Nonlinear Programming and Inventory Control Multiple Items

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Nonlinear Programmingand Inventory Control
(Multiple Items)
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Multiple Items

• Materials management involves many items and
  transactions
• Uneconomical to apply detailed inventory control
  analysis to all items
• Because a small percentage of inventory items
  accounts for most of the inventory value
• Must focus on important items
• Isolate those items requiring precise control
• ABC analysis indicates where managers should
  concentrate

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ABC Analysis

• Divides inventory into three classes according to
  dollar volume (dollar volumeAnnual demand unit
  purchase cost)
• A class is high value items whose dollar volume
  accounts for 75-80 of the total inventory value,
  while representing 15-20 of the inventory items
• B class is lesser value items whose dollar volume
  accounts for 10-15 of the total inventory value,
  while representing 20-25 of the inventory items
• C class is low value items whose volume accounts
  for 5-10 of the total inventory value, while
  representing 60-65 of the inventory items
• Thus, the same degree of control is not justified
  for all items

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Example
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Example-Cont.
B10-15 B20-25
A75-80 A15-20
C5-10 C60-65
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EOQ and Multiple Items

• Most inventory systems stock many items
• May treat each item individually and then add
  them up
• Restrictions may be imposed (limited warehouse
  capacity, upper limit on the maximum dollar
  investment, number of orders per year, and so on)
• Will be covered after covering NLP

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Overview of NLP

• Many realistic problems have nonlinear functions
• When LP problems contain nonlinear functions,
  they are referred NLP
• Have a separate name, because they are solved
  differently
• In LP, solutions are found at the intersections
• In NLP, there may no corner point
• Solution space can be undulating line or surface
• Like a mountain range with many peaks and valleys
• Optimal point at the top of any peak or at the
  bottom of any valley
• In NLP, we have local and global points

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Local and Global Optimal Point

• Solution techniques generally search for high
  points or low points
• Difficulty of NLP is to determine whether the
  identified is a local or global optimal point
• Global optimal point can be found by the very
  complex mathematical techniques

Global optimal point
Local optimal points
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Constrained and UnconstrainedOptimization

• Constrained optimization
• Profit function Z vp-cf-vcv, v1500-24.6p,
  cf10,000 and cv8
• vp creates a curvilinear relationship
• Results in quadratic function
• Z 1696.8p-24.6p2-22,000
• dZ/dp0, 1696.8-49.2p0, p34.49
• Called classical unconstrained optimization

profit
price
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Constrained Optimization

• Max Z 1696.8p-24.6p2-22,000
• s.t.
• plt20
• Referred NLP
• Solution is on the boundary formed by constraint
• Change RHS from 20 to 40
• Solution is no longer on the boundary
• Makes process of finding optimal solution
  difficult, particularly, with more variables and
  constraints

Feasible space
P20
price
Feasible space
price
P40
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Single Facility Location Problem

• Locate a centralized facility that serves several
  customers
• Minimize the total miles traveled between the
  facility and all customers
• Locations of the cities and the number of trips
  are (20,20,75), (10,35,105), (25,9,135), (32,15,
  60), (18,8,90)
• d(xi-x)2(yi-y)21/2
• Min Sditi

city1
city3
city5
city4
city2
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NLP with Multiple Constraints

• Consider a problem with two constraints
• A company produces two products and the
  production is subject to resource constraints
• Demand of each product is dependent on the price
  (x11500-24.6p1 and x22700-63.8p2)
• Cost of producing x1 and x2 are 12 and 9
• Production resources are (2x12.7 x2lt6000,
  3.6x12.9x2lt8500, 7.2x18.5 x2lt15,000)
• Model Max z(p1-12)x1(p2-9)x2
• s.t.
• 2x12.7 x2lt6000
• 3.6x12.9x2lt8500
• 7.2x18.5 x2lt15,000
• Where x11500-24.6p1 and x22700-63.8p2
• Decision variables are?

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Solution Techniques

• Very complex
• Two method
• Least or Substitution method
• Transferring a constraint optimization to an
  unconstraint optimization
• Lagrange multiplier

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Substitution Method

• Restricted to models containing only equality
  constraints
• Involves solving the constraint for one variable
  for another
• New expression will be substituted into the
  objective function to eliminate it completely

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Example

• Max Z vp-cf-vcv
• s.t.
• v15,00 -24.6 p
• cf10,000 and cv8
• vp creates a curvilinear relationship
• Constraint has been solved v for p
• Substitute it in objective function
• Z 1500p-24.6p2-cf-1500cv24.6 pcv
• Z 1696.8p-24.6p2-22,000
• Differentiating and setting it equal to zero
• 01696.8-49.2p
• p34.49

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Example

• Consider the Furniture Company
• Assume contribution of each declines as the
  quantity increases
• Relationship for x1 4-0.1x1
• Relation for x2 5-0.2 x2
• Profit earned form each(4-0.1x1)x1 and (5-0.2
  x2)x2
• Total profit z4x15x2-0.1x12-0.2x22
• Consider just one constraint x12x240 or
  x140-2x2
• z4(40-2x2)5x2-0.1 (40-2x2)-0.2x22
• z13x2-0.6x22
• x118.4, z70.42

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Limitation of Substitution

• Highest order of decision variable was a power of
  two
• Dealt only with two decision variables and single
  constraint

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Lagrange Multiplier

• Used for constraint optimization consisting of
  nonlinear objective function and constraints
• Transform objective function into a Lagrangian
  function
• Constraints as multiples of Lagrange multipliers
  are subtracted from the objective function
• Consider an example
• Max z4x15x2-0.1x12-0.2x22
• s.t.
• x12x240
• Forming Lagrangian function
• L4x15x2-0.1x12-0.2x22 -? (x12x2-40)

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Lagrange Multiplier-Cont.

• Partial derivatives of L with respect to each of
  variables
• dL/dx10, dL/dx20, dL/d?0
• 4-0.2x10, 5-0.4-2?0, x1-2x2400
• X118.3, x210.8, ?0.33, z70.42

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Sensitivity Analysis

• Lagrangian multiplier, ?, is analogous to dual
  variables in LP
• Shows changes in objective function value by
  changing the RHS
• Positive value of ? shows the increase in
  objective function
• Example
• Max z4x15x2-0.1x12-0.2x22
• s.t.
• x12x241
• x118.8, x211.2, ?0.27, z70.75

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Mathematical Notations

• DiAnnual demand for item i in units
• CiUnit purchase cost of item i in dollar
• Ai ordering cost of item i in dollar
• fiRequired storage space for item i in square
  foot
• FMaximum total storage space available
• TCTotal average annual costs in dollar
• Hi Annual holding cost per unit i per year in
  dollar

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Application of NLP and Inventory Modeling
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Example-Limited Working Capital
Total17120
Available budget15,000 Inventory carrying
charge0.2
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Solution

• Solved by the method of Lagrange-multiplier
• Before applying the method, we should solve the
  total cost function by ignoring the constraint
• Otherwise, we form the Lagrangian expression

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Lagrangian Expression

• Total budget17120gt available budget15000
• Form the Lagrangian Expression
• Derivative with respect to Qi and lambda

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Final Result

• From the first equation
• From the second equation

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Problem Solution
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HW Assignment 1

• Assume that the annual inventory carrying charge
  is 10
• and that 15000 sq.ft of floor space are
  available.
• What is the optimal inventory policy for these
  items.
• Determine the cost of having only 15,000 sq.ft of
  floor space?

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HW2
Classify these items into A, B, and C.
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HW3

• The Furniture Company has developed the following
  NLP model to determine the optimal number of
  chairs and tables to produce daily.
• Max z7x1-0.3x12 8x2-0.4x22
• S.t.
• 4x15x2100 hr
• Determine the optimal solution to this NLP using
  the substitution method.
• Determine the optimal solution to this NLP using
  the Lagrange multipliers.

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HW4

• Consider the following NLP model to determine
  solution.
• Max z30x1-2x12 25x2-0.5x22
• S.t.
• 3x16x2300
• Determine the optimal solution to this NLP using
  the substitution method.
• Determine the optimal solution to this NLP using
  the Lagrange multipliers.

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HW5

• Section 11.2c, Problems 1 and 4