Deals with multi-period planning.

1. PRODUCTION SCHEDULING Meenu S Babu

5. Formulating Linear Programming model : Let X im denote production volume in units for product i in month m. i=1,2 m=1,2,3 i=1 means component 322A m=1 means April If 322A costs $20 per unit produced and 802B costs $10 per unit produced, then Total production cost = 20X 11 + 20X 12 +20X 13 +10X 21 +10X 22 +10X 23 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

6. Let S im denotes Inventory level for product i at the end of month m. If Inventory Holding costa are 1.5 percentage of cost of product (i.e 0.015 * $20 = $0.30 per unit for 322A and .015 * $10 = $0.15 per unit for 802B) Inventory Holding Cost = 0.30s 11 +0.30s 12 +0.30s 13 +0.15S 21 + 0.15S 22 +0.15S 23 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

7. 2 additional variables could be defined to incorporate fluctuations in production levels from month to month. I m = increase in the total production level necessary during month m. D m = decrease in the total production level necessary during month m. Let I m for any month be 0.50 and D m be 0.20 Change-in-production-level costs = 0.50I 1 + 0.50I 2 + 0.50I 3 + 0.20D 1 + 0.20D 2 + 0.20D 3 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

8. Combinig all the three costs, Objective function becomes : Min 20X 11 + 20X 12 +20X 13 +10X 21 +10X 22 +10X 23 +0.30s 11 +0.30s 12 +0.30s 13 +0.15S 21 + 0.15S 22 +0.15S 23 +0.50I 1 + 0.50I 2 + 0.50I 3 + 0.20D 1 + 0.20D 2 + 0.20D 3 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

9. Making the constraints : To guarantee schedule meets customer demand, it must me in the form : This Month's Demand = Ending Inventory From Previous Month + Current Production - Ending Inventory for This Month

10. Let the inventories at the beginning of 3 month scheduling period were 500 units for component 322A and 200 units for component 802B. Demand in first month : 500 + X 11 -S 11 = 1000 200 + X 21 -S 21 =1000 => X 11 – S 11 = 500 X 21 – S 21 = 800 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

11. Demand in second month : S 11 + X 12 - S 12 = 3000 S 21 + X 22 - S 22 = 500 Demand in third month : S 12 + X 13 - S 13 = 5000 S 22 + X 23 - S 23 = 3000 Component April May June 322A 1000 3000 5000 802B 1000 500 3000

12. If the company specifies a minimum inventory level at the end of the 3 month period of at least 400 units of component 322A and at least 200 units of component 802B, then S 13 >= 400 S 23 >= 200

13. Following table shows machine,labor and storage capacity available: Following table shows machine,labor and storage space requirement needed: MONTH MACHNE CAPACITY (HOURS) LABOR CAPACITY (HOURS) STORAGE CAPACITY (HOURS) APRIL 400 300 10,000 MAY 500 300 10,000 JUNE 600 300 10,000 COMPONENT MACHINE (HOURS/UNIT) LABOR (HOURS/UNIT) STORAGE (SQUARE FEET/UNIT) 322A 0.10 0.05 2 802B 0.08 0.07 3

14. Machine Capacity : 0.10X 11 + 0.08X 21 <= 400 Month 1 0.10X 12 + 0.08X 22 <= 500 Month 2 0.10X 13 + 0.08X 23 <= 600 Month 3 MONTH MACHNE CAPACITY (HOURS) LABOR CAPACITY (HOURS) STORAGE CAPACITY (HOURS) APRIL 400 300 10,000 MAY 500 300 10,000 JUNE 600 300 10,000 COMPONENT MACHINE (HOURS/UNIT) LABOR (HOURS/UNIT) STORAGE (SQUARE FEET/UNIT) 322A 0.10 0.05 2 802B 0.08 0.07 3

15. Labor Capacity : 0.05X 11 + 0.07X 21 <= 300 Month 1 0.05X 12 + 0.07X 22 <= 300 Month 2 0.05X 13 + 0.07X 23 <= 300 Month 3 MONTH MACHNE CAPACITY (HOURS) LABOR CAPACITY (HOURS) STORAGE CAPACITY (HOURS) APRIL 400 300 10,000 MAY 500 300 10,000 JUNE 600 300 10,000 COMPONENT MACHINE (HOURS/UNIT) LABOR (HOURS/UNIT) STORAGE (SQUARE FEET/UNIT) 322A 0.10 0.05 2 802B 0.08 0.07 3

16. Storage Capacity : 2X 11 + 3S 21 <= 10,000 Month 1 2X 12 + 3S 22 <= 10,000 Month 2 2X 13 + 3S 23 <= 10,000 Month 3 MONTH MACHNE CAPACITY (HOURS) LABOR CAPACITY (HOURS) STORAGE CAPACITY (HOURS) APRIL 400 300 10,000 MAY 500 300 10,000 JUNE 600 300 10,000 COMPONENT MACHINE (HOURS/UNIT) LABOR (HOURS/UNIT) STORAGE (SQUARE FEET/UNIT) 322A 0.10 0.05 2 802B 0.08 0.07 3

17. Suppose the production levels for march had been 1500 units of component 322A and 1000 units of component 802B for a total production level of 1500 + 1000 = 2500 units Amount of change in production of April is : April production – March production = Change (X 11 + X 21 ) -2500 = Change Change can be positive or negative (X 11 + X 21 ) -2500 = I 1 - D 1

18. Similarly, (X 12 + X 22 ) - (X 11 + X 21 ) = I 2 - D 2 MAY (X 13 + X 23 ) - (X 12 + X 22 ) = I 3 - D 3 JUNE Complete Set : (X 11 + X 21 ) -2500 - I 1 + D 1 = 2500 (X 12 + X 22 ) - (X 11 + X 21 ) - I 2 + D 2 = 0 (X 13 + X 23 ) - (X 12 + X 22 ) - I 3 + D 3 = 0

19. Optimal Solution to the Bollinger Electronics Production Scheduling Problem : Variable Value Reduced Costs X 11 500 0 X 12 3200 0 X 13 5200 0 X 21 2500 0 X 22 2000 0 X 23 0 0.128 S 11 0 0.172 S 12 200 0 S 13 400 0

20. Variable Value Reduced Costs S 21 1700 0 S 22 3200 0 S 23 200 0 I 1 500 0 I 2 2200 0 I 3 0 0.072 D 1 0 0.700 D 2 0 0.700 D 3 0 0.628

21. Constraint Slack/Surplus Dual Prices 1 0 -20 2 0 -10 3 0 -20.128 4 0 -10.150 5 0 -20.428 6 0 -10.300 7 0 -20.728 8 0 -10.450 9 150 0

22. Constraint Slack/Surplus Dual Prices 10 20 0 11 80 0 12 100 0 13 0 1.111 14 40 0 15 4900 0 16 0 0 17 8600 0 18 0 0.500 19 0 0.500 20 0 0.428

23. Total Production,Inventory & production-smoothing cost = $225,295 Activity April May June PRODUCTION 322A 500 3200 5200 802B 2500 2000 0 TOTALS 3000 5200 5200 ENDING INVENTORY 322A 0 200 400 802B 1700 3200 200 MACHINE USAGE SCHEDULED HOURS 250 480 520 SLACK CAPACITY HOURS 150 20 80 LABOR USAGE SCHEDULED HOURS 200 300 260 SLACK CAPACITY HOURS 100 0 40 STORAGE USAGE SCHEDULED HOURS 5100 10,000 1400 SLACK CAPACITY 4900 0 8600