Equipment Replacement

Four Really Great Models1.Deterministic Age Replacement2.Minimal Repair Model3.Repair versus Replace4.Block Replacement

Applications Continuous operating equipment

Transportation Systems Radar sites Power generating equipment Computers and communications systems

Intermittent operating equipment Vehicles Aircraft Appliances and entertainment components Lighting Systems

Deterministic Age Replacement

Section 12.6 with extensions

When to replace?When to replace?

Some Very Good Assumptions

Equipment is used continuously or time is measured in operating units

Negligible downtime for repair and maintenance

Infinite planning horizon Identical replacement equipment Only maintenance and replacement costs are

considered Objective is to minimize long-run costs Time value of money not addressed

A Simple Model Let t = the decision variable, age of equipment at

replacement time (length of the replacement cycle) K = replacement cost C(u) = maintenance cost rate ($/unit time) at age u Cost per replacement cycle of length t:

2

0 0

let ( )

( )2

t t

C u au

atK C u du K au du K

Solving a simple model: Average cost per unit time:

To find the value of t that minimizes G(t):

21( )

2 2

at K atG t K

t t

2

*

'( ) 02

2

K aG t

t

Kt

a

The necessary first example

Maintenance cost on the first year of a power generator was $1200. Replacement cost is $64,000.

2 2

*

(1)1200

2 22400

2 2(64,000)7.3 yr.

2400

at a

a

kt

a

A more general replacement model

S(u) = salvage value at age u

0 0

1 ( )( ) ( ); and ( ) ( )

t tK S tK C u du S t G t C u du

t t t

0 02 2 2

02 2 2

( ) ( )1 ( ) '( )

Solve: '( )

( )( ) ( ) '( )

0

t t

t

d C u du C u duK S t S t

G tt t dt t t t

C u duK C t S t S t

t t t t t

Still a more general model

02 2 2

0

0

( )( ) ( ) '( )

0

( ) ( ) ( ) '( ) 0

( ) ( ) ( ) '( )

t

t

t

C u duK C t S t S t

t t t t t

K tC t C u du S t tS t

t C t S t K C u du t S t

A Specific Case

0

0

let ( ) , , 0 and ( ) , , 0

( ) ( ) ( ) '( )

1

bu dt

t

dttbt dt bu

bt dt bt dt

C u ae a b S t ce c d

t C t S t K C u du t S t

d ceate ce K ae du t

dt

aate ce K e tcde

b

Rearranging terms:

11bt dt a

ae t ce dt Kb b

The necessary second example

Replacement cost of an automobile is $10,000. The car loses 15% of its value each year.

(0)

(1)

(0) 10,000

(1) .85(10,000) 8500 10,000

.85 or ln(0.85) .1625

d

d

d

S ce c

S e

e d

0.1625( ) 10,000 tS t e

More of the necessary second example

First year maintenance cost was $200 and is increasing at the rate of 40 percent per year.

0

(1)

1

1 200

tbu bt

b

aae du e

b

ae

b

( 1)

( )1.4

( 1)

1.4 or ln(1.4) .3365bt

bb t

C t

C t

aee b

ae

.3365.3365

(200)(.3365)1 200 168.25

.3365 1

ae or a

e

0.3365( ) 168.25 tC t e

Bringing it home…

0.3365 0.1625

11

168.25 2.972 10,000 1 0.1625 500 10,000

bt dt

t t

aae t ce dt K

b b

e t e t

4 9778.426

4.1 9812.65

4.2 9851.775

4.3 9896.092

4.4 9945.903

4.5 10001.52

4.6 10063.27

4.7 10131.48

4.8 10206.51

4.9 10288.73

5 10378.5

Revised maintenance costFirst year maintenance cost was $200 and is increasing at the

rate of 20 percent per year. LetM = rate of increase in maintenance costI0 = first year maintenance cost

D = yearly rate of depreciation as a fractionThenb = ln(1 + M) = ln(1 + .2) = .18232a = I0 b / M = 200 (.18232) / .2 = 182.322

c = K = 10,000d = -ln(1 – D) = -ln(1-.15) = .1625

( ) ; (0)

(1) (1 )

dt

d

S t ce S c K

S Ke D K

( 1)

1

0

0

( )1 ; 1

( 1)

1

btb

b t

b

bt

C t aeM e M

C t ae

a eae dt I

b

Our new solution

8.5 9572.033

8.6 9650.86

8.7 9734.041

8.8 9821.688

8.9 9913.913

9 10010.83

9.1 10112.57

9.2 10219.23

9.3 10330.96

9.4 10447.86

9.5 10570.08

0.18232 0.1625

11

182.322 1/ 0.18232 10,000 1 0.1625 1000 10,000

bt dt

t t

aae t ce dt K

b b

e t e t

Yet another general case

let ( ) ; , 0 and ( ) ; , 0b dC t at a b S t ct c d

Then the average cost per unit time is given by:

0

1

1 ( )( ) ( )

1

1

t db

b d

K S t K a ctG t C u du t

t t t t b t

K at ct

t b

Minimize G(t) Directly

The Die Forge Company operates several die forges die forging pounds or presses metal between two dies (called

tooling) that contain a precut profile of the desired part. Parts from a few ounces to 60,000 lbs. can be made using this process.

Each unit cost $150,000 with C(t) = 200t1.5 and S(t) = 500t-.15

1

0

0.151.5 1.5 1.15

0

1 ( )( )

1

150,000 1 500 150,000 200200 500

2.5

tb b d

t

K S t K aG t au du t ct

t t t t b

tu du t t

t t t t

The Solution

1.5 1.15150,000 200Min ( ) 500

2.5G t t t

t Using Solver

t* = 17.31128 yr.

G(t)

15 14625.3715.5 14537.92

16 14474.3816.5 14432.87

17 14411.7217.31 14408.18

18 14424.7318.5 14456.38

19 14503.3419.5 14564.66

20 14639.47

Let’s take derivatives…

1.5 1.15

.5 2.152

2.5 0.15

150,000 200Min ( ) 500

2.5200 1.5150,000

'( ) 500 1.15 02.5

150,000 120 575 0

G t t tt

G t t tt

t t

t G'(t)15 -200.206

15.5 -150.32316 -104.456

16.5 -62.134917 -22.9577

17.31128 -0.0002818 47.30427

18.5 78.9479519 108.5795

19.5 136.396820 162.5735

A MINIMAL REPAIR MODEL

Replacement Based upon random failures

Minimal repair model

• Equipment is restored upon failure• Restoration results in minimal repair • Equipment continues to deteriorate over time• Define (t) = failure rate at time t (an increasing function)• N(t) = number of failures in time t

E N t t dtt

[ ( )] ( ' ) 'z0

Power Law Process (t) = a b tb-1 , a,b > 0

A six year old regional transit bus experiences minimal repairupon failure. It was found to have an intensity function given by

(t) = .5 t1.5 with t measured in years.

The expected number of failures during the first two years is given by

The expected number of failures during the first 5 years is given by55 2.5

1.5

0 0

.5[ (5)] .5 11.18

2.5

tE N t dt

22 2.51.5

0 0

.5[ (2)] .5 1.13

2.5

tE N t dt

Replacement Model

Cu = unit cost,Cf = cost of a failure t = replacement time (decision variable)

0( ) ( )

t

u fC t C C t dt

0( )

tfuCC

C t dtt t

Replacement Model – Power Law process

dC

dt

C

tb C atu

fb

221 0( )

tC

C a bu

f

b

*( )

/

LNMM

OQPP1

1

what if b <= 1?

1 1

0

tf b bu u

f

CC CC abu du C at

t t t

A Really Good Example

A repairable machine has a NHPP with an intensity functionof (t) = 2 x 10-6 t with t measured in operating hours. If the cost of a failure (repair) is $500 and the unit cost is $21000,

then

Replace at 6,481 hr or 810 days if used 8 hr/day

1/ 1/2

6

21,000* 6481

( 1) 500(10 )

b

u

f

Ct

C a b

REPAIR VERSUS REPLACE

Decisions…Decisions

Repair vs. Replacement

f = the number of part failures over the life of the system,c = unit cost of the part,ar = fixed cost of repairad = fixed cost of discarding where ad < ar

br = cost to repair a failurebd = cost to remove and replace a part where bd < br ,k = condemnation fraction

A Cost Trade-off Model

repair cost = ar + br f + c k fdiscard cost = ad + (c + bd ) f

Repair vs. Replacement

ad + (c + bd ) f <= ar + br f + c k fIf Then discard

ca a

f k

b b

kr d r d

( )1 1

c

f

discard

repair

br-bd

1-k

Example - circuit board

ar = $ 20,000 (primarily test equipment and facilities) ,ad = $ 1200 (warehouse overhead for the spares),br = $ 768 / failure ($48 / hr labor x 8 hr MTTR x 2 crew members),bd = $ 24 / failure ( $48 / hr labor x .5 hr R&R x 1 crew member),k = .05

0

5000

10000

15000

20000

25000

UNIT COST

NBR FAILURES

DISCARD VS REPAIR

c = 783.2 + 19789.5 / f .

BLOCK REPLACEMENT

Block replacement for a group of items (pages 736-738)When is it more economical to replace a group of items that will eventually fail at the same time rather than one at a time?

Block Replacement

Let pk = fraction of items that fail in period kn0 = number of items placed in service at time 0

Period Number of failures1 n1 = n0p1

2 n2 = n0p2 + n1p1

3 n3 = n0p3 + n1p2 + n2p1

k nk = n0pk + n1pk-1 + …+ nk-1p1

...

Group Replacement Continues

Let a1 = cost of an individual replacementa2 = cost of replacing all n0 items

number periods for block replacement average cost

1 a2 + a1n1

2 [a2 + a1(n1 + n2)]/2

k

2 11

0 1 1 1 1( ) where ...

k

jj

j j j j

a a n

G k n n p n p n pk

nj is the expected number of failures in period j

Expected Cost without Block Replacement

1

( ) kk

E T k p

The expected lifetime of a single item:

Therefore the expected (steady-state) cost for the entire block of items: 0 1Cost per time period

( )

n a

E T

Example: 200 items cost $5 each to replace and have anexpected lifetime of 25 months. Then (200)(5)/25 = $40 per month(on the average, there will be 8 failures per month).

The mandatory example A large sign us lit by 8,000 bulbs. The bulbs cost $2 each to

replace as they fail but can be replaced for 30 cents each when they are replaced all at once. Based upon past history:

months fraction failing1 0.022 0.033 0.034 0.055 0.086 0.097 0.078 0.19 0.11

10 0.1311 0.1512 0.14

1

n0 = 8,000n1 = no p1 = (8,000) (.02) = 160n2 = no p2 + n1 p1 = (8,000) (.03) + (160)(.02) = 243etc.

Use Excel

a1 = 2

a2 = 2400

n0 = 8000

From Excel…nbr mo.

fraction failing 1 2 3 4 5 6 7 8 9 10 11 12 nj Sum nj G(k)

1 0.02 160 160.0 160.0 2720.02 0.03 240 3.2 243.2 403.2 1603.23 0.03 240 4.8 4.9 249.7 652.9 1235.24 0.05 400 4.8 7.3 5.0 417.1 1070.0 1135.05 0.08 640 8 7.3 7.5 8.3 671.1 1741.1 1176.46 0.09 720 12.8 12.2 7.5 12.5 13.4 778.4 2519.5 1239.87 0.07 560 14.4 19.5 12.5 12.5 20.1 15.6 654.6 3174.0 1249.78 0.1 800 11.2 21.9 20.0 20.9 20.1 23.4 13.1 930.5 4104.5 1326.19 0.11 880 16 17.0 22.5 33.4 33.6 23.4 19.6 18.6 1064.0 5168.5 1415.2

10 0.13 1040 17.6 24.3 17.5 37.5 53.7 38.9 19.6 27.9 21.3 1298.4 6466.9 1533.411 0.15 1200 20.8 26.8 25.0 29.2 60.4 62.3 32.7 27.9 31.9 26.0 1542.9 8009.8 1674.512 0.14 1120 24 31.6 27.5 41.7 47.0 70.1 52.4 46.5 31.9 39.0 30.9 1562.4 9572.3 1795.4

sums 1 8000

1

2400 2

( )

k

jj

n

G kk

12

1

2 8,000( ) 8.22 mo. cost per mo. $1,946.47

8.22kk

E T k p

This is the End