No of items to be replaced:

N0 = 1000

N1 = N0 X P1 = 1000 X 0.03 = 30

N2 = N0 X P2 + N1 X P1 = 1000 X 0.12 + 30 X 0.03 = 121

N3 = 97

N4 = 220

N5 = 265

N6 = 254

N7 = 212

N8 = 202

Individual cost / resistor = Rs 1.30

Group replacement of 1000 resistors = Rs 0.32

Determining cost of replacement:

Period Total Cost

Avg cost /

week

0 1000 x 0.32 NA

1 1000 x 0.32 + 30 x 1.30 = 359 359

2 1000 x 0.32 + 30 x 1.30 +121 x 1.30 = 516.90 258.15

3 1000 x 0.32 + 30 x 1.30 +121 x 1.30 + 97 x 1.30 =

642.40

214.13

4 642.40 + 220 x 1.30 = 928.40 232.10

Optimum

SEQUENCING PROBLEMS

Introduction

To determine the order in which jobs or activities are

to be performed.

Types:

Sequencing

Processing n jobs through 2 machines


Processing n jobs with m

machines

Travelling salesman problems


Developed by S.M, Johnson

Two machines A and B are involved

Jobs are processed in the order AB

Steps Involved:

Examine Process times on machine A and B together and

determine the smallest processing time.

If the selected time is on machine A, note the job from first

If the selected time is on machine B, note the job in last

column

Strikeout the assigned job and continue the process.

Problems

There are six jobs each of which must go through

the two machine A and B in the order AB. Processing

time in hours are given in table. Determine the

sequence of six jobs which will minimize the elapse

time and idle time. JOB

PROCESSING TIME

MACHINE – A MACHINE - B

1 3 2

2 6 5

3 4 6

4 7 3

5 5 2

6 8 8

Solution JOB

PROCESSING TIME

MACHINE –

A

MACHINE -

B

1 3 2

2 6 5

3 4 6

4 7 3

5 5 2

6 8 8

Sequencing Display

Min time on machine

A

Min time on machine

B

1 5 4 3 2 6

Sequence order is : 3 – 6 – 2 – 4 – 5 – 1

To Calculate elapse time and Idle time

Optimal

Sequence

Machine – A Machine - B

Time In Time out Time In Time out

3 0 0 + 4 = 4 4 4 + 6 =10

6 4 4 + 8 = 12 12 12 + 8 = 20

2 12 12 + 6 = 18 20 20 + 5 = 25

4 18 18 + 7 = 25 25 25 + 3 = 28

5 25 25 + 5 = 30 30 30 + 2 = 32

1 30 30 + 3 = 33 33 33 + 2 = 35

JOB

PROCESSING TIME

MACHINE –

A

MACHINE -

B

1 3 2

2 6 5

3 4 6

4 7 3

5 5 2

6 8 8

Elapse time = 35 hrs

Idle time of machine A = 35 – 33 = 2 hrs

Idle time of machine B = (4-0) + (12-10) + (20-20) + (25-25) + (30 – 28)

+ (33-32) = 9 hrs

replacement problems

Documents