Proof of Theorem 1: We use the 3-Satisfiability (3SAT) problem (Garey and
Johnson 1979) for our reduction. 3-Satisfiability (3SAT)
Instance: A set Z = {z1, z2, . . . , zp} of boolean variables and a collection W =
{W1 ∩ W2 ∩ . . . ∩ Wk} of clauses over Z, each of which is a disjunction of literals,
z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp such that|Wi| = 3 for 1 ≤ i ≤ k and ¯zj = 1− zj for 1≤ j ≤ p.
Solution: Find an assignment of either a true (1) or a false (0) value to each variable in {z1, z2, . . . , zp}, such that the expression W evaluates to true (1).
Given an arbitrary instance of 3SAT, we construct the following instance of Prob- lem P.
• The parameter values related to estimating the transportation cost are as fol- lows: cu = 0, ct = 0, ca = 1, fa = 4, fu = 0, ba = 1, bt = 1, ljU = p, lIUj = 2p,
and lIM j = 4p.
• The capacity limit for the retail vault is cℓ= pL− 1 and h = 1. The unit cost
of incremental capacity at the vaults is co = 0. The fixed cost of upgrading the
retail vaults zi, (or ¯zi) to a regional vault, Fi = 1, i = 1, 2, . . . , 2p.
• If the distance is less than 1 then the transportation mode is urban (U), oth- erwise it is inter-urban (IU) or inter-modal (IM).
• Each variable in the set {z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp} corresponds to a distinct retail
vault, i.e., N = 2p. There is one big vault BV and one regional vault RGV . That is, MB = 1, Mrv = 1, and Nv = N + Mbr + Mrv = 2p + 2.
• A branch is attached to each retail vault in {z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp}. See Ta-
• Each clause in {W1, W2, . . . Wk} corresponds to a distinct branch. Currently
branch Wj, where j = 1, 2, . . . , k, is assigned to the regional vault RGV since
there is not enough capacity at any of the retail vaults in{z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp}.
The distance between Wj and RGV is 1. The transportation mode between
Wj and RGV is inter-urban and the transportation cost for a period is L. See
Table A.3 for details.
• There are p dummy branches, {U1, U2, . . . Up} and one big vault, BV . Currently
branch Uj, where j = 1, 2, . . . , p, is assigned to big vault BV since there is not
enough capacity at any of the retail vaults in {z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp}. The
distance between Uj and BV is 1. The transportation mode between Uj and
BV is inter-urban and the transportation cost for a period is L. See Table A.3 for details.
• There is one branch near BV that is served by BV . There is one branch near RGV that is served by RGV . See Table A.3 for details. There is no other financial intuition present in this problem instance.
• The number of branches is Mbr = k + 3p + 2.
• The distance between the clause node, Wj, where j = 1, 2, . . . , k and a variable
node in Wj is 0, and other distances are 1. The transportation cost per period
(urban mode) from a clause node, Wj, j = 1, 2, . . . , k to a variable node in
Wj is therefore 0; from Wj to other nodes it is L (the cost corresponds to the
inter-urban mode).
• The distance between node Uj, where j = 1, 2, . . . , p and the node zj, (or ¯zj) is
0, and other distances are 1. The transportation cost per period (urban mode) from Uj, j = 1, 2, . . . , p to node zj (or ¯zj) is therefore 0; from Uj to other nodes
Table A.3: Current Transportation Cost. If the Distance is Less Than 1, Then the Transportation Mode is U, and Otherwise is IU or IM.
Bank Current Distance from Load Withdrawals Transportation Branch j Supply the Current Size per Period (Dj+) Cost (IU)
j Point i Supply Point Load Size Deposits per Period (Dj−) Wj RV dtij= 1, l IU j = 2p D + j = pL, 1 lIU j bt.Dj.(2dij) daij= 1 D−j = 0 = L Uj BV dtij= 1, ljIU= 2p D+j = pL 1 lIU j bt.Dj.(2dij) da ij= 1 D−j = 0 = L branches D+j = pL− 1, at zj, ¯zj zj, ¯zj dtij= 0 - D−j = 0 0 Bank Branch D+j = pL, near BV BV dt ij= 0 - D−j = 0 0 Bank Branch D+j = pL, near RV RV dt ij= 0 - D−j = 0 0
• Before proceeding with the proof, we provide the construction of the instance of Problem P using the following example.
• Example: p = 3 and k = 4. Z = {z1, z2, z3}; W = {W1 ∩ W2 ∩ W3 ∩ W4}, where W1 = (z1 ∪ z2 ∪ ¯z3), W2 = (¯z1 ∪ ¯z2 ∪ ¯z3), W3 = (¯z1 ∪ z2 ∪ z3), and W4 = (¯z1∪ ¯z2∪z3). Each variable in the set{z1, ¯z1, z2, ¯z2, z3, ¯z3} corresponds to a retail vault. The current total transportation cost is (p + k)L. The distance is 1 for all edges that are not shown in the network. Refer to Figure A.1. Let L = 2p. For the instance of Problem P constructed above, we consider the following question:
Decision Problem: Does there exist a feasible solution with total cost Φ ≤ p? The decision problem is clearly in NP. It can also be easily verified that the construction of our decision problem from the 3SAT instance can be performed in polynomial time. We now show that the decision problem has an affirmative answer if and only if the 3SAT instance is satisfiable.
Figure A.1: Problem Instance Corresponding to the Example Problem with p = 3, k = 4, W1 = {z1, z2, ¯z3}, W2 = {¯z1, ¯z2, ¯z3}, W3 = {¯z1, z2, z3}, W4 = {¯z1, ¯z2, z3}, Solution z1 = 1, ¯z2 = 1, and z3 = 1. The Distance is 1 for All Edges That Are Not Shown in the Network.
BV
U1 U2 U3
L L L
Big Vault (1), Bank Branch (1) Bank Branches (p) Z1 Z1 Z2 Z2 Z3 Z3 L 0 0 0 0 0 0 Bank Branches (p)
Retail Vaults (2p), Bank Branches(2p)
Z1 Z1 Z2 Z2 Z3 Z3 L 0 0 0 0 0
Retail Vaults (2p), Bank Branches(2p)
0 W1 W2 W3 W4 0 0 0 0 0 0 0 Bank Branches (k) L L L L 0 0 0 RGV
Regional Vault (1), Bank Branch (1)
L L L L
true variables in a truth assignment. These vaults then correspond to the truth assignment, zi (or ¯zi), where i = 1, 2, . . . , p, will be upgraded. A total of p retail
vaults will therefore be upgraded with the total cost of p. branch Wj is assigned to a
variable node (a retail vault) corresponding the truth assignment in Wj. The trans-
portation cost for Wj accordingly corresponds with this truth assignment which is 0
(the urban mode). Similarly, branch Uj is assigned to a variable node (a retail vault)
corresponding to the truth assignment of variable zj (or ¯zj). The transportation cost
for Uj therefore corresponds this truth assignment which is 0 (the urban mode), and
Φ = p.
Only if part: Suppose there exists a solution to the decision problem with Φ ≤ p. The current total transportation cost, (p + k)L must be eliminated since L = 2p. branches Uj, where j = 1, 2, . . . , p and Wj, j = 1, 2, . . . , k must therefore be assigned
to one of the upgraded vaults in{z1, ¯z1, z2, ¯z2, . . . , zp, ¯zp}. Since the currency demand
exactly one vault form {zj, ¯zj} must be upgraded. The total upgrade cost of retail
vaults is therefore p and Φ = p. Since Φ = p, branch Wj must be assigned to
a variable node (a retail vault) with the transportation cost 0 (urban mode), and otherwise Φ > L > p. A satisfiable assignment corresponding to Φ = p for the 3SAT instance is now immediate. This completes the proof.