Alternative lot sizing schemes

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

The three maor control phases o the productive system Inormation requirements or each end item over the planning horizon Master Production Schedule Phase 1 Lot sizing rules and capacity planning Materials Requirements Planning Phase 2 Requirements or raw material Detailed shop loor schedule Phase 3

The explosion calculus Explosion calculus is as term that reers to the set o rules by which gross requirements at one level o product structure are translated into a production schedule at that level and requirements at lower level. End item level End item Parent level (level 1) A(2) 1 week B(1) 2 weeks Child level (level2) C(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 o wind instruments at its plant in Joliet, Illinois. Because the company relatively small, it would like to minimize the amount o 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 o the instruments produced is the model 85C trumpet. The trumpet retails or $800 and has been a reasonably proitable item or the company. Figure gives the product structure diagram or the construction o the trumpet. Bell Assembly (1) Lead Time: 2 weeks Trumpet (end item) Valve casing (1) Lead times 4 weeks Slide assemblies (3) Lead Time : 2 weeks Valves (3) Lead times : 3 weeks

Explosion o trumpet BOM Week 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Demand 77 42 38 21 26 112 45 14 76 38 Week 8 9 10 11 Scheduled receipts 12 6 9 week 8 9 10 11 12 13 14 15 16 17 net predicted demand 42 42 32 12 26 112 45 14 76 38 Week 6 7 8 9 10 11 12 13 14 15 16 17 Gross requirements 42 42 32 12 26 112 45 14 76 38 Net requirements 42 42 32 12 26 112 45 14 76 38 Time-phased net requirements 42 42 32 12 26 112 45 14 76 38 planned order relase (lot-or-lot) 42 42 32 12 26 112 45 14 76 38 Time-phased net requirements 42 42 32 12 26 112 45 14 76 38 planned order relase (lot-or-lot) 42 42 32 12 26 112 45 14 76 38 Trumpet 23 trumpet at the end o the week 7 Bell Assembly Valve Casing Assembly Week 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Gross requirements 126 126 96 36 78 336 135 42 228 114 Scheduled receipts 96 On-han inventory 186 60 30 Net requirements 0 0 66 36 78 336 135 42 228 114 Time-phased net requirements 66 36 78 336 135 42 228 114 planned order relase (lot-or-lot) 66 36 78 336 135 42 228 114 valves

Alternative Lot-Sizing Schemes EOQ lot sizing The silver-meal heuristic Least Unit Cost Part Period Balancing

EOQ Lot Sizing To apply the EOQ ormula, we need three inputs: The average demand rate, The holding cost rate, h Setup cost, K Q 2K h

EOQ Lot Sizing- Valve casing assembly in Ex.7.1 Suppose that the setup operation or 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 or labor, Holding cost, h = (141.82)(0.22) / 52 = 0.60 Lot-or-lot policy requires Total holding cost or Total setup cost = 0 = (132)(10) = $1320

EOQ Lot Sizing- Valve casing assembly in Ex.7.1 Q 2K h (2)(132)(43.9) 0.6 139 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Gross requirements 42 42 32 12 26 112 45 14 76 38 Net requirements 42 42 32 12 26 112 45 14 76 38 Time-phased net requirements 42 42 32 12 26 112 45 14 76 38 Planned order relase (EOQ) 139 0 0 0 139 0 139 0 0 139 Planned deliveries 139 0 0 0 139 0 139 0 0 139 Ending inventory 97 55 23 11 124 12 106 92 16 117 Lot-or-lot policy requires Cumulative ending inventory 97 +55 +23 +11 +124 +12 +106 +92 +16 +117 = 653 Total holding cost or = (0.6)(653) = $391.8 Total setup cost = (132)(4) )= $528 The total cost o EOQ policy )= $528 + $391.8 = $ 919.8

The silver meal heuristic The silver meal heuristic named or Harlan Meal and Edward Silver is a orward method that requires determining the average cost per period as a unction o the number o periods the current order is to span, and stopping the computation when this unction irst increases. C(T) as the average holding and setup cost per period i the current orders spans the next T periods.

The silver meal heuristic Let (r 1,,r n ) be the requirements over n-period horizon. Consider period 1. I we produce ust enough in period 1 to meet the demand in period 1, then ust incur the order cost K. C(1) = K I we order enough in period 1 to satisy the demand both periods 1 and 2 then we must hold r 2 or one period. C(2)= (K + hr 2 )/2 Similarly C(3)= (K + hr 2 + 2hr 3 )/3 and in general, C()= (K + hr 2 + 2hr 3 + + (-1)hr 3 ) / Once C()>C(-1), we stop and set y 1 = r 1 + r 2 + + r n, and begin the process again starting at period.

The silver meal heuristic Example 7.2 A machine shop uses the Silver-Meal heuristic to schedule production lot sizes or computer casing. Over the next ive weeks the demands or 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 y 1 = r 1 + r 2 = 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 y 3 = r 3 + r 4 = 42 + 5 = 47 Because period 5 is inal 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 o dividing the cost over periods by the number o periods,, we divide it by the total number o units demanded through period, r 1 + r 2 + + r. Deine C(T) as the average holding and setup cost per unit or a T- period order horizon. Then, C(1) = K/r 1 C(2)= (K + hr 2 )/(r 1 + r 2 ) C()= [K + hr 2 + 2hr 3 + + (-1)hr 3 ] /(r 1 + r 2 + + r ) As with the Silver-Meal heuristic, this computation is stopped when C()>C(-1), and production level is set to r 1 + r 2 + + r -1. The process is then repeated, starting at period and continuing until the end o 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 y 1 = r 1 + r 2 = 18 + 30 = 48 Starting in period 3 C(1) = 80 /42 = 1.90 C(2) =[80 + (2)(5)] /(42+5) = 1.92 stop Set y 3 = r 3 = 42 Starting in period 4 C(1) = 80/5 = 16 C(2) =[80 + (2)(20)] /(5+20) = 4.8 As we reached the en o the horizon, we set y 4 = r 4 + r 5 = 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 o cases, part period balancing seems to be more popular in practice. The method is set the order horizon equal to the number o 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 o periods.

Part Period Balancing- Example 7.5 Consider Example 7.2. Starting in period 1 Order horizon Total holding cost 1 0 2 60 (30)(2) 3 228 (30)(2)+(2)(2)(42) Because 228 exceeds the setup cost o 80, we stop. As 80 is closer to 60 than to 228, the irst order horizon is two periods. y 1 = r 1 + r 2 = 18 + 30 = 48 We start again in period 3 Order horizon Total holding cost 1 0 2 10 (5)(2) 3 90 (5)(2)+(2)(2)(20) We have exceeded the setup cost o 80, so we stop. Because 90 is closer to 80 than 10, the order horizon is three periods. y 3 = r 3 + r 4 + r 5 = 42+ 5 + 20= 67 y = (48,0,67,0,0)

Comparison o results Silver-Meal LUC Part Period Balancing Demads 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 or valve casing are 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Time-phased net requirements 42 42 32 12 26 112 45 14 76 38 planned order relase (lot-or-lot) 42 42 32 12 26 112 45 14 76 38 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 y 4 = 42+ 42+ 32 + 12= 128 Valve Casing Assembly

Incorporating lot-sizing algorithms into Starting in week 8 C(1) = 132 the explosion calculus 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 y 8 = 26+ 112+ 45 + 14= 197 y 12 = 76 + 38 Week 4 5 6 7 8 9 10 11 12 13 14 15 16 17 Gross requirements 42 42 32 12 26 112 45 14 76 38 Net requirements 42 42 32 12 26 112 45 14 76 38 Time-phased net requirements 42 42 32 12 26 112 45 14 76 38 Planned order relase (S-M) 128 0 0 0 197 0 0 0 114 0 Planned deliveries 128 0 0 0 197 0 0 0 114 0 Ending inventory 86 44 12 0 171 59 14 0 38 0

S-M policy requires Total holding cost or = (0.6)(424) = $254.4 Total setup cost = (132)(3)= $396 The total cost o EOQ policy )= $396 + $254.4 = $ 650.50 Lot or lot : TC L4L =$1320 EOQ : TC EOQ =$919.8

Lot sizing with capacity constraints Capacity constraint clearly makes the problem ar more realistic. However it also makes the problem more complex. Even inding a easible solution may not be obvious. Consider the Example: r=(52, 87, 23, 56) and capacity or each period c=(60, 60, 60, 60). First we must determine i the problem is easible On the surace the problem looks solvable, 4*60 =240>218 But problem is ineasible.! 60+60=120<139 Feasibility condition i1 c i i1 r i or 1,, n. Even easibility condition is satisied, it is not obvious how to ind easible solution.

Lot sizing with capacity constraints-example 7.7 r= (20, 40, 100, 35, 80, 75, 25) c= (60, 60, 60, 60, 60, 60, 60) Checking or easibility: r 1 = 20 c 1 = 60 r 1 + r 2 = 60, c 1 + c 2 = 120, r 1 + r 2 + r 3 = 160, c 1 + c 2 + c 3 = 180, r 1 + r 2 + r 3 + r 4 = 195, c 1 + c 2 + c 3 + c 4 = 240, r 1 + r 2 + r 3 + r 4 + r 5 = 275, c 1 + c 2 + c 3 + c 4 + c 5 = 300, r 1 + r 2 + r 3 + r 4 + r 5 + r 6 = 350, c 1 + c 2 + c 3 + c 4 + c 5 + c 6 = 360, r 1 + r 2 + r 3 + r 4 + r 5 + r 6 + r 7 = 375, c 1 + c 2 + c 3 + c 4 + c 5 + c 6 + c 7 = 420, 50 60 40 60 60 55 60 60 r = ( 20, 40, 100, 35, 80, 75, 25 ) c =( 60, 60, 60, 60, 60, 60, 60 ) Setting y=r gives a easible solution.

Improvement step Modiied requirements schedule r is easible. Is there another easible policy that has lower cost. Variety o reasonable improvement rules can be used For each lot that is scheduled Start rom the last and work backward to the beginning Determine whether it is cheaper to produce the units composing that lot by shiting production to prior periods o excess capacity By eliminating a lot, one reduces setup const in that period to zero, but shiting production to prior periods increases the holding cost.

Example 7.8 Assume that K=$450 and h=$2 r=(100, 79, 230, 105, 3, 10, 99, 126, 40) c=(120, 200, 200, 400, 300, 50, 120, 50, 30) Feasibility check cum. c 100 179 409 514 517 527 626 752 792 easible. cum. r 120 320 520 920 1220 1270 1390 1440 1470 r'=(100, 109, 200, 105, 28, 50, 120, 50, 30) 1 2 3 4 5 6 7 8 9 r' 100 109 200 105 28 50 120 50 30 c 120 200 200 400 300 50 120 50 30 y 100 109 200 105 28 50 120 50 30 Excess capacity 20 91 0 295 272 0 0 0 0

Example 7.8 1 2 3 4 5 6 7 8 9 r' 100 109 200 105 28 50 120 50 30 c 120 200 200 400 300 50 120 50 30 y 100 109 200 Excess capacity 20 91 0 263 105 137 295 y=(100, 109, 200, 263, 0, 0, 120, 0, 0) r=(100, 79, 230, 105, 3, 10, 99, 126, 40) Setup cost = 5 x 450 = 2250 Holding cost = 2 x (0+30+0+158+155+145+166+40+0) = 2 x 694 = $ 1388 Total cost o this policy $3638, ($4482 or initial easible policy). 0 158 108 58 28 4 3 0 50 120 0 50 0 30 300 142 192 242 272 0 0 0 0 (4 period) ($2 per unit per period) (30 unit) =$240 <K=$450 2 1

Optimal Lot-Sizing or the time Varying Demand Heuristic techniques (EOQ lot sizing, The silver-meal heuristic, Least Unit Cost, Part Period Balancing) are easy to use and give lot sizes with costs that are near the true optimum. Lot size problem can be expressed as a shortest path problem. The dynamic programming can be used to ind shortest path. Assume that 1. Forecasted demands over the next n periods are known and given by vector r=(r 1,r 2,,r n ). 2. Costs are charged against holding at $h per unit per period and $K per setup.

Example 7A.1 The orecast demand or an electronic assembly produced at Hi-tech, a local semiconductor abrication shop, over the next our weeks is 52, 87, 23, 56. There is only one setup each week or production o these assemblies, and there is no back-ordering o excess demand. Assume that the shop has the capacity to produce any number o assemblies in a week. Let y 1, y 2, y 3, y 4 be the order quantities in each week. Clearly y 1 52, i we assume that ending inventory in week 4 is zero, then y 1 218 (52+87+23+56) y 1 can take any one o 167 possible values It is thus clear that or even moderately sized problems the number o easible solution is enormous.

Wagner-Whitin Algorithm The Wagner Whitin algorithm is based on the ollowing observation: Result. An optimality policy has the property that each value o y is exactly the sum o a set o uture demands(exact requirement policy). y 1 =r 1, or y 1 =r 1 +r 2 +, or y 1 =r 1 +r 2 + +r n y 2 =0, or y 2 =r 2 or y 2 =r 2 +r 3 +, or y 2 =r 2 +r 3 + +r n y n =0, or y n =r n. The number o exact requirement policy is much smaller than the total number o easible policies.s

Wagner-Whitin Algorithm Because y 1 must satisy exact requirements, it can take only values : 52, 139, 162, 218 Ignoring value o y1, y2 can take 0, 87, 110, 166. Every exact policy is the orm (i 1,i 2,,i n ), where i are either 0 or 1. For example the policy (1,0,1,0) means that production occurs in period 1 and 3 only, that is y= (139, 0, 79, 0). For this example, there are exactly 2 3 = 8 distinct exact requirement policies.

Wagner-Whitin Algorithm A convenient way to look at the problem is as a one-way network with the number o nodes equal to exactly one more than the number o periods. Every path through to network corresponds to a speciic exact requirements policy. c 13 c 14 c 15 c 12 c 23 c 34 c 45 1 2 3 4 5 c 24 c 25 For any pair (i,) with i<, i the arc (i,) is one path, it means that ordering takes place in period i and the order size is equal to the sum o requirements in periods i, i+1, i+2,, -1. Period is the next period o ordering. All paths end at period n+1, The value o each arc, c i, is deined as setup and holding cost o ordering in period i to meet requirements through -1. c 35

Example 7A.2 r = (52, 87, 23, 56), h=$1, K=$75 The irst step compute c i or 1i4 and 25 c 12 =75 (setup cost) c 13 =75 + 87 = 162 c 14 =75 + (23 x 2) + 87 = 208 c 14 =75 + (56 x 3) + (23 x 2) + 87 = 376 c 23 =75 c 24 =75 + 23 = 98 c 25 =75 + (56 x 2) + 23 = 210 c 34 =75 c 35 =75 + 56= 131 i \ 1 2 3 4 5 c 23 =75 (setup cost) 1 75 162 208 376 c 45 =75 2 75 98 210 3 75 131 4 75 Path Cost 1-2-3-4-5 300 1-2-4-5 248 1-2-5 285 1-2-3-5 281 1-3-4-5 312 1-3-5 293 1-4-5 283 1-5 376

Solution by Dynamic Programming The total number o exact policies or a problem o n periods is 2 n-1. As n gets large, total enumeration is not eicient. Dynamic programming is a recursive solution technique that can signiicantly reduce the number o computations required. Dynamic programming is based on the principle o optimality. Deine k as the minimum cost starting at node k. Initial condition is n+1 =0 min ( ck ) or k 1,2,, n k k

Solution by Dynamic Programming 5 0 5 75 ) ( min 4 4 4 at c 5 131 131 150 min 0 131 75 75 min min ) min( 5 35 4 34 3 3 3 at c c c 4 173 210 173 206 min 0 210 75 98 131 75 min min ) min( 5 25 4 24 3 23 2 2 2 at c c c c 2 248 376 283 293 248 min 0 376 75 208 131 162 173 75 min min ) min( 5 15 4 14 3 13 2 12 1 1 1 at c c c c c