1

Material ManagementClass Note #3-B

~ Multiple product EPQ analysis ~

Prof. Yuan-Shyi Peter Chiu Feb. 2011

2

T1 T2

T

t

H

I(t)

§ . Inventory Management for Finite Production Rate

• Inventory Levels for Finite Production Rate Model

Slope=P- λ

Slope= - λ

3

P : production rate (per unit time)λ: demand rate (per unit time)

T1 T2t

H

I(t) T

Q

1

QQ T T

Q P T1

H QH P- T1 T1 P- P

QT

H Q 1-P

P

P >λ

4 where h' h (1- )P

2 k Q*h'

………... [Eq.1]

k HG(Q) h

Q 2

Q 1-k P Q 2

Let h' h 1- Pk h'Q

G(Q) Q 2

C

hC

C

§. Finite Production Rate Model

5

~ EPQ (Finite Production Rate ) Models for Production Planning ~

◆ Producing n products on a single machine (or line)

◆ Goals： (1) To determine the optimal procedure for producing n products on the machine to minimize the cost of holding & setups.

(2) To guarantee that no stock-outs occur during the production cycle.

j

j

j

j

P

h

k

Demand rate for product jProduction rate for product j Holding cost per unit per unit time for product j

Cost of setting up the production facility to produce product j

‧

6

◆ Assumption (required)

◇ To ensure that the facility has sufficient capacity to satisfy the demand for all products.

1

1

nj

jj

~ Multiple items production planning ~

………... [Eq.2]

11 = Q P an d QT

from page 3Q

T PT Q

T P

,

7

◆ Rotation cycle policy assumption :

means that in each cycle there is exactly one setup foreach product, and products are produced in the samesequence in each production cycle .

◇ Recall “the finite production rate “ solution

'

2 j jj

j

KQ

h

where ' 1 j

j jj

h h

H

T

21

~ Multiple items production planning ~

8

◆ Let T be the cycle time ( new, for the rotation cycle policy)

During time T , we assume that exactly one lot of each

product is produced.

◇ In order that the lot for product j be large enough to

meet the demand occurring during time T , it follows that the lot size should be

◇ The average annual cost ( of setup & holding costs only )

j jQ T jjQ

(e.g. if T=0.5 year , =12,000 then =6,000)

'

2

j j j jj

j

K h QG Q

Q

………... [Eq.3]

9

◇ The average annual cost for all products

◇ The avg. annual cost for n products in terms of the cycle time T

◇ To find T, minimize G(T) ; (check G”(T)>0 for minimal)

'

1 1

2

n nj j j j

jj j j

K h QG Q

Q

j

j

Q

'

1 2

nj j j

j

K hG

0

d

dG

'

21

0

2

nj j j

j

K h

………... [Eq.4]

10

'1

21

0

2

n

j nj j j

j

Kh

'2

1 12

n nj j

jj j

hK

12

'

1

2

n

jj

n

j jj

K

h

1

'

1

*2

n

jj

n

j jj

K

h

where ' 1 j

j jj

h h

………... [Eq.5]

11

◆ If setup times are a factor, we must check that if there is enough time each cycle to account for both setup times and production time of the n products. Let be the setup time for product jjs

1

nj

jjj

Qs P

12

3

4

T 1 year0

Idle time

fits into idle timejs

~ Multiple items production planning ~

………... [Eq.6]

12

then [Eq.6] become

1

n

jj

jj

S P

1 1

n n

jj

jj j

S p

j jQ j

j

Q

(or )

◆ The optimal solution is to choose the cycle time T

( Let )

T = max { T* , } if considering

1

1

1

n

jjn

j

jj

S

P

1

min

1

1

n

jjn

j

jj

S

P

min

min jS

………... [Eq.7]

13

Example 4.7

Bali produces several styles of men’s and women’s shoes at a single facility near Bergamo, Italy. The leather for both the uppers and the soles of the shoes is cut on a single machine, This Bergamo plant is responsible for seven styles and several colors in each style. (The colors are not considered different products for our purposes, because no setup is required when switching colors.)

Bali would like to schedule cutting for the shoes using a rotation policy that meets all demand and minimizes setup and holding costs. Setup costs are proportional to setup times. The firm estimates that setup costs amount to an average of $110 per hour, based on the cost of worker time and the cost of forced machine idle time during setups. Holding costs are based on a 22% annual interest charge. The relevant data for this problem appear in Table 4-1.

14

7

1

2695jj

K

7

'

1

230,490j jj

h

7

1*7

'

1

2 2 2695

0.1529 (years)230,490

jj

j jj

K

h

◆ Example 4.7

# Style

Production Rate

Setup Time (hr)

Variable Cost

Assuming$110 per hr

Assumingί =22%

1 WP 4,520 35,800 3.2 40 352 8.80 7.69 0.1263

644 34,759

2 WL 6,600 62,600 2.5 26 275 5.72 5.12 0.1054

842 33,792

3 WB 2,340 41,000 4.4 52 484 11.44 10.79 0.0571

458 25,249

4 WS 2,600 71,000 1.8 18 198 3.96 3.81 0.0366

520 9,906

5 MW 8,800 46,800 5.1 38 561 8.36 6.79 0.1880

1206 59,752

6 ML 6,200 71,200 3.1 28 341 6.16 5.62 0.0871

868 34,844

7 MO 5,200 56,000 4.4 31 484 6.82 6.19 0.0929

902 32,188

∵ $110 * (Setup Time)

Holding Cost= ί*c=(22%)*c

Σ=

0.6934

j j jC jK jh'

1

j

jj

j

h

h

j

j

*'

2 j jj

j

KQ

h

'

j jh

'

230,490

j jh

2695jK

[Eq.2]

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◆ Assuming 250 working days a year

(0.1529 year ) * 250 = 38.2 days

1041.07

1

*1

jj

7*'

11

0.1060jj

*'

2 j j

j

KQ

h

*

j * *1j *'

jQ*'

1j

** jj

j

Q

*

**1

j

jj P

Q *' *

j jQ *'

*'1 *

jj

j

Q

P

# style

1 WP 644 0.1425 0.1529 0.0180 691 0.0193

2 WL 842 0.1276 0.1529 0.0135 1009 0.0161

3 WB 458 0.1962 0.1529 0.0112 358 0.0087

4 WS 520 0.2000 0.1529 0.0073 398 0.0056

5 MW 1206 0.1370 0.1529 0.0258 1346 0.0288

6 ML 868 0.1400 0.1529 0.0122 948 0.0133

7 MO 902 0.1735 0.1529 0.0161 795 0.0142

◆ Example 4.7

16

12

34

56

7

0 10 20 30 40 50 60 70 80

3.4

9.9

13

17.5

21.5

24.3

26.1

31.9

37.7

44.9

48.660.9

74.3

Original – by EPQ single item Not feasible31.9

34.3

35

35.6

43.4

49.1

50

3.4

6.5

3.1

4.5

4.0

2.8

1.8

◆ Example 4.7

in SPT

?

70.6

272.0

17

*jQ *

j *1j * *'

jQ•

*'1j# style

1 WP 644 35.6 4.5 38.2 691 4.8

2 WL 842 31.9 3.4 38.2 1009 4.0

3 WB 459 49.1 2.8 38.2 358 2.2

4 WS 520 50 1.8 38.2 398 1.4

5 MW 1206 34.3 6.5 38.2 1346 7.2

6 ML 868 35 3.1 38.2 948 3.3

7 MO 902 43.4 4.0 38.2 795 3.6

Σ=26.1 Σ=26.5

◆ 1 year = 250 days

◆ Example 4.7

Original by Rotation cycle policy

18

12

34

56

7

0 10 20 30 40 50 60 70 80

3.4

9.9

13

17.5

21.5

24.3

26.1

31.9

37.7

44.9

48.660.9

74.3

Original31.9

34.3

35

35.6

43.4

49.1

50

3.4

6.5

3.1

4.5

4.0

2.8

1.8

◆ Example 4.7

in SPT

19

◆ Example 4.7

by Rotation cycle policy

0

5 10 15 20 25 30 40 50 60

4.88.8

1112.4

19.6

22.9

26.5

38.243

4749.2

50.657.8

61.1

T* = 38.2

in SPT

# 1

# 2# 3

# 4# 5 # 6

# 7

20

◆ The optimal solution is to choose the cycle time T

( Let )

T = max { T* , } if considering

1

1

1

n

jjn

j

jj

S

P

1

min

1

1

n

jjn

j

jj

S

P

min

min jS

………... [Eq.7]

If including setup time sj :

1min

1

24.5 / 8 / 2500.03995

1 0.69341

n

jj

nj

jj

S

P

T = max { 0.1529 , 0.03995 } = 0.1529

Setup Time (hr)

3.2

2.5

4.4

1.8

5.1

3.1

4.4

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Class NoteClass Note # 3b# 3b

§.§. I.4.9: I.4.9: Home Work Problems Home Work Problems

Chapter 4 :Chapter 4 :( # 29( # 29, , 3030 ) ) pp.219-220pp.219-220

(a) using Rotation Cycle(a) using Rotation Cycle

( ( # 45# 45 ) ) pp.227pp.227

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Discussion : 10 ~ 15 minutes ■

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§.§. I.eg.4.7.1: I.eg.4.7.1: Class ExerciseClass Exercise

“IF there are only 2 products”

Holding costs are based on a 22% annual interest charge

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Discussion : 10 ~ 15 minutes

Setup Time

#2 WL 6,600 62,600 2.5 26 275

4* WS 2,600 71,000 3.6 18 396

j jP jC jK

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Setup Time

#2 WL 6,600 62,600 2.5 26 275 0.1054 $5.72 5.12 842

4* WS 2,600 71,000 3.6 18 396 0.0366 3.96 3.82 734

j jP jC jKj

j

jh

'jh *

'

2 j jj

j

KQ

h

1 year = 250 days

◆ Example “IF there are only 2 products”

Original by Rotation cycle policy

#2 WL 0.1276 = 31.9 33,792 3.4 28.5 43.8 1,156 4.6 40.2

4* WS 0.2824 = 70.6 9,932 2.6 68 43.8 456 1.6 43.2

*j daysT

'j jh **

1Q

jT *2jT

*T *Q 1jT 2jT

* Data being modified

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• By rotation cycle policy T*

• Costs - when using T*

2

1*2

'

1

22 275 396 1342

0.175233,792 9,932 43,724

43.8 ( )j

j

j jj

K

daysh

' *2*

*1

275 6600 5.12 1156 396 2600 3.82 456

2 1156 2 456 2j j j j

jj j

K h QG Q

Q

= (1570+2959) + (2258+871)= (4529) + (3129) = $7,658

◆ Example “IF there are only 2 products”

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#2 WL 0.1276 = 31.9 33,792 3.4 28.5 43.8 1,156 4.6 40.2

4* WS 0.2824 = 70.6 9,932 2.6 68 43.8 456 1.6 43.2

*j daysT

'j jh **

1Q

jT *2jT

*T *Q 1jT 2jT

◆ To Think about - when producing To Think about - when producing 2 products2 products

Item “WL” has T1 = 31.9 days

Item “WS” has T2 = 70.6 days

Why using R-C-P ?

To allow producing 1 run

per item per period ?

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Class NoteClass Note # 3b# 3b

§.§. I.eg.4.7.2.I.eg.4.7.2.AA: : Class ExerciseClass Exercise

From eg.4.7.1

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Discussion : 10 ~ 15 minutes

Setup Time

#2 WL 6,600 62,600 2.5 26 275

4* WS 2,600 71,000 3.6 18 396

j jP jC jK

If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or 0.2824 years)

(a) What will the annual costs be?(b) Draw a chart for production schedule.

27

Solution to eg.4.7.2.AA :

When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 70.6 days (or 0.2824 years)

(a)(a)

∴ T1 = 0.2824 / 2 = 0.1412 years = 35.3 days

∴ Q1 = (λ1) ( T1) = (6600) (0.1412) ≒ 932

∴ T11 = Q1 / P1 = 932 / 62,600 0.01488 years 3.72 days≒ ≒ [ up-time]

'

1 1 1 11

1

2K h Q

G QQ

275 6600 5.12 932

932 2

= (1947.42) + (2385.92) = $4,333

'

2 2 2 22

2

2K h Q

G QQ

= $2,805

28

' *2

**

1

2

j j j jj

j j

K h QG Q

Q

= $4,333 + $2,805 = $7,138

Solution to eg.4.7.2.AA :

Comparing to R-C-P: $7658Comparing to R-C-P: $7658

There are ($7658 - There are ($7658 - $7138$7138) / ) / $7658$7658 = = 6.8 %6.8 % cost saved. cost saved.

29

0

3.72

35.3

39.02

70.6

T* = 70.6 days

#WL-1# WS

Solution to eg.4.7.2.AA :

(b)(b)

6.31

#WL-2

74.32

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Class NoteClass Note # 3b# 3b

§.§. I.eg.4.7.2.B:I.eg.4.7.2.B: Class ExerciseClass Exercise

From eg.4.7.1

Preparation Time : 15 ~ 20 minutesPreparation Time : 15 ~ 20 minutes Discussion : 10 ~ 15 minutes Discussion : 10 ~ 15 minutes

Setup Time

#2 WL 6,600 62,600 2.5 26 275

4* WS 2,600 71,000 3.6 18 396

j jP jC jK

If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or 2*0.1276 = 0.2552 years)

(a) What will the annual costs be?(b) Draw a chart for production schedule.

◆G-s-32

31

Class NoteClass Note # 3b# 3b

§.§. I.eg.4.7.2.CI.eg.4.7.2.C: : Class ExerciseClass Exercise

From eg.4.7.1From eg.4.7.1

Setup Time

#2 WL 6,600 62,600 2.5 26 275

4* WS 2,600 71,000 3.6 18 396

j jP jC jK

If we producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or 0.2656 years) – i.e. the optimal length for period – i.e. the optimal length for period

What will the annual costs be?What will the annual costs be?

◆G-s-35

32

The EndThe End

33

Solution to eg.4.7.2.BB :

When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 63.8 days (or 0.2552 years)

(a)(a)

∴ T2 = 0.2552 years = 63.8 days

∴ Q2 = (λ2) ( T2) = (2600) (0.2552) 664 ≒

∴ T21 = Q2 / P2 = 664 / 71,000 0.00935 years 2.34 days ≒ ≒ [ up-time]

'

2 2 2 22

2

2K h Q

G QQ 664

6

396 2600 3

4

.82

6 2

= (1550.6) + (1268.24) = $2,819

'

1 1 1 11

1

2K h Q

G QQ = $4,311

34

' *2

**

1

2

j j j jj

j j

K h QG Q

Q

= $4,311 + $2,819 = $7,130

Solution to eg.4.7.2.BB :

Comparing to R-C-P: $7658Comparing to R-C-P: $7658

There are ($7658 - $7130) / $7658 =There are ($7658 - $7130) / $7658 = 6.9 %6.9 % costcost saved saved..

35

0

3.4

31.9

35.3

63.8

T* = 63.8 days

#WL-1 # WS

Solution to eg.4.7.2.BB :

(b)(b)

5.74

#WL-2

◆G-b-29

67.2

36

Solution to eg.4.7.2.CSolution to eg.4.7.2.C :

When producing 2 runs of item “WL” and 1 run of item “WS” in a period of 66.4 days (or 0.2656 years) – the optimal period length– the optimal period length

∴ for item “WL”, each run T1 = 0.2656 / 2 = 0.1328 years = 33.2 days

∴ for item “WS”, T2 = 0.2656 years = 66.4 days

∴ Q1 = (λ1) ( T1) = (6600) (0.1328) 876 ≒

∴ Q2 = (λ2) ( T2) = (2600) (0.2656) 691 ≒

∴ T11 = Q1 / P1 = 876 / 62,600 0.01399 years 3.50 days ≒ ≒ [ up-time]

∴ T21 = Q2 / P2 = 691 / 71,000 0.00973 years 2.43 days ≒ ≒ [ up-time]

'

1 1 1 11

1

2K h Q

G QQ

876

8

275 6600 5

6

.12

7 2

= (2072) + (2243) = $4,315

37

' *2

**

1

2

j j j jj

j j

K h QG Q

Q

= $4,315 + $2,810 = $7,125= $7,125

Solution to eg.4.7.2.C :Solution to eg.4.7.2.C :

Comparing to R-C-P: $7658Comparing to R-C-P: $7658

There are ($7658 - $7125) / $7658 = There are ($7658 - $7125) / $7658 = 7.5 %7.5 % cost saved. cost saved.

'

2 2 2 22

2

2K h Q

G QQ 691

6

396 2600 3

1

.82

9 2

= (1490) + (1320) = $2,810

38

0

3.5

33.2

36.7

66.4

T* = 66.4 days

#WL-1

# WS

Solution to eg.4.7.2.C :Solution to eg.4.7.2.C :

(b)(b)

5.93

#WL-2

69.9

39

Alternative # Period

Length

Cost SavedCost Saved

Vs. R-C-PVs. R-C-P

Using Rotation

Cycle Policy44.8 days $7,658 ——

Using original

“WL” * 2 63.8 days $7,130 6.9 %6.9 %

OptimalOptimal

LengthLength 66.4 days66.4 days $7,125$7,125 7.0%7.0%

Using original

“WS” period 70.6 days $7,138 6.8%6.8%

2

*

1j

j

G Q

§.§. I.eg.4.7.1: I.eg.4.7.1: 2 items production planning2 items production planning comparison comparison

40

The EndThe End