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Chapter 7 ~ HOMEWORK SOLUTION ~. By: Prof. Y. Peter Chiu. What is the MRP Calculations for the slide assemblies ? ( 3 slide assemblies for each valve casing ). Lead Time = 2 weeks. Assume On-hand inventory of 270 slide assemblies at the end of week

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By: Prof. Y. Peter Chiu

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Chapter 7

~ HOMEWORK SOLUTION ~

By: Prof. Y. Peter Chiu

• What is the MRP Calculations for the slide assemblies ?

( 3 slide assemblies for each valve casing )

• Lead Time = 2 weeks

• Assume On-hand inventory of 270 slide assemblies at the end of week

3 & Scheduled receipts of 78 & 63 at the beginning of week 5 & 7

• Solution to MRP Calculations for the slide assemblies

2 3 4 5 6 7 8 9 10 11 12 13

Week

Gross

Requirements

126 126 96 36 78 336 135 42 228 114

78 63

Scheduled Receipts

On-hand inventory

270 144 96 0 27

Net

Requirements

0 0 0 051336 135 42 228 114

Time-Phased

Net Requirements

51 336 135 42 228 114

Planned Order

Release (lot for lot)

51 336 135 42 228 114

#4(a) MPS for the computers

#5

#6

# 9

(b)

Week

27 28 29 30 31 32 33 34 35

MPS-end item

165 180 300 220 200 240

Component B (P.O.R)

330 360 600 440 400 480

Component F

330 360 600 440 400 480

-Net. Req.

Time Phased Net. Req.

330 360 600 440 400 480

Ans. →Planned Order Release

330 360 600 440 400 480

# 9

(c)

Week

21 22 23 24 25 26 27 28 29 30 31 32 33 34 35

165 180 300 220 200 240

MPS-end item

P.O.R –Comp. B

330 360 600 440 400 480

330 360 600 440 400 480

Net Req. –Comp. E

330 360 600 440 400 480

Time Phased –Net Req.

P.O.R – Comp. E

330 360 600 440 400 480

660 720 1200 880 800 960

Net Req. –Comp. G

660 720 1200 880 800 960

Time Phased –Net Req.

P.O.R –Comp. G

660 720 1200 880 800 960

660 720 1200 880 800 960

Net Req. –Comp. I

660 720 1200 880 800 960

Time Phased –Net Req.

Ans. → P.O.R –Comp. I

660 720 1200 880 800 960

(d)

1980 2160 3600 2640 2400 2880

Net Req. –Comp. H

Time Phased –Net Req.

1980 2160 36002640 2400 2880

Ans. → P.O.R –comp. H

1980 2160 36002640 2400 2880

# 14

Month

1 2 3 4 5 6 7 8 9 10 11 12

Demand

6 12 4 8 15 25 20 5 10 20 5 12

• Current Inventory : 4

• An ending Inventory should be : 8

• h = \$ 1

• k = \$ 40

Month

1 2 3 4 5 6 7 8 9 10 11 12

Net. Demand

212 4 8 15 25 20 5 10 20 5 20

(a) Silver-Meal

• Starting in Period 1 :

C(1) = 40

C(2) = (40+12)/2 = 26

C(3) = [40+12+2(4)] /3 = 20

C(4) = [40+12+2(4)+3(8)] /4= 21 <stop>

# 14

(a) Silver-Meal

C(1) = 40

C(2) = [40+15]/2 = 27.5

C(3) = [40+15+2(25)] /3 = 35 <stop>

• Starting in Period 4 :

• Starting in Period 6 :

C(1) = 40

C(2) = [40+20]/2 = 30

C(3) = [40+20+2(5)] /3 = 23.3

C(4) = [40+20+2(5)+3(10)] /4 = 25 <stop>

• Starting in Period 9 :

C(1) = 40

C(2) = [40+20]/2 = 30

C(3) = [40+20+2(5)] /3 =23.3

C(4) = [40+20+2(5)+3(20)] /4 = 32.5 <stop>

∴ Using Silver- Meal ; y = [ 18 , 0 , 0 , 23 , 0 , 50 , 0 , 0 , 35 , 0 , 0 , 20 ]

# 14

(b) LUC

• Starting in Period 1 :

C(1) = 40 /2 = 20

C(2) = 52 /14 = 3.71

C(3) = 60 /18 = 3.33

C(4) = 84 /26 = 3.23

C(5) = (84+60) /41 = 3.51 <stop>

• Starting in Period 5 :

C(1) = 40 /15 = 2.67

C(2) = 65 /40 = 1.63

C(3) = [65+2(20)] /60 = 1.75 <stop>

• Starting in Period 7 :

C(1) = 40 /20 = 2

C(2) = (40+5) /25 = 1.80

C(3) = [40+5+2(10)] /35 = 1.86 <stop>

# 14

(b) LUC

• Starting in Period 9 :

C(1) = 40 /10 = 4

C(2) = 60 /30 = 2

C(3) = [60+2(5)] /35 = 2

C(4) = [70+60] /55 = 2.36 <stop>

∴ Using LUC ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 25 , 0 , 35 , 0 , 0 , 20 ]

(C) PPB

Period

Holding cost

• Starting in Period 1 :

1* (12) = 12

12+2(4) = 20

20+3(8) = 44

2

3

4

K = \$40

Closer to period 4

(C) PPB

Period

Holding cost

• Starting in Period 5 :

1

2

3

0

25

25+2(20) = 65

K = \$40

Closer to period 2

Period

Holding cost

• Starting in Period 7 :

2

3

4

5

5+2(10) = 25

25+3(20) = 85

K = \$40

Closer to period 3

Period

Holding cost

• Starting in Period 10 :

2

3

5

5+2(12) = 29

K = \$40

∴ Using PPB ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 35 , 0 , 0 , 45 , 0 , 0 ]

# 14 (d)

1 2 3 4 5 6 7 8 9 10 11 12

SM

18 0 0 23 0 50 0 0 35 0 0 20

LUC

26 0 0 0 40 0 25 0 35 0 0 20

PPB

26 0 0 0 40 0 35 0 0 45 0 0

Demand

2 12 4 8 15 25 20 5 10 20 5 20

Inv. SM

Σ = 95

16 4 0 15 0 25 5 0 25 5 0 0

24 12 8 0 25 0 5 0 25 5 0 0

Inv. LUC

Σ = 104

Inv. PPB

24 12 8 0 25 0 15 10 0 25 20 0

Σ = 139

Cost of S.M. (\$40*5)+(\$1*95) = \$295

Cost of LUC (\$40*5)+(\$1*104) = \$304

Cost of PPB (\$40*4)+(\$1*139) = \$299

∴ Silver Meal Method is the least expensive one !

K=200; h=0.3

#17

#17

#17

#18

#18 (cont’d)

The End