IP modeling techniques II In this handout Modeling

IP modeling techniques II In this handout, Modeling techniques: ØEither-Or Constraints ØBig M method ØBalance constraints ØFixed Charges Applications: ØMulti-period production planning ØInventory management

Modeling technique: Either-Or Constraints • In some situations, a choice can be made between two constraints, so that only one (either one) must hold whereas the other one can hold but is not required to do so. • E. g. , recall the capacity constraints for the furniture manufacturer example: pine: 5 xt + 1 xc + 9 xd 1500 (1) oak: 2 xt + 3 xc + 4 xd 1000 (2) Suppose the furniture can be made from either pine or oak but we don’t need both. How to achieve that in the model?

Either-Or Constraints • Introduce new binary variables. For i=1, 2 • Only one of (1) and (2) must hold. Thus, add a constraint: y 1+y 2 = 1 • We also need: y 1=1 implies 5 xt+1 xc+9 xd 1500 y 2=1 implies 2 xt+3 xc+4 xd 1000 How to express these implications by linear constraints?

Either-Or Constraints Ø New idea: use the big number method. • Select a huge positive number M. • Note that 5 xt+1 xc+9 xd 1500+M holds for any reasonable choices of xt, xc, xd. It is equivalent of not putting any restriction on xt, xc, xd at all. • Consider constraint 5 xt+1 xc+9 xd 1500+M(1 -y 1) (3) – If y 1=1 then (3) is equivalent to (1) – If y 1=0 then (3) doesn’t impose any restriction on xt, xc, xd • Thus, the set of constraints 5 xt+1 xc+9 xd 1500+M(1 -y 1) 2 xt+3 xc+4 xd 1000+M(1 -y 2) y 1+y 2 = 1 provides that only one of 5 xt+1 xc+9 xd 1500 and 2 xt+3 xc+4 xd 1000 must hold.

k out of p constraints must hold • Suppose the model includes a set of p constraints f 1(x 1, x 2, …, xn) d 1 f 2(x 1, x 2, …, xn) d 2. . fp(x 1, x 2, …, xn) dp such that only some k of these constraints must hold. • Generalizing the big M method of the previous slide, that condition is achieved by the following set of constraints: f 1(x 1, x 2, …, xn) d 1+My 1 f 2(x 1, x 2, …, xn) d 2+My 2 …. fp(x 1, x 2, …, xn) dp+Myp y 1+y 2+…+yp = p – k y 1, y 2, …, yp binary

IP modeling: Multi-period production planning • A manufacturer wishes to schedule production for K periods in advance to meet known monthly demands for a given product. • Demand for period i is Di. • In period i, at most Ci items can be produced at cost $pi per item. • The demand of the current period can be satisfied by the items produced in earlier production periods (aka inventory). • The cost of holding an item in inventory from period i to period i+1 is $hi. • No inventory at the beginning of the first period. At most Hi items are allowed in inventory at the beginning of period i. • Goal: Formulate an IP which will minimize the total cost while satisfying the demands.

IP modeling: Multi-period production planning • What variables should we have? Define xi = number of items produced in period i , for i=1, . . . , k. wi = number of items in inventory at the beginning of period i , for i=1, . . . , k+1. (xi and wi are nonnegative integers) • What is the objective function? Minimize the total production and inventory cost: • Some obvious constraints. Production limit: xi Ci , for i=1, . . . , k Inventory limit: wi Hi , for i=1, . . . , k+1 No inventory before period 1: w 1 = 0

Multi-period production planning: Balance Constraints • Also need constraints satisfying the demands, and relating xi and wi. • Consider the following diagram for period i: wi INPUT xi Di Period i wi+1 OUTPUT • The corresponding constraint for period i (i=1, …, k): wi + xi = Di + wi+1 This is known as balance constraint, and is often used in multi-period problems. Note that the balance constraints provide that wi + xi ≥ Di (since wi+1 ≥ 0), and thus the demands are satisfied.

Multi-period production planning: Complete IP model s. t. xi Ci , wi Hi , wi + xi = Di + wi+1 , w 1 = 0 xi ≥ 0 integer wi ≥ 0 integer for i=1, . . . , k+1 for i=1, …, k for i=1, . . . , k+1

Multi-period production planning: Fixed Setup Cost • Note that the demand of the current period can be totally satisfied by the inventory carried from the previous period. • Suppose there is a setup cost $si for period i if we decide to have any production for that period (and there is no setup cost if there is no production). • How to take the setup costs into account? • We need a new entry in the objective function: if xi>0 then si for each period i. But this is not a linear function (because no conditions are allowed on variables).

Multi-period production planning with setup cost • To overcome the problem, introduce new variables. For i=1, . . . , k, • Then the setup cost is siyi for period i. • But we also need new constraints relating xi and yi. • Idea: Use the big M method. For each i=1, . . . , k , add a constraint xi Myi. Why does it work? § If yi=0 then xi must be 0; § If yi=1 then there is no restriction on xi. • Note that we can take M=Ci for period i. Thus, we don’t need new constraints. Simply, replace xi Ci with xi Ciyi. • Question: Can we have yi=1 but xi =0 for period i ?

Multi-period production planning with setup cost • Summarizing, the modified IP model is: s. t. xi Ciyi , wi Hi , wi + xi = Di + wi+1 , w 1 = 0 xi ≥ 0 integer wi ≥ 0 integer yi binary for i=1, . . . , k+1 for i=1, …, k