Alternative lot sizing schemes Anadolu University Industrial Engineering Department Source: Nahmias, S., Production And Operations Analysis, McGraw-Hill /Irwin, Fifth Edition, 2005, ISBN: 0073018651

The three major control phases of the productive system Information requirements for each end item over the planning horizon

Phase 1 Master Production Schedule

Lot sizing rules and capacity planning

Materials Requirements Planning

Requirements for raw material

Detailed shop floor schedule

Phase 2

Phase 3

The explosion calculus • Explosion calculus is as term that refers to the set of rules by which gross requirements at one level of product structure are translated into a production schedule at that level and requirements at lower level.

End item level

Parent level (level 1) Child level (level2)

End item

A(2) 1 week C(1) 2 weeks

B(1) 2 weeks D(2) 1 week

C(2) 2 weeks

E(3) 2 weeks

Example 7.1 • The Harmon Music Company produces a variety of wind instruments at its plant in Joliet, Illinois. Because the company relatively small, it would like to minimize the amount of money tied to inventory. For that reason production levels are set to match predicted demand as closely as possible. In order to achieve this goal, the company has adopted an MRP system to determine production quantities. • One of the instruments produced is the model 85C trumpet. The trumpet retails for $800 and has been a reasonably profitable item for the company. • Figure gives the product structure diagram for the construction of the trumpet.

Trumpet (end item)

Bell Assembly (1) Lead Time: 2 weeks

Valve casing (1) Lead times 4 weeks

Slide assemblies (3) Lead Time : 2 weeks

Valves (3) Lead times : 3 weeks

Explosion of trumpet BOM Week Demand

2

3

4

5

6

7

8 9 10 11 12 13 14 15 16 17 77 42 38 21 26 112 45 14 76 38

Week Scheduled receipts

8 12

week net predicted demand

8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38

Week Gross requirements Net requirements Time-phased net requirements planned order relase (lot-for-lot)

6

8 42 42 42 42 32 42 42 32

Time-phased net requirements planned order relase (lot-for-lot)

Week Gross requirements Scheduled receipts On-han inventory Net requirements Time-phased net requirements planned order relase (lot-for-lot)

7

9 10 11 6 9

9 42 42 12 12

10 11 32 12 32 12 26 112 26 112

23 trumpet at the end of the week 7

12 13 26 112 26 112 45 14 45 14

42 42 32 12 26 112 45 14 76 38 42 42 32 12 26 112 45 14 76 38

2

3

4 5 6 7 126 126 96 36 96 186 60 30 0 0 66 36 66 36 78 336 135 66 36 78 336 135

Trumpet

14 45 45 76 76

15 16 17 Bell Assembly 14 76 38 14 76 38 38 38

Valve Casing Assembly

8 9 10 11 12 13 14 15 16 17 valves 78 336 135 42 228 114

78 336 135 42 228 114 42 228 114 42 228 114

Alternative Lot-Sizing Schemes • • • •

EOQ lot sizing The silver-meal heuristic Least Unit Cost Part Period Balancing

EOQ Lot Sizing • To apply the EOQ formula, we need three inputs: – The average demand rate,  – The holding cost rate, h – Setup cost, K

2 K Q h

EOQ Lot Sizing- Valve casing assembly in Ex.7.1 • Suppose that the setup operation for the machinery used in this assembly operation takes two workers about three hours. – The worker average cost: $22 per hour – K= (22)(2)(3) = $132

• The company uses a holding cost on a 22 percent annual interest rate. – Each valve casing assembly costs $141.82 in materials and value added for labor, – Holding cost, h = (141.82)(0.22) / 52 = 0.60

• Lot-for-lot policy requires – Total holding cost for – Total setup cost

=0 = (132)(10) = $1320

EOQ Lot Sizing- Valve casing assembly in Ex.7.1 2 K (2)(132)(43.9) Q   139 h 0.6 Week Gross requirements Net requirements Time-phased net requirements Planned order relase (EOQ) Planned deliveries Ending inventory

4

5

6

7

8 9 10 42 42 32 42 42 32 42 42 32 12 26 112 45 139 0 0 0 139 0 139 139 0 0 97 55 23

11 12 13 14 12 26 112 45 12 26 112 45 14 76 38 0 0 139 0 139 0 139 11 124 12 106

15 16 17 14 76 38 14 76 38

0 0 139 92 16 117

Lot-for-lot policy requires Cumulative ending inventory 97

+55

+23

+11 +124 +12 +106 +92

+16 +117

= 653

Total holding cost for = (0.6)(653) = $391.8 Total setup cost = (132)(4) )= $528 The total cost of EOQ policy )= $528 + $391.8 = $ 919.8

The silver meal heuristic • The silver meal heuristic – named for Harlan Meal and Edward Silver – is a forward method that requires determining the average cost per period as a function of the number of periods the current order is to span, – and stopping the computation when this function first increases.

• C(T) as the average holding and setup cost per period if the current orders spans the next T periods.

The silver meal heuristic • Let (r1,…,rn) be the requirements over n-period horizon. • Consider period 1. – If we produce just enough in period 1 to meet the demand in period 1, then just incur the order cost K. C(1) = K – If we order enough in period 1 to satisfy the demand both periods 1 and 2 then we must hold r2 for one period. C(2)= (K + hr2)/2 – Similarly C(3)= (K + hr2 + 2hr3)/3 – and in general, C(j)= (K + hr2 + 2hr3 + … + (j-1)hr3) /j – Once C(j)>C(j-1), we stop and set y1 = r1 + r2 + … + rn , and begin the process again starting at period j.

The silver meal heuristic – Example 7.2 • A machine shop uses the Silver-Meal heuristic to schedule production lot sizes for computer casing. 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. • 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 because C(3)>C(2). Set y1 = r1 + r2 = 18 + 30 = 48

• 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 Set y3 = r3 + r4 = 42 + 5 = 47

• Because period 5 is final period in the horizon, we do not need to start the process again. Set y5 = r5 = 20 y = (48, 0, 47, 0, 20)

Least unit cost • The least unit cost (LUC) heurist – Similar to Silver – Meal heuristic except instead of dividing the cost over j periods by the number of periods, j, we divide it by the total number of units demanded through period j, r1 + r2 + … + rj .

• Define C(T) as the average holding and setup cost per unit for a Tperiod order horizon. Then, C(1) = K/r1 C(2)= (K + hr2)/(r1 + r2) … C(j)= [K + hr2 + 2hr3 + … + (j-1)hr3] /(r1 + r2 + … + rj ) – As with the Silver-Meal heuristic, this computation is stopped when C(j)>C(j-1), and production level is set to r1 + r2 + … + rj-1 . – The process is then repeated, starting at period j and continuing until the end of the planning horizon is reached.

Least Unit Cost – Example 7.4 • Assume the same requirements schedule and costs as given in Ex.7.2 • 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 because C(3)>C(2). Set y1 = r1 + r2 = 18 + 30 = 48

• Starting in period 3 – C(1) = 80 /42 = 1.90 – C(2) =[80 + (2)(5)] /(42+5) = 1.92 stop – Set y3 = r3 = 42

• Starting in period 4 – C(1) = 80/5 = 16 – C(2) =[80 + (2)(20)] /(5+20) = 4.8

• As we reached the en of the horizon, we set y4 = r4 + r5 = 5 + 20 =25. The solution obtain by LUC heuristic y = (48, 0, 42, 25, 0)

Part Period Balancing • Although Silver-Meal heuristic seems to give better results in a greater number of cases, part period balancing seems to be more popular in practice. • The method is 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. • The order horizon that exactly equates holding and setup costs will rarely be an integer number of periods.

Part Period Balancing- Example 7.5 • Consider Example 7.2. • Starting in period 1 Order horizon 1 2 3

Total holding cost 0 60 (30)(2) 228 (30)(2)+(2)(2)(42)

– 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

• We start again in period 3 Order horizon 1 2 3

– – – –

Total holding cost 0 10 90

(5)(2) (5)(2)+(2)(2)(20)

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)

Comparison of results Silver-Meal Demads

LUC

Part Period Balancing

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

Incorporating lot-sizing algorithms into the explosion calculus • Consider valve casing assembly in the Ex.7.1 • Time phased net requirements for valve casing are Time-phased net requirements planned order relase (lot-for-lot)

4 5 6 7 8 9 10 11 12 13 14 15 16 17 42 42 32 12 26 112 45 14 76 38 42 42 32 12 26 112 45 14 76 38

Valve Casing Assembly

• The setup cost the valve casing is K=$132, and holding cost h=$0.60 per assembly per week. Silver Meal heuristic. • Starting in week 4 – – – – – –

C(1) = 132 C(2) =[132 + (0.6)(42)] /2 = 78.6 C(3) = [132 + (0.6)[42+ (2)(32)]] /3 = 65.2 C(4) = [132 + (0.6)[42+ (2)(32)+(3)(12)]] /4 = 54.3 C(5) = [132 + (0.6)[42+ (2)(32) +(3)(12) +(4)(26)]] /5 = 55.92 , stop. Set y4 = 42+ 42+ 32 + 12= 128

Incorporating lot-sizing algorithms into the explosion calculus • Starting in week 8 – – – – – –

C(1) = 132 C(2) =[132 + (0.6)(112)] /2 = 99.6 C(3) = [132 + (0.6)[112+ (2)(45)]] /3 = 84.4 C(4) = [132 + (0.6)[112+ (2)(45)+(3)(14)]] /4 = 69.6 C(5) = [132 + (0.6)[112+ (2)(45)+(3)(14)+(4)(76)]] /5 = 92.12 , stop. Set y8 = 26+ 112+ 45 + 14= 197

• y12 = 76 + 38 Week Gross requirements Net requirements Time-phased net requirements Planned order relase (S-M) Planned deliveries Ending inventory

4

5

6

7

8 42 42 42 42 32 12 26 128 0 0 0 197 128 86

9 42 42 112 0 0 44

10 32 32 45 0 0 12

11 12 12 14 0 0 0

12 26 26 76 114 197 171

13 112 112 38 0 0 59

14 15 16 17 45 14 76 38 45 14 76 38

0 14

0 114 0 38

0 0

• S-M policy requires – Total holding cost for = (0.6)(424) = $254.4 – Total setup cost = (132)(3)= $396 – The total cost of EOQ policy )= $396 + $254.4 = $ 650.50

• Lot for lot : TCL4L=$1320 • EOQ : TCEOQ=$919.8

Lot sizing with capacity constraints • Capacity constraint clearly makes the problem far more realistic. • However it also makes the problem more complex. • Even finding a feasible solution may not be obvious. Consider the Example: r=(52, 87, 23, 56) and capacity for each period c=(60, 60, 60, 60). • First we must determine if the problem is feasible • On the surface the problem looks solvable, 4*60 =240>218 • But problem is infeasible. ! 60+60=120