bilinear approximation techniques to solve network flow...

Supply Chain ProblemsDynamic Traffic Assignment Problems

Conclusion

Bilinear Approximation Techniques toSolve Network Flow Problems with

Nonlinear Arc Cost Functions

Artyom Nahapetyan

University of FloridaDepartment of Industrial and Systems Engineering

July 18, 2006

Artyom Nahapetyan Dissertation Defense


Conclusion

1 Supply Chain ProblemsConcave Piecewise Linear Network Flow ProblemFixed Charge Network Flow ProblemCapacitated Multi-Item Dynamic Pricing Problems

2 Dynamic Traffic Assignment ProblemsDiscrete-Time DTA with Periodic Planning HorizonA Heuristic Algorithm for DTDTA-UDynamic Toll Pricing Framework

3 Conclusion



Conclusion

Concave Piecewise Linear Network Flow ProblemFixed Charge Network Flow ProblemCapacitated Multi-Item Dynamic Pricing Problems

Description of the Problem

CPLNF Problem

Let

G (N,A) represent a network, and

fa(xa) denote a cost function of arc a.

minx

∑a∈A

fa(xa)

s.t. Bx = b

xa ∈ [λ0a, λ

naa ] ∀a ∈ A

B - node-arc incident matrix of the network Gfa(xa) - concave piecewise linear functions



Conclusion


Arc Cost Function

Function fa(xa)

fa(xa) =

c1a xa + s1

a xa ∈ [λ0a, λ

1a)

c2a xa + s2

a xa ∈ [λ1a, λ

2a)

· · · · · ·cnaa xa + sna

a xa ∈ [λna−1a , λna

a ]

c1a > c2

a > · · · > cnaa



Conclusion


Arc Cost Function

Function fa(xa)

fa(xa) =

c1a xa + 0 xa ∈ [0, λ1

a)c2a xa + s2

a xa ∈ [λ1a, λ

2a)

· · · · · ·cnaa xa + sna

a xa ∈ [λna−1a , λna

a ]

c1a > c2

a > · · · > cnaa



Conclusion


Arc Cost Function

Function fa(xa)



Conclusion


Mixed Integer Formulation

CPLNF-IP Problem

minx ,y

∑a∈A

∑k∈Ka

cka xk

a +∑a∈A

∑k∈Ka

ska yk

a

Bx = b∑k∈Ka

xka = xa

∑k∈Ka

λk−1a yk

a ≤ xa ≤∑k∈Ka

λkayk

a ,∑k∈Ka

yka = 1

xka ≤ Myk

a , xka ≥ 0, yk

a ∈ 0, 1

where xka denotes the portion of the total flow that has cost

according to the linear function f ka (xa) = ck

a xa + ska



Conclusion


Relaxation of the Binary Variables

CPLNF-R Problem

minx ,y

∑a∈A

∑k∈Ka

cka xk

a +∑a∈A

∑k∈Ka

ska yk

a

Bx = b∑k∈Ka

xka = xa

∑k∈Ka

λk−1a yk


λkayk

a ,∑k∈Ka

yka = 1

xka = xay

ka , xk

a ≥ 0, yka ≥ 0



Conclusion



CPLNF-R Problem

minx ,y

∑a∈A

∑k∈Ka

cka xk

a +∑a∈A

∑k∈Ka

ska yk

a

Bx = b

∑k∈Ka

λk−1a yk


λkayk

a ,∑k∈Ka

yka = 1

xka = xay

ka , yk

a ≥ 0



Conclusion



CPLNF-R Problem

minx ,y

∑a∈A

∑k∈Ka

cka yk

a

xa +∑a∈A

∑k∈Ka

ska yk

a

Bx = b

∑k∈Ka

λk−1a yk


λkayk

a ,∑k∈Ka

yka = 1

yka ≥ 0



Conclusion


Theoretical Results

Lemma

Any feasible vector of the CPLNF-IP problem is feasible to theCPLNF-R

Lemma

Any local optimum of the CPLNF-R problem is either feasible tothe CPLNF-IP or leads to a feasible vector of CPLNF-IP with thesame objective function value.

Theorem

A global optimum of the CPLNF-R problem is a solution or leadsto a solution of the CPLNF-IP .



Conclusion


Concave Piecewise Linear Problem with a SeparableObjective Function

CPLPwSOF Problem

minx

n∑i=1

fi (xi )

x ∈ X ⊆ Rn

xi ∈ [λ0i , λ

nii ]



Conclusion


Relaxation

CPLPwSOF-R Problem

minx ,y

n∑i=1

∑k∈Ki

f ki (xi )y

ki

x ∈ X ⊆ Rn∑k∈Ki

λk−1i yk

i ≤ xi ≤∑k∈Ki

λki yk

i

∑k∈Ki

yki = 1

yki ≥ 0,



Conclusion


Theoretical Results

Theorem

A concave minimization problem with a separable piecewise linearobjective function, CPLPwSOF, is equivalent to a bilinear program,CPLPwSOF-R.



Conclusion


Two Problems

CPLNF-R Problem

minx ,y

∑a∈A

∑k∈Ka

cka yk

a

xa +∑a∈A

∑k∈Ka

ska yk

a

Bx = b∑k∈Ka

λk−1a yk


λkayk

a

∑k∈Ka

yka = 1

yka ≥ 0



Conclusion


Two Problems

LP(y)Problem (y is fixed)

minx

∑a∈A

∑k∈Ka

cka yk

a

xa

Bx = b

xa ∈ [0, λnaa ]



Conclusion


Two Problems

CPLNF-R Problem

minx ,y

∑a∈A

∑k∈Ka

[cka xa + sk

a

]yka

Bx = b

∑k∈Ka

λk−1a yk


λkayk

a

∑k∈Ka

yka = 1

yka ≥ 0



Conclusion


Two Problems

LP(x) Problem (x is fixed)

miny

∑a∈A

∑k∈Ka

[cka xa + sk

a ]yka

∑k∈Ka

λk−1a yk


λkayk

a

∑k∈Ka

yka = 1, yk

a ≥ 0

A binary variable that satisfies the inequality is a solution ofthe problem.



Conclusion


Dynamic Cost Updating Procedure (DCUP)

DCUP: Iteratively Solves LP(x) and LP(y)

Step 1: Let y0 denote the initial vector of yk0a , where y10

a = 1 andyk0a = 0, ∀k ∈ Ka, k 6= 1. m← 1.

Step 2: Let xm = argminLP(ym−1), andym = argminLP(xm).

Step 3: If ym = ym−1 then stop. Otherwise, m← m + 1 and goto Step 2.



Conclusion


Theoretical Results

Theorem

Given any initial binary vector y0, DCUP converges in a finitenumber of iterations.

Theorem

Let (x∗, y∗) be the solution returned by DCUP. If y∗ is a uniquesolution of the LP(x∗) problem then (x∗, y∗) is a local minimum ofCPLNF-R.



Conclusion


Theoretical Results

y∗ is not unique only if ∃k ∈ K , k 6= 0, k 6= na, and a ∈ A,such that x∗a = λk

a .∑k∈Ka

λk−1a yk


λkayk

a

It is highly unlikely that x∗a takes those values because it is asolution of the LP(y) problem.

minx

∑a∈A

∑k∈Ka

cka yk

a

xa

Bx = b, xa ∈ [0, λnaa ]



Conclusion


DSSP



Conclusion


DSSP

Kim D., Pardalos P., “A Solution Approach to the FixedCharged Network Flow Problems Using a Dynamic SlopeScaling Procedure”, Operations Research Letters, 24, pp.195-203, 1999.

Kim D., Pardalos P., “Dynamic Slope Scaling and TrustInterval Techniques for Solving Concave Piecewise LinearNetwork Flow Problems”, Networks, 35(3), pp. 216-222,2000.



Conclusion


An Alternative Formulation

NFPwFDCF Problem

minx

FT (x)x

s.t. Bx = b,

xa ∈ [0, λnaa ],

where F (x) is a vector function with components

Fa(xa) =

fa(xa)

xaxa > 0

M xa = 0=

c1a xa ∈ (0, λ1

a]

c2a + s2

axa

xa ∈ (λ1a, λ

2a]

· · · · · ·cnaa + sna

axa

xa ∈ (λna−1a , λna

a ]

M xa = 0



Conclusion


Theoretical Results

Theorem

The NFPwFDCF problem is equivalent to the CPLNF problem.

Theorem

The solution of the DSSP is a user equilibrium solution of thenetwork flow problem with the flow dependent cost functionsFa(xa), i.e

“find feasible x∗ such that FT (x∗)(x − x∗) ≥ 0, ∀xa ∈ [0, λnaa ],

Bx = b”,

We need to find a system optimum solution of NFPwFDCF.

It is well known that system optimum and user equilibriumsolutions are not the same.



Conclusion


Test Problems

Parameters of the Problems

Network size (nodes-arcs-supply/demand nodes): 12-35-2,20-100-3, 40-300-4, 100-2000-20, and 200-5000-50.

Demand: U[10,20], U[20,30], or U[30,40]

Number of linear pieces: 5 or 10.

Total number of problems sets is 30.

There are 30 problems per problem set. (In total 900problems)



Conclusion


Test Problems

Computational Results

DCUP provides a better solution than DSSP in about 50% ofthe test problems.

Both algorithms provide the same solution in about 30% ofthe test problems.

DCUP spends 2-5 times less CPU time than DSSP.

DCUP converges using less number of iterations than DSSP.



Conclusion



FCNF Problem

minx

f (x) =∑a∈A

fa(xa)

s.t. Bx = b,

xa ∈ [0, λa],

where fa(xa) =

caxa + sa xa ∈ (0, λa]0 xa = 0



Conclusion


ε-Approximation



Conclusion


ε-Approximation

CPLNF(ε) Problem

minx

φε(x) =∑a∈A

φεaa (xa)

s.t. Bx = b,

xa ∈ [0, λa],

where ε is a vector of εa.



Conclusion


Theoretical Results

Let

x∗ = argmin(FCNF ),

xε = argmin (CPLNF (ε)), and

δ = minxva |xv ∈ V , a ∈ A, xv

a > 0, where V represents theset of vertices of the feasible region.

Theorem

For all ε such that εa ∈ (0, λa], ∀a ∈ A, φε(xε) ≤ f (x∗).

Theorem

For all ε such that εa ∈ (0, δ], ∀a ∈ A, φε(xε) = f (x∗).



Conclusion


Adaptive Dynamic Cost Updating Procedure



Conclusion


Test Problems


Network size (nodes-arcs-supply/demand nodes): 20-100-3,40-300-4, 100-1000-10, and 150-3000-15.

Setup cost: U[50,100], U[100,200], or U[200,400]

Slope: U[1,5], U[10,20], or U[30,40]

Demand: U[30,50]

Total number of problems sets is 36.

There are 30 problems per problem set. (In total 1080problems)



Conclusion


Test Problems


ADCUP provides a better solution than DSSP

Small problems - 62%, andLarge problems - 98%.

Both algorithms provide the same solution

Small problems - 28%, andLarge problems - 0%.

ADCUP spends 2-8 times less CPU time than DSSP.

ADCUP converges using less number of iterations than DSSP.



Conclusion


Problem Description

Given Data:

Set of products.

Unit and setup costs.

Inventory costs.

Production capacities.

Demand is a function of the price.

Problem

Find a production and pricing policy, which maximizes the profit.



Conclusion


Price-Demand and Profit-Demand Relationships

g(d) = f (d)d



Conclusion


Formulation

CMDP Problem

maxx ,y ,d

∑p∈P

∑j∈∆

g(p,j)

∑i∈∆|i≤j

x(p,i ,j)

−

∑p∈P

∑i ,j∈∆|i≤j

[c in(p,i ,j) + cpr

(p,i)

]x(p,i ,j) −

∑p∈p

∑i∈∆

cst(p,i)y(p,i)

s.t.∑p∈P

∑j∈∆|i≤j

x(p,i ,j) ≤ Ci ,∑

j∈∆|i≤j

x(p,i ,j) ≤ Ciy(p,i),

x(p,i ,j) ≥ 0, y(p,i) ∈ 0, 1



Conclusion


Approximation



Conclusion


Approximate Formulation

ACMDP Problem

maxx ,y ,λ

∑p∈P

∑j∈∆

∑k∈K

gk(p,j)λ

k(p,j)

−∑p∈P

∑i ,j∈∆|i≤j

[c in(p,i ,j) + cpr

(p,i)

]x(p,i ,j) −

∑p∈p

∑i∈∆

cst(p,i)y(p,i),

s.t.∑p∈P

∑j∈∆|i≤j

x(p,i ,j) ≤ Ci ,∑

j∈∆|i≤j

x(p,i ,j) ≤ Ciy(p,i),

∑i∈∆|i≤j

x(p,i ,j) =∑k∈K

dk(p,j)λ

k(p,j),

N∑k=0

λk(p,j) = 1,

x(p,i ,j) ≥ 0, λk(p,j) ≥ 0, y(p,i) ∈ 0, 1.



Conclusion


Approximate Formulation

ACMDP Problem

maxx ,y

∑p∈P

∑i∈∆

∑j∈∆|i≤j

∑k∈K

qk(p,i ,j)x

k(p,i ,j) − cst

(p,i)y(p,i)

s.t.

∑p∈P

∑j∈∆|i≤j

∑k∈K

xk(p,i ,j) ≤ Ci ,

∑j∈∆|i≤j

∑k∈K

xk(p,i ,j) ≤ Ciy(p,i),

∑k∈K

∑i∈∆|i≤j

xk(p,i ,j)

dk(p,j)

≤ 1, xk(p,i ,j) ≥ 0, y(p,i) ∈ 0, 1,

where qk(p,i ,j) = f k

(p,j) − c in(p,i ,j) − cpr

(p,i)



Conclusion


Objective Function



Conclusion


Bilinear Reduction

ACMDP-B Problem

maxx ,y

∑p∈P

∑i∈∆

∑j∈∆|i≤j

∑k∈K

qk(p,i ,j)x

k(p,i ,j) − cst

(p,i)

y(p,i) = ϕ(x , y)

x ∈ X and y ∈ Y ,

where Y = [0, 1]|P||∆| and

X = x |xk(p,i ,j) ≥ 0,

∑p,k,j |i≤j xk

(p,i ,j) ≤ Ci ,∑

k,i |i≤j

xk(p,i,j)

dk(p,j)

≤ 1



Conclusion


Theoretical Results

Theorem

Any local maximum of the ACMDP-B problem is feasible or leadsto a feasible solution of the ACMDP problem with the sameobjective function value.

Theorem

A global maximum of the ACMDP-B problem is a solution or leadsto a solution of the ACMDP problem.



Conclusion


Procedure 1

Tow LPs

LP(x) :

maxy∈Y

∑p∈P

∑i∈∆

[∑j∈∆|i≤j

∑k∈K qk

(p,i ,j)xk(p,i ,j) − cst

(p,i)

]y(p,i)

LP(y) :

maxx∈X

∑p∈P

∑i∈∆

∑j∈∆|i≤j

∑k∈K

[qk(p,i ,j)y(p,i)

]xk(p,i ,j)



Conclusion


Procedure 1

Procedure 1

Step 1: Let y0 denote an initial binary vector, where y(p,i) = 1.m← 1.

Step 2: Let xm = argmaxLP(ym−1), andym = argmaxLP(xm).

Step 3: If ym = ym−1 then stop. Otherwise, m← m + 1 and goto Step 2.



Conclusion


Procedure 1

Disadvantage

The quality of the solution is not good enough.

Procedure 1 converges to a local maximum of the problem.

If ym(p,i) = 0 then in all following iterations xk

(p,i ,j) = 0, ∀j ∈ ∆,i ≤ j , and p ∈ P.

As a result, the local maximum can be far from being a globalone.



Conclusion


Approximation of the Objective Function



Conclusion


Procedure 2

Procedure 2

Step 1: Let ε(p,i) be a sufficiently large number, and y0 be suchthat y0

(p,i) = 1, ∀p ∈ P and i ∈ ∆. m← 0.

Step 2: Construct the approximation problem and run Procedure1 to find a local maximum of the problem, where ym is an initialbinary vector. Let (xm+1, ym+1) denote the local maximum.

Step 3: If ∃p ∈ P and i ∈ ∆ such that∑j∈∆|i≤j

∑k∈K qk

(p,i ,j)x(m+1)k(p,i ,j) − cst

(p,i) ≤ εm(p,i) and∑

j∈∆|i≤j

∑k∈K x

(m+1)k(p,i ,j) > 0 then ε← αε, m← m + 1 and go to

Step 2. Otherwise, stop.



Conclusion


How Big is ε(p,i)?

Procedure 3

For all i ∈ ∆ and p ∈ P assign the available capacity first to thevariables with a higher value of qk

(p,i ,j)



Conclusion


Test Problems


Number of products: |P| =5, 10, or 20.

Planning horizon: |∆| =12, or 52.

Capacity: Ci = |P|U[10, 100], |P|U[50, 150], |P|U[100, 200],or |P|U[150, 250].

Costs: cpr(p,i) = U[20, 40], cst

(p,i) = U[600, 1000], and

c in(p,i) = U[4, 8].

Maximum price/demand: U[70, 90]/U[500, 1000].

Profit per unit of investment: β ∈ [0.7, 1.3]

The value of the parameter α: 1/2, 2/3, 9/10.

Total number of problem sets is 24.

There are 10 problems per problem set. In total 240 problems.



Conclusion


Test Problems


Quality of the solution:

Procedure 1 - 1-8.3%, andProcedure 2 - <1.2%.

CPU time:

Procedure 1 - 0.2-35 sec, andProcedure 2

α = 1/2 - 0.5-58 sec,α = 2/3 - 0.8-78 sec, andα = 9/10 - 1.8-257 sec.

A higher value of α provides a slightly better solution.



Conclusion

Discrete-Time DTA with Periodic Planning HorizonA Heuristic Algorithm for DTDTA-UDynamic Toll Pricing Framework


Given Data:

Network - G (N,A).

Travel time - φa(·).Set of OD pairs -C = (1, 4), (3, 4).Demand - hk(t).

Problem

Find traffic flows which minimize the total travel time of allvehicles during the time period [0,T ].



Conclusion



Variables:

xa(t) - number of cars on arc a at time t,

ua(t) - inflow rate into arc a at time t,

va(t) - outflow rate from arc a at time t.

Assumptions:

xa(0) = 0 and xa(T ) = 0,

If t is the time to enter arc a then t + φa(xa(t)) is the time toleave the arc.



Conclusion


Disadvantage

Requires the network to be empty at time 0 and T

It is not necessarily valid to ignore xa(0) even when it is small.



Conclusion


Cyclic Time Period

How to model DTA problem with positive xa(0) and xa(T )?



Conclusion


Discrete-Time Formulation

[0,T ) =⇒ 0, 1, 2, . . . , (T − 1).

Γa = a set of possible discrete travel times on arc a.

Use time-expanded network



Conclusion


Time-Expanded Network

T = 5,

∆ = 0, 1, 2, 3, 4,2 ≤ φa(·) < 5, and

Γa = 2, 3, 4



Conclusion


Variables

Index a(t, s) uniquely identifies an arc in TE network.

Arc flow on TE network: Ya(t,s) =∑

k∈C yka(t,s)

za(t,s) =

1 if the arc is used0 o/w



Conclusion


Set Ωa(t)

It is necessary to compute xa(t)

Ωa(t) = (τ, s) : τ = [t − 1]T , · · · , [t − s]T , s ∈ Γa

xa(t) =∑

(τ,s)∈Ωa(t)Ya(τ,s)



Conclusion


Discrete Formulation

Periodic DTA Problem

min(x ,y ,z,g)

∑t∈∆

∑a∈A

[φa(xa(t))

∑s∈Γa

Ya(t,s)

]= Φ(Y )TY

s.t. Byk + gk = bk ,∑t∈∆

gkd(k)t

=∑t∈∆

hkt

Ya(t,s) =∑

k∈C yka(t,s), xa(t) =

∑(τ,s)∈Ωa(t)

Ya(τ,s)∑s∈Γa

za(t,s) = 1∑s∈Γa

(s − δ)za(t,s) ≤ φa(xa(t)) ≤∑

s∈Γasza(t,s)

Ya(t,s) ≤ Maza(t,s)

yka(t,s) ≥ 0, gk

d(k)t≥ 0, xa(t) ≥ 0, za(t,s) ∈ 0, 1



Conclusion


Linearization of the Constraint

Denote φ−1a (1) = 0

∑s∈Γa

(s − δ)za(t,s) ≤ φa(xa(t)) ≤∑

s∈Γasza(t,s)

m∑s∈Γa

φ−1a (s − δ)za(t,s) ≤ xa(t) ≤

∑s∈Γa

φ−1a (s)za(t,s)



Conclusion


Bounds on the Problem

The upper bound on the objective

Φ(Y )TY ≤∑t∈∆

∑a∈A

[ ∑s∈Γa

sza(t,s)

] [ ∑s∈Γa

Ya(t,s)

]=

∑t∈∆

∑a∈A

∑s∈Γa

sYa(t,s) = qTu Y

The lower bound on the objectiveΦ(Y )TY ≥

∑t∈∆

∑a∈A

∑s∈Γa

(s − δ)Ya(t,s) = qTl Y



Conclusion


Three Problems

Let S(δ) represent the feasible region.

Upper Bounding Problem(Y u,Zu) = argminqT

u Y : (Y ,Z ) ∈ S(δ)

Original Problem(Y ∗,Z ∗) = argminΦ(Y )TY : (Y ,Z ) ∈ S(δ)

Lower Bounding Problem(Y l ,Z l) = argminqT

l Y : (Y ,Z ) ∈ S(δ)



Conclusion


Theoretical Results

Lemma

For any δ > 0, qTl Y l ≤ Φ(Y ∗)TY ∗ ≤ qT

u Y u

Theorem

Given ε > 0, there exist δ > 0 such that

Φ(Y ∗)TY ∗ − qTl Y l ≤ ε, and

qTu Y u − Φ(Y ∗)TY ∗ ≤ ε



Conclusion


Solution Improvement

Refinement Problem

Find (Y u,Zu)⇓

Y u is the solution of the original problem with Z = Zu

⇓The refined solution is (Y u,Zu).

Corollary

Φ(Y ∗)TY ∗ ≤ Φ(Y u)T Y u ≤ Φ(Y u)TY u ≤ qTu Y u



Conclusion


Test Problems


Time horizon - [0,T ) = [0, 10),

Travel time function

linear: φa(xa(t)) = 1.5 + 2.5(

xa(t)100

), or

quadratic: φa(xa(t)) = 1.5 + 2.5(

xa(t)100

)2

DemandTime

Traffic Intensity 0 1 2 3 4 5 6 7 8 9 TotalLow 20 25 30 35 40 40 35 30 25 20 300

Medium 30 35 40 45 50 50 45 40 35 30 400High 40 45 50 55 60 60 55 50 45 40 500



Conclusion


Test Problems

Computational Results: Linear Travel Time Function

(Y∗, Z∗) (Y u , Zu) (Y l , Z l ) Rel. cpu

Traffic cpu∗ cpuu cpul Err Ratio

Intensity Delay (sec) Delay (sec) Delay (sec) (%)cpu∗cpuu

cpu∗

cpul

Low 1337.50 27.42 1385.00 2.57 1392.50 2.38 3.55 10.7 11.5Medium 1800.00 15.92 1866.30 2.66 1815.30 2.90 0.85 6.0 5.5

High 2290.00 95.02 2327.50 4.07 2315.00 1.25 1.09 23.3 76.0



Conclusion


Test Problems

Computational Results: Quadratic Travel Time Function

(Y∗, Z∗) (Y u , Zu) (Y l , Z l ) Rel. cpu

Traffic cpu∗ cpuu cpul Err Ratio

Intensity Delay (sec) Delay (sec) Delay (sec) (%)cpu∗cpuu

cpu∗

cpul

Low 1054.50 0.88 1054.50 0.09 1054.50 0.08 0.00 9.8 11.0Medium 1543.80 6.62 1543.80 0.14 1543.80 0.34 0.00 47.3 19.5

High 2129.80 501.17 2128.60 0.10 2128.60 0.13 -0.06 5011.7 3855.2



Conclusion


The Upper Bounding Problem

DTDTA-U Problem

min(x ,y ,z,g)

qTu Y


gkd(k)t

=∑t∈∆

hkt

Ya(t,s) =∑

k∈C yka(t,s), xa(t) =

∑(τ,s)∈Ωa(t)

Ya(τ,s)∑s∈Γa

za(t,s) = 1∑s∈Γa

φ−1a (s − δ)za(t,s) ≤ xa(t) ≤

∑s∈Γa

φ−1a (s)za(t,s)

Ya(t,s) ≤ Maza(t,s)

yka(t,s) ≥ 0, gk

d(k)t≥ 0, xa(t) ≥ 0, za(t,s) ∈ 0, 1



Conclusion


Difficulties

For practical problems, general IP solvers (e.g., Cplex) are noteffective.

LP relaxation does not provide a good lower bound.

LP relaxation problem decomposes into shortest path problems.

The neighborhood search we implemented was slow and didnot produce a good solution.

Many infeasible neighbors.

Optimal solutions may be isolated.

Difficult to find feasible vectors z .



Conclusion


Nonlinear Relaxation (DTDTA-R)

Replace za(t,s) byYa(t,s)P

r∈ΓaYa(t,r)

.

DTDTA-R Problem

min(x ,y ,z,g)

qTu Y


gkd(k)t

=∑t∈∆

hkt

xa(t) =∑

(τ,s)∈Ωa(t)

∑k∈C yk

a(t,s)

∑s∈Γa

φ−1a (s − δ)Ya(t,s) ≤ xa(t)

∑s∈Γa

Ya(t,s) ≤∑s∈Γa

φ−1a (s)Ya(t,s)

yka(t,s) ≥ 0, gk

d(k)t≥ 0, xa(t) ≥ 0



Conclusion


Theoretical Results

Theorem

DTDTA-R provides a tighter lower bound for the DTDTA-U thanthe LP relaxation.

Equivalent Objective Functions

Objective Function 1 Objective Function 2

min(x ,y ,z,g)

qTu Y min

(x ,y ,z,g)δeT x



Conclusion


Heuristic Algorithm

Main Procedure

Step 1: Solve the DTDTA-R problem and let (yR , xR) denote thesolution of the problem.

Step 2: Find values of binary variables za(t,s) based on the

solution xRa(t,s).

Step 3: In the DTDTA-U problem, fix binary variables to thevalues of za(t,s) and solve the resulting LP.

Step 4: If the LP problem is feasible, stop and return the solution.Otherwise, run the UpSet procedure and go to Step 1.



Conclusion


Heuristic Algorithm

UpSet Procedure: Example 1

s = 1 2 3 4 5Ya(t,s)P

r∈ΓaYa(t,r)

= 0.8 0.0 0.0 0.0 0.2

Force Ya(t,5) to be 0



Conclusion


Heuristic Algorithm


s = 1 2 3 4 5Ya(t,s)P

r∈ΓaYa(t,r)

= 0.0 0.3 0.7 0.0 0.0

Allow Ya(t,4) to be positive



Conclusion


Heuristic Algorithm


s = 1 2 3 4 5Ya(t,s)P

r∈ΓaYa(t,r)

= 0.0 1.0 0.0 0.0 0.0

Allow Ya(t,3) to be positive



Conclusion


Heuristic Algorithm

UpSet Procedure: Example 4∑r∈Γa

Ya(t,r) = 0

Allow Ya(t,s) to be positive ∀s ∈ Γa



Conclusion


Heuristic Algorithm

UpSet procedure potentially cuts away the current localminima.

In the algorithm it is not require to find a global solution ofthe DTTDTA-R problem.



Conclusion


Test Problems


Two networks

Travel time, φa(xa(t)) = αa + βa(xa(t)/ca)γa

αa βa ca γa

U(1,2) U(2,8) U(200,300) random (1 or 2)

Demand=U(20,100)



Conclusion


Test Problems

Computational Results: Separate Mode

Heuristic Approach Cplex Rel.A CPU Iter. CPU Err. B

(%) (sec.) (sec.) (%) (ratio)4-Nodes, [0,10), OBJ1 12% 0.81 10.23 0.84 4.09% 1.744-Nodes, [0,10), OBJ2 8% 1.03 10.35 1.00 4.13% 1.414-Nodes, [0,30), OBJ1 20% 19.76 34.47 376.58 3.92% 22.624-Nodes, [0,30), OBJ2 22% 27.05 38.64 396.51 3.32% 9.559-Nodes, [0,10), OBJ1 28% 633.00 40.58 1,515.44 4.16% 7.809-Nodes, [0,10), OBJ2 10% 111.92 9.98 1,758.65 2.09% 38.919-Nodes, [0,30), OBJ1 30% 1338.37 7.56 5,000.30 -3.33% 12.659-Nodes, [0,30), OBJ2 27% 1029.79 4.53 4,945.14 -4.22% 14.28

A - percentage of the problems not solved by the heuristic

B - CPUCplex/CPUheur.



Conclusion


Test Problems

Computational Results: Combined Mode

Heuristic Approach Cplex Rel.A CPU CPU Err. B

(%) (sec.) (sec.) (%) (ratio)4-Nodes, [0,10) 4% 0.59 0.97 3.63% 1.854-Nodes, [0,30) 10% 19.91 543.00 3.70% 24.229-Nodes, [0,10) 6% 166.86 1,359.16 2.50% 30.279-Nodes, [0,30) 14% 706.03 4,952.60 -3.62% 16.97

A - percentage of the problems not solved by the heuristicB - CPUCplex/CPUheur.



Conclusion


Toll Pricing Framework in the Static DTA

Problem Description: 4-Node Network



Conclusion



Problem Description: User Equilibrium Solution



Conclusion



Problem Description: System Optimum Solution



Conclusion



Problem Description: Toll Pricing Problem



Conclusion


Reduced TE Networks

(y , z) feasible



Conclusion


Reduced TE Networks

RTE(z)



Conclusion


Reduced TE Networks

(p, t)m

p(t) in RTE (y , x , z)



Conclusion


Theoretical Results

SP(y , z)-A Problem

miny ,g

ΦT∑k∈C

yk

s.t. Bzyk + gk = bk ,

∑t∈∆

gkd(k)t

=∑t∈∆

hkt

yka(t,sa(t))

≥ 0, gkd(k)t

≥ 0

Theorem

A feasible solution (y , z) is an user equilibrium solution if and onlyif yk

a(t,sa(t))is an optimal solution of the corresponding SP(y , z)-A

problem.



Conclusion


Valid Toll Vector

Let β denote an arc toll vector.

Definition

β is a valid toll vector with respect to a feasible solution (y , z) if(y , z) is a solution of the tolled user equilibrium problem.

SP(y , z , β)-A Problem

miny ,g

(Φ + β)T∑k∈C

yk


∑t∈∆

gkd(k)t

=∑t∈∆

hkt

yka(t,sa(t))

≥ 0, gkd(k)t

≥ 0



Conclusion


Theoretical Results

SP(y , z , β)-A Problem

miny ,g

(Φ + β)T∑k∈C

yk


∑t∈∆

gkd(k)t

=∑t∈∆

hkt

yka(t,sa(t))

≥ 0, gkd(k)t

≥ 0

Corollary

Given a feasible vector (y , z), vector β is a valid toll with respectto (y , z) if and only if y is an optimal solution of the correspondingSP(y , z , β)-A problem.



Conclusion


Dynamic Toll Set

Theorem

Given a feasible vector (y , z), β is a valid toll if and only if thereare vectors ρk and γ such that

∑k∈C

[(bk)T ρk + γk

∑t∈∆

hkt

]= (Φ + β)T

∑k∈C

yk

BTz ρk ≤ Φ + β

ρkd(k)t

≤ γk

β = −Φ is a valid toll vector given any feasible solution (y , z)



Conclusion


What if (y , z) is a System Optimum Solution?

Let

(y∗, z∗) denote a system optimum solution,

L∗a(t) =∑

s∈Γaφ−1

a (s − δ)z∗a(t,s), and

U∗a(t) =

∑s∈Γa

φ−1a (s)z∗a(t,s)

DTDTA(z∗) Problem

min(y ,g)

ΦT (y)y

s.t. Bz∗yk + gk = bk ,

∑t∈∆

gkd(k)t

=∑t∈∆

hkt

L∗a(t) ≤∑

(τ,s∗a (τ))∈Θa(t)

∑k∈C

yka(τ,s∗a (τ)) ≤ U∗

a(t), yka(t,s∗a (t)) ≥ 0, gk

d(k)t≥ 0



Conclusion


What if (y , z) is a System Optimum Solution?

Theorem

Given a system optimum solution (y∗, g∗, z∗), vector βMSC withcomponents

βMSCa(t,s∗a (t)) =

∑r∈∆|(t,s∗a (t))∈Θa(r)

[∇xa(r)

φa(r)(x∗a(r))

∑k∈C

y∗ka(r ,s∗a (r))

]

+∑

r∈∆|(t,s∗a (t))∈Θa(r)

[λ∗a(r) − µ∗a(r)

]is a valid toll vector with respect to (y∗, z∗), where x∗a(t) representsthe optimal number of cars on arc a at time t, λ∗ and µ∗ are thedual multipliers of the bounding restrictions, and λ∗a(t)µ

∗a(t) = 0,

∀a ∈ A and t ∈ ∆.



Conclusion


Similarities with Marginal Cost.

βMSCa(t,s∗a (t)) =

∑r∈∆|(t,s∗a (t))∈Θa(r)

[∇xa(r)

φa(r)(x∗a(r))

∑k∈C

y∗ka(r ,s∗a (r))

]

+∑

r∈∆|(t,s∗a (t))∈Θa(r)

[λ∗a(r) − µ∗a(r)

]

The first component of the toll vector measures the totalbenefit that is experienced by drivers who enter the arc aftertime t

The second component of the toll vector measures the cost tomaintain the (optimal) network structure of RTE (z∗)



Conclusion


Toll Pricing Problem

Let =(y∗, z∗) denote the set of valid tolls with respect to systemoptimum solution (y∗, z∗).

MinRev Problem

min(β,ρ,γ)∈=(y∗,g∗,z∗)

(y∗)Tβ

MinCost Problem

min(β,ρ,γ,κop ,κinf )

(cop)Tκop + (c inf )Tκinf

s.t. (β, ρ, γ) ∈ =(y∗, g∗, z∗)

|βa(t,s∗a (t))| ≤ Mκopa(t), κ

opa(t) ≤ κinf

a ,

κopa(t) and κinf

a ∈ 0, 1,



Conclusion


Toll Pricing Problem

Additional Constraints

Name ConstraintNonnegativity β(a,t,s∗a (t)) ≥ 0, ∀t ∈ [0,T ]Maximum toll β(a,t,s∗a (t)) ≤ βmax

a ,∀t ∈ [0,T ]Maximum subsidy −βmin

a ≤ β(a,t,s∗a (t)), ∀t ∈ [0,T ]Road restriction β(a,t,s∗a (t)) = 0, ∀t ∈ [0,T ]Time restriction β(a,t,s∗a (t)) = 0, ∀t ∈ [t1

a , t2a ]

Variability |β(a,t,s∗a (t)) − β(a,t+δ,s∗a (t+δ))| ≤ εa, ∀t ∈ [0,T ]



Conclusion


Illustrative Examples

Parameters of the Problems: Network

Time horizon= [0, 30)



Conclusion



Parameters of the Problems: Travel Time Function

φa(xa(t)) = Aa + Ba(xa(t)/Ca)Da ,

where parameters Aa, Ba, Ca and Da, are random numbers.

Aa Ba Ca Da

U[1,2] U[2,8] U[200,300] random (1 or 2)



Conclusion



Parameters of the Problems: Demand Function

hkt = 1.1αtζ,

where the value of αt depends on time t, and ζ is a numberuniformly generated from the interval [20, 30].



Conclusion



Total Collected Toll and Total Cost

Total Collected Toll Total CostA εa = 1 εa = 0.5 εa = 0.1 A εa = 1 εa = 0.5 εa = 0.1

MinRev(ε) 3854 4015 5157 No Sol. 2670 3298 3870 No Sol.MinCost(ε) 5433 5563 6978 No Sol. 1249 1624 2418 No Sol.

Number of Toll Collecting Centers

Number of Toll Collecting CentersA εa = 1 εa = 0.5 εa = 0.1

MinRev(ε) 17 16 17 N/AMinCost(ε) 8 9 12 N/A.



Conclusion



Toll Vector for different values of εa: MinRev Problem



Conclusion



Toll Vector for different values of εa: MinCost Problem



Conclusion

Conclusion

Supply Chain Problems

MIP formulations are equivalent to continuous bilinearproblems.Bilinear reduction problems can be used in heuristic algorithms.By applying a cutting plane method to the bilinear problems itis possible to find an exact solution of the problem.

Dynamic Traffic Assignment Problems

Initial SO formulation belongs to the class of nonlinear mixedinteger problems.Theoretical results suggest solving upper bounding MIPproblem.Bilinear relaxation provides a tighter bound than the LPrelaxation and can be used in heuristic algorithms.A toll pricing framework can be constructed based on anapproximate or exact solution of the SO problem.



Conclusion

Questions?


bilinear approximation techniques to solve network flow...

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