Disaggregate Planning

1

• Disaggregate Planning

2
Background

• The aggregate plan specifies work-force size and
  production quantities for the firm as a whole or
  for large families of related items.
• For production planning, aggregate plan must be
  disaggregated into item by item production
  quantities remember aggregate plan in surrogate
  units.
• We will examine an approach to disaggregation
  developed by Bitran and Hax.

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Disaggregation - Product Families

• The first step in the Bitran-Hax approach to
  disaggregation is to identify families of items
  that will be produced in each planning period.
• A family of items is a group of products so
  similar that it is economically or
  technologically wise to produce them together.
  Typically, within family setup change cost is
  small and between family setup change cost is
  large.

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Disaggregation - Product Families (Cont.)

• We will consider producing some units of each
  item in a family whenever it is necessary to
  schedule production of any item in family.
• Looking at each item produced by the firm, in
  each period in the planning horizon, determine
  the quantity expected to be in inventory if no
  production is scheduled for the period.

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Disaggregation - Product Families (Cont.)

• If any item in a particular planning period will
  have an ending inventory less than or equal to
  that items safety stock level zero if no safety
  stock is held, then all members of that items
  family are candidates for production in that
  planning period.

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Disaggregation - Product Families (Cont.)

• Formally, for any item j in each family i, if I
  ij,t-1 is the amount of the item held in
  inventory at the end of period t-1, the demand
  for item j in period t is Dij, t and the safety
  stock is SSij, then
• if min Iij,t-1 - Dij, t - SSij ? 0 for
  all j e i
• then all items j in family i are considered for
  production in period t

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Disaggregation - Product Families (Cont.)

• Let zt be the set of families identified for
  production in period t.
• Find amounts xit (in aggregate units) for each
  family i e zt to be produced in period t by
  solving the following knapsack problem

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Disaggregation - Product Families (Cont.)

• Min ? (hixit)/2 (Si/xit ) ? Kij Dij, t
• for all i e zt for all j e i
• Subject to ? xit xt
• for all i e zt
• xit ? LBit
• xit ? UBit
• where Si set up cost to produce family i
• xt aggregate plan production for period t

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Disaggregation - Product Families (Cont.)

• xit amount of family i to be produced in
  period t
• Kij conversion factor for item j in family i
  expressing item in
• aggregate units
• hi holding cost per aggregate unit for
  family i
• LBit lower bound on production of family i in
  period t
• UBit upper bound on production of family i in
  period t

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Disaggregation - Product Families (Cont.)

• Note that xit 0 for each family i that is
  not an element of zt i.e., the optimal
  production quantity is zero.
• Bounds LBit and UBit are calculated as follows

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Disaggregation - Product Families (Cont.)

• where n specifies the maximum number of
  periods of demand held in stock in any planning
  period

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Disaggregation - Product Families (Cont.)

• Note that a feasible solution to the knapsack
  problem will exist when
• ? LBit ? xt ? ? UBit
• for all i e z for all i e zt
• If xt gt ? UBit
• for all i e zt
• the upper bound constraint will have to be
  violated in order to meet the aggregate plan

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Disaggregation - Product Families (Cont.)

• In this case, allocate the excess production to
  reflect the relative cost of inventory. If that
  cost is same over each family, then set the
  production level as
• yit xit (xt) UBit / ? UBit
• for all i e zt
• If xt lt ? LBit
• for all i e zt
• The aggregate plan does not produce enough
  inventory to meet safety stock requirements for
  period t.

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Disaggregation - Product Families (Cont.)

• In this case, should share planned production
  among families to reflect the relative costs of
  stock outs.
• If relative costs of stock outs is same over each
  family, then set the production level as
• yit (xit) (xt) LBit / ? LBit
• for all i e zt

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Family Disaggregation Algorithm

• If feasible solution to the knapsack problem
  exists, that is
• ? LBit ? xt ? ? UBit
• for all i e zt for all i e zt
• Set ß 1, P1 xt, and z1 zt

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Family Disaggregation Algorithm (Cont.)

• Step 1. Compute, for all i e z ß
• Si ? (KijDij, t)1/2
• for all j e i
• yiß _________________ P ß
• ? Si ? (KijDij,
  t)1/2
• for all i e z ß for
  all j e i

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Family Disaggregation Algorithm (Cont.)

• Step 2. For each i e ?zß
• If LBit ? yi ß ? UBit, set yit yi ß
• For families i (if any) where yi ß does not
  satisfy LBit or UBit,
• go to Step 3.

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Family Disaggregation Algorithm (Cont.)

• Step 3. Divide families for which yiß does not
  satisfy bounds into
• two groups
• z ß i e z ß yi ß gt UBit
• z- ß i e z ß yi ß lt LBit
• Compute ? ? (yi ß - UBit)
• i e z ß
• ? - ? (LBit - yi ß)
• i e z- ß

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Family Disaggregation Algorithm (Cont.)

• Step 4. If ? ? ? -, let yit UBit for all i
  e zß
• if ? lt ? -, let yit LBit for
  all i e z- ß
• Let ß ß 1,
• z ß 1 z ß - all families for which yi has
  been found
• and let P ß 1 P ß - ? yit
• all i found in iteration ß
• If z ß 1 F, stop. Otherwise return
  to Step 1.

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Disaggregating - Example Problem

• Consider the specialty automobile manufacturer we
  discussed in our examination of aggregate
  planning.
• Suppose that we have decided to disaggregate
  aggregate plan Alpha 3500 production hours in
  each of four quarters that we developed with the
  graphical/tabular approach.
• Handout page 8 shows the demand data for the four
  types of specialty vehicles hardtops A1 and
  A2, and convertibles B1 and B2. The handout also
  identifies the distribution of beginning
  inventory among the four vehicle types.

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Disaggregating - Example Problem

• We will use the Bitran-Hax approach to
  disaggregate aggregate plan Alpha.
• Assumptions relevant to the disaggregation
  process are shown on handout page 8. Note, in
  particular, that we have assumed that 5th Qtr
  demands are equal to 4th Qtr demands for all
  items in all families
• Note that by management policy, no more than n
  2 quarters of demand will be held in inventory
  for any car type.

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Disaggregating - Example Problem

• To disaggregate Alphas 1st Qtr specification of
  3500 production hours into the firms two
  families A and B we first determine which
  families should be scheduled for production?
  With no production in 1st Qtr we have
• for A1 8 - 50 - 42 lt 0 for B1 5 - 30 -
  25 lt 0
• for A2 3 - 30 - 27 lt 0 for B2 9 - 40
  - 31 lt 0
• Hence z1 A, B.

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Disaggregating - Example Problem

• Next compute upper and lower bounds for families
  in z1
• LBA1 max 0, 20(50-80) max 0,
  20(30-30) 1380
• UBA1 20(5060)-8020(3040)-30 3380
• LBB1 max 0, 20(30-50) max 0,
  20(40-90) 1120
• UBB1 20(3070)-5020(4080)-90 4120

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Disaggregating - Example Problem

• Now calculate
• yA1 10,00020(50)20(30)1/
  2 3,500
• 10,00020(50)20(30)1/2
  15,00020(30)20(40)1/2
• 1631.21
• yB1 15,00020(30)20(40)1/
  23,500
• 10,00020(50)20(30)1/2
  15,00020(30)20(40)1/2
• 1868.79

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Disaggregating - Example Problem

• Since 1380 ? 1631.21 lt 3380 and 1120 lt 1869.79 lt
  4120
• knapsack solution is within upper and lower
  bounds, hence
• yA1 1631.21
• and yB1 1869.79
• Now we must disaggregate family specifications in
  production hours to item specifications in terms
  of cars.
• We will use the Bitran-Hax approach to item
  disaggregation.

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Item Disaggregation

• In item disaggregation attempt to allocate family
  production among family member items so that all
  members of family reach their safety stock levels
  at same time.
• Note We might as well do that because Bitran-Hax
  considers producing all members of a family when
  any one item falls below safety stock level.

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Item Disaggregation (Cont.)

• For each family i e zt
• Step 1. Find smallest integer Nit such that
• Nit-1
• yit ? ? Kij ? Dij, tk SSij - Iij, t-1
• k0
• for all j e i

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Item Disaggregation (Cont.)

• Note that Nit specifies the number of periods of
  demand Dij,t, Dij, t1, ...,Dij, t N it-1 this
  is Nit time periods of demand for all items j in
  family i such that the sum of these demands just
  exceeds yit.
• Hence, if we summed only Nit-1 periods of demand,
  that total would be strictly less than yit.

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Item Disaggregation (Cont.)

• Step 2. Calculate
• Eit is the amount that producing Nit- 1
  periods of demand the exceeds aggregate plan
  quantity of yit

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Item Disaggregation (Cont.)

• Step 3. Calculate for each item j in family i
• The last term in this expression is reduction
  in production so that yit will be met.

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Item Disaggregation (Cont.)

• If for some item j say item g, the calculation
  of yigt gives
• yigt lt 0, then set yigt 0
• then recalculate

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Item Disaggregation (Cont.)

• Recalculate
• Now we will return to the specialty auto firm to
  disaggregate family production into production of
  cars in the 1st Quarter.
• Then, we look at family and item disaggregations
  for the 2nd, 3rd, and 4th Quarters.