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Production Management Application by Aparna Asha. v Saritha Jinto Antony Kurian.

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Presentation on theme: "Production Management Application by Aparna Asha. v Saritha Jinto Antony Kurian."— Presentation transcript:

1 Production Management Application by Aparna Asha. v Saritha Jinto Antony Kurian

2 Contents A Make-or-Buy Decision Production Scheduling Workforce Assignment

3 A Make-or-Buy Decision

4 QUESTION A company markets various business and engineering products. Currently it is preparing to introduce two new calculators: one for the business market called the Financial Manager and one for the engineering market called the Technician. Each calculator has three components: a base, an electronic cartridge, and a face plate or top. The same base is used for both calculators, but the cartridges and tops are different. All components can be manufactured by the company or purchased from the outside suppliers. 3000 Financial Manager calculators and 2000 Technician calculators will be needed.

5 Contd: Manufacturing capacity is limited. The company has 200 hours of regular manufacturing time and 50 hours of overtime that can be scheduled for the calculators. Overtime involves a premium at the additional cost of $9 per hour. The problem for the company- to determine how many units of each component to manufacture and how many units of each component to purchase.

6 OBJECTIVE FUNCTION Decision Variables BM No. of bases manufactured BP No. of bases purchased FCM No. of Financial manager cartridges mfrd FCP No. of Financial manager cartridges prchd TCM No. of technician cartridges manufactured TCP No. of technician cartridges purchased FTM No. of financial manager tops manufactured FTP No. of financial manager tops purchased TTM No. of technician tops manufactured TTP No. of technician tops purchased OT No. of hours of overtime to be scheduled.

7 Objective of the decision maker To minimize the total cost, including manufacturing costs, purchase costs and overtime costs. Hence the objective function is: Min 0.5BM+0.6BP+3.75FCM+4FCP+3.3TCM+3.9TCP+0.6F TM+0.65FTP+0.75TTM+0.78TTP+9 OT

8 CONSTRAINTS Number of each component needed to satisfy the demand for 3000 FM calculators and 2000 Technician calculators. The five demand constraints are BM+BP =5000 Bases FCM+FCP=3000 FM cartridges TCM+TCP=2000 Technician cartridges FTM+FTP=3000 FM tops TTM+TTP=2000 Technician tops

9 Manufacturing capacities for regular time and overtime cannot be exceeded 1) Limits overtime capacity to 50 hours OT <= 50 2) Total manufacturing time required for all components must be less than or equal to the total manufacturing capacity ( regular time + overtime) BM+3FCM+2.5TCM+FTM+1.5TTM<=12,000+60OT ie, BM+3FCM+2.5TCM+FTM+1.5TTM-60OT<=12,000

10 Complete Formulation Min 0.5BM+0.6BP+3.75FCM+4FCP+3.3TCM+3.9TCP+0.6FTM+ 0.65FTP+0.75TTM+0.78TTP+9OT S.t. BM + BP = 5000 Bases FCM + FCP =3000 FC TCM + TCP = 2000 TC FTM + FTP = 3000 FT TTM + TTP = 2000 TT OT <= 50 Overtime hours BM+3FCM+2.5TCM+FTM+1.5TTM-60 OT <= 12,000 Manuftrng cpcty

11 Production Scheduling

12 One of the most important applications of linear programming deals with multiperiod planning such as production scheduling. Let us consider the case of Bollinger Electronics Company, which produces two different electronic components for a major airplane engine manufacturer. The airplane engine manufacturer notifies the Bollinger sales office each quarter of their monthly requirements for components for each of the next 3 months. The monthly requirements for the components may vary considerably, depending on the type of engine the airplane engine manufacturer is producing.

13 The table below shows the order that has been received for the next 3 month period. Three month demand schedule for Bollinger Electronics Company ComponentAprilMayJune 322A 802B 1000 3000 500 5000 3000

14 After the order is processed, a demand statement is sent to the production control department. The production control department must then develop a 3 month production plan for the components. In arriving at the desired schedules, the production manager will identify: Total production cost Inventory holding cost Change in production schedule cost

15 Let x im denote the production volume in units for product i in month m. Here i=1,2 and m=1,2,3; i=1 refers to component 322 A, i=2 refers to component 802B, m=1 refers to April, m=2 refers to May and m=3 refers to June.

16 If component 322A costs $20 per unit produced and component 802B costs $10 per unit produced, the total production cost part of the objective function is Total production cost= 20x11+20x12+20x13+10x21+10x22+10x23

17 In order to incorporate the relevant inventory holding cost into the model, let ‘Sim’denote the inventory level for the product ‘i ’ at the end of the month ‘m’. Bollinger has determined that the monthly inventory holding costs are 1.5% of the cost of the product,ie., (0.015)($20)=$0.30 per unit for the component 322A and (0.015)($10)=$0.15 per unit of component 802B..

18 A common assumption made in production scheduling is that the monthly inventories are an acceptable approximation of the average inventory levels throughout the month. Making this assumption the inventory holding cost portion of the objective function will be as follows: Inventory holding cost= 0.30s11 +0.30s12 +0.30s13+0.15s21+0.15s22+0.15s23

19 To incorporate the costs of fluctuations in production levels from month to month, two more variables are defined: Im =increase in the total production levels during the month m Dm=decrease in the total production level during the month m After estimating the effects of employee layoffs,turnovers,reassignment training costs and other costs, Bollinger estimates that the cost of increase in production level for any month is $.0.50 per unit increase.Similarly the cost of decrease in production level for any month is estimated as $0.20 per unit decrease.

20 The third portion of the objective function will be: Change in production level costs- 0.50I1 + 0.50I2 + 0.50I3 + 0.20D1 + 0.20D2 + 0.20D3 Combining all the costs, the complete objective function becomes: Min 20x11 + 20x12 +20x13 +10x21 + 10x22 + 10x23 + 0.30s11 + 0.30s12+ 0.30s13 + 0.15s21 + 0.15s22 + 0.15s23 + 0.50I1 + 0.50I2 + 0.50I3 + 0.20D1 + 0.20D2 + 0.20D3

21 Constraints The units demanded can be expressed as: Ending inventory from previous month + Current production-Ending inventory for this month=This month’s demand Suppose the inventories at the beginning of the 3- month scheduling period were 500 units for component 322A and 200 units for component 802B. The demand for both products in the first month (April) was 1000 units, so the constraints for meeting demand in the first month becomes

22 500+x11-s11=1000 200+x21-s21=1000 Moving the constants to the right side we have: X11-s11=500 X21-s21=800

23 Similarly demand constraints for second and third month: Month 2 S11+x12-s12=3000 S21+x22-s22=500 Month 3 S12+x13-s13=5000 S22+x23-s23=3000

24 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 S13>=400 S23>=200

25 Suppose if additional information on machine, labour and storage capacity is given: Month Machine Labour Storage capacity capacity capacity (hours) (hours) (square feet) April 400 300 10,000 May 500 300 10,000 June 600 300 10,000

26 Machine, Labour and storage requirements Component Machine Labour Storage (hours/unit) (hours/unit) (square feet) 322A 0.10 0.05 2 802B 0.08 0.07 3

27 Then the constraints are: Machine capacity 0.10x11+0.08x21<=400 Month 1 0.10x12+0.08x22<=500 Month 2 0.10x13+0.08x23<=600 Month 3 Labour capacity 0.05x11+0.07x21<=300 Month 1 0.05x12+0.07x22<=300 Month 2 0.05x13+0.07x23<=300 Month 3 Storage capacity 2s11+3s21<=10,000 Month 1 2s12+3s22<=10,000 Month 2 2s13+3s23<=10,000 Month 3

28 One final set of constraints must be added to guarantee that Im and Dm will reflect the increase or decrease in the the total production level for month m. Suppose that the production level for March, the month before the start of the current production scheduling period, had been 1500 units of component 322A and 1000 units of component 802B for a total production level of 1500+1000=2500 units. We can find the change in production for April from the relationship April production-March production= Change Using the April production variables, x11 and x21 and the March production of 2500 units, we have (x11 +x21) -2500 = Change

29 A positive change reflects an increase in the total production level, and a negative change reflects a decrease in the total production level. We can use the increase in production for April, I1, and the decrease in production for April, D1, to specify the constraint for the change in total production for the month of April: (x11 +x21) – 2500= I1 – D1 We cannot have an increase in production and a decrease in production during the same month; thus either I1 OR D1 will be zero. This approach of denoting the change in production level as the difference between two non negative variables, I1 and D1, permits both positive and negative changes in the production level. If a single variable (like cm) had been used, only positive changes would be possible because of the nonnegativity requirement.

30 Using the same approach in May and June, we obtain the constraints for the second and third months of the production scheduling period: (x12 + x22) – (x11 + x21) = I2 –D2 (x13 +x23) – (x12 + x22) = I3 – D3 Placing the variables on the left-hand side and the constraints on the right-hand side yields the complete set of what is commonly referred to as production-smoothing constraints. x11 +x12 – I1 + D1 = 2500 -x11 –x21 +x12 +x22 –I2 +D2 = 0 -x12 – x22 + x13 + x23 – I3 + D3 =0

31 Minimum-cost production schedule information for ActivityAprilMayJune Production50032005200 Comp 322A Comp 802B 250020000 Ending inventory-322A 802B 0 1700 200 3200 400 200 Machine usage- scheduled hrs Slack capacity 250 150 480 20 520 80 Labour usage-scheduled hrs Slack capacity 200 100 300 0 260 40 Storage usage-scheduled storage Slack capacity 5100 4900 10000 0 1400 8600

32 Workforce Assignment

33 Workforce assignment Important for managers to make decisions regarding staffing. Reduces the cost of labor if employees can be cross trained in two or more jobs. Not only optimal product mix, but also optimal workforce assignment.

34 Case overview McCormick manufacturing company produces two products with contributions to profit per unit of $10 and $9 respectively. The labor requirement per unit produced and the total hours of labor available from personnel assigned to each of four department are shown

35 Labor-Hours per unit DepartmentProduct1Product2Total Available Hours 10.650.956500 20.450.856000 31.000.707000 40.150.301400

36 Decision variable P 1 = units of products 1 P 2 = units of products 2 Objective function : To maximize profit Max Z = 10P 1 +9P 2 Subjected to constraints 0.65P 1 +0.95P 2 <= 6500 0.45P 1 +0.85P 2 <= 6000 1.00P 1 +0.70P 2 <= 7000 0.15P 1 +0.30P 2 <= 1400

37 Non-negativity constraints P 1,P 2 >= 0 Optimal solution P 1 = 5744 P 2 = 1795 Z = $ 73,590

38 Slack/Surplus Department 3 & 4 are on max operation at capacity. Department 1 & 2 have a slack of appx 1062 and 1890 hours. Feasible solution Transfer labor hours from the departments which have slack to one’s which need more hours to increase the profit.

39 Cross training ability and capacity information From Department 1234Max hours transferable 1--Yes --400 2-- Yes 800 3-- Yes100 4Yes -- 200

40 Additional variables b i = the labor hours allocated to department i (i = 1,2,3,4) t ij = the labor hours transferred from department i to department j Writing the capacity constraints in terms of b 0.65P 1 + 0.95P 2 <= b 1 0.45P 1 + 0.85P 2 <= b 2 1.00P 1 + 0.70P 2 <= b 3 0.15P 1 + 0.30P 3 <= b 4 Bringing b i on to the left side of the inequalities for all the equations. 0.65P 1 + 0.95P 2 – b 1 <= 0 …………

41 b 1 = Hours initially in department 1 + hours transferred into department 1 – hours transferred out of department 2 b 1 = 6500+t 41 -t 12 -t 13 rewriting ….. b 1 -t 41 +t 12 +t 13 = 6500 b 2 -t 12 +t 42 +t 23 +t 24 = 6000 b 3 -t 13 -t 23 +t 34 = 7000 b 4 -t 24 -t 34 +t 41 +t 42 = 1400

42 Transfer capacity t 12 +t 13 <= 400 t 23 +t 24 <= 800 t 34 <= 100 t 41 +t 42 <= 200 Profits maximized from $73,590 to $ 84,011 P 1 = 6825 P 2 = 1751 Conclusion Not only the profits are maximized, but also the labor workforce is also optimized.

43 THANK YOU


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