Aggregate Planning Production Planning and Control Anadolu Üniversitesi Mühendislik Mimarlık Fakültesi Endüstri Mühendisliğ Bölümü
Example 3.1 Washing Machine Plant Model Number Number of Worker-Hours Required to Produce Selling Price The ratio of selling price divided by the worker-hours Percentages of total number of sales A 5532 4.2 285 K 4242 4.9 345 L 9898 5.1 395 L 3800 5.2 425 M 2624 5.4 525 M 3880 5.8 725 67.86 70.41 77.45 81.73 97.22 125.00 32 21 17 14 10 6 M FICTIOUS = (4.2)(.32)+ (4.9)(.21)+ (5.1)(.17)+ (5.2)(.14)+ (5.4)(.10)+ (5.8)(.06) = 4.856
Example 3.2 Densepack is to plan workforce and production levels for six-month period January to June. The firm produces a line of disk drives for mainframe computers that are plug compatible with several computers produced by major manufacturers. Forecast demand over the next six months for a particular line of drives product in Milpitas, California, plant are 1280, 640, 900, 1200, 2000 and 1400. There are currently (end of December) 300 workers employed in the Milpitas plant. Ending inventory in December is expected to by 500 units, and the firm would like to have 600 units on hand at the end of June.
Problem parameters Planning horizon Current number of workers 300 Starting inventory Ending Inventory Cost of hiring one worker C H =$500 Cost of firing one worker C F =$1000 Cost of holding one unit of inventory for one month C I =$80 Number of aggregate units produced by one worker in one day January-June, six months {January, February, March, April, May, June} {1,2,3,4,5,6} I 0 =500 (Ending inventory in December) I 6 =600 (Inventory level end of June) It is known that in the past for 22 working days with workforce level 76 workers, the firm produced 245 disk drivers. 245/22 = 11.1364 drivers per day when 76 workers employed K = 11.1364 / 76 = 0.14653 drive per worker
Problem Parameters Month Net Predicted Demand Net Cumulative Demand January 780 (1280-500) 780 February 640 1420 March 900 2320 April 1200 3520 May 2000 5520 June 2000 (1400+600) 7520
Evaluation of chase strategy (Zero Inventory Plan ) A B C D E Month Number of working Days Number of Unitst Produced per worker (Bx0.14653) Forecast Demand Minimum Number of Workers Required (D/C rounded up) January 20 2.931 780 267 February 24 3.517 640 182 March 18 2.638 900 342 April 26 3.810 1200 315 May 22 3.224 2000 621 June 15 2.198 2000 910
Evaluation of chase strategy (Zero Inventory Plan ) continue A B C D E F G H I Number of Cumulati Number of units ve Cumulati Number of Number of Number of units per produced Producti ve workers hired fired worker (BxE) on Demand Total cost of hiring, firing and holding is: (755)(500)+(145)(1000)+(30)(80) = $524900 + the cost of holding for ending inventory $524900 + (600)(80) = $572900 Ending Inventor y (G-H) Month January 267 33 2.9306 783 783 780 3 February 182 85 3.51672 640 1423 1420 3 March 342 160 2.63754 902 2325 2320 5 April 315 27 3.80978 1200 3525 3520 5 May 621 306 3.22366 2002 5527 5520 7 June 910 289 2.19795 2000 7527 7520 7 755 145 30
Evaluation of the Constant Workforce Plan A B C D Month Cumulative Demand Cumulative Number of Ratio B/C units Produced per (round up) worker January 780 2.931 267 February 1420 6.448 221 March 2320 9.086 256 April 3520 12.896 273 May 5520 16.120 343 June 7520 18.318 411
Evaluation of the Constant Workforce Plan A B C D E F Number of units Monthly Cumulative Cumulative net Month produced per Production Production Demand worker (B x 411) Total cost of hiring and holding is: (111)(500)+(5962+600)(80) = $580460 Ending Inventory (D-E) January 2.931 1205 1205 780 425 February 3.517 1445 2650 1420 1230 March 2.638 1084 3734 2320 1414 April 3.81 1566 5300 3520 1780 May 3.224 1325 6625 5520 1105 June 2.198 903 7528 7520 8 Total 5962
Mixed Strategies and Additional With more flexibility, small modifications can result in dramatically lower cost. Aggregate planning problem can be formulated and solved optimally* by linear programming. 8000 7000 6000 5000 4000 3000 2000 1000 0 1 2 3 4 5 6 7 Constraints The graphical method can also be used when additional constraints are present. Suppose that capacity of the plant is only 1800 units per month. Maximum change from one month to the nest be no more than 750 units. As the constraints become more complex finding good solutions graphically becomes more difficult. Most constraint of this nature can be incorporated easily into linear programming formulations.
Solution of Aggregate Planning Problems by Linear Programming Cost parameters and given information The following values are assumed to be known: Cost of hiring one worker, c H c F c I c R c O c U c s Cost of firing one worker, Cost of holding one unit of stock for one period, Cost of producing one unit on regular time, Incremental cost of producing one unit on overtime, Idle cost per unit of production Cost to subcontract one unit of production, n t Number of production days in period t, K Number of aggregate units produced by one worker in one day, Initial inventory on hand at the start of the planning horizon, I 0 W 0 Initial workforce at the start of the planning horizon D t Forecast of demand in period t.
Solution of Aggregate Planning Problems by Linear Programming Problem variables The following values are assumed to be known: W t Workforce level in period t, P t Production level in period t, I t Inventory level in period t, H t Number of workers hired in period t, F t Number of workers fired in period t, O t U t S t Overtime production in units, Worker idle time units ( undertime ) Number of units subcontracted from outside.
Solution of Aggregate Planning Problems by Linear Programming Problem Constraints 1. Conservation of workforce constraints. W t = W t-1 + H t - F t, for 1 t T 2. Conservation of units constraints. I t = I t-1 + P t + S t - D t, for 1 t T 3. Constraints relating production levels to workforce levels. P t = Kn t W t + O t - U t, for 1 t T
Solution of Aggregate Planning Problems by Linear Programming Linear Programming Formulation Minimize t (c H H t + c F F t +c I I t + c R P t + c O O t + c U U t + c S S t ) subject to Conservation of workforce constraints. W t = W t-1 + H t - F t, for 1 t T I t = I t-1 + P t + S t - D t, for 1 t T Conservation of units constraints. P t = Kn t W t + O t - U t, for 1 t T Constraints relating production levels to workforce levels. H t,f t, I t, O t, U t, S t,w t,p t 0
Rounding the Variables In general optimal values of the problem will not integers. Fractional values for many variables do not make sense Size of workforce, the number of workers hired each period, etc. All variables assumes integer (Integer linear programming problem) Solution algorithm considerably more complex. Rounding Closets Integer? Inconsistency. Round up Wt to the next integer.
Extentions Minimum buffer inventory Bt each period. I t B t, for 1 t T Capacity constraints on the amount of production each period P t C t, for 1 t T Backlogging I t = I t+ + I - $Cost t I t+ 0, I t- 0 Slope = c 2H convex-piecewise linear cost Slope = c 1H t (c H1 H 1t +c H2 H 2t ) H H t = H 1t + H * 2t 0 H 1t H* 0 H 2t
Solving Aggregate Planning Problems By Linear Programming: an example Minimize t (500H t + 1000F t +80I t ) subject to W t = W t-1 + H t - F t, for 1 t T W 1 - W 0 - H 1 + F 1 =0 W 2 - W 1 - H 2 + F 2 =0 W 3 - W 2 - H 3 + F 3 =0 W 4 - W 3 - H 4 + F 4 =0 W 5 - W 4 - H 5 + F 5 =0 W 6 - W 5 - H 6 + F 6 =0 Conservation of workforce constraints. I t = I t-1 + P t + S t - D t, for 1 t T P 1 - I 1 + I 0 = 1280, P 2 - I 2 + I 1 = 640, P 3 - I 3 + I 2 = 900, P 4 - I 4 + I 3 = 1200, P 5 - I 5 + I 4 = 2000, P 6 - I 6 + I 5 =1400, Conservation of units constraints. P t = Kn t W t + O t - U t, for 1 t T P 1-2.931 W 1 =0 P 2-3.517 W 2 =0 P 3-2.638 W 3 =0 P 4-3.810 W 4 =0 P 5-3.224W 5 =0 P 6-2.198 W 6 =0 Constraints relating production levels to workforce levels. H t,f t, I t,w t,p t 0 W 0 =300 I 0 =500 I 6 =600