math models of or: the boxes problem

Math Models of OR:The Boxes Problem

John E. Mitchell

Department of Mathematical SciencesRPI, Troy, NY 12180 USA

December 2018

Mitchell The Boxes Problem 1 / 22

Mass production

Outline

1 Mass production

2 The Boxes Problem

3 Backward recursion equations

4 Solving the problem

5 Modeling as a shortest path problem


Mass production

Mass production

When we mass-produce something (eg, cars, telephones, computers,shirts, etc.), we are not usually able to exactly meet the demand ofevery customer.

We typically have to limit the number of different types of models weproduce.

For example, a particular line of cars comes with certain combinationsof color, number of cylinders, different configurations of theentertainment system, etc.

This saves money in production, but it has a cost in either potential lostdemand or in reduced prices to entice customers who wantedsomething slightly different.


The Boxes Problem

Outline

1 Mass production

2 The Boxes Problem





The Boxes Problem

The boxes problem

The boxes problem is a variant on this.

We want to ship 6 packages of different sizes, but we can onlymanufacture 3 different sizes of box.

We assume item 1 is bigger than item 2, which in turn is bigger thanitem 3, and so on.

For example, a box that is large enough to contain item 3 would alsobe large enough to ship item 4 or item 5 or item 6.

The costs of manufacturing a single box of each size are in the table.

size k 1 2 3 4 5 6manufacturing cost ck 25 20 18 15 11 8


S i x items : I 7 2 7 3 7 4 > 5 7 6 .

Three b o x sires.

Boxofsi.c2.b.no/iiu4,boxcfsica#2 boxerof s i n 5 , 2 boxer c i r c 5

2 boxers i r . l , --

✓ item, item,i tems 5 8 6l o l 3 5 4

-

I box s i n 1 , I box i i i . 2 , 4 boxes i n 3

:.

The Boxes Problem

Stages and states

Our decision variables are

xk = number of boxes of size k that are manufactured, k = 1, . . . , 6

The STAGES are the indices k , so at stage k we decide the valueof xk .The STATE is the number of box sizes still available. Initially, we haves = 3. Every time we choose a positive value for xk , the value of sdecreases by one.


Backward recursion equations

Outline

1 Mass production

2 The Boxes Problem






Backward recursion

We use backward recursion to minimize the manufacturing costs. Wehave

fk (s) = manufacturing cost of optimal policy to ship items k , . . . , 6,given that we can use at most s different sizes

fk (s, xk ) = manufacturing cost of optimal policy to ship items k , . . . , 6,given that we can use at most s different sizeswith the decision xk


Eventually find f , (3).

falsixu) = c a n . + f (s-l)kexy

W ✓cost t o make

a boxer of futureS e c k :

cost

c o s t c µ e µEg: take k=3, s = 2 .I f X u = 3 : the 3 boxesof i c e 3 ship

item, 3 , 4,5.

Next consider boxesof s i n . 6.


Setting up the recursion

The maximum number of boxes we need to build at size k is equal tothe number of items still to be shipped, which is 6 � k + 1. We cannottake xk = 0 in the recursion, since we have to ship item k . Then

fk (s) = maxxk=1,...,6�k+1

{fk (s, xk )}

If we make the decision xk > 0 then the state decreases by one. Theimmediate cost is ckxk .

Further, these k boxes are used to ship items k , . . . , k + xk � 1, so thefuture costs depend on fk+xk (s � 1).

Notice in this example that the stage has changed by an amountthat may well be different from one.


m i n A l lremaining item, shipped:← 6=ktxu.- I

K k i ka...tt#?kexu?X u : I 2 . . . x ,


Setting up the recursion

The maximum number of boxes we need to build at size k is equal tothe number of items still to be shipped, which is 6 � k + 1. We cannottake xk = 0 in the recursion, since we have to ship item k . Then

fk (s) = maxxk=1,...,6�k+1

{fk (s, xk )}

If we make the decision xk > 0 then the state decreases by one. Theimmediate cost is ckxk .

Further, these k boxes are used to ship items k , . . . , k + xk � 1, so thefuture costs depend on fk+xk (s � 1).

Notice in this example that the stage has changed by an amountthat may well be different from one.



The recursion

Combining the immediate costs with the future costs, we obtain theformula

fk (s, xk ) =

⇢ckxk + fk+xk (s � 1) if 1 k + xk 6ckxk if k + xk > 6

In the end, we need to determine f1(3).


Solving the problem

Outline

1 Mass production

2 The Boxes Problem





Solving the problem

The posible states and decisions

At optimality, we are going to build three different sizes of boxes.

Thus, the only state we need to consider when k = 6 is s = 1.

We must build at least one of size 1.

Thus, the only possible states for stages 3, . . . , 5 are s = 1 or s = 2.

At stage 2, the only possible state is s = 2: we used up one size onitem 1.

When s = 1, we must choose xk = 6 � k + 1 so we ship all theremaining items.

When s = 2, we can choose xk = 1, . . . , 6 � k , so we don’t ship item 6in a box of size k .


l e t 4 - s e ei f n o tenough^ earlierboxer:I ✓ build1 boxo f a i r . G

'ai:÷÷÷÷:)

1st.".".im.1 o r 2

Must have5 = 2 :could only have used u p 1 r i c e

previously

m a t eD D D 0 D o✓ ÷¥¥÷¥...

I f s = 1 : choose a ⇐ 4

Solving the problem

Find f6(s)

We have f6(1) = c6 = 8, since the optimal solution to the subproblemwould be to build one box to ship item 6.


Solving the problem

Find f5(s)

We have f6(1) = 8 and c5 = 11.

state s f5(s, x5) f5(s) x⇤5

x5 = 1 x5 = 21 — 2c5 = 22 22 22 c5 + f6(1) = 19 — 19 1


I F h e

0

Solving the problem

Find f4(s)

We have f5(1) = 22, f6(1) = 8 and c4 = 15.


x4 = 1 x4 = 2 x4 = 31 — — 3c4 = 45 45 32 c4 + f5(1) = 37 2c4 + f6(1) = 38 — 37 1


ME i n D 8

-

I!÷:{''I:#think.I box ofair. G .

Solving the problem

Find f3(s)

We have f4(1) = 45, f5(1) = 22, f6(1) = 8 and c3 = 18.


x3 = 1 x3 = 2 x3 = 3 x3 = 41 — — — 4c3 = 72 72 42 c3 + f4(1) = 63 2c3 + f5(1) = 58 3c3 + f6(1) = 62 — 58 2


T I F I I I I 183

3! I21¥'s' + HE's

Solving the problem

Find f2(s)

We only need to consider s = 2 for f2(s), since we can only use onesize that is larger than size 2.

We have f3(1) = 72, f4(1) = 45, f5(1) = 22, f6(1) = 8 and c3 = 20.


x2 = 1 x2 = 2 x2 = 3 x2 = 42 c2 + f3(1) = 92 2c2 + f4(1) = 85 3c2 + f5(1) = 82 4c2 + f6(1) = 88 82 3

Hence f2(2) = 82 with x⇤2 = 3.


§ MyFit I D D8

O'boi:{+ !!:::'s

Solving the problem

Find f1(3)

It is not optimal to take x1 � 5, since then we’d only make at most 2different sizes.

We have f2(2) = 82, f3(2) = 58, f4(2) = 37, f5(2) = 19, and c1 = 25.


x1 = 1 x1 = 2 x1 = 3 x1 = 43 c1 + f2(2) = 107 2c1 + f3(2) = 108 3c1 + f4(2) = 112 4c1 + f5(2) = 119 107 1

Thus, f1(3) = 107 and x⇤1 = 1.


M¥122128115 D i d

a

Solving the problem

Backtracking to get the optimal solution

The optimal value is 107.

Backtracking through the tables, we see that x1 = 1, x2 = 3, andx5 = 2.

Checking the cost, we have 1 ⇥ 25 + 3 ⇥ 20 + 2 ⇥ 11 = 107,confirming the value we found in the tables.


Modeling as a shortest path problem

Outline

1 Mass production

2 The Boxes Problem







Each (stage,state) combination is a node.

An edge goes from one node (k , s) to another if there is a decision xkthat leads from one node to the other;

the edge is labelled with the immediate cost of that decision, namelyckxk .



The network

1, 3

2, 2 3, 2 4, 2 5, 2

3, 1 4, 1 5, 1 6, 1

end

25

5075

100

20

40

60

80

18

36 54

15 30

11

72

4522

8

The optimal solution x1 = 1, x2 = 3, x5 = 2 has value 107 andcorresponds to the path (1, 3) ! (2, 2) ! (5, 1) ! end.


0 ¥ ,O O O O

states] 1 2 30 ok E § e n d

0

state= 2

state= I


The network

1, 3

2, 2 3, 2 4, 2 5, 2

3, 1 4, 1 5, 1 6, 1

end

25

5075

100

20

40

60

80

18

36 54

15 30

11

72

4522

825

60

22

The optimal solution x1 = 1, x2 = 3, x5 = 2 has value 107 andcorresponds to the path (1, 3) ! (2, 2) ! (5, 1) ! end.


math models of or: the boxes problem

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