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Terminology. Tight vs. Slack constraints Sign / Nonnegativity vs. normal constraints Shadow prices Reduced costs Agenda: Shadow prices Reduced costs Duality. Shadow Prices. (Change in objective) / (Change in constraint right hand side) 0 for loose constraints
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Terminology • Tight vs. Slack constraints • Sign / Nonnegativity vs. normal constraints • Shadow prices • Reduced costs • Agenda: • Shadow prices • Reduced costs • Duality
Shadow Prices • (Change in objective) / (Change in constraint right hand side) • 0 for loose constraints • Applies to normal constraints • Interpretation: • Price for additional capacity
Reduced Cost • Shadow price for x >= 0 constraints • 0 for variables that are positive • Change in objective when variable set to 1 • Making constraint tighter • Objective becomes worse • Contribution (profit from setting variable to 1) - opportunity cost (loss from reallocating resources)
Recreational Vehicles Example • Shadow prices: • E (engine shop) = $140/hr • B (body shop) = $420/hr • Reduced cost for L (luxury car) • Contribution $1200 • Opportunity cost,needs 1hr in engine shop + 3hr in body shop= 1E+3B=$1400 • Reduced cost = 1200-1400= $ -200
Excel • Solver -> options -> “assume linear model” • Solver -> Solve -> Sensitivity Report • Concepts apply to nonlinear problems too • Shadow price = “Lagrange multiplier” • Reduced cost = “Reduced gradient”
Dual Problem • Theory is works particularly for LPs • Also an LP (when original is an LP) • Another way of understanding problem • Normal constraints become variables • Variables become constraints • Max becomes min • …
Dual Problem Dual: min cost to buy all capacity s.t. willing to sell capacity instead of produce variables are prices Original: max profit from running plant s.t. capacity not exceeded variables are production quantities
Dual Problem Dual: min price E * 120 hr engine shop capacity + … s.t. 3hr * E + 1hr *B + 2hr * SF >= $840 (car profit) … variables E, B, SF, FF, FL are prices Original: max $840 profit * S cars + … s.t. 3hr * S + 2hr * F + 1hr * L <= 120hr engine shop capacity … variables S, F, L are production quantities
Dual Problem maxx pTx s.t. Ax <= c x >= 0 equivalent to miny cTy s.t. ATy >= p y >= 0