ALTERNATIVES LOT-SIZING SCHEMES

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ALTERNATIVES LOT-SIZING SCHEMES
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Alternatives Lot-Sizing Schemes

• The silver-meal heuristic
• Least Unit Cost
• Past Period Balancing

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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

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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) gt C(j-1)
• Set y1 r1 r2 rj-1
• Start over at period j, repeat step (I) (III)

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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.

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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) gt C(2)
• y1 r1 r2
• 18 30
• 48

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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

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• Since period 5 is the final period, thus no need
  to start the process again.
• Set y5 r5
• 20
• Thus y (48, 0, 47, 0, 20)

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Least Unit Cost

• 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

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Method

• 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) gt C(j-1)
• Set y1 r1 r2 rj-1
• Start over at period j, repeat step (I) (III)

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Step I, II III

• 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) / (183042)
• 3.42
• Stop the calculation as the C(3) gt C(2)
• y1 r1 r2
• 18 30
• 48

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Step IV

• Starting in period 3
• C(1) 80 / 42
• 1.9
• C(2) 80 (2)(5) / (42 5)
• 1.92 Stop
• y3 r3
• 42
• 42

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Step IV

• 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)

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Part Period Balancing

• 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.

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Example

• 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

Order horizon Total holding cost
1 0
2 2(30) 60
3 2(30) 2(2)(42) 228
closest
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• Starting in period 3
• 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)

Order horizon Total holding cost
1 0
2 2(5) 10
3 2(5) 2(2)(20) 90
closest
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Comparison of Results
Silver Meal Least Unit Cost Part Period Balancing
Demand r (18, 30, 42, 5, 20) r (18, 30, 42, 5, 20) r (18, 30, 42, 5, 20)
Solution y (48, 0, 47, 0, 20) y (48, 0, 42, 25, 0) y (48, 0, 67, 0, 0)
Holding inventory 30535 302050 305(2)(20)75
Holding cost 35(2)70 50(2)100 75(2)150
Setup cost 3(80)240 3(80)240 2(80)160
Total Cost 310 340 310
  The Silver Meal and Part Period Balancing
heuristics resulted in the same least expensive
costs.
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Exercise 14 pg 381

• 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?

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Exercise 17 pg 381

• 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?

Week 1 2 3 4 5 6
Requirements 335 200 140 440 300 200