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					       Chapter 3
Aggregate Planning




     By: Prof. Y. Peter Chiu
          9 / 1 / 2011
                               1
§. A1: Introduction
          ~ Aggregate Planning ~
 Macro production planning
 The problem of deciding how many
  employees the firm should retain
  — for a manufacturing firm to decide the
     quantity and mix of products to be produced
 e.g. In Service organizations:
 — Airlines plans staffing levels for flight
    attendants & pilots
 — Hospitals plans staffing levels for nurses

                                                   2
§. A1: Introduction               ( page 2 )



 Macro planning begins with the forecast of
  demand
 The aggregate planning methodology we
  discussed later, requires the assumption that
  〝demand is deterministic, or known in
  advance〞.
 Ways to satisfy demand
  — in house production
  — out-sourced / subcontracting

                                               3
§. A1: Introduction                     ( page 3 )

■   Competing objectives of Aggregate planning:
— To react quickly to anticipated changes in
   demand(i.e. making frequent and potentially
   large changes in the size of the labor force
   – A chase strategy, may be cost effective
   short term.)
— To retain a stable workforce.
— To develop a production plan for the firm that
   maximizes profit over the planning horizon
   subject to constraints on capacity.


                                                     4
§. A1: Introduction                  ( page 4 )

 Aggregate planning methodology is
  designed to translate demand forecasts into
  a blueprint for planning staffing and
  production levels for the firm over a
  predetermined planning horizon
 Production planning may be viewed as a
  hierarchical process in
   — purchasing
   — production
   — staffing
 decisions must be made at several levels in
 the firm
                                            5
Fig. 3-1   Fig.1 The hierarchy of prodution planning decision
p.128
                     Forcast of aggregate demand
                     for t period planning horizon


                     Aggregate production plan
                Determination of aggregate production
                & Workforce levels for t period planning
                               horizon


                     Master Production Schedule
                   Production levels by item by time
                                period


                    Materials Req. Planning System
                   Detailed timetable for production &
                     assembly of components and
                              subassemblies
                                                                6
§. A1: Introduction                      ( page 6 )



 Aggregate Units of Production
  — amount of work required(worker-years)
  — weight(tons of steel)
  — volume(gallons of gasoline)
  — dollar value(value of inventory in dollars)
  ~ Not Always Obvious




                                                  7
§. A2: Example ~ Aggregate Planning ~
 Example 3-1                              ( p.127 )
    A plant manager working for a large national
  appliance firm is considering implementing an
  aggregate planning system to determine the
  workforce and production levels in his plant. This
  particular plant produces 6 models of TVs. The
  characteristics of the TVs are :
   Model #       Number of Worker - Hours Required    Selling Price
                  to produce
      1                 4.2                          $285
      2                 4.9                          $345
      3                 5.1                          $395
      4                 5.2                          $425
      5                 5.4                          $525
      6                 5.8                          $725
                                                                 8
§. A2: Example                                          ( page 2 )
             ~ Aggregate Planning ~

 Example 3-1 (continued)
     The manger notices that the percentages of the
  total number of sales for these six models have been
  fairly constant:
 Model   #            % of the total numbers of sales


    1                    32%
    2                    21%
    3                    17%
    4                    14%
    5                    10%
    6                     6%
                                                              9
 §. A2: Example                                 ( page 3 )
                  ~ Aggregate Planning ~
Eg. 3-1 / Solution:
   To find the particular aggregation scheme
   (1) Selling price / Number of worker-hours required
         = $ per Input-Hour

        Model #                     $/hr


           1               $285/4.2 = $67.86
           2               $345/4.9 = $70.41
           3               $395/5.1 = $77.45
           4               $425/5.2 = $81.73
           5               $525/5.4 = $97.22
           6               $725/5.8 = $125.00
                                                      10
§. A2: Example          ~ Aggregate Planning ~         ( page 4 )

Eg. 3-1 / Solution:
      then
              Model #           $/hr * % of Sales

                1              $67.86* 0.32 = $21.72
                2              $70.41* 0.21 = $14.79
                3              $77.45* 0.17 = $13.17
                4              $81.73* 0.14 = $11.44
                5              $97.22* 0.10 = $9.72
                6              $125.00* 0.06 = $7.50


     what is $78.34?
                                         $78.34
     ―Average dollars of output / worker-hour input‖
        in this particular production plant
                                                             11
 §. A2: Example                                          ( page 5 )
                 ~ Aggregate Planning ~
 Eg. 3-1 / Solution: (continued)
  (2) The manager decides to define an aggregate unit of
      production as a fictitious TV

       Model #         Number of worker-Hours Required

         1                 4.2*0.32 = 1.34
         2                 4.9*0.21 = 1.03
         3                 5.1*0.17 = 0.87
         4                 5.2*0.14 = 0.73
         5                 5.4*0.10 = 0.54
         6                 5.8*0.06 = 0.35

                                     4 4.86
                                  Sum$=.86
                                                               12
§. A2: Example                                        ( page 6 )
              ~ Aggregate Planning ~
  ◆ what is 4.86 hours?
  ― Average worker-hours required to produce
      a fictitious TV‖
 What if we like to know
 ― how many fictitious TV can one worker – one
 day (8hrs) produce ? ‖
 [ 1 / 4.86 ] x 8 = 1.646

— Applications of this Avg. worker-hours/ TV :
~ If the manager can obtain sales forecast of overall models, then
he can use this to plan workforce ~

                                                             13
§. A3:     Hierarchical production planning (HPP)

 Hierarchy for Aggregate planning ( by Hax & Meal 1975 )
    (1) Items:Final products to be delivered to the customer.
               (SKU-stock keeping unit)
    (2) Families:A group of items that share a common
                 manufacturing setup cost
    (3) Types:Groups of families with production quantities
               that are determined by a single aggregate
               production plan
■   In our previous example, different models of TVs are families,
        while type might be large appliances.
■   Hax & Meal aggregation scheme will not necessarily work
        in every situation


                                                                 14
§. A4: Overview of the
         Aggregate planning problem

 The goal of aggregate planning is to determine
  aggregate production quantities and the levels of
  resources required to achieve these production goals

 The primary issues related to the aggregate planning
  problem include:
  ◆ Smoothing
     — 2 key components of smoothing costs are the
       costs that result from hiring and firing workers
  ◆ Bottleneck problems
     — System unable to respond to sudden changes
        in demand as a result of capacity restrictions
                                                    15
§. A4: Overview of the
        Aggregate planning problem             ( page 2 )


 The primary issues related to the aggregate planning
  problem (continued) :

  ◆ Planning horizon — Not too small T, Not too large T
                     — End-of-horizon effect
  ◆ Treatment of demand
     — Assumption of deterministic or known, ignores
         the possibility of forecast errors
     — Needs a buffer for forecast errors
     — incorporates the effects of seasonal
         fluctuations & business cycles
                                                     16
§. A5: Costs in Aggregate planning

 ◆ Smoothing Costs — laid off (firing) workers
                   — hiring workers
 ◆ Holding Costs — capital tied up in inventory
 ◆ Shortage Costs — excess demand normally
                    assumes backlogged
 ◆ Regular time Costs
          — the cost of producing one unit of output
              during regular working hours
 ◆ Overtime and subcontracting costs


                                                   17
§. A6: A Prototype problem
Example 3.2:      ( p.133)
    Densepack company is to plan workforce and
 production levels for the 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 demands for the next 6 months for a
 particular line of drives produced in their plant 1, are
 1280, 640, 900,1200, 2000, and 1400. There are
 currently (end of December) 300 workers employed in
 plant 1. Ending inventory in December is expected to
 be 500 units, and the firm would like to have 600 units
 on hand at the end of June. And It is estimated that:
                                                      18
§. A6: A Prototype problem                        ( page 2 )

Example 3.2:(continued)

 Cost of hiring 1 worker, C H  $500
 Cost of firing 1 worker, C F  $1000
 Cost of holding 1 unit of Inventory for 1month, C I  $80
   The plant manager observed that in the past, over 22
   working days, with the workforce level constant at 76
   workers, the firm produced 245 disk drives.
 What is ‗‘Number of aggregate units produced by 1
  worker in 1 day ?‘‘
  Evaluate : (1) the chase strategy (zero inventory plan)
        and (2) the constant workforce plan
                                                         19
§. A6: A Prototype problem                       ( page 3 )

Example 3.2 : Solution
Starting:500 units   300 workers                   300 workers

          1   1280                 1   780(-)    780
          2    640                 2   640      1420
          3    900                 3   900      2320

          4   1200                 4 1200       3520
          5   2000                 5 2000       5520
          6   1400                 6 2000(+) 7520

Ending:600 units     x worker?                   x worker?
                                                         20
§. A6: A Prototype problem                  ( page 4 )

Example 3.2 : Solution


   K= # of aggregate units produced by one worker in one
       day
   76 workers work 22 days producing 245 disk drives

       245
   K          0.14653
      76 (22)



                                                     21
§. A6: A Prototype problem                                          ( page 5 )

 Example 3.2 : Solution
(1) Evaluation of chase strategy
  Table 1: Initial Calculation for Chase Strategy Zero Inventory Plan for Densepack

   A                 B                  C                 D                  E
                                   Number of                            Minimum
                                Units Produced                         Number of
               Number of          per Worker        Forecasted      Workers Required
 month       Working Days         (B ×.14653)      Net Demand      (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


                                                                                 22
    §. A6: A Prototype problem                                   ( page 6 )

    Example 3.2 : Solution
    (1) Evaluation of chase strategy
    Table 2: Chase Strategy Zero Inventory Plan for Densepack

A         B         C         D         E      F        G       H          I
                                            Number of
                               Number of    Units                        Ending
       Number of Number Number  unit     Produced Cumulative Cumulative Inventory
month    Worker   Hired Fired per Worker (B ×E) Production Demand        (G-H)
January 267               33    2.931        783       783     780          3
February 182              85    3.517        640      1423    1420          3
March     342      160          2.638        902      2325    2320          5
April     315             27    3.810       1200      3525    3520          5
May       621      306          3.224       2002      5527    5520          7
June      910      289          2.198       2000      7527    7520          7

Total              755     145                                             30

                                                                                23
§. A6: A Prototype problem                                            ( page 7 )

 Example 3.2 : Solution
(1) Evaluation of chase strategy
    Table 2: Chase Strategy Zero Inventory Plan for Densepack

A         B         C         D         E      F        G       H          I
                                            Number of
                               Number of    Units                        Ending
       Number of Number Number  unit     Produced Cumulative Cumulative Inventory
month    Worker   Hired Fired per Worker (B ×E) Production Demand        (G-H)
January 267               33    2.931        782       782     780          2
February 182              85    3.517        640      1422    1420          2
March     342      160          2.638        902      2324    2320          4
April     315             27    3.810       1200      3525    3520          5
May       621      306          3.224       2002      5527    5520          7
June      910      289          2.198       2000      7527    7520          7

Total              755     145                                              30
                   753     144                                              13

    CH  $500                 755*(500)+145*(1000)+30*(80)=$524,900
                               600*(80)=4800               + 48,000
    C F  $1000
    C I  $ 80                                              572,900
                                  $569,540/910 workers                        24
               753 & 144 & 13 
A6: A Prototype problem                                                                        (page 7)
 Example 3.2 : Solution
 (1)Evaluation of chase strategy
Table 2:Chase Strategy Zero Inventory Plan for Densepack


    A              B             C             D              E            F              G           H            I
                                                                       Number of
                                                           Number of     Units
               Number 0f      Number        Number          unit per   Produced       Cumulative   Cumulativ    Ending
  month         Worker         Hired         Fired          Worker      (B × E)       Production   e Demand    Inventory
  January         267                          33            2.931        783            783         780          3
 February         182                          85            3.517        640           1423         1420         3
  March           341           159                          2.683        900           2323         2320         3
   April          315                          26            3.810       1200           3523         3520         3
   May            620           305                          3.224       1999           5522         5520         2
   June           910           290                          2.198       2000           7522         7520         2


   Total                        754           144                                                                 16


CH=$500                                    745*(500)+(144)*(1000)+16*(80)= $522,280
CF=$1000                                   600*(80)=48,000                 + 48,000
CI=$80                                                                     $570,280
                                                                                                                 25
§. A6: A Prototype problem                                          ( page 8 )

Example 3.2 : Solution

(2) Evaluation of Constant workforce plan
 Table 3: Computation of the Minimum Work Force Required by Densepack

     A                B                 C                           D
                                     Cumulative               Ratio
                  Cumulative     Number of units               B/C
   month          Net Demand    Produced per Worker        (Rounded Up)

   January           780            2.931                         267
   February         1420            6.448                         221
   March            2320            9.086                         256
   April            3520           12.896                         273
   May              5520           16.120   Hire to max. 411
                                                                  343
   June             7520           18.318   workers initially    411


                                                                          26
§. A6: A Prototype problem                                        ( page 9 )

Example 3.2 : Solution
(2) Evaluation of Constant workforce plan
     Table 4: Inventory Levels for Constant Work Force Schedule

 A            B            C             D             E              F
         Number of      Monthly                                    Ending
       unit Produced   Production    Cumulative    Cumulative     Inventory
 month   per Worker    (B ×411)     Production    Net Demand      (D - E)
 January    2.931       1,205          1,205          780            425
 February   3.517       1,445          2,650        1,420          1,230
 March      2.638       1,084          3,734        2,320          1,414
 April      3.810       1,566          5,300        3,520          1,780
 May        3.224       1,325          6,625        5,520          1,105
 June       2.198         903          7,528        7,520               8

 Total                                                             5,962

         Constant Work Force Plan hires 111 workers in January
         111*(500)+5962*(80)+600*(80)=$580,460
                                                                           27
§. A6: A Prototype problem                      ( page 10 )

Example 3.2 : Solution
  (3) Comparison of 2 plans
    Chase Strategy ( Zero Inventory Plan )
    ■ Hiring & Firing Constantly Appropriate?
      H:755(-2) $524,900
      F:145(-1)
    ■ Minimum Inventory Level, total I = 30 or 13 (-17)
                              $48,000
    ■ Ending at desirable work force level ? 910
          Total costs = $569,540 & ending 910 workers

                                                        28
§. A6: A Prototype problem                      ( page 11 )

Example 3.2 : Solution

 (3) Comparison of 2 plans (continued)
     Constant Work Force Plan
      ■ Minimum hiring & firing (one time) $55,500
         H : 111 
        F :0       55,500
         one time 
                  
                  
      ■ More carryovers units, total I = 5962     $476,960
      ■ Ending at better work force level . 411
           Total costs = 580,460 & ending 411 workers
                                                        29
§. A6: A Prototype problem                                   ( page 12 )

Example 3.2 : Solution

(4) Other Suggestions:(A) CHIU’s – Suggestion Ⅰ
  A          B            C              D            E              F
             Di            Ki
                                  '
                                         B/C         300           Inv
 January     780        2.931           267          300            99
 February   1420        6.448           221          300           514
 March      2320        9.086           256          300           405
 April      3520       12.896           273          300           348
 May        5520 2000 16.120 3.224    343          513             1
 June       7520 2000 18.318 2.198    411          910             1

                                                     H:610       I:1368
                              610*(500)+1368*(80)=$414,440       910

                   $ 414,440                         910 workers
                       +600*($80)=$462,440                           30
§. A6: A Prototype problem                                   ( page 13 )

Example 3.2 : Solution
(4) Other Suggestions: (B) CHIU’s – Suggestion II
   A         B            C              D            E              F
             Di           Ki
                                  '
                                         B/C          300          Inv
 January     780        2.931           267          300            99
 February   1420        6.448           221          300           514
 March      2320        9.086           256          300           405
 April      3520       12.896           273          300           348
 May        5520 2000 16.120 3.224    343          674           520
 June       7520 2000 18.318 2.198    411          674             1

                                                   H:374        I:1887
                              374*(500)+1887*(80)=$337,960         674


                   $ 337,960                              674 workers
                      +600*($80)=$385,960                            31
(4) Other Suggestions: (C) CHIU’s – Suggestion III
   A            B               C              D              E              F
              D                K
                                         '
                    i                i         B/C           300           Inv
 J anuar y     780        2. 931              267            300
                                                             273           99
                                                                           20
 Febr uar y   1420        6. 448              221            300
                                                             273          514
                                                                          340
   ar
 M ch         2320        9. 086              256            300
                                                             273          405
                                                                          160
 Apr i l      3520       12. 896              273            300
                                                             273          348
                                                                            1
 M ay         5520 2000 16. 120 3.224       343            674
                                                             738          520
                                                                          380
 J une        7520 2000 18. 318 2.198       411            674
                                                             738            1
                                                                            2
                              F:27                       H: 465
                                                          H:374       Inv.: 903
                                                                       I :1887
                                 374*( 500) +1887*( 80) =$337, 960
                             465*($500)+903*($80)= $304,740                674


                        $ 304,740 +$27000                        738 workers
                          +600*($80)=$379,740



                                                                                  32
§. A6.1: Class
          Problems Discussion

  Chapter 3 :
                 ( # 12, 14 )          pp.139-140



      Preparation Time : 10 ~ 15 minutes
         Discussion    : 10 minutes


                                                    33
The End


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