Production Management Application by Aparna Asha v Saritha

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

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

A Make-or-Buy Decision

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.

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.

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

• 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. 5 BM+0. 6 BP+3. 75 FCM+4 FCP+3. 3 TCM+3. 9 TCP+0. 6 F TM+0. 65 FTP+0. 75 TTM+0. 78 TTP+9 OT

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

• 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+3 FCM+2. 5 TCM+FTM+1. 5 TTM<=12, 000+60 OT ie, BM+3 FCM+2. 5 TCM+FTM+1. 5 TTM-60 OT<=12, 000

Complete Formulation • Min 0. 5 BM+0. 6 BP+3. 75 FCM+4 FCP+3. 3 TCM+3. 9 TCP+0. 6 FTM+ 0. 65 FTP+0. 75 TTM+0. 78 TTP+9 OT 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+3 FCM+2. 5 TCM+FTM+1. 5 TTM-60 OT <= 12, 000 Manuftrng cpcty

Production Scheduling

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.

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

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

Let xim 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 802 B, m=1 refers to April, m=2 refers to May and m=3 refers to June.

If component 322 A costs $20 per unit produced and component 802 B costs $10 per unit produced, the total production cost part of the objective function is Total production cost= 20 x 11+20 x 12+20 x 13+10 x 21+10 x 22+10 x 23

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 322 A and (0. 015)($10)=$0. 15 per unit of component 802 B. .

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. 30 s 11 +0. 30 s 12 +0. 30 s 13+0. 15 s 21+0. 15 s 22+0. 15 s 23

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.

The third portion of the objective function will be: Change in production level costs- 0. 50 I 1 + 0. 50 I 2 + 0. 50 I 3 + 0. 20 D 1 + 0. 20 D 2 + 0. 20 D 3 Combining all the costs, the complete objective function becomes: Min 20 x 11 + 20 x 12 +20 x 13 +10 x 21 + 10 x 22 + 10 x 23 + 0. 30 s 11 + 0. 30 s 12+ 0. 30 s 13 + 0. 15 s 21 + 0. 15 s 22 + 0. 15 s 23 + 0. 50 I 1 + 0. 50 I 2 + 0. 50 I 3 + 0. 20 D 1 + 0. 20 D 2 + 0. 20 D 3

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 322 A and 200 units for component 802 B. The demand for both products in the first month (April) was 1000 units, so the constraints for meeting demand in the first month becomes

500+x 11 -s 11=1000 200+x 21 -s 21=1000 Moving the constants to the right side we have: X 11 -s 11=500 X 21 -s 21=800

Similarly demand constraints for second and third month: Month 2 S 11+x 12 -s 12=3000 S 21+x 22 -s 22=500 Month 3 S 12+x 13 -s 13=5000 S 22+x 23 -s 23=3000

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

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

Machine, Labour and storage requirements Component 322 A 802 B Machine Labour Storage (hours/unit) (square feet) 0. 10 0. 05 2 0. 08 0. 07 3

Then the constraints are: Machine capacity 0. 10 x 11+0. 08 x 21<=400 0. 10 x 12+0. 08 x 22<=500 0. 10 x 13+0. 08 x 23<=600 Month 1 Month 2 Month 3 Labour capacity 0. 05 x 11+0. 07 x 21<=300 0. 05 x 12+0. 07 x 22<=300 0. 05 x 13+0. 07 x 23<=300 Month 1 Month 2 Month 3 Storage capacity 2 s 11+3 s 21<=10, 000 2 s 12+3 s 22<=10, 000 2 s 13+3 s 23<=10, 000 Month 1 Month 2 Month 3

One final set of constraints must be added to guarantee that Im and Dm will reflect the increase or decrease in 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 322 A and 1000 units of component 802 B 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, x 11 and x 21 and the March production of 2500 units, we have (x 11 +x 21) -2500 = Change

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, I 1, and the decrease in production for April, D 1, to specify the constraint for the change in total production for the month of April: (x 11 +x 21) – 2500= I 1 – D 1 We cannot have an increase in production and a decrease in production during the same month; thus either I 1 OR D 1 will be zero. This approach of denoting the change in production level as the difference between two non negative variables, I 1 and D 1, 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.

Using the same approach in May and June, we obtain the constraints for the second and third months of the production scheduling period: (x 12 + x 22) – (x 11 + x 21) = I 2 –D 2 (x 13 +x 23) – (x 12 + x 22) = I 3 – D 3 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. x 11 +x 12 – I 1 + D 1 = 2500 -x 11 –x 21 +x 12 +x 22 –I 2 +D 2 = 0 -x 12 – x 22 + x 13 + x 23 – I 3 + D 3 =0

Minimum-cost production schedule information for Activity April May June Production 500 3200 5200 Comp 322 A Comp 802 B 2500 2000 0 Ending inventory-322 A 802 B 0 1700 200 3200 400 200 Machine usagescheduled hrs Slack capacity 250 150 480 20 520 80 Labour usage-scheduled hrs Slack capacity 200 100 300 0 260 40 10000 0 1400 8600 Storage usage-scheduled 5100 storage 4900 Slack capacity

Workforce Assignment

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.

Case overview • Mc. Cormick 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

Labor-Hours per unit Department Product 1 Product 2 Total Available Hours 1 0. 65 0. 95 6500 2 0. 45 0. 85 6000 3 1. 00 0. 70 7000 4 0. 15 0. 30 1400

• Decision variable P 1 = units of products 1 P 2 = units of products 2 • Objective function : To maximize profit Max Z = 10 P 1+9 P 2 • Subjected to constraints 0. 65 P 1+0. 95 P 2 <= 6500 0. 45 P 1+0. 85 P 2 <= 6000 1. 00 P 1+0. 70 P 2 <= 7000 0. 15 P 1+0. 30 P 2 <= 1400

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

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.

Cross training ability and capacity information From Department 1 2 3 4 Max hours transferable 1 -- Yes -- 400 2 -- -- Yes 800 3 -- -- -- Yes 100 4 Yes -- -- 200

Additional variables • bi= the labor hours allocated to department i (i = 1, 2, 3, 4) • tij = the labor hours transferred from department i to department j Writing the capacity constraints in terms of b 0. 65 P 1 + 0. 95 P 2 <= b 1 0. 45 P 1 + 0. 85 P 2 <= b 2 1. 00 P 1 + 0. 70 P 2 <= b 3 0. 15 P 1 + 0. 30 P 3 <= b 4 Bringing bi on to the left side of the inequalities for all the equations. 0. 65 P 1 + 0. 95 P 2 – b 1<= 0 …………

• 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

• 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.

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