Review and Samples:

1
Review and Samples

• LP Models
• Solution MethodsThe Graphical Method
• The Simplex Method
• Duality and Sensitivity AnalysisDual (Marginal,
  Shadow) PricesRange for Objective
  CoefficientsRange for Right-hand Side DataThe
  Dual Problem

2
LP Model Standard (Inequality) Form
3
Duality Theory

• Standard (Inequality) Primal Form
• Dual Form

4
LP Model Standard (Inequality) Matrix Form
5
Primal-Dual in Matrix Form

• Standard (Inequality) Primal Form
• Dual Form

6
Primal-Dual in Matrix Form Equality

• Standard (Equality) Primal Form
• Dual Form

7
Relations Between Primal and Dual

• 1. The dual of the dual problem is again the
  primal problem.
• 2. Either of the two problems has an optimal
  solution if and only if the other does if one
  problem is feasible but unbounded, then the other
  is infeasible if one is infeasible, then the
  other is either infeasible or feasible/unbounded.
• 3. Weak Duality Theorem The objective function
  value of the primal (dual) to be maximized
  evaluated at any primal (dual) feasible solution
  cannot exceed the dual (primal) objective
  function value evaluated at a dual (primal)
  feasible solution.
• cTx gt bTy (in the standard equality
  form)

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Relations between Primal and Dual (continued)

• 4. Strong Duality Theorem When there is an
  optimal solution, the optimal objective value of
  the primal is the same as the optimal objective
  value of the dual.
• cTx bTy
• 5. Complementary Slackness Theorem Consider an
  inequality constraint in any LP problem. If that
  constraint is inactive for an optimal solution to
  the problem, the corresponding dual variable will
  be zero in any optimal solution to the dual of
  that problem.
• xj (c-ATy)j 0, j1,,n.

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Optimality Conditions

• Primal Feasibility
• Dual Feasibility
• Strong Duality
• or Complementary Slackness

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Optimality Conditions

• Primal Feasibility
• Dual Feasibility
• Strong Duality
• or Complementary Slackness

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Theory of Linear Programming

• An LP problem falls in one of three cases
• Problem is infeasible Feasible region is empty.
• Problem is unbounded Feasible region is
  unbounded towards the optimizing direction.
• Problem is feasible and bounded then there
  exists an optimal point an optimal point is on
  the boundary of the feasible region and there is
  always at least one optimal corner point (if the
  feasible region has a corner point).
• When the problem is feasible and bounded,
• There may be a unique optimal point or multiple
  optima (alternative optima).
• If a corner point is not worse than all its
  neighbor corners, then it is optimal.

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Graphical Solution Seeking

• Plot the feasible region.
• If the region is empty, stop the problem is
  infeasible there must be conflicting constraints
  in the model.
• Plot the objective function contour and choose
  the optimizing direction.
• Determine whether the objective value is bounded
  or not. If not, stop the problem is unbounded
  there must be mistakes in model formulation.
• Determine an optimal corner point.
• Identify active constraints at this corner.
• Solve simultaneous linear equations for the
  optimal solution.
• Evaluate the objective function at the optimal
  solution to obtain the optimal value of the
  problem.

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Summary of the Simplex Method for Max (Min)

• 1. Select a basic feasible solution
• 2. Express the basic variables in terms of the
  nonbasic variables, and express the objective
  function in terms of only nonbasic variables.
• If all objective coefficients are
  non-positive (non-negative), then stop the
  current basic feasible solution is optimal.
• Otherwise, select the entering variable such
  that its coefficient is the greatest (least), and
  among basic variables select the leaving variable
  such that it becomes zero first when increasing
  the entering variable and keeping the other
  nonbasic variables at zero. If no basic variable
  becomes zero, then stop the objective function
  is unbounded.
• 3. Goto Step 2.

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Complementary Slackness Conditions in the Primal
Simplex Method

• The simplex method maintain the complementary
  slackness condition, and
• moving toward

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Matrix form of the initial tableau for the
inequality standard form

• Basic
• Variable x xs RHS
• Z -c 0 0
• xs A I b
• Matrix form of the tableau with a selected basis
  B
• Basic
• Variable x xs RHS
• Z cBB-1A -c cBB-1 cBB-1b
• xB B-1A B-1 B-1b

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The Primal Simplex Method in Tableau

• 1. Initialization A(A, I) and c(c, 0)
  xBB-1b? 0.
• 2. Calculate ycBB-1 and r yA - c ? 0. If r ?
  0, then optimal and stop otherwise, goto next
  step.
• Determine the entering basic variable say select
  the basic variable with the lowest value in r
  determine the leaving basic variable whose
  coefficient in xB reaches zero first as the
  entering variable increases (use the ratio test
  of the entering column of B-1A against B-1b). If
  the increase is unlimited (the column contains
  all non-positive numbers), then stop, the primal
  problem is unbounded. Otherwise, using the
  pivoting procedure to update B, B-1A and B-1b and
  return to Step 2.

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The Dual Simplex Method in Tableau

• 1. Initialization A(A, I) and c(c, 0)
  ycBB-1 such that r yA - c ? 0
• 2. Calculate xBB-1b. If xB ? 0, then optimal and
  stop otherwise, goto next step.
• Determine the leaving basic variable say select
  the basic variable with the lowest value in xB
  determine the entering basic variable whose
  coefficient in r reaches zero first as the dual
  variable of the leaving row increases (use the
  ratio test of the leaving row of B-1A against r).
  If the increase is unlimited (the row contains
  all non-negative numbers), then stop, the primal
  problem is infeasible or dual is unbounded.
  Otherwise, using the pivoting procedure to update
  B, B-1A, ycBB-1 and r yA c, and goto Step 2.

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Reduced Cost and Objective Coefficient Range

• All positive variables have zero reduced cost
• In general, the reduced cost of any zero variable
  is the amount the objective coefficient of that
  variable would have to change, with all other
  data held fixed, in order for it to become a
  positive variable in the OS. If the OS is
  degenerate, the objective coefficient of a zero
  variable would have to change by at least, and
  possibly more that, the reduced cost in order to
  become a positive variable in the OS.
• The objective coefficient ranges give the ranges
  of the objective function over which no change in
  the OS will occur. If the OS is degenerate, any
  objective coefficient must be changed by at
  least, and possibly more than, the indicated
  allowable amounts in order to produce a new
  optimal solution.
• One of the allowable increase and decrease for
  a zero variable is infinite and the other is the
  reduced cost. If a zero variable has zero reduced
  cost, then there exist an alternative optimal
  solution.

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Dual (Shadow) Price and Constraint RHS Ranges

• All inactive constraint have zero dual price
• In general, the dual price on a given active
  constraint is the rate of improvement in the OV
  as the RHS of the constraint increases with all
  other data held fixed. If the RHS is decreased,
  it is the rate at which the OV is impaired.
• The constraint RHS ranges give the ranges of the
  constraint RHS over which no change in the dual
  price will occur. If the solution is degenerate,
  the dual price may be valid for one-sided changes
  in the RHS.
• One of the allowable increase and decrease for
  an inactive constraint is infinite and the other
  equals to the slack or surplus.
• In general, when the RHS of an active constraint
  changes, both the OV and OS will change

20
Samples

• The Concrete Products Corporation has the
  capability of producing four types of concrete
  blocks. Each block must be subjected to three
  processes batch mixing, mold vibrating and
  inspection. The plant manager desires to maximize
  the profit during the next month. During the
  upcoming thirty days, he has 800 machine hours
  available on the batch mixer, 1000 hours on the
  mold vibrator and 340 man-hours of inspection
  time. The problem is formulated as follows max
  8x1 14x2 30x3 50x4 S.t.
  x1 2 x2 10x3 16x4 lt 800
  1.5x1 2 x2 3 x3 5 x4 lt 1000
  0.5x1 0.6x2 x3 2 x4 lt 340 where
  xi is the production level of the ith product i
  1, 2, 3, 4.

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After solving by the simplex method, the final
tableau is

• Basic x1 x2 x3 x4 x5 x6 x7 Right
• Z 0 0 28 40 5 2 0 6000
• 2 0 1 11 19 1.5 -1 0 200
• 1 1 0 -12 -22 -2 2 0 400
• 7 0 0 -4 1.6 .1 -.4 1 20

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• In answering the following question provide
  a short explanation and
• computation when needed.
• 1. By how much must the unit profit on block 3
  be increased before it would be profitable to
  manufacture it ?
• 2. What must the minimum unit profit on block 2
  be so that it remains in the production schedule?
• 3. If the 800 machine-hours on the batch mixer is
  uncertain, for what range of machine hours will
  it remain feasible to produce blocks 1 and 2 ?
• 4. A competitor located next door has offered the
  manager additional batch-mixing time at a rate
  4.50 per hour. Should he accept his offer?
• 5. Suppose instead that the competitor offers the
  manager 250 hours of batch-mixing time for a
  total 1,100, Should he accept his offer? ( The
  manager can only accept or reject the extra 250
  hours.)
• 6. The owner has approached the manager with a
  thought about producing a new type of concrete
  block that would require 4 hours of batch mixing,
  4 hours of molding and 1 hour of inspection per
  pallet. What should be the profit per pallet if
  block 5 is to be manufactured?
• 7. If the next door competitor would like to buy
  the resources from Concrete Products next month,
  what would be the fair prices? Formulate it as a
  linear Program.

23
The following tableau associated with a LP problem

• Basic x1 x2 x3 x4 x5 x6 x7 Right
• Z 0 0 0 g 3 h i 0
• 2 0 1 0 d 1 0 3 e
• 3 0 0 1 -2 2 f -1 2
• 1 1 0 0 0 -1 2 1 3
• The entries d, e, f, g, h, i in the tableau are
  parameters. Each of the following
• questions is independent, and they all refer to
  this tableau. Clearly state the range
• of values of the various parameters that will
  make the conclusions in
• the following questions true.
• 1. The present tableau has a basic feasible
  solution
• 2. Row 1 in the present tableau indicates that
  the problem is infeasible.
• 3. The current basic solution is feasible, and
  the present tableau indicates that the problem is
  unbounded.
• 4. The current basic solution is feasible, x6 is
  the entering basic variable, and x3 is the
  leaving basic variable.
• 5. The current basic solution is optimal.