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Material Management Class Note # 1-A MRP – Capacity Constraints. Prof. Yuan-Shyi Peter Chiu Feb. 2011. § M1: Push & Pull Production Control System MRP : Materials Requirements Planning (MRP) ~ PUSH JIT : Just-in-time (JIT) ~ PULL Definition (by Karmarkar, 1989)

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slide1

Material Management

Class Note #1-A

MRP – Capacity Constraints

Prof. Yuan-Shyi Peter Chiu

Feb. 2011

slide2
§ M1: Push & Pull Production Control System
  • MRP: Materials Requirements Planning (MRP) ~ PUSH
  • JIT:Just-in-time (JIT) ~ PULL
  • Definition(by Karmarkar, 1989)

A pull system initiates production as a reaction to present demand, while

A push system initiates production in anticipation of future demand

Thus, MRP incorporates forecasts of future demand while JIT does not.

slide3
§ M2: MRP ~ Push Production Control System

We determine lot sizes based on forecasts of future demandsand possibly on cost considerations

  • Atop-downplanning system in that all production quantity decisions are derived from demand forecasts.
  • Lot-sizing decisionsare found for every level of the production system. Item are produced based on this plan and pushed to the next level.
slide4
§ M2: MRP ~ Push Production Control System ( p.2 )
  • A production plan is a complete spec. of
    • The amounts of final product produced
    • The exact timing of the production lot sizes
    • The final schedule of completion
  • The production plan may be broken down into several component parts
    • The master production schedule (MPS)
    • The materials requirements planning (MRP)
    • The detailed Job Shop schedule
  • MPS - a spec. of the exact amounts and timing of production of each of the end items in a production system.
slide6
§ M2: MRP ~ Push Production Control System ( p.4 )
  • The data sources for determining the MPS include
    • Firm customer orders
    • Forecasts of future demand by item
    • Safety stock requirements
    • Seasonal plans
    • Internal orders
  • Three phases in controlling of the production system

Phase 1: gathering & coordinating info to develop MPS

Phase 2: development of MRP

Phase 3: development of detailed shop floor and resource requirements from MRP

slide7
§ M2: MRP ~ Push Production Control System ( p.5 )
  • How MRP Calculus works:
    • Parent-Child relationships
    • Lead times into Time-Phased requirements
    • Lot-sizing methods result in specific schedules
slide8
§ M3: JIT ~ Pull Production Control System

Basics :

  • WIP is minimum.
  • A Pull system ~ production at each stage is initiated only when requested.
  • JIT extends beyond the plant boundaries.
  • The benefits of JIT extend beyond savings of inventory-related costs.
  • Serious commitment from Top mgmt to workers.

Lean Production ≈ JIT

slide9
§ M4: The Explosion Calculus (BOM Explosion)

Gross Requirements of one level

Push down

Lower levels

slide10

Fig.7-5 p.353

Trumpet

( End Item )

§ M4: The Explosion Calculus (page 2)

Eg. 7-1

Bell assembly (1)

Lead time = 2 weeks

Valve casing

assembly (1)

Lead time = 4 weeks

b-t-13

b-t-14

Slide assemblies (3)

Lead time = 2 weeks

Valves (3)

Lead time = 3 weeks

b-t-15

slide11
§ M4: The Explosion Calculus (page 3)

=>Steps

  • Predicted Demand (Final Items)
  • Net demand (or MPS)
    • Forecasts
    • Schedule of Receipts
    • Initial Inventory
  • Push Down to the next level (MRP)
    • Lot-for-lot production rule (lot-sizing algorithm) – no inventory carried over.
    • Time-phased requirements
    • May have scheduled receipts for different parts.
  • Push all the way down
slide12

Eg. 7-1

1 Trumpet

1 Bell Assembly

1 Valve casing Assembly

  • Slide Assemblies
  • 3 Valves
  • 7 weeks to produce a Trumpet ?
  • To plan 7 weeks ahead
  • The Predicted Demands:

Week

8 9 10 11 12 13 14 15 16 17

Demand

77 42 38 21 26 112 45 14 76 38

  • Expected schedule of receipts

Week

8 9 10 11

12 0 6 9

Scheduled receipts

slide13

Beginning inventory = 23, at the end of week 7

  • Accordingly the net predicted demands become

8 9 10 11 12 13 14 15 16 17

Week

Net Predicted

Demands

42 42 32 12 26 112 45 14 76 38

Master Production Schedule (MPS) for the end product (i.e. Trumpet)

  • MRP calculations for the Bell assembly (one bell assembly

for each Trumpet) & Lead time = 2 weeksgo-see-10

6 7 8 9 10 11 12 13 14 15 16 17

Week

Gross

Requirements

42 42 32 12 26 112 45 14 76 38

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

Planned Order

Release (lot for lot)

42 42 32 12 26 112 45 14 76 38

slide14

MRP Calculations for the valve casingassembly (1 valves

casing assembly for each Trumpet) & Lead time = 4 weeksgo-see-10

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Gross

Requirements

42 42 32 12 26 112 45 14 76 38

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

Planned Order

Release (lot for lot)

42 42 32 12 26 112 45 14 76 38

b-t-20

b-t-38

slide15

MRP Calculations for the valves( 3 valves for each valve casing

  • assembly)go-see-10
  • Lead Time = 3 weeks
  • On-hand inventory of 186 valves at the end of week 3
  • Receipt from an outside supplier of 96 valves at the start of week 5
  • MRP Calculations for the valves

Week

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

Gross

Requirements

126 126 96 36 78 336 135 42 228 114

Scheduled Receipts

96

On-hand inventory

18660 30

Net

Requirements

0 0 66 36 78 336 135 42 228 114

Time-Phased

Net Requirements

66 36 78 336 135 42 228 114

Planned Order

Release (lot for lot)

66 36 78 336 135 42 228 114

slide16

§. M4.1: Class Work # CW.1

  • 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

Show the MRP Calculations for the slide assemblies !

◆1g-s-62

Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

slide17

To Think about …

  • Lot-for-Lot may not be feasible ?!

e.g. 336 Slide assemblies required at week 9 may exceeds

plant’s capacity of let’s say 200 per week.

  • Lot-for-Lot may not be the best way in production !?
  • Why do we have to produce certain items (parts) every week?
  • why not in batch ? To minimize the production costs.
slide18

§. M4.2: Class Problems Discussion

Chapter 7 : ( # 4, 5,6 ) p.356-7

( # 9 (b,c,d) ) p.357

Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

slide19
§ M5: Alternative Lot-sizing schemes
  • Log-for-log : in general,not optimal
  • If we have a known set of time-varying demands and costs of setup & holding, what production quantities will minimize the total holding & setup costs over the planning horizon?
slide20

(1)EOQ Lot sizing

(page 2)

  • MRP Calculation for the valve casing assembly when applying E.O.Q.
  • lot sizing Technique instead of lot-for-lot (g-s-14)

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

Planned order

release (EOQ)

139 0 0 0 139 0 139 0 0 139

Planned deliveries

139 0 0 0 139 0 139 0 0 139

Ending inventory

97 55 23 11 124 12 106 92 16 117

slide21

Ending

  • Inventory

Beginning

Inventory

Planning

Deliveries

Net

Requirements

=

+

-

  • Total ordering ( times ) = 4 ; cost = $132 * 4 = $528
  • Total ending inventory = = 653 ;

cost = ($0.6) (653) = $391.80

Total Costs

= Setup costs + holding costs

= 4*132+$0.6*653 = $919.80

vs. lot-for-lot 10*132 = $1320 (setup costs)

g-b-41

slide22
§ M5: Alternative Lot-sizing schemes (page 3)

(2) The Silver-Meal Heuristic (S-M)

  • Forward method ~ avg. cost per period (to span)
  • Stop when avg. costs increases.
  • i.e. Once c(j) > c(j-1) stop

Them let y1 = r1+r2+…+rj-1 and begin again starting at period j

slide23

§ M5: Alternative Lot-sizing schemes

  • The silver-meal heuristicWill Not Alwaysresult in an optimal solution (see eg.7.3; p.360)
  • Computing Technology enables heuristic solution

● S-M example 1 :

  • Suppose demands for the casings are r = (18, 30, 42, 5, 20)
  • Holding cost = $2 per case per week
  • Production setup cost = $80

Starting in Period 1 :

C(1) = $80

C(2) = [$80+$2(30)] /2 = $70

C(3) = [$80+$2(30)+$2(2)(42)] /3 =308/3 = $102.7

∵ C(3) >C(2) ∴ STOP ; Set

slide24

Starting in Period 3 :

  • r = (18, 30, 42, 5, 20)

C(1) = 80

C(2) = [80+2(5)] /2 = 45

C(3) = [80+2(5)+$2(2)(20)] /3 = 170/3 = 56.7

∵ C(3) >C(2) ∴ STOP ; Set

∴ Solution = (48, 0, 47, 0, 20) cost = $310

● S-M example 2 : (counterexample)

Let r = (10, 40, 30) , k=50 & h=1

Silver-Meal heuristic gives the solution y=(50,0,30)

but the optimal solution is (10,70,0)

Conclusion of Silver-Meal heuristic

  • It will not always result in an optimal solution
  • The higher the variance (in demand) , the better the
  • improvement the heuristic gives (versus EOQ)
slide25
§ M5: Alternative Lot-sizing schemes(page 4)

(3) Least Unit Cost (LUC)

  • Similar to the S-M except it divided by total demanded quantities.
  • Once c(j) > c(j-1) stop and so on.
slide26

● LUC example:

r = (18, 30, 42, 5, 20)

h = $2

K = $80

Solution :in period 1

C(1) = $80 /18 = $4.44

C(2) = [80+2(30)] /(18+30) = 140/48 = $2.92

C(3) = [80+2(30)+2(2)(42)] /(18+30+42) = 308/90 = $3.42

∵ C(3) >C(2) ∴ STOP ; Set

Starting in period 3

C(1) = $80 /42 = 1.90

C(2) = [80+2(5)] /(42+5) = 90 /47 = 1.91

∵ C(3) >C(2) ∴ STOP ; Set

slide27

r = (18, 30, 42, 5, 20)

Starting in period 5

C(1) = $80 /5 = 16

C(2) = [80+2(20)] /(5+20) = 120 /25 = 4.8

∴ Set

∴ Solution = ( 48, 0, 42, 25, 0) cost = 3(80)+2(30)+2(20) = $340

slide28
§ M5:Alternative Lot-sizing schemes(page 5)

(4) Part Period Balancing (PPB)

  • More popular in practice
  • Set the order horizon equal to “# of periods”

~ closely matches total holding cost closely with the setup cost over that period.

  • Closer rule

Eg. 80 vs. (0, 10, 90) then choose 90

Last three : S-M, LUC, and PPB are heuristic methods ~means reasonable but not necessarily give the optimal solution.

slide29

● PPB example :

r = (18, 30, 42, 5, 20)

h = $2

K = $80

Starting in Period 1

Order

Horizon

Total Holding

cost

1

2

3

0

60 (2*30)

228 (2*30+2*2*42)

K=80

∵ K is closer to period 2

slide30

r = (18, 30, 42, 5, 20)

h = $2

K = $80

Starting in Period 3 :

Order

Horizon

Total Holding

cost

1

2

3

0

10 (2*5)

90 (2*5+2*2*20)

K=80

∵ K is closer to period 3

∴ Solution = (48, 0, 67, 0, 0)

cost = 2(80)+2(30)+2(5)+2(2)(20) = $310 #

slide31

§. M5.1: Class Problems Discussion

Chapter 7 : ( # 14,17 ) p.363

Preparation Time : 25 ~ 40 minutes Discussion : 15 minutes

slide32
§ M6:Wagner – Whitin Algorithm ~guarantees an optimal solution to the production planning problem with time-varying demands.

Eq.

slide33
§ M6:Wagner – Whitin Algorithm (page 2)

Eg.A four periods planning

◆2g-t-63

slide34
§ M6:Wagner – Whitin Algorithm (page 3)
  • Enumerating vs. dynamic programming

◆ Dynamic Programming

slide35

§ M6:Wagner – Whitin Algorithm (page 4)

See ‘ PM00c6-2 ‘ for Example

slide36
§ M6.1: Dynamic Programming

Eq 7.2 r =(18,30,42,5,20) h=$2 k=$80

slide37

§. M6.2: Class Problems Discussion

#1: Inventory model when demand rate λ is not constant

K=$20

C=$0.1

h=$0.02

300 200 300 200

#2: ( Chapter 7: # 18(a),(b) ) p.363

Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes

slide38
§ M7: Incorporating Lot-sizing Algorithms into the Explosion calculus

▓ From Time-phased net requirements applies algorithmp.364

  • Example 7.6

g-s-14

from the time-phased net requirements for the valve casing assembly :

Week

4 5 6 7 8 9 10 11 12 13

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

  • Setup cost = $132 ; h= $0.60 per assembly per week
  • Silver-Meal heuristic :
slide39

Starting in week 4 :

C(1) = $132

C(2) = [132+(0.6)(42)] /2 = 157.2/2 = $78.6

C(3) = [132+(0.6)(42)+(0.6)(2)(32)] /3= 195.6/3 =$65.2

C(4) = [195.6+(0.6)(3)(12)] /4 = 217.2/4 = $54.3

C(5) = [217.2+(0.6)(4)(26)] /5 = 279.6/5 = $55.9 (STOP)

Starting in week 8 :

C(1) = $132

C(2) = [132+(0.6)(112)] /2 = 199.2/2 = $99.6

C(3) = [199.2+(0.6)(2)(45)] /3= 253.2/3 =$84.4

C(4) = [253.2+(0.6)(3)(14)] /4 = 278.4/4 = $69.6

C(5) = [278.4+(0.6)(4)(76)] /5 = 460.8/5 = $92.2 (STOP)

∵ C(5) >C(4)∴

slide40

Starting in week 12 :

C(1) = $132

C(2) = [132+(0.6)(38)] /2 = $77.4

∴ y = (128 , 0 , 0 , 0 , 197 , 0 , 0 , 0 , 114 , 0)

  • MRP Calculation using Silver-Meal lot-sizing algorithm :

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

Planned Order

Release (S-M)

128 0 0 0 197 0 0 0 114 0

Planned deliveries

128 0 0 0 197 0 0 0 114 0

Ending inventory

86 44 12 0 171 59 14 0 38 0

slide41

▓ Compute the total costs

  • S-M : Total cost = 132(3)+(0.6)(86+44+12+171+59+14+38) = $650.4
  • Lot-For-Lot : $132*10 = $1320

g-s-14

  • E.O.Q : 4(132)+(0.6)(653) = $919.80

g-t-20

  • for optimal schedule by Wagner-Whitin algorithm it is
  • y4=154 , y9=171 , y12=114 ; Total costs= $610.20

▓ push down to lower level…

slide42

§. M7.1: Class Work# CW.2

  • Applies Least Unit Cost in MRP Calculation for
  • the valve casing assembly.

  • Applies Part Period Balancing in MRP Calculation
  • for the valve casing assembly.

◆3g-t-64

  • Applies Wagner-Whitin algorithm in MRP for the

valve casing assembly.

Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

slide43

§. M 7.2: Class Problems Discussion

Chapter 7 : ( # 24, 25 ) p.365-6

( # 49 ) p.393

Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes

slide44

§ M8:Lot sizing with Capacity Constraints

▓ Requirements vs production capacities.

’’realistic’’~more complex.

▓ True optimal is difficult, time-consuming and probably not practical.

▓ Even finding a feasible solution may not be obvious.

▓ Feasibility condition must be satisfied

e.g. Demand r = ( 52 , 87 , 23 , 56 ) Total demands = 218

Capacity C= ( 60 , 60 , 60 , 60 ) Total capacity = 240

though total capacity > total demands ;

but it is still infeasible (why?)

slide45

§ M8:Lot sizing with Capacity Constraints(page 2)

▓ Lot-shifting technique to find initial solution

▓ Eg. #7.7 (p.376)γ=(20,40,100,35,80,75,25)

C =(60,60,60,60,60,60,60)

◆ First tests for Feasibility condition → satisfied

◆ Lot-shifting

C = (60,60, 60,60,60,60,60)

γ = (20,40,100,35,80,75,25)  demand

(C-γ) = (40,20,-40,…)

(C-γ)’ =(20, 0, 0,…)

(production plan)γ’= (40,60,60,35,80,75,25) [γ’=C- (C- γ)’]

(C-γ’)’ = (20,0,0,25,-20,…)

(C-γ’)’ = (20,0,0,5,0,…)

γ’ = (40,60,60,55,60,75,25) [γ’=C- (C- γ’)’]

slide46

§ M8:Lot sizing with Capacity Constraints(page 3)

(C-γ’)’ = (20,0,0,5,0,-15,…)

(C-γ’)’ = (10,0,0,0,0,0,…)

γ’ = (50,60,60,60,60,60,25) [γ’=C- (C- γ’)’]

(C-γ’)’ = (10,0,0,0,0,0,35)

(production plan)γ’= (50,60,60,60,60,60,25)

∴ lot-shifting technique solution (backtracking)gives a feasible solution.

▓ Reasonable improvement rules for capacity constraints

◆ Backward lot-elimination rule

slide47

§ M8:Lot sizing with Capacity Constraints(page 4)

◆ Eg. 7.8

Assume k=$450 , h=$2

C = (120,200,200,400,300,50,120, 50,30)

γ= (100, 79,230,105, 3,10, 99,126,40)

from lot-shifting γ’=?

γ’ = (100,109,200,105,28,50,120,50,30) [ How ? ]

costs = (9*$450)+2*(216)=$4482

◆ Improvement

Find Excess capacity first.

C = (120,200,200,400,300,50,120, 50,30)

γ’ = (100,109,200, 105, 28, 50,120, 50,30)

(C - γ’) = ( 20, 91, 0, 295,272, 0, 0, 0, 0)

slide48

§ M8:Lot sizing with Capacity Constraints(page 5)

◆ Is there enough excess capacity in prior periods to consider shifting this lot?

  • excess capacity:(C –γ’) = (20,91,0,295,272,0,0,0,0)
  • γ’ = (100,109,200,105,28,50,120,50,30)
  • ∵ 30 units shifts from the 9th period to the 5th period

242

192

142

58

108

158

slide49

§ M8:Lot sizing with Capacity Constraints(page 6)

  • ∵ 50 units shifts from the 8th period to the 5th
  • ∵ 120 units shifts from the 7th period to the 5th [not Okay]
  • ∵ okay to shift 50 from the 6th period to the 5th
  • Result :

→ γ’ = (100,109,200,105,158,0,120,0,0)

slide50

∵ Furthermore, it is okay to shift 158 from the 5th period to the 4th period

  • 263 0
  • → γ’ = (100,109,200,105,158,0,120,0,0)
  • (C-γ’) = (20,91,0,295,142,50,0,50,30)
  • 137 300

Excess capacity

∵ 158 units shifts from the 5th period to the 4th increase holding cost by $2*158=$316 < $K “ okay ’’

→ final γ’ = (100,109,200,263,0,0,120,0,0)

slide51

§ M8:Lot sizing with Capacity Constraints(page 7)

◆ after improvement; total cost = [ 5*$450+ $2*(694) ] = $3638

vs { $4482 (before improvement)

where γ’ = (100,109,200,105,28,50,120,50,30) }

◆ improvement save 20% of costs

slide52

§. M 8.1: Class Problems Discussion

Chapter 7 : # CW.3 ; # 28(a)(b)p.369

# CW.5 ; #CW.4

Preparation Time : 25 ~ 30 minutes Discussion : 15 minutes

slide53

# CW.5

Consider problem #28 (a), suppose the setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week, and the time-phased net requirements and production capacity for the base assembly in a table lamp over the next 6 weeks are:

Week

1 2 3 4 5 6

Time-Phased Net

Requirements r =

335 200 140 440 300 200

Production c=

Capacity

600 600 600 400 200 200

  • Determine the feasible planned order release
  • Determine the optimal production plan
slide54
§ M 9: Shortcoming of MRP

■ Uncertainty◆ forecasts for future sales

◆ lead time from one level to another

■ Two implication in MRP

  • all of the lot-sizing decisions could be incorrect.
  • former decisions that are currently being implemented
  • in the production process may be incorrect.

■Safety stockto protect against the uncertainty of demand

◆ not recommended for all levels

◆ recommended for end products only, they will be transmitted down thru the explosion calculus.

slide55

§ M 9: Shortcoming of MRP ( page 2 )

■ Applies the coefficient variationσ/μ

◆ obtain σ, find →ratio = ∴ σ=μxratio

◆ obtainsafety stockσx z(e.g. z = 1.28 → 90%)

◆ obtain (μ+σ*z ) as planned production schedule.

slide56

Example 7.9(p.381)[ Using a Type 1 service lever of 90 %]

  • Consider example 7.1 (p.362) Demands for Trumpets
  • If analyst finds that the ratio σ/μ (coefficient of variation) is0.3
  • Harmon co. decided to produce enough Trumpets to meet
  • all weekly demand with probability 0.90
  • 0.90 for Normally Distributed demand has a Z = 1.28

Week

8 9 10 11 12 13 14 15 16 17

Predicted

Demand ( μ)

77 42 38 21 26 112 45 14 76 38

Standard

Deviation ( σ= μ*0.3 )

23.1 12.6 11.4 6.3 7.8 33.6 13.5 4.2 22.8 11.4

Mean demand

Plus safety stock

( μ+ z σ )

107 58 53 29 36 155 62 19 105 53

[ i.e. μ+(1.28) σ ]

slide57

§ M 9: Shortcoming of MRP (page 3)

■ Capacity Planning

◆ Feasible solution at one level may result in an

‘’ infeasible ’’ requirements schedule at a lower level.

◆ CRP – Capacity requirements planning by using MRP

planned order releases.

~ If CRP results in an ‘’ infeasible ’’ case then to correct it by

◇ schedule overtime, outsourcing

◇ revise the MPS

~ Trial & Error between CRP and MRP until fitted.

slide58

§ M 9: Shortcoming of MRP (page 3)

▓ Rolling Horizons and System Nervousness

◆ MRP is not always treated as a static system.

~ may need to rerun each period for 1st period decision

▓ Other considerations

◆ Lead times is not always dependent on lot sizes

~ sometimes lead time increases when lot size increases

◆ MRP Ⅱ:Manufacturing Resource Planning

◇ MRP converts an MPS into planned order releases.

◇ MRP Ⅱ:Incorporate Financial , Accounting ,

& Marketing functions into the production

planning process

slide59

§ M 9: Shortcoming of MRP (page 4)

Ultimately, all divisions of the company would work together to find a production schedule consistent with the overall business plan and long-term financial strategy of the firm.

◇ MRP Ⅱ:~ incorporation of CRP

◆ Imperfect production Process

◆ Data Integrity

slide60

§. M 9.1: Class Problems Discussion

Chapter 7 : ( # 33 ) p.376

Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes

slide61
§ M 10:J I T

◆Kanban

◆ SMED (Single minute exchange of dies)

‧IED (inside exchange of dies )

‧OID (out side exchange of dies )

◆ Advantages vs. Disadvanges (See Table 6-1)

§ M 11: MRP & JIT

36 distinct factors to compare JIT, MRP, & ROP (reorder point) [Krajewski et al 1987]

slide63

# CW.1

◆1

  • Solution: MRP Calculations for the Slide assemblies ( 3 )
  • Lead Time = 2 weeks
  • On-hand inventory of 270 valves at the end of week 3
  • Receipt from an outside supplier of78 & 63 at the start of week 5 & 7
  • MRP Calculations for the valves

Week

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

Gross

Requirements

126 126 96 36 78 336 135 42 228 114

63

Scheduled Receipts

78

On-hand inventory

270144 96 27

Net

Requirements

0 0 0 0 51 336 135 42 228 114

Time-Phased

Net Requirements

51 336 135 42 228 114

g-b-16

Planned Order

Release (lot for lot)

51 336 135 42 228 114

slide64

◆2

g-b-33

slide65

# CW.2

◆3

  • Solution: Applies Part Period Balancing in MRP Calculation
  • for the valve casing assembly.
  • MRP Calculation using Part Period Balancing lot-sizing algorithm :

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

Planned Order

Release (PPB)

?

Starting in Period 4:

Order Horizon

Total Holding cost

1

2

3

4

5

0

$25.2 (0.6)*(42)

$63.6 $25.2+2(0.6)(32)

$85.2 $63.6+3(0.6)(12)

$147.6 $85.2+4(0.6)(26)

∵ K is closer to period 5

K=132

slide66

# CW.2

◆3

Starting in Period 9:

Order Horizon

Total Holding cost

1

2

3

4

0

$27 (0.6)*(45)

$43.8 $27+2(0.6)(14)

$180.6 $43.8+3(0.6)(76)

∵ K is closer to period 4

K=132

  • MRP Calculation using Part Period Balancing lot-sizing algorithm :

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

Time-Phased

Net Requirements

42 42 32 12 26 112 45 14 76 38

P.O.R. (PPB)

154 0 0 0 0 247 0 0 0 38

Planned deliveries

154 0 0 0 0 247 0 0 0 38

Ending inventory

112 70 38 26 0 135 90 76 0 0

slide67

# CW.2

◆3

▓ Compute the total costs

  • PPB : Total cost = $132(3)+(0.6)(547) = $724.2
  • S-M : $650.4
  • Lot-For-Lot : $132*10 = $1320
  • E.O.Q : $919.80
  • foroptimal schedule by Wagner-Whitin algorithm it is
  • y4=154 , y9=171 , y12=114 ; Total costs= $610.20

g-s-42

slide68

# 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

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

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

slide69

# CW.3

[1]

First test for:

It is okay!

[2] Lot-shifting technique (back-shift demand from rj > cj):

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

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

excess (c-r) =

Capacity

8 8 18 38 24 (62) 5 36 (26) 12

(c-r)’ =

8 8 18 38 24 (62) 5 36 (26) 12

(c-r)’ =

8 8 18 0 0 0 5 10 0 12

final r ’ =

42 42 32 50 50 50 45 40 50 38

slide70

# CW.4 ( 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.

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

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

Production c=

Capacity (O-T)

120 120 120 120 120 120 120 120 120 120

slide71

# CW.4 ( 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.

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

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)

Ending Inventories =

0 0 0 38 62 0 0 26 0 0 Σ= 126

[1] First, the cost for using regular shift is$100(10) + $0.65 (126)

= $1,081.9 [ lot for lot ]

slide72

# CW.4

[1] First, the cost for using regular shift is$100(10) + $0.65 (126)

= $1,081.9 [ lot for lot ]

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Time-Phased Net

Requirements r =

42 42 32 12 26 112 45 14 76 38

Production c=

Capacity (O-T)

120 120 120 120 120 120 120 120 120 120

excess (c-r) =

Capacity

78 78 88 108 94 8 75 106 44 82

r =

42 42 32 12 26 112 45 14 76 38

excess (c-r)’=

Capacity

78 78 88 108 94 8 75 106 44

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

final r ’ =

85 0 120 0 0 120 0 0 114 0

slide73

# CW.4

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Time-Phased Net

Requirements r =

42 42 32 12 26 112 45 14 76 38

final r ’ =

85 0 120 0 0 120 0 0 114 0

Ending Inventories =

43 1 89 77 51 59 14 0 38 0

[2] The cost for using Overtime shift is$205(4) + $0.65(372) = $1061.8

Less than the cost for using regular shift $1,081.9, Saved $ 20.10

slide74

# CW.4

[3] To think about the following solution:

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Time-Phased Net

Requirements r =

42 42 32 12 26 112 45 14 76 38

Suppose r ’=

116 0 050 50 50 45 40 50 38[ One OT, 7 regular ]

Ending Inventories =

74 32 0 38 62 0 0 26 0 0 Σ= 232

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 $ 26.1

Less than the cost for using all Overtime shift $1061.8 Saved $ 6.0

WHY ?

slide75

# CW.4

[4] A Better Solution :

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Time-Phased Net

Requirements r =

42 42 32 12 26 112 45 14 76 38

Suppose r ’=

116 0 0 39 0 120 50 0 114 0

Ending Inventories =

74 32 0 27 1 9 14 0 38 0 Σ= 195

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 !

42 113 0 0 0 120 50 0 114 0 $944.35

slide76

§. M4.2: Class Problems Discussion

Chapter 7 : ( # 4, 5,6 ) p.356-7

( # 9 (b,c,d) ) p.357

Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

p.18

slide77

§. M5.1: Class Problems Discussion

Chapter 7 : ( # 14,17 ) p.363

Preparation Time : 25 ~ 40 minutes Discussion : 15 minutes

p.31

slide78

§. M6.2: Class Problems Discussion

#1: Inventory model when demand rate λ is not constant

K=$20

C=$0.1

h=$0.02

300 200 300 200

#2: ( Chapter 7: # 18(a),(b) ) p.363

Preparation Time : 10 ~ 15 minutes Discussion : 10 minutes

p.37

slide79

§. M7.1: Class Work# CW.2

  • Applies Least Unit Cost in MRP Calculation for
  • the valve casing assembly.

  • Applies Part Period Balancing in MRP Calculation
  • for the valve casing assembly.

◆3g-t-64

  • Applies Wagner-Whitin algorithm in MRP for the

valve casing assembly.

Preparation Time : 25 ~ 35 minutes Discussion : 20 minutes

p.42

slide80

§. M 7.2: Class Problems Discussion

Chapter 7 : ( # 24, 25 ) p.365-6

( # 49 ) p.393

Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes

p.43

slide81

§. M 8.1: Class Problems Discussion

Chapter 7 : # CW.3 ; #CW.5 ; #CW.4

Preparation Time : 25 ~ 30 minutes Discussion : 15 minutes

slide82

# CW.5

Consider problem #28 (a), suppose the setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week, and the time-phased net requirements and production capacity for the base assembly in a table lamp over the next 6 weeks are:

Week

1 2 3 4 5 6

Time-Phased Net

Requirements r =

335 200 140 440 300 200

Production c=

Capacity

600 600 600 400 200 200

  • Determine the feasible planned order release
  • Determine the optimal production plan

p.53

slide83

# CW.5Solution

Week

1 2 3 4 5 6

Time-Phased Net

Requirements r =

335 200 140 440 300 200

Production c=

Capacity

600 600 600 400 200 200

  • Determine the feasible planned order release

(c-r) =

265 400 460 -40 -100 0

Adj.(c-r) =

265 400 320 0 0 0

r’ =

335 200 280 400 200 200

(b) Determine the optimal production plan

r’ =

335 200 280 400 200 200

p.53

(c-r’) =

265 400 320 0 0 0

slide84

# CW.5Solution (b)

Production c=

Capacity

600 600 600 400 200 200

(b) Determine the optimal production plan

r’ =

335 200 280 400 200 200

(c-r’) =

265 400 320 0 0 0

65 600 120 0 200 0

r’’ =

535 0 480 400 0 200

Increase holding cost = $0.3*(200) + $0.3*2*(200)=$180

Saving setup cost = 2*K = 2*$200= $400

Overall savings = $220

Final production plan r’’ =

535 0 480 400 0 200

slide85

§. M 9.1: Class Problems Discussion

Chapter 7 : ( # 33 ) p.376

Preparation Time : 15 ~ 20 minutes Discussion : 10 minutes

p.60

slide86

# 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

4 5 6 7 8 9 10 11 12 13 14 15 16 17

Net

Requirements

42 42 32 12 26 112 45 14 76 38

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

p.68

slide87

# CW.4 ( 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.

Week

4 5 6 7 8 9 10 11 12 13 14 15 16 17

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

Production c=

Capacity (O-T)

120 120 120 120 120 120 120 120 120 120

p.70

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