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Aggregate Production Planning (APP)

 Matches market demand to company resources  Plans production 6 months to 12 months in advance  Expresses demand, resources, and capacity in general terms  Develops a strategy for economically meeting demand  Establishes a companywide game plan for allocating resources

Inputs and Outputs to Aggregate Production Planning

Capacity Constraints Strategic Objectives Company Policies Demand Forecasts Aggregate Production Planning Size of Workforce Production per month (in units or $) Inventory Levels Financial Constraints Units or dollars subcontracted, backordered, or lost

Strategies for Meeting Demand

1. Use inventory to absorb fluctuations in demand (level production) 2. Hire and fire workers to match demand (chase demand) 3. Maintain resources for high demand levels 4. Increase or decrease working hours (over & undertime) 5. Subcontract work to other firms 6. Use part-time workers 7. Provide the service or product at a later time period (backordering)

Strategy Details

 Level production - produce at constant rate & use inventory as needed to meet demand  Chase demand - change workforce levels so that production matches demand  Maintaining resources for high demand levels ensures high levels of customer service  Overtime & undertime - common when demand fluctuations are not extreme

Strategy Details

 Subcontracting - useful if supplier meets quality & time requirements  Part-time workers - feasible for unskilled jobs or if labor pool exists  Backordering - only works if customer is willing to wait for product/services

Units

Level Production

Demand Production Time

Units

Chase Demand

Demand Time Production

APP Example

 The Bavarian Candy Company (BCC) makes a variety of candies in three factories worldwide. Its line of chocolate candies exhibits a highly seasonal pattern with peaks in winter months and valleys during the summer months. Given the costs and quarterly sales forecasts, determine whether a level production or chase demand production strategy would be more economically meet the demand for chocolate candies.

APP Using Pure Strategies

Quarter Spring Summer Fall Winter Sales Forecast (kg) 80,000 50,000 120,000 150,000 Hiring cost = $100 per worker Firing cost = $500 per worker Inventory carrying cost = $0.50 per kilogram per quarter Production per employee = 1,000 kilograms per quarter Beginning work force = 100 workers

Level Production Strategy

Quarter Spring Summer Fall Winter Sales Forecast 80,000 50,000 120,000 150,000 400,000 Production Plan 100,000 100,000 100,000 100,000 Inventory 20,000 70,000 50,000 0 140,000 Cost = 140,000 kilograms x $0.50 per kilogram = $70,000

Chase Demand Strategy

Quarter Spring Summer Fall Winter Sales Forecast 80,000 50,000 120,000 150,000 Production Plan 80,000 50,000 120,000 150,000 Workers Needed 80 50 120 150 Workers Hired 70 30 100 Workers Fired 20 30 50 Cost = (100 workers hired x $100) + (50 workers fired x $500) = $10,000 + 25,000 = $35,000

LP Formulation

Define H t F t I t P t = inventory at end of period t = Production in period t W t = # hired for period t = # fired for period t = Workforce in period t

Min Z = $100 (H 1 $0.50 (I 1 + I 2 + I + H 3 + I 2 4 + H ) 3 + H 4 ) + $500 (F 1 + F 2 + F 3 + F 4 ) +

Min Z = $100 (H 1 + H 2 + H 3 + H 4 ) + $500 (F 1 + F 2 + F 3 + F 4 )+ $0.50 (I 1 + I 2 + I 3 + I 4 )

Subject to P 1 - I 1 = 80,000 I 1 + P 2 - I 2 = 50,000 I 2 + P 3 - I 3 = 120,000 I 3 + P 4 - I 4 = 150,000 (1) (2) (3) (4) Demand constraints

Min Z = $100 (H 1 Subject to + H 2 + H 3 + H 4 ) + $500 (F 1 I 1 I 2 I 3 P 1 + P 2 + P 3 + P 4 - I 1 - I 2 - I 3 - I 4 = 80,000 = 50,000 = 120,000 = 150,000 + F 2 + F 3 + F 4 )+ $0.50 (I 1 + I 2 + I 3 + I 4 ) (1) (2) (3) (4) Demand constraints

P 1 - 1,000 W 1 = 0 P 2 - 1,000 W 2 = 0 P 3 - 1,000 W 3 = 0 P 4 - 1,000 W 4 = 0 (5) (6) (7) (8) Production constraints

Min Z = $100 (H 1 Subject to + H 2 + H 3 + H 4 ) + $500 (F I 1 I 2 I 3 P 1 + P + P 2 3 - I 1 - I 2 - I 3 = 80,000 = 50,000 = 120,000 + P 4 P 1 P 2 P 3 P 4 - I 4 = 150,000 - 1,000 W - 1,000 W - 1,000 W - 1,000 W 1 2 3 4 = 0 = 0 = 0 = 0 1 + F 2 + F 3 + F 4 )+ $0.50 (I 1 + I 2 (1) (2) (3) (4) (5) (6) (7) (8) Demand constraints Production constraints + I 3 + I 4 )

W 1 - H 1 + F 1 = 100 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 (9) (10) (11) (12) Work force constraints

Min Z = $100 (H 1 + H 2 + H 3 + H 4 ) + $500 (F 1 + F 2 + F 3 + F 4 )+ $0.50 (I 1 + I 2 + I 3 + I 4 ) Subject to P 1 - I 1 = 80,000 I 1 + P 2 - I 2 = 50,000 I 2 + P 3 - I 3 = 120,000 I 3 + P 4 - I 4 = 150,000 P 1 - 1,000 W 1 = 0 P 2 - 1,000 W 2 = 0 P 3 - 1,000 W 3 = 0 P 4 - 1,000 W 4 = 0 W 1 - H 1 + F 1 = 100 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 (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) Demand constraints Production constraints Work force constraints

LP Solution: Z = $32,000 H 1 = 0, F 1 = 20, I 1 = 0, P 1 =80000; H 2 = 0, F 2 = 0, I 2 = 30000, P 2 =80000; H 3 = 10, F 3 = 0, I 3 = 0, P 3 =90000; H 4 = 60, F 4 = 0, I 4 = 0, P 4 =150000;

Summary: APP By Linear Programming

Min Z = $100 (H 1 + H 2 + H 3 + H 4 ) + $500 (F 1 + F 2 + F 3 + F 4 )+ $0.50 (I 1 + I 2 + I 3 + I 4 ) Subject to I 1 I 2 I 3 P + P 2 + P 3 + P 4 1 - I - I 2 - I 3 - I 4 1 = 80,000 = 50,000 = 120,000 = 150,000 P 1 - 1,000 W 1 = 0 P 2 - 1,000 W 2 = 0 P 3 - 1,000 W 3 = 0 P 4 - 1,000 W 4 = 0 W 2 W 3 W 1 - W 1 - H - H 1 - W 2 - H 3 2 + F 1 + F + F 2 3 = 100 = 0 = 0 W 4 - W 3 - H 4 + F 4 = 0 (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) Demand constraints Production constraints Work force constraints where H t F t I t = # hired for period t = # fired for period t = inventory at end of period t

LP Solution: Z = $32,000 H 1 = 0, F 1 = 20, I 1 = 0, P 1 =80000; H 2 = 0, F 2 = 0, I 2 = 30000, P 2 =80000; H 3 = 10, F 3 = 0, I 3 = 0, P 3 =90000; H 4 = 60, F 4 = 0, I 4 = 0, P 4 =150000;

Quarter 1 2 3 4

APP By The Transportation Method

Expected Demand 900 1500 1600 3000 Regular Capacity 1000 1200 1300 1300 Overtime Subcontract Capacity 100 150 200 200 Capacity 500 500 500 500 Regular production cost per unit = $20 Overtime production cost per unit = $25 Subcontracting cost per unit = $28 Inventory carrying cost per unit per period = $3 Beginning inventory = 300 units

Initial Setup of the Transportation Tableau

Period of Production Period of 1 Use Beginning Inventory 1 Regular

300 20

Un used Capa -city

1000

Overtime

25 28 31 34

Subcontract

28 31 34 37

2 Regular

20 23 26

Overtime

25 28 31

Subcontract

28 31 34

3 Regular

20 23 100 500 1200 150 500 1300

Overtime

25 28 200

Subcontract

28 31 500

4 Regular

20 1300

Overtime

25 200

Subcontract

28 500

Demand

900 1500 1600 3000

Solution: Step 1

Period of Production Period of 1 Use Beginning Inventory 1 Regular Overtime 300 20

600

25 2 23 28 3 4 26 29 31 34 Subcontract 28 31 34 37 2 Regular 20 23 26 Overtime 25 28 31 Subcontract 28 31 34 3 Regular 20 23 Overtime 25 28 Subcontract 28 31 4 Regular 20 Overtime 25 Subcontract 28 Un used Capa -city 1000 100 500 1200 150 500 1300 200 500 1300 200 500 Demand 900 1500 1600 3000

Solution: Step 2

Period of Production Period of 1 Use Beginning Inventory 1 Regular Overtime 300 20 600 25 2 23

300

28 3 4 26 29 31 34 Subcontract 28 31 34 37 2 Regular Overtime 20

1200

25 23 26 28 31 Subcontract 28 31 34 3 Regular 20 23 Overtime 25 28 Subcontract 28 31 4 Regular 20 Overtime 25 Subcontract 28 Un used Capa -city 1000 100 500 1200 150 500 1300 200 500 1300 200 500 Demand 900 1500 1600 3000

Final Transportation Tableau

Period of Production Period of 1 Use Beginning Inventory 1 Regular Overtime

300

600

25 2 23

300

28 3 26

100

31 Subcontract 28 31 34 4 29 ---- 34

100

37 2 Regular Overtime Subcontract 3 Regular Overtime Subcontract 4 Regular Overtime Subcontract Demand 900 20

1200

25 28 ---- 28

200

23 31 20

1300

25 28 26 ---- 31

150

250

23 ---- 28 ---- 31

500

1300

200

500

1500 1600 3000 Un used Capa -city 0 0 500 500 0 0 1000 100 1200 150 250 500 0 0 0 0 0 0 1300 200 500 1300 200 500

Production Plan

Period 1 2 3 4 Demand 900 1500 1600 3000 Total Strategy Variable Reg Prodn Overtime Sub End Inv 1000 1200 1300 1300 7000 4800 100 0 150 200 250 500 200 500 650 1250 500 600 1000 0 2100

 Regular Production Cost = (4,800 * $20)=$96,000  Overtime Production Cost = (650 * $25)= $16,250  Subcontracting Cost = (1,250 * $28) = $35,000  Inventory Cost = (2,100 * $3) = $ 6,300  The Total Cost of the Plan = $153,550

Linear Programming Formulation

Let:  D t = units required in period t, (t = 1,…,T)      m = number of sources of product in any period P it = capacity, in units of product, of source i in period t, (i = 1,…,m) X it = planned quantity to be obtained from source i in period t c it = variable cost per unit from source i in period t h t = cost to store a unit from period t to period t+1  I t = inventory level at the end of period t, after satisfying the requirement in period t

Minimize z



t T

   1

i m

 1

c it X it



h t I t

]

ST X it



P it

, (

 1 ,...,

;

 1 ,...,

)

I t X it I t

  

I t

0 , 0 ,  (

1 (



i m

  1

X it

   1 ,...,

1 ,...,

) ;

t D t

,  (

 1 ,...,

) 1 ,...,

)

Minimize z

  28 ( 20 (

X R

X S

1  

X S

X R

2  

X S

X R

3  

X S

X R

4 ) )   3 (

25 (

X O

1 1 

2  

I X O

2 3  

4 )

X O

3 

X O

4 )

ST X X O

X X R

S X R

X R

4 1

1        1000 ,

X R

2  1200 ,

X R

3  1300 ,

X R

4  1300 ; 100 500 ,

X X X X O

O O O

2 3 4 ,    

X O X S

X S

2  

2   600 ,

1 

X X

S S

3 4     150 500 ,

I I

3 4 ,  

X O X S

I I

3 2 3     200 , 500 ,

X O

X S

4   200 500 ; ; 1500 1600 3000 , , ;

Optimal Value (Z) = $153,550

            XR1 XR2 XR3 XR4 XO1 XO2 XO3 XO4 XS1 XS2 XS3 XS4 = 1000 = 1200 = 1300 = 1300 = 0 = 150 = 200 = 200 = 0 = 350 = 500 = 500

Strategies for Managing Demand

 Shift demand into other periods incentives, sales promotions, advertising campaigns  Offer product or services with counter cyclical demand patterns create demand for idle resources

Aggregate Planning for Services

1. Most services can’t be inventoried 2. Demand for services is difficult to predict 3. Capacity is also difficult to predict 4. Service capacity must be provided at the appropriate place and time 5. Labor is usually the most constraining resource for services

Services Example

 The central terminal at the Deutsche Cargo receives airfreight from aircraft arriving from all over Europe and redistributes it to aircraft for shipment to all European destinations. The company guarantees overnight shipment of all parcels, so enough personnel must be available to process all cargo as it arrives. The company now has 24 employees working in the terminal. The forecasted demand for warehouse workers for the next 7 months is 24, 26, 30, 28, 28, 24, and 24. It costs $2,000 to hire and $3,500 to lay off each worker. If overtime is used to supply labor beyond the present work force, it will cost the equivalent of $2,600 more for each additional worker. Should the company use a level capacity with overtime or a matching demand plan for the next six month?

The Level Capacity with Overtime Plan

(1) Month

5 6 1 2 3 4

(2) Number of Workers Forecasted

24 26 30 28 28 24

(3) Overtime Labor Cost [(2)-24]*$2,600

0 $5,200 15,600 10,400 10,400 0

The Matching Demand Plan

0 1 2 3 4 5 6 7 (1)

Month

(2)

Number of Workers Required

24 24 26 30 28 28 24 24 (3)

Number of Workers Hired

0 2 4 0 (4)

Number of Workers Laid Off

(5)

Cost of Hired Workers [(3)*$2,000]

(6)

Cost of Laid-Off Workers [(4)*$3,500]

2 0 4 0 $4,000 8,000 0 $7,000 0 14,000

 The cost of the Level Capacity with Overtime = $ 41,600  The total cost of the Matching Demand plan = $12,000 + $21,000 = $33,000  Hence, since the cost of matching demand plan is less than the level capacity plan with overtime and would be the preferred plan

Aggregate Planning Example 1

 A manufacturer produces a line of household products fabricated from sheet metal.

To illustrate his production planning problem, suppose that he makes only four products and that his production system consists of five production centers: stamping, drilling, assembly, finishing (painting and printing), and packaging. For a given month, he must decide how much of each product to manufacture, and to aid in this decision, he has assembled the data shown in Tables 1 and 2. Furthermore, he knows that only 2000 square feet of the type of sheet metal used for products 2 and 4 will be available during the month.

Product 2 requires 2.0 square feet per unit and product 4 uses 1.2 square feet per unit.

 

TABLE 1

Production Data for Example 1

DEPARTMENT Stamping PRODUCTION RATES IN HOURS PER UNIT PRODUCT 1 PRODUCT 2 PRODUCT 3 PRODUCT 4 Production Hours Available 0.03 0.15 0.05 0.10 400



Drilling



Assembly 0.06 0.12 ---- 0.10 400 0.05 0.10 0.05 0.12 500

 

Finishing Packaging 0.04 0.20 0.03 0.12 450 0.02 0.06 0.02 0.05 400



TABLE 2

Product Data for Example 1     NET SELLING VARIABLE SALES POTENTIAL PRODUCT PRICE/UNIT COST/UNIT MINIMUM MAXIMUM 1 2 3 4 $10 $6 1000 6000 25 15 ---- 500 16 11 500 3000 20 14 100 1000

A Linear Program of Example 1:  Define x i 3, and 4.

be the number of units of Product i to be produced per month, i = 1, 2,

Maximize

1  10

2  5

3  6

0 0

x x x x x

0 .

1 2 3 3 4 4 03 06     

x x

1 1    500 500 100 0 0 3000 1000 .

15 12

x x

2 2 0 .

2 0

x x

0 .

1 2 .

2   0 .

06 1 .

2  1 1 04  

1  0 1000 6000 .

2      0 0 0 0 0 .

05 10 05 03 02 2000

x x x x

3 4 3

3 3      0 .

4 400 0 0 0 .

12 12 05

x x x

4 4 4     400 500 450 400



Solution

of Example 1 using

LINGO Software Package

copy of this package from the web site at www.lindo.com): (get a free 

Objective value

42600.00

   

Variable X1 X2 X3 X4 Value Reduced Cost 5500.000 0.0000000

500.0000 0.0000000

3000.000 0.0000000

100.0000 0.0000000

             

Row PROFIT STAMPING DRILLING ASSEMBLY FINISHING PACKAGING SHEETMETAL MINPROD1 MAXPROD1 MAXPROD2 MINPROD3 MAXPROD3 MINPROD4 MAXPROD4 Slack or Surplus Dual Price 42600.00 1.0000000

0.0000000 0.0000000

0.0000000 66.66666

13.00000 0.0000000

28.00000 0.0000000

195.0000 0.0000000

880.0000 0.0000000

4500.000 0.0000000

500.0000 0.0000000

0.0000000 2.0000000

2500.000 0.0000000

0.0000000 5.000000

0.0000000 -0.6666667

900.0000 0.0000000

Ranges in which the basis is unchanged

Objective Coefficient Ranges

   

Variable X1 X2 X3 X4 Current Allowable Allowable Coefficient Increase Decrease 4.000000 INFINITY INFINITY 10.00000 INFINITY INFINITY 5.000000 INFINITY INFINITY 6.000000 INFINITY INFINITY

Righthand Side Ranges:

            

Row STAMPING DRILLING ASSEMBLY FINISHING PACKAGING SHEETMETAL MINPROD1 MAXPROD1 MAXPROD2 MINPROD3 MAXPROD3 MINPROD4 MAXPROD4 Current Allowable Allowable RHS Increase Decrease 400.0000 INFINITY 100.0000

400.0000 INFINITY 3000.000

500.0000 INFINITY 500.0000

450.0000 INFINITY 5500.000

400.0000 INFINITY 0.0

2000.000 INFINITY 13.00000

1000.000 INFINITY 28.00000

6000.000 INFINITY 195.0000

500.0000 INFINITY 880.0000

500.0000 INFINITY 4500.000

3000.000 INFINITY 500.0000

100.0000 INFINITY 2500.000

1000.000 INFINITY 900.0000

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Aggregate Production Planning (APP)