ALTERNATIVES LOT-SIZING SCHEMES

Alternatives Lot-Sizing SchemesThe silver-meal heuristicLeast Unit CostPast Period Balancing

The Silver-Meal HeuristicForward 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 periodFormula : C(j) = (K + hr2 + 2hr3 + … + (j-

1)hrj) / jC(j) average holding cost and setup cost per

periodk order cost or setup costh holding costr demand

MethodI. 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-1IV. Start over at period j, repeat step (I) – (III)

ExampleA 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 IVStarting 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 = 20Thus 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

MethodI. 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-1IV. 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 IVStarting 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 BalancingSet the order horizon equal to the number

of periods that most closely matches the total holding cost with the setup cost over that period.

Exampler = (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 02 2(30) = 603 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 02 2(5) = 103 2(5) + 2(2)(20) = 90

closest

Comparison of ResultsSilver – Meal Least Unit

CostPart 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)=75Holding cost 35(2)=70 50(2)=100 75(2)=150Setup cost 3(80)=240 3(80)=240 2(80)=160Total 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 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 6Requirements 335 200 140 440 300 200