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ALTERNATIVES LOT-SIZING SCHEMES. Alternatives Lot-Sizing Schemes. The silver-meal heuristic Least Unit Cost Past Period Balancing. The Silver-Meal Heuristic. Forward method that requires determining the average cost per period as a function of the number of periods the current order to span.

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Alternatives lot sizing schemes1
Alternatives Lot-Sizing Schemes

  • The silver-meal heuristic

  • Least Unit Cost

  • Past Period Balancing


The silver meal heuristic
The Silver-Meal Heuristic

  • Forward method that requires determining the average cost per period as a function of the number of periods the current order to span.

  • Minimize the cost per period

  • Formula : C(j) = (K + hr2 + 2hr3 + … + (j-1)hrj) / j

  • C(j)  average holding cost and setup cost per period

  • k  order cost or setup cost

  • h  holding cost

  • r  demand


Method
Method

  • Start the calculation from period 1 to next period

    • C(1) = K

    • C(2) = (K + hr2) / 2

    • C(3) = (K + hr2 + 2hr3) / 3

  • Stop the calculation when C(j) > C(j-1)

  • Set y1 = r1 + r2 + … + rj-1

  • Start over at period j, repeat step (I) – (III)


Example
Example

  • A machine shop uses the Silver-Meal heuristic to schedule production lot sizes for computer casings. Over the next five weeks the demands for the casing are r = (18, 30, 42, 5, 20). The holding cost is $2 per case per week, and the production setup cost is $80. Find the recommended lot sizing.


Step i ii iii
Step I, II & III

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

k = $80

h = $2

Starting in period 1

  • C(1) = 80

  • C(2) = [80 + (2)(30)] / 2

    = 70

  • C(3) = [80 + (2)(30)+ (2)(2)(42)] / 3

    = 102.67

    Stop the calculation as the C(3) > C(2)

  • y1 = r1 + r2

    = 18 +30

    = 48


Step iv
Step IV

Starting in period 3

  • C(1) = 80

  • C(2) = [80 + (2)(5)] / 2

    = 45

  • C(3) = [80 + (2)(5)+ (2)(2)(20)] / 3

    = 56.67. Stop

  • y3 = r3+ r4

    = 42 + 5

    = 47



Least unit cost
Least Unit Cost the process again.

  • Similar to Silver-Meal method

  • Minimize cost per unit of demand

  • Formula : C(j) = (K + hr2 + 2hr3 + … + (j-1)hrj) / (r1 + r2 + … + rj

  • C(j)  average holding cost and setup cost per period

  • k  order cost or setup cost

  • h  holding cost

  • r  demand


Method1
Method the process again.

  • Start the calculation from period 1 to next period

    • C(1) = K / r1

    • C(2) = (K + hr2) / (r1 + r2)

    • C(3) = (K + hr2 + 2hr3) / (r1 + r2 + r3 )

  • Stop the calculation when C(j) > C(j-1)

  • Set y1 = r1 + r2 + … + rj-1

  • Start over at period j, repeat step (I) – (III)


Step i ii iii1
Step I, II & III the process again.

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

k = $80

h = $2

Starting in period 1

  • C(1) = 80 / 18

    = 4.44

  • C(2) = [80 + (2)(30)] / (18 + 30)

    = 2.92

  • C(3) = [80 + (2)(30)+ (2)(2)(42)] / (18+30+42)

    = 3.42

    Stop the calculation as the C(3) > C(2)

  • y1 = r1 + r2

    = 18 +30

    = 48


Step iv1
Step IV the process again.

  • Starting in period 3

  • C(1) = 80 / 42

    = 1.9

  • C(2) = [80 + (2)(5)] / (42 + 5)

    = 1.92 Stop

  • y3 = r3

    = 42

    = 42


Step iv2
Step IV the process again.

Starting in period 4

  • C(1) = 80 / 5

    = 16

  • C(2) = [80 + (2)(20)] / (5 + 20)

    = 4.8

  • y4 = r4 + r5

    = 5 + 20

    = 25

  • Thus y = (48, 0, 42, 25, 0)


Part period balancing
Part Period Balancing the process again.

  • Set the order horizon equal to the number of periods that most closely matches the total holding cost with the setup cost over that period.


Example1
Example the process again.

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

Holding cost = $2 per case per week

Setup cost = $80

Starting in period 1

  • Because 228 exceeds the setup cost of 80, we stop. As 80 is closer to 60 than to 228, the first order horizon is two periods,

  • y1 = r1 +r2 = 18 + 30 = 48

closest


Starting in period 3 the process again.

  • We have exceeded the setup cost of 80, so we stop.

  • Because 90 is closer to 80 than 10, the order horizon is three periods.

  • y3 = r3 + r4 + r5

    = 42 + 5 + 20

    = 67

  • y = (48, 0, 67, 0, 0)

closest


Comparison of results
Comparison of Results the process again.

The Silver Meal and Part Period Balancing heuristics resulted in the same least expensive costs.


Exercise 14 pg 381
Exercise 14 – pg 381 the process again.

  • A single inventory item is ordered from an outside supplier. The anticipated demand for this item over the next 12 months is 6, 12, 4, 8, 15, 25, 20, 5, 10, 20, 5, 12. Current inventory of this item is 4, and ending inventory should be 8. Assume a holding cost of $1 per period and a setup cost of $40. Determine the order policy for this item based on

    • Silver-Meal

    • Least unit cost

    • Part period balancing

    • Which lot-sizing method resulted in the lowest cost for the 12 periods?


Exercise 14 pg 3811
Exercise 14 - pg 381 the process again.

Demand = (6, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 12)

Starting inventory = 4

Ending inventory = 8

h = 1

K = 40

Net out starting and ending inventories to obtain

r = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)

a) Silver Meal

Start 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.

y1 = r1 + r2 + r3 = 2 + 12 + 4 = 18


Start in period 4 the process again.:

C(1) = 40

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

C(3) = [40 + 15 + (2)(25)]/3 = 35 Stop.

y4 = r4 + r5 = 8+15 = 23

Start in period 6:

C(1) = 40

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

C(3) = [40 + 20 + (2)(5)]/3 = 23.3333

C(4) = [40 + 20 + (2)(5) + (3)(10)]/4 = 25 Stop.

y6 = r6 + r7 + r8 = 25+20+5 = 50

Start in period 9:

C(1) = 40

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

C(3) = [40 + 20 + (2)(5)]/3 = 23.3333

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

y9 = r9 + r10 + r11 = 10+20+5=35

y12 = r12 = 20

y= (18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20)


b) Least unit cost the process again.

Start in period 1:

C(1) = 40/2 = 20

C(2) = (40 + 12)/(2 + 12) = 3.71

C(3) = (40 + 12 + 8) /(2 + 12 + 4) = 3.33

C(4) = (40 + 12 + 8 + 24) /(2 + 12 + 4 + 8) = 3.23

C(5) = (40 + 12 + 8 + 24 + 60) /(2 + 12 + 4 + 8 + 15) = 3.51 Stop.

y1 = r1 + r2 + r3 + r4 = 2 + 12 + 4 + 8= 26

Start in period 5:

C(1) = 40/15 = 2.67

C(2) = (40 + 25)/(15 + 25) = 1.625

C(3) = (40 + 25 + 40)/(15 + 25 + 20) = 1.75 Stop.

y5 = r5 + r6 = 15+25 = 40

Start in period 7:

C(1) = 40/20 = 2

C(2) = (40 + 5)/(20 + 5) = 1.8

C(3) = (40 + 5 + 20)/(20 + 5 + 10) = 1.86 Stop.

y7 = r7 + r8 = 20+5= 25


Start in period 9 the process again.:

C(1) = 40/10 = 4

C(2) = (40 + 20)/(10 + 20) = 2

C(3) = (40 + 20 + 10)/(10 + 20 + 5) = 2

C(4) = (40 + 20 + 10 + 60)/(10 + 20 + 5 + 20) = 2.3636

y9 = r9 + r10 + r11 = 10+20+5=35

y12 = r12 = 20

y= (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20)


c) Part period balancing the process again.

r = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)

h = 1

K = 40

Starting in period 1

 y1 = r1 + r2 + r3 + r4 = 2 + 12 + 4 + 8= 26

closest


closest

closest


  • Start in period 10 the process again.

     y10 = r10 + r11 + r12

    = 20+5+20

    = 45

    y = (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0, 0)

closest


Comparison of results1
Comparison of results the process again.

The Silver Meal resulted the least expensive cost.


Exercise 17 pg 381
Exercise 17 – pg 381 the process again.

  • The time-phased net requirements for the base assembly in a table lamp over the next six weeks are

  • The setup cost for the construction of the base assembly is $200, and the holding cost is $0.30 per assembly per week

    • Determine the lot sizes using the Silver-Meal heuristic

    • Determine the lot sizes using the least unit cost heuristic

    • Determine the lot sizes using part period balancing

    • Which lot-sizing method resulted in the lowest cost for the 6 periods?


Exercise 17 pg 3811
Exercise 17 pg 381 the process again.

K = $200

h = $0.30

a) Silver Meal

Start in period 1:

C(1) = 200

C(2) = [200 + (200)(0.3)]/2 = 130

C(3) = [(2)(130) + (2)(140)(0.3)]/3 = 114.67

C(4) = [(3)(114.67) + (3)(440)(0.3)]/4 = 185 Stop.

y1= = r1 + r2 + r3 = 335 + 200 + 140 = 675

Start in period 4:

C(1) = 200

C(2) = [200 + (300)(0.3)]/2 = 145

C(3) = [(2)(145) + (2)(200)(0.3)]/3 = 136.67 Stop.

y4=r4 + r5 + r6 =440 + 300 + 200 = 940

y = (675, 0, 0, 940, 0, 0)


b) Least unit cost the process again.

Start in period 1:

C(1) = 200/335 = 0.597

C(2) = [200 + (200)(0.3)]/(335 + 200) = 0.486

C(3) = [200 + (200)(0.3) + (140)(2)(0.3)]/(335 + 200 + 140) = 0.51 Stop.

y1= r1 + r2 = 335 + 200 = 535

Start in period 3:

C(1) = 200/140 = 1.428

C(2) = [200 + (440)(0.3)]/(140 + 440) = 0.572

C(3) = [200 + (440)(0.3) + (300)(2)(0.3)]/(140 + 440 + 300) = 0.58 Stop.

y3= r3 + r4 = 140 + 440 = 580

Start in period 5:

C(1) = 200/300 = 0.67

C(2) = [200 + (200)(0.3)]/(300 + 200) = 0.52 Stop.

 y5 = r5 + r6 = 300 + 200 = 500

y = (535, 0, 580, 0, 500, 0)


C part period balancing
c) Part period balancing the process again.

K = $200

h = $0.30

  • Starting in period 1

    y1= r1 + r2 + r3 = 335 + 200 + 140 = 675

closest


  • Starting in period 4 the process again.

    y4=r4 + r5 + r6 =440 + 300 + 200 = 940

    y = (675, 0, 0, 940, 0, 0)

closest


Comparison of results2
Comparison of results the process again.

The Silver Meal and Part Period Balancing heuristics resulted in the same least expensive costs.


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