A business dinner problem

A business dinner problem

Fr ´ed ´eric Meunier

December 9th, 2016

Joint work with Alejandra Estanislao

A “scheduling” problem

A problem submitted to the Junior company of the Ecole des Ponts a few years ago.

•

A pool of providers and a pool of customers.

•

Each provider and each customer have to meet once.

•

Two providers have to meet at most once.

A series of dinners has to be planned. Tables are available, with a given capacity.

Two people meet if they take their dinner at a same table.

Objective. Minimize the number of dinners in order to achieve

the constraints above.

An example

Feasible schedule in 6 dinners for an instance with

• 2 tables, 5 suppliers, 6 customers

• at most 2 suppliers per table, at most 3 customers per table

dinner 1: table 1: suppliers: 1,2

customers: 1,2,3 table 2: suppliers: 3,4 customers: 4,5 dinner 2: table 1: suppliers: 3

customers: 2 table 2: suppliers: 5 customers: 5 dinner 3: table 1: suppliers: 2

customers: 4,5 table 2: suppliers: 4 customers: 2,6 dinner 4: table 1: suppliers: 3

customers: 1,3 table 2: suppliers: 5 customers: 2,6 dinner 5: table 1: suppliers: 1,5

customers: 4 table 2: suppliers: 2,3 customers: 6 dinner 6: table 1: suppliers: 4,5

customers: 1,3 table 2: suppliers: 1 customers: 5,6

Parameters and Results

Parameters:

•

Number of tables: t

•

Number of suppliers: s

•

Number of customers: c

•

At most σ suppliers per table

•

At most γ customers per table Contribution:

•

Feasible schedule with not too many dinners for any value of the parameters

•

Lower bound for any value of the parameters

•

Special cases with closed formula

•

Open questions

Related works: combinatorial design

Kirkman Schoolgirl problem

15schoolgirls must be split into five groupe of three girls each day of the week so that no two girls are in the same group twice.

Similarity with our problem: suppliers = schoolgirls, tables = groups, etc.

Social Golfer problem Howell designs

Oberworlfach problem(Ringel, 1967)

Is it possible to seat an odd number n of mathematicians at m round tables of sizes(`1, . . . , `m)withP

i`i =n in(n−1)/2dinners so that each mathematician sits next to everyone else exactly once?

Conjecture: ‘yes’ (except for exceptional values of the`i’s) Bipartite Analogue of the Oberwolfach problem

Oberwolfach Rectangular Table Negotiation problem

None of these are optimization problems.

Lower bounds

• Number of tables:t

• Number of suppliers:s

• Number of customers:c

• ^{At mostσ}suppliers per table

• ^{At mostγ}customers per table

Two suppliers at most once together; each supplier and each customer exactly once together.

Lower bounds on the number of dinners:

lb1=ls σ m

lb2= c

γ

lb3= s

tσ c

γ

lb4=











√s tγ





(c−γ)max r γ

c−γ,1

+ γ

maxq _γ

c−γ,1















 lb5= max

j∈{2,...,σ}

s t

2 j

c γ

− s−1 j(j−1)

Lower bounds lb

and lb

• Number of tables:t

• Number of suppliers:s

• Number of customers:c

• ^{At mostσ}suppliers per table

• ^{At mostγ}customers per table

During a dinner, any customer sits at a table with at mostσsuppliers. To meet all suppliers, he needs at leastds/σedinners.

During a dinner, any supplier sits at a table with at mostγcustomers. To meet all customers, he needs at leastdc/γedinners.

Lower bounds lb

and lb

xij =number of times supplierisits at a table withjsuppliers (him included)

min 1 t

s

X

i=1 σ

X

j=1

1 jxij

s.t.

σ

X

j=1

xij ≥ l

c γ

m

i∈[s]

σ

X

j=1

(j−1)x_ij ≤ s−1 i∈[s]

xij∈R⁺ i∈[s],j∈[σ].

Optimal value equal to s t max

1 σ

c γ

, max

j∈{2,...,σ}

2 j

c γ

− s−1 j(j−1)

.

Proof.Strong duality. Maximization of a piecewise linear concave function R→R.

Exact solutions

Ifσ=1, then the optimal number of dinners is

max

s, c

γ

, s

t c

γ

.

Proof.Ifγ=1, proper edge-coloring ofKs,cwith each color present at most ttimes andlb1,lb2,lb3. Forγ >1, make groups.

t=s= 3, c= 4, σ=γ= 1

customers providers

4 colors ⇒4 dinners

Howell designs

A square array is aHowell designif

1. every cell is either empty or contains an unordered pair of elements (symbols) chosen from a set of size 2n,

2. every symbol occurs exactly once in each row and each column, 3. every unordered pair of symbols occurs at most once in the array.

t=3 tables,s=4 suppliers, dc/γe=3 groups of customers (the columns),

at mostσ=2 suppliers per table, in 3 dinners (the rows)

1,2 3,4

3 1 2,4

4 2 1,3

t=5 tables,s=6 suppliers,dc/γe=5 groups of customers (the columns), at mostσ=2 suppliers per table, in 5 dinners (the rows)

6 2,3 4,5 1

6 3,4 1 2 5

3,5 1,2 6 4

2,4 5 1 3,6

1 5 4,6 3 2

Exact solutions

Ifσ=2, s>c/γ, t ≥min(c/γ,s/2)and(dc/γe,s)6= (2,4), the optimal number of dinners is

max c

γ

,ls 2

m . Proof.Almost directly from Howell design andlb1,lb2.

If there are4tables,7suppliers,26customers, at most2suppliers per table, at most4 customers per table, the optimal number of dinners is7.

Exact solutions

If t=ds/2e,σ=2, anddc/γe ≥³₂s, then the optimal number of dinners is

2 c

γ

−s+1.

Proof.Combination of the previous two exact solutions andlb5.

If there are5tables,9suppliers,41customers, at most2suppliers per table, at most3 customers per table, the optimal number of dinners is20.

Feasible schedules

r(t,s,c, σ, γ) :=optimal value

r(t1,s,c, σ, γ)≤ t2

t1

r(t2,s,c, σ, γ) r(t,s,c, σ1, γ)≤

σ2

σ1

r(t,s,c, σ2, γ)

r(t,s,c, σ, γ1)≤r(t,s,dc/γ2e, σ,dγ1/γ2e) ifγ1≥γ2

General upper bounds onr(t,s,c, σ, γ)

ub1= 2

σ 1 t min

c γ

,s

max c

γ

,ls 2

m

ub2= 1

t ls

σ m

1−σ+σmax c

γ

,2ls σ

m

ifds/σe ≤ dc/γe

Inefficient whenσ≥3.

Open question: making groups of customers

Example from the beginning:

t=2 tables,s=5 suppliers,c=6 customers, at mostσ=2 suppliers per table, at mostγ=3 customers per table

dinner 1: table 1: suppliers: 1,2

customers: 1,2,3 table 2: suppliers: 3,4 customers: 4,5 dinner 2: table 1: suppliers: 3

customers: 2 table 2: suppliers: 5 customers: 5 dinner 3: table 1: suppliers: 2

customers: 4,5 table 2: suppliers: 4 customers: 2,6 dinner 4: table 1: suppliers: 3

customers: 1,3 table 2: suppliers: 5 customers: 2,6 dinner 5: table 1: suppliers: 1,5

customers: 4 table 2: suppliers: 2,3 customers: 6 dinner 6: table 1: suppliers: 4,5

customers: 1,3 table 2: suppliers: 1 customers: 5,6

Open question: making groups of customers

t=2 tables,s=5 suppliers,c=6 customers, at mostσ=2 suppliers per table, at mostγ=3 customers per table

Optimal solution:

dinner 1: table 1: suppliers: 1,2

customers: 1,2,3 table 2: suppliers: 3,4 customers: 4,5,6 dinner 2: table 1: suppliers: 3

customers: 1,2,3 table 2: suppliers: 2,5 customers: 4,5,6 dinner 3: table 1: suppliers: 4,5

customers: 1,2,3 table 2: suppliers: 1 customers: 4,5,6

Is there always an optimal solution in which the customers are

split into groups, the members of each group staying together

for all dinners?

Open question: t = γ = 1

What is the optimal solution when there is only one table and

exactly one customer per table?

Thank you.