By: Prof. Y. Peter Chiu
35
1 By: Prof. Y. Pe ter Chiu MRP & JIT ~ HOMEWORK SOLUTION ~
description
MRP & JIT ~ 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 - PowerPoint PPT Presentation
Transcript of By: Prof. Y. Peter Chiu
1 By: Prof. Y. Peter Chiu
MRP & JIT~ HOMEWORK SOLUTION ~
2
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
Week
Gross Requirements
Scheduled Receipts
On-hand inventory
Net RequirementsTime-Phased Net Requirements
Planned Order Release (lot for lot)
2 3 4 5 6 7 8 9 10 11 12 13
126 126 96 36 78 336 135 42 228 114
78 63
270 144 96 0 27
0 0 0 0 51 336 135 42 228 114
51 336 135 42 228 114
51 336 135 42 228 114
3
#4 (a) MPS for the computersweek 1 2 3 4 5 6 7 8 9 10 11
Gross requirements 220 165 180 120 75 300
Scheduled receipts 30 10
On-hand inventory 75
(a)Net requirements-MPS 145 165 150 120 65 300
MRP:Mother board (3 weeks, 1)Gross requirements 145 165 150 120 65 300
Net requirements 145 165 150 120 65 300
Time phased net reqt. 145 165 150 120 65 300
(b) Planned Order Rel. 145 165 150 120 65 300
Disk drivers: outside orders schedule (1 week, 2)Gross requirements 290 330 300 240 130 600
Net requirements 290 330 300 240 130 600
Time phased net reqt. 290 330 300 240 130 600
(c)Planned order release 290 330 300 240 130 600
4
#5MRP: DRAM Chips [Devired from Motherboard 6-4 (b)] (2weeks, 90)
1 2 3 4 5 6 7 8145 165 150 120 65 300
Gross Reqts. (x 90) 13050 14850 13500 10800 5850 27000
Scheduled Recipts 3000
On-Hand Inv 23000
Net Reqts 0 1900 13500 10800 5850 27000
Time-phased net reqt. 0 1900 13500 10800 5850 27000
(a) Planned Order
Release (VendorA) 1900 10000 10000 5850 10000
Release (VendorB) 0 3500 800 0 17000
(b) Planned order release (A)2066 10000 10000 6359 10000
Vendor A 92%(yield) 9200 9200 9200
Vendor B 96%(yield) 0 4300 1600 0 17800 (yield)
Planned rorder release (B) 0 4480 1667 0 18542
5
#6(a) POR for component A
6 7 8 9 10 11 12 13 14 15 16 17Gross Reqts. 200 200 80 80 200 400 400 400Net Reqts 200 200 80 80 200 400 400 400Time-phased net reqt. 200 200 80 80 200 400 400 400for component A Planned order release 200 200 80 80 200 400 400 400(b) POR for component B (2 weeks, 1)Gross Reqts. 100 100 40 40 100 200 200 200Net Reqts 100 100 40 40 100 200 200 200Time-phased net reqt. 100 100 40 40 100 200 200 200P.O.R 100 100 40 40 100 200 200 200(c) POR for Component C (2 weeks, 1*A+2*B)Gross Reqts. For A 200 200 80 80 200 400 400 400Gross Repts. For B 200 200 80 80 200 400 400 400Net Reqts A+B 200 400 280 160 280 600 800 800 400Time-phased net reqt. 200 400 280 160 280 600 800 800 400POR for C 200 400 280 160 280 600 800 800 400
6
# 9
(b)
Week
MPS-end item
Component B (P.O.R)
Component F
-Net. Req.
Time Phased Net. Req.
Ans. →Planned Order Release
27 28 29 30 31 32 33 34 35
165 180 300 220 200 240
330 360 600 440 400 480
330 360 600 440 400 480
330 360 600 440 400 480
330 360 600 440 400 480
7
# 9(c) Week
MPS-end item
P.O.R –Comp. B
Net Req. –Comp. ETime Phased –Net Req.P.O.R – Comp. E
P.O.R –Comp. GTime Phased –Net Req.Net Req. –Comp. G
Net Req. –Comp. I
Net Req. –Comp. H
Time Phased –Net Req.
Time Phased –Net Req.
Ans. → P.O.R –Comp. I
Ans. → P.O.R –comp. H
21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
165 180 300 220 200 240
330 360 600 440 400 480
330 360 600 440 400 480
330 360 600 440 400 480
330 360 600 440 400 480
660 720 1200 880 800 960
660 720 1200 880 800 960
660 720 1200 880 800 960
660 720 1200 880 800 960
660 720 1200 880 800 960
660 720 1200 880 800 960
1980 2160 3600 2640 2400 2880
1980 2160 36002640 2400 2880
1980 2160 36002640 2400 2880
(d)
8
Month
Demand
1 2 3 4 5 6 7 8 9 10 11 12
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
Net. Demand
# 14
1 2 3 4 5 6 7 8 9 10 11 12
2 12 4 8 15 25 20 5 10 20 5 20
• 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>
1841221 y∴
(a) Silver-Meal
9
• Starting in Period 4 :
231584 y∴
• Starting in Period 9 :
• Starting in Period 6 :
C(1) = 40
C(2) = [40+15]/2 = 27.5
C(3) = [40+15+2(25)] /3 = 35 <stop>
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>
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>
35520109 y
50520256 y∴
∴∴ 2012 y
∴ Using Silver- Meal ; y = [ 18 , 0 , 0 , 23 , 0 , 50 , 0 , 0 , 35 , 0 , 0 , 20 ]
# 14 (a) Silver-Meal
10
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>
(b) LUC
• Starting in Period 1 :
• Starting in Period 7 :
• Starting in Period 5 :
C(1) = 40 /20 = 2
C(2) = (40+5) /25 = 1.80
C(3) = [40+5+2(10)] /35 = 1.86 <stop>
C(1) = 40 /15 = 2.67
C(2) = 65 /40 = 1.63
C(3) = [65+2(20)] /60 = 1.75 <stop>
261 y
257 y
405 y
∴
∴
∴
# 14
11
(C) PPB
• Starting in Period 1 : Period Holding cost
234
1* (12) = 1212+2(4) = 2020+3(8) = 44
Closer to period 4
K = $40
26841221 y∴
• 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>
2012 y359 y
∴ Using LUC ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 25 , 0 , 35 , 0 , 0 , 20 ]∴∴
(b) LUC# 14
12
• Starting in Period 5 : Period Holding cost
123
0 2525+2(20) = 65
Closer to period 2
35105207 y
∴
• Starting in Period 7 :
• Starting in Period 10 :23
Period Holding cost
55+2(12) = 29
Period Holding cost
234
5 5+2(10) = 2525+3(20) = 85
Closer to period 3
4025155 y
452052010 y∴
∴
∴ Using PPB ; y = [ 26 , 0 , 0 , 0 ,40 , 0 , 35 , 0 , 0 , 45 , 0 , 0 ]
(C) PPB
K = $40
K = $40
K = $40
13
SM
LUC
PPB
Demand
Inv. SM
Inv. LUC
Inv. PPB
1 2 3 4 5 6 7 8 9 10 11 12
18 0 0 23 0 50 0 0 35 0 0 20
26 0 0 0 40 0 25 0 35 0 0 20
26 0 0 0 40 0 35 0 0 45 0 0
2 12 4 8 15 25 20 5 10 20 5 20
16 4 0 15 0 25 5 0 25 5 0 0
24 12 8 0 25 0 5 0 25 5 0 0
24 12 8 0 25 0 15 10 0 25 20 0
Σ = 95
Σ = 104
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 !
# 14 (d)
14
#17 K=200; h=0.3
Week 1 2 3 4 5 6Time-phased net reqts335 200 140 440 300 200(a) EOQ 599 0 599 0 599 0
Ending Inventory 264 64 523 83 382 182
(b) S-M 675 0 0 940 0 0
Ending Inventory 340 140 0 500 200 0
2( ) 599
ka EOQ
h
( )
(1) 200
(2) [200 (0.3)200]/ 2 130
(3) [200 (0.3)(200) 2(0.3)(140)] / 3 114.7
(4) [200 60 84 3(0.3)(440)] / 4 185
4
(1) 200
(2) [200 (0.3)(300)] / 2 145
(3) [200 90 2(0.3)(200)] / 3 136.7
( )
b
c
c
c
c
Start Week
c
c
top
c
S
He
1 1 2 3
54 4 6
y 675
940
r r rnce
y r r r
15
#17
5
1 3 5
Starting Week 5
200(1) 0.667
300200 60
(2) 0
300 200 500
535 ; 580 ; 50
.520300 200
0
y
Hence y y
c
y
c
1
2
1 2
200(1) 0.597
335
260 (2) 0.486
535
200 60 2(0.3)140 (3) 0.509
335 200 140Start Week 3
200(1) 1.
( ) LUC
( )
4286140200 132
(2) 0.572140 440
kc
r
k hrc
r r
c
c
S
c
c
top
1 3
4
200 132 180(3) 0
335 200 535 y 140 440 58
.582( )140 440 300
0
c Stop
y
O
(
r
)
d
er
Part
ing
-Period
Comparing
Holdi
setup cost ($200)
ng
Horizo
to Holding c
n Costs
_______
____
osts
_____
__
__
__
d
1 0
2 60
3 60+84=144
4 144+396=540
_____________
*
______
1 21
______
Starting week 4
1 0
2 90
3 90+120=210
________________
*
__
y
________
r r 3
4 5 64
675
y 940
r
r r r
16
#17Week 1 2 3 4 5 6Time-phased net reqts 335 200 140 440 300 200(a) EOQ 599 0 599 0 599 0
Ending Inventory 264 64 523 83 382 182(b) S-M 675 0 0 940 0 0
Ending Inventory 340 140 0 500 200 0
(c) LUC 535 0 580 0 500 0
Ending Inventory 200 0 440 0 200 0(d) Part-Period 675 0 0 940 0 0
Ending Inventory 340 140 0 500 200 0
( ) : ($200*3) + (0.3)
(264+64+523+83+382+182) =
( ) S-M : ($200*2) + (0.3)
:
$200*6=
(340+1
$1049.4
$740+500+200) =
$1
( ) L
20
54
0
UC :
Lot for
a EOQ
b
L t
c
o
($200*3) + (0.3)(840) =
( ) Part-Period : the same as S-M =
$852
$754d
17
#18
4 1 3
(1)(1,1,1,1) (23,86,40,21) $1200
(2)(1,1,1,0) (23,86,52,0) $900+3(12)=$936
(3)(1,1,0,1) (23,126,0,12) $900
$300; $3 (23
+3(40)=$1020
(4)(1,
(
,86,40,12
1,0,0) (23,138,0,0)
) 2
$600+3(
2
)
8
k
a
h r
52+12)=$792
(5)(1,0,1,1) (109,0,40,12) $900+3(86)=$1158
(6)(1,0,1,0) (109,0,52,0) $600+3(86+12)=$894
(7)(1,0,0,1) (149,0,0,12) $600+3(126+40)=$1098
(8)(1,0,0,0) (161,0,0,0) $300+3(138+52+12)=$906
Hence
the optimal costs = $792 (23,138,0,0)
18
#18 (cont’d)
4
33 4
3 43
22 3
32 2 4
42
11 2
21
1
300
300 300 600min
300 3(12)
300 336 636
mi
$300; $3 (23,86,40,12)
( )
336#
492#
79
n 300 3(40) 300 720
300 3(40 24)
300 492
min
2#
k h r
C
C CC
C
C C
C C C
C
C C
C
b
CC 3
31 4
1 2 2 3 4
41
300 3(86) 336 894
300 3(86 80) 300 1098
300 3(86 80 3
y 23; 138
Hence the optimal cost $792
6) 906
y r r r
C C
C
19
h = $0.4 K = $180
Starting Inventory Week 6 is 75 Receiving: 30 & 10 in week 8 & 10
# 24
Week
Net Demand
Demand
6 7 8 9 10 11
220 165 180 120 75 300
145 165 150 120 65 300
(a) PPB :
• Starting in Period 1 :
• Starting in Period 4 : Holding costs
Holding costsWeek
Period
123
0 165*0.4= 66 66+2(0.4)(150)= 186
Closer to period 34601501651451 y
485300651204 y∴
∴
123
0 65*0.4= 26 26+2(300)(0.4)= 266
Closer to period 3
K = $180
K = $180
20
# 24
MRP - Mother Boards
Week
Net. Req.
Time-Phased Net Req.
P.O.R. (lot-for-lot)
Ans. → ( PPB ) P.O.R
For DRAMAns. → Gross Req.
For DRAMTime-Phased Net Req.For DRAM P.O.R.
1 2 3 4 5 6 7 8 9 10 11
145 165 150 120 65 300
145 165 150 120 65 300
145 165 150 120 65 300
460 0 0 485 0 0
41,400 0 0 43,650 0 0
41,400 0 0 43,650 0 0
41,400 0 0 43,650 0 0
21
#25Week 8 9 10 11 12 13 14 15 16 17End Products 100 100 40 40 100 200 200 200L-F-L Planned OrderRelease ~ A 200 200 80 80 200 400 400 400S-M (A) 560 0 0 0 600 0 800 0L-F-L Planne OrderRelease ~ B 100 100 40 40 100 200 200 200S-M (B) 280 0 0 0 300 0 400 0Gross Reqts. ( C)C=(1*A+2*B) 560 560 0 0 600 600 800 800 0Gross Reqts. ( D)D=2*A 1120 0 0 0 1200 0 1600 0Gross Reqts. (E)E=3*B 840 0 0 0 900 0 1200 0 0
22
#25 (cont’d)
1 1 2 3 4
(1) 100
(2) (100 30) / 2 65
(3) (100 30 24) / 3 51.33
(4) (100 30 24 36) / 4 47
: K=100, h=0.15
(
.5
(5) (100 30 24 36 60) / 5 50
(1) 100
(2) (100 60) / 2 80
(3) (1
)
56
0
0
0 60 120) /
S M A
Stop
y r
S M
c
c
c
c
c
c
c
r r r
c
5 5 6
7 7 8
3 93
(1) 100
(2) (100 60)
( )
200 400 600
400 40
/ 2
0 8 0
80
0
c
Stop
y r r
y r
c
r
1 1 2 3 4
5 5 6
7 7 8
(1) 100
(2) (100 25) / 2 62.5
(3) (100 25 20) / 3 48.3
(4) (145 30) / 4 43.75
(5) 43.75
(1) 100
(2) 1
: K=100, h=0.25
( )
280
( )
1
50 / 2 75
(3
00 200 300
20
0
) 75
c
S M B
Stop
y r r r r
Stop
c
c
c
c
c
c
y r r
y r r
c
200 400
23
#28(a) K=$200 ; h=$0.30r 335 200 140 440 300 200Capacity 600 600 600 400 200 200 335 200 140 400 200 200 180 280 r' 335 200 280 400 200 200 Initial Solution 1,272
Excess 265 400 320 0 0 0
320 1200 0 0
r'' 335 280 600 400 0 0 1,196535 0 480 400 200 0 1,112
Cost =4*200+(0.3)(80+540+500+200)=$1196
r'' 535 0 480 400 0 200 1,052Week 2 -> Week 1 : 200 units
0.3*200 = $60
Cost =4*200+(0.3)(200+340+300)=$1052
24
#28(b)
17(b) S-M: ($1052 - $754) / ($754) = 40 %17(c) LUC: ($1052 - $852) / ($852) = 23 %17(d) PPB: ($1052 - $754) / ($754) = 40 %
25
# CW.3# CW.3 Consider the example presented previously for the scheduling of the valve casing assembly. Suppose that the production capacity in any week is 50 valve casings. Determine the feasible planned order release for the valve casings. Recall that the time phased net requirements for the valve casings as followed:
Week
Net Requirements
Time-Phased Net
Requirements r =
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
26
Week
Net Requirements
Time-Phased Net
Requirements r =
4 5 6 7 8 9 10 11 12 13 14 15 16 17
42 42 32 12 26 112 45 14 76 38
42 42 32 12 26 112 45 14 76 38
Production c=Capacity
# CW.3# CW.3
excess (c-r) =Capacity
8 8 18 38 24 (62) 5 36 (26) 12
(c-r)’ =
[2] Lot-shifting technique (back-shift demand from rj > cj):
50 50 50 50 50 50 50 50 50 50
8 8 18 38 24 (62) 5 36 (26) 12
8 8 18 0 0 0 5 10 0 12 (c-r)’ =
final r ’ = 42 42 32 50 50 50 45 40 50 38
1 1
1,2, ,j j
i ii i
C for j n
[1]First test for: It is okay!
27
# CW.4 # CW.4 ( continues on #CW.3)( continues on #CW.3)
Suppose that with overtime work on 2nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan.Determine the optimal production plan.
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Production c=Capacity (O-T)
120 120 120 120 120 120 120 120 120 120
28
# CW.4 # CW.4 ( continues on #CW.3)( continues on #CW.3)
Suppose that with overtime work on 2nd shifts, the company could increase the weekly production capacity to 120 valve casings, however, extra cost per week = $105. Where K=$100, is the regular setup cost. The holding cost per valve casing per week is estimated to be $0.65. Determine the optimal production plan.Determine the optimal production plan.
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity
50 50 50 50 50 50 50 50 50 50
final r ’ = 42 42 32 50 50 50 45 40 50 38 (using regular shift)
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
0 0 0 38 62 0 0 26 0 0 Σ= 126 Ending Inventories =
29
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Production c=Capacity (O-T)
excess (c-r) =Capacity
120 120 120 120 120 120 120 120 120 120
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
# CW.4# CW.4[1] First, the cost for using regular shift is $100(10) + $0.65 (126)
= $1,081.9 [ lot for lot ][ lot for lot ]
78 78 88 108 94 8 75 106 44 82
42 42 32 12 26 112 45 14 76 38r = 78 78 88 108 94 8 75 106 44 excess (c-r)’=
Capacity 35 77 0 31 43 0 6
42 42 32 12 26 112 45 14 76 38
85 43 120 89 77 120 59 0 114 0
0 0 0 0
85 0 120 0 0 120 0 0 114 0 final r ’ =
30
# CW.4# CW.4
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
85 0 120 0 0 120 0 0 114 0 final r ’ =
[2] The cost for using Overtime shift is $205(4) + $0.65(372) = $1061.8
43 1 89 77 51 59 14 0 38 0 Ending Inventories =
Less than the cost for using regular shift $1,081.9, Saved $ 20.10
31
# CW.4# CW.4
Ending Inventories =
[3] To think about the following solution:
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Suppose r ’=
74 32 0 38 62 0 0 26 0 0 Σ= 232
116 0 0 50 50 50 45 40 50 38 [ One OT, 7 regular ]
The cost for using only one Overtime shift on week 4
is $205(1) + $100(7) + $0.65(232) = $1055.8
Less than the cost for using regular shift $1,081.9, Saved $ 26.1Less than the cost for using all Overtime shift $1061.8 Saved $ 6.0
WHY ?
32
# CW.4# CW.4
Ending Inventories =
[4] A Better Solution :
Time-Phased Net
Requirements r =42 42 32 12 26 112 45 14 76 38
Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17
Suppose r ’=
74 32 0 27 1 9 14 0 38 0 Σ= 195
116 0 0 39 0 120 50 0 114 0
The cost for using the above solution
is $205 (3) + $100 (2) + $0.65(195) = $ 941.75
Less than the cost for using regular shift $1,081.9, Saved $ 140.05
Wow !
33
#33Week 6 7 8 9 10 11 12 13 14 15 16POR A 200 200 80 80 200 400 400 400POR B 100 100 40 40 100 200 200 200POR C 200 400 280 100 280 600 800 800 400A(0.84) 239 239 96 96 239 477 477 477B(0.92) 109 109 44 44 109 218 218 218C(0.70) 286 572 400 143 400 858 1143 1143 572
34
The EndThe End
35
End Item
A(2)1 week
B(1)2 weeks
C(1)2 weeks
D(2)1 week
C(2)2 weeks
E(3)2 weeks
