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 14 - pg 381

• 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

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• Start in period 4
• C(1) 40
• C(2) (40 15)/2 27.5
• C(3) 40 15 (2)(25)/3 35 Stop.
• y4 r4 r5 815 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 25205 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
  

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• b) Least unit cost
• 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 1525 40
• Start in period 7
• C(1) 40/20 2
• C(2) (40 5)/(20 5) 1.8

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• Start in period 9
• 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 1020535
• y12 r12 20
• y (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20)

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• c) Part period balancing
• 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

Order horizon Total holding cost
1 0
2 1(12) 12
3 1(12) 2(1)(4) 20
4 1(12) 2(1)(4) 3(1)(8) 44
closest
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• We start again in period 5
• y5 r5 r6 1525 40
• Start in period 7
•  y7 r7 r8 r9 20510 35

Order horizon Total holding cost
1 0
2 1(25) 25
3 1(25) 2(1)(20) 65
closest
Order horizon Total holding cost
1 0
2 1(5) 5
3 1(5) 2(1)(10) 25
4 1(5) 2(1)(10) 3(1)(20) 85
closest
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• Start in period 10
•  y10 r10 r11 r12
• 20520
• 45
• y (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0,
  0)

Order horizon Total holding cost
1 0
2 1(5) 5
3 1(5) 2(1)(20) 45
closest
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Comparison of results
Silver Meal Least Unit Cost Part Period Balancing
Demand r (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20) r (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20) r (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)
Solution y (18, 0, 0, 23, 0, 50, 0, 0, 35, 0, 0, 20) y (26, 0, 0, 0, 40, 0, 25, 0, 35, 0, 0, 20) y (26, 0, 0, 0, 40, 0, 35, 0, 0, 45, 0, 0)
Holding inventory 122(4)15202(5)202(5) 95 122(4)3(8)255 202(5)104 122(4)3(8)255 2(10)52(20) 139
Holding cost 95(1)95 104(1)104 139(1)139
Setup cost 5(40)200 5(40)200 4(40)160
Total Cost 295 304 299
  The Silver Meal resulted the least expensive
cost.
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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
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Exercise 17 pg 381

• 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

Week 1 2 3 4 5 6
Requirements 335 200 140 440 300 200
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• b) Least unit cost
• 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

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c) Part period balancing
Week 1 2 3 4 5 6
Requirements 335 200 140 440 300 200

• K 200
• h 0.30
• Starting in period 1
• y1 r1 r2 r3 335 200 140 675

Order horizon Total holding cost
1 0
2 0.3(200) 60
3 0.3(200) 2(0.3)(140) 144
4 0.3(200) 2(0.3)(140) 3(0.3)(440) 540
closest
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• Starting in period 4
• y4 r4 r5 r6 440 300 200 940
• y (675, 0, 0, 940, 0, 0)

Order horizon Total holding cost
1 0
2 0.3(300) 90
3 0.3(300) 2(0.3)(200) 210
closest
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Comparison of results
Silver Meal Least Unit Cost Part Period Balancing
Demand r (335, 200, 140, 440, 300, 200) r (335, 200, 140, 440, 300, 200) r (335, 200, 140, 440, 300, 200)
Solution y (675, 0, 0, 940, 0, 0) y (535, 0, 580, 0, 500, 0) y (675, 0, 0, 940, 0, 0)
Holding inventory 2002(140)300 2(200) 1180 200440200 840 2002(140)300 2(200) 1180
Holding cost 1180(0.3)354 840(0.3)252 1180(0.3)354
Setup cost 2(200)400 3(200)600 2(200)400
Total Cost 754 852 754
  The Silver Meal and Part Period Balancing
heuristics resulted in the same least expensive
costs.