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.

• 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

I. Start the calculation from period 1 to next period C(1) = K C(2) = (K + hr2) / 2 C(3) = (K + hr2 + 2hr3) / 3

II. Stop the calculation when C(j) > C(j-1)III. Set y1 = r1 + r2 + … + rj-1

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

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 & IIIr = (18, 30, 42, 5, 20)k = $80h = $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

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

• 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)

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

Method

I. 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 )

II. Stop the calculation when C(j) > C(j-1)III. Set y1 = r1 + r2 + … + rj-1

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

Step I, II & IIIr = (18, 30, 42, 5, 20)k = $80h = $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 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

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)

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.

Example

r = (18, 30, 42, 5, 20) Holding cost = $2 per case per weekSetup 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

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

Comparison of Results

Silver – Meal Least Unit Cost Part Period Balancing

Demand 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 30+5=35 30+20=50 30+5+(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.

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

a) Silver-Mealb) Least unit costc) Part period balancingd) Which lot-sizing method resulted in the lowest cost for the 12

periods?

Exercise 14 - pg 381

Demand = (6, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 12)Starting inventory = 4Ending inventory = 8h = 1 K = 40

Net out starting and ending inventories to obtainr = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)

a) Silver Meal Start in period 1:C(1) = 40C(2) = (40 + 12)/2 = 26C(3) = [40 + 12 + (2)(4)]/3 = 20C(4) = [40 + 12 + (2)(4) + (3)(8)]/4 = 21 stop. y1 = r1 + r2 + r3 = 2 + 12 + 4 = 18

Start in period 4:C(1) = 40C(2) = (40 + 15)/2 = 27.5C(3) = [40 + 15 + (2)(25)]/3 = 35 Stop.y4 = r4 + r5 = 8+15 = 23

Start in period 6:C(1) = 40C(2) = (40 + 20)/2 = 30C(3) = [40 + 20 + (2)(5)]/3 = 23.3333C(4) = [40 + 20 + (2)(5) + (3)(10)]/4 = 25 Stop.y6 = r6 + r7 + r8 = 25+20+5 = 50 Start in period 9:C(1) = 40C(2) = (40 + 20)/2 = 30C(3) = [40 + 20 + (2)(5)]/3 = 23.3333C(4) = [40 + 20 + (2)(5) + (3)(20)]/4 = 32.5 y9 = r9 + r10 + r11 = 10+20+5=35y12 = r12 = 20

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

b) Least unit cost

Start in period 1:C(1) = 40/2 = 20C(2) = (40 + 12)/(2 + 12) = 3.71C(3) = (40 + 12 + 8) /(2 + 12 + 4) = 3.33C(4) = (40 + 12 + 8 + 24) /(2 + 12 + 4 + 8) = 3.23C(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 = 2.67C(2) = (40 + 25)/(15 + 25) = 1.625C(3) = (40 + 25 + 40)/(15 + 25 + 20) = 1.75 Stop. y5 = r5 + r6 = 15+25 = 40

Start in period 7:C(1) = 40/20 = 2C(2) = (40 + 5)/(20 + 5) = 1.8C(3) = (40 + 5 + 20)/(20 + 5 + 10) = 1.86 Stop. y7 = r7 + r8 = 20+5= 25

Start in period 9:C(1) = 40/10 = 4C(2) = (40 + 20)/(10 + 20) = 2C(3) = (40 + 20 + 10)/(10 + 20 + 5) = 2C(4) = (40 + 20 + 10 + 60)/(10 + 20 + 5 + 20) = 2.3636y9 = r9 + r10 + r11 = 10+20+5=35y12 = r12 = 20

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

c) Part period balancingr = (2, 12, 4, 8,15, 25, 20, 5, 10, 20, 5, 20)h = 1 K = 40Starting 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

• We start again in period 5

y5 = r5 + r6 = 15+25 = 40

• Start in period 7

y7 = r7 + r8 + r9= 20+5+10= 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

• Start in period 10

y10 = r10 + r11 + r12

= 20+5+20= 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

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)

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 12+2(4)+15+20+2(5)+20+2(5)= 95

12+2(4)+3(8)+25+5+20+2(5)=104

12+2(4)+3(8)+25+5+2(10)+5+2(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.

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 weeka) Determine the lot sizes using the Silver-Meal heuristicb) Determine the lot sizes using the least unit cost heuristicc) Determine the lot sizes using part period balancingd) 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

Exercise 17 pg 381

K = $200h = $0.30

a) Silver Meal Start in period 1:C(1) = 200C(2) = [200 + (200)(0.3)]/2 = 130C(3) = [(2)(130) + (2)(140)(0.3)]/3 = 114.67C(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) = 200C(2) = [200 + (300)(0.3)]/2 = 145C(3) = [(2)(145) + (2)(200)(0.3)]/3 = 136.67 Stop.

y4= r4 + r5 + r6 = 440 + 300 + 200 = 940y = (675, 0, 0, 940, 0, 0)

Week 1 2 3 4 5 6

Requirements 335 200 140 440 300 200

b) Least unit costStart in period 1:C(1) = 200/335 = 0.597C(2) = [200 + (200)(0.3)]/(335 + 200) = 0.486C(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 = 1.428C(2) = [200 + (440)(0.3)]/(140 + 440) = 0.572C(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 = 0.67C(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

K = $200h = $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

Week 1 2 3 4 5 6

Requirements 335 200 140 440 300 200

closest

• 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

Comparison of results

Silver – Meal Least Unit Cost Part Period Balancing

Demand 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 200+2(140)+300+ 2(200) = 1180 200+440+200 = 840 200+2(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.