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THE TIME LIMIT ASSIGNMENT PROBLEMS
Guozhong Bai1and Xiao-Xiong Gan2
1Department of Mathematics
Guangdong University of Business Studies
Guangzhou, China
baiguozhong@163.com
2Department of Mathematics
Morgan State University
Baltimore, MD 21251, US
gxiaoxio@morgan.edu
ABSTRACT
This paper establishes a new assignment problem called the time limit assignment problem
(TLAP) which pursues the best overall efficiency only if all tasks have been completed in
time or very close to the deadline. The time limit assignment problem contains one discrete
goal constraint, and therefore it is a special goal programming problem. This paper
provides a mathematical model for such a time limit assignment problem and introduces a
method to find its optimal solutions and the deviation of the optimal solution to the warning
time. We also provide an example that illustrates how to solve the time limit assignment
problem.
Keywords: assignment problem, goal programming, time limit, mathematical model, optimal
solution.
2000 Mathematics Subject Classification: 90B50, 90C29, 68Q05, 90B06, 90B35.
1 Introduction
A traditional assignment problem concentrates on the usual assignment problems and bottle-
neck assignment problems. Many new assignment problems, such as “one task - multi per-
sons” (or one person - multi tasks) assignment problem [1,2], B-assignment problems[3] (optimal
efficiency and optimal time assignment problem), C-assignment problems [4,5,6] (the number of
the assignments is less than or equal to the number of human resources and the number of the
tasks), and gray assignment problems[7,8] (efficiency is indeterminate) have been discussed
recently. Each of these assignment problems considers only the efficiency with constrains
and provides ways to obtain the best efficiency.
International Journal of Applied Mathematics & Statistics,
Int. J. Appl. Math. Stat.; Vol. 13; No. S08; September 2008; 31-40
ISSN 0973-1377 (Print), ISSN 0973-7545 (Online)
Copyright © 2008 by IJAMAS, CESER
The due-date assignment problems received a lot of attention recently. In many practical
scheduling environment, the promised due-date is to be determined during sales negotiations
with the customer. The due-date assignment problem approaches investigate the maximum
earliness cost, tardiness cost and due-date cost and then minimized those factors. A very
helpful survey about some due-date assignment problems is provided in [9] and some more
recent developments can be found in [10].
In practice, however, we frequently meet such an assignment problem that pursues the best
overall efficiency only if all the tasks have been completed in time or very close to the deadline,
we call such an assignment problem the time limit assignment problem (TLAP), and call the
period of time required to complete all tasks the warning time. In such an assignment problem
the due-date is not negotiable and the jobs must be completed in time. The following is a
typical example.
Example. A city has four projects and plans to accomplish all of them within twenty months.
There are four companies bidding for these projects. Table 1 and 2 below show us how many
months each bidder needs to accomplish each project and how many dollars each bidder asks
for. The city wants all four projects to be completed within 20 months with the least cost, and
the priority is given to the time of completion. Which bidder(s) should the city select?
Table 1. Efficiency. Table 2. Cost.
proj
time
bidder @
@
@
X
X
X
X
X
XABCD
I M 25 20 21
II 20 21 24 25
III 19 24 18 23
IV 17 18 20 19
proj
bidding
bidder @
@
@
X
X
X
X
X
XABCD
I M 1.15 1.11 1.04
II 1.10 1.05 1.14 1.15
III 1.10 1.15 1.08 1.13
IV 1.08 1.09 1.15 1.09
The number Min these tables is a sufficiently large number indicating that the company did
not bid the corresponding project, the unit in Table 1 is the month and the unit in Table 2 is
$1,000,000.
The assignment problems with a fixed contract period are such problems and its contract period
is the warning time provided that the project starts at the time zero. The assignment prob-
lems for all emergency preparations or preventing projects for the natural disasters such as
hurricanes, earthquakes and winter storms are also such problems where the warning times
for such disasters are their forecast times. It is more often for us to meet such assignment
problems in military affairs.
2 Mathematical Model for Time Limit Assignment Problem
Given a time limit assignment problem, it usually has ntasks that need to be accomplished
by m(human) resources. Let eij and cij denote the time efficiency and job efficiency (cost or
32 International Journal of Applied Mathematics & Statistics
others) for the ith resource to accomplish the jth task, we call the matrices E=(eij)m×nand
C=(cij)m×nthe time (efficiency) matrix and cost (efficiency) matrix, respectively. Define
F(x) = max{eij |xij =0,1≤i≤m, 1≤j≤n},and
f(x)=
m
i=1
n
j=1
cijxij
for any x=(x11,x
12,···,x
21,x
22,···,x
mn)∈{0,1}mn. A standard time limit assignment
problem with the warning time e0, time matrix Eand cost matrix Cis defined to be
min z=W
1p1+W
2p2
s.t. F (x)+n1−p1=e0
f(x)+n2−p2=0
n
j=1 xij =1,i=1,2,···,m
m
i=1 xij =1,j=1,2,···,n
xij =0or1,1≤i≤m, 1≤j≤n
n1,n
2,p
1,p
2≥0,
(2.1)
where Wkis a given priority factor (priority classification), nkis the negative deviational variable
and pkis the positive deviational variable with nkpk=0,k=1,2.W1≺W2means that W1
has priority over W2, that is, W2-class goal will not be considered until the W1-class goal has
been achieved. The constrain without deviational variables is called the absolute constrain
and the other is called the goal constrain. We only discuss the problem (2.1) with nk=0
because our goal is to minimize the positive deviations. As usual, eij ≥0,c
ij ≥0,1≤i≤
m, 1≤j≤n.
Time limit assignment problem could be regarded as a goal programming problem with a dis-
crete goal constrain[11]. The definitions we use in this paper related to the goal programming
such as the definitions of absolute constrains, goal constrains, deviational variables, imple-
mentable solutions, and optimal solutions are the same as the definitions in [11].
It is not difficult to convert the problem of “one task - multi persons” (or one person - multi tasks)
into the standard form [1,2] and that is why we only introduce the case that xij = 0 or 1 .Itis
then clear that the time limit assignment problem (2.1) has implementable solution if and only
if m=n. We suppose that m=nand E=0 and C=0 from now on.
3 Solving Time Limit Assignment Problems
Time limit assignment problems are special goal programming problems. We can not solve
the time limit assignment problem by the traditional goal programming method because such
a TLAP has discrete goal constrains. We introduce a method of solving time limit assignment
problem based on its particular property.
Int. J. Appl. Math. Stat.; Vol. 13, No. S08, September 2008 33
Definition 1.Ifx∈{0,1}n2satisfies all absolute constrains in (2.1), then xis called a
implementable solution of the problem.
If we denote Sthe set of all implementable solutions of the problem (2.1), then
S={xT|n
j=1 xij =1,n
i=1 xij =1,x
ij =0or1,1≤i, j ≤n}
where x=(x11,x
12,···,x
21,x
22,···,x
nn)∈Rnn .
Definition 2. A unit step function Ue:R−→ Ris defined to be
Ue(t)=U(t−e)=0t≤e
1t>e (3.1)
Let e>0, the following problem (3.2) is called the g(e, E)assignment problem of (2.1), or
simply the g(e, E)problem.
min g(e, x)=
n
i=1
n
j=1
Ue(eij)eij xij
s.t. n
j=1 xij =1,i=1,2,···,n
n
i=1 xij =1,j=1,2,··· ,n
xij = 0 or 1,1≤i, j ≤n.
(3.2)
Denote
B(e)={(i, j)|eij ≤e, 1≤i, j ≤n}⊂(N×N)n.(3.3)
Let Mbe a sufficiently large positive number ( Me), we call the problem (3.4) bellow the
G(e, C)assignment problem of the problem (2.1), or call it the G(e, C)problem.
min G(e, x)=
(i,j)∈B(e)
cijxij +M
(i,j)/∈B(e)
xij
s.t. n
j=1 xij =1,i=1,2,···,n
n
i=1 xij =1,j=1,2,··· ,n
xij = 0 or 1,1≤i, j ≤n.
(3.4)
Lemma 1. With the notations as above, we have
(i) if e1≤e2, then g(e1,x)≥g(e2,x)for every x∈S,
(ii) if e=F(x)for some x∈S, then g(e, x)= 0,
(iii) if e1≤e2, then B(e1)⊂B(e2). Moreover, if e1≤e2and xij=0 for some
(i,j)∈B(e2)\B(e1), then G(e1,x)>G(e2,x).
Proof. (i) and (ii) can be easily derived from the formula (3.2). For (iii), it is obvious that
B(e1)⊂B(e2)∀e1≤e2. Since xij=0 for some (i,j)∈B(e2)\B(e1)we have
M
(i,j)/∈B(e1)
xij −M
(i,j)/∈B(e2)
xij ≥Mxij=M>0.
34 International Journal of Applied Mathematics & Statistics
Because Mis a sufficiently large number, we have G(e1,x)>G(e2,x).
Theorem 1. Given a time limit assignment problem (2.1). If there exists a x0∈Ssuch
that
g(e0,x0)=
n
i=1
n
j=1
Ue0(eij)eij x0
ij =0,
then the optimal solution of the assignment problem G(e0,C)is the optimal solution of the time
limit assignment problem (2.1).
Proof. For any x∈S, the corresponding deviational variable p1is determined by the
value max{0,F(x)−e0}, that is, p1is also a function of x,orp1=p1(x)=F(x)−e0if
F(x)−e0≥0.
Let x∗be an optimal solution of the G(e0,C)assignment problem (3.4), then x∗is also a
implementable solution of the problem (2.1) because both problems have the same constrains
for x∗. We first show that g(e0,x∗)=0.
In fact, if g(e0,x∗)>0, then there exist some i1and j1such that ei1j1>e
0and ei1j1x∗
i1j1=0,
and hence (i1,j
1)/∈B(e0). Then
G(e0,x∗)=
(i,j)∈B(e0)
cijx∗
ij +M
(i,j)/∈B(e0)
x∗
ij ≥
(i,j)∈B(e0)
cijx∗
ij +M≥M,
Since g(e0,x0)=0, it follows that eij≤e0if x0
ij=0, and hence (i,j)∈B(e0). Then
G(e0,x0)=
(i,j)∈B(e0)
cijx0
ij +M
(i,j)/∈B(e0)
x0
ij =
(i,j)∈B(e0)
cijx0
ij M≤G(e0,x∗).
It is in contradiction with that x∗is a optimal solution of the G(e0,C)assignment problem.
Thus, g(e0,x∗)=0.
By the definition of the g(e0,E)and that g(e0,x∗)=0,wehaveF(x∗)≤e0and then
p1(x∗)=0. Assume that x∗is not an optimal solution of the time limit assignment problem
(2.1), then there must exist a solution y0=(y0
11,··· ,y
0
nn)T∈Ssuch that p1(y0)=0 and
n
i=1
n
j=1
cijy0
ij <
n
i=1
n
j=1
cijx∗
ij.
Since p1(y0)=0, it follows that F(y0)≤e0, and hence (i,j)/∈B(e0)y0
ij =0, then we have
G(e0,y0)=
(i,j)∈B(e0)
cijy0
ij +M
(i,j)/∈B(e0)
y0
ij =
n
i=1
n
j=1
cijy0
ij <
n
i=1
n
j=1
cijx∗
ij ≤G(e0,x∗).
This contradicts that x∗is an optimal solution of the assignment problem G(e0,C). Thus, x∗
is a optimal solution of the time limit assignment problem (2.1).
Theorem 2.Ifminx∈Sg(e0,x)>0, then
(i) there exists some time efficiency ei0j0, called the critical time efficiency for e0, such that
ei0j0>e
0,minx∈Sg(ei0j0,x)=0, and g(ei1j1,x)>0for all ei1j1<e
i0j0;
Int. J. Appl. Math. Stat.; Vol. 13, No. S08, September 2008 35
(ii) we can obtain a critical time efficiency within n2−n+1 steps;
(iii) if ei0j0is a critical time efficiency and x∗is an optimal solution of the assignment problem
G(ei0j0,C), then F(x∗)=ei0j0,
(iv) for any critical time efficiency ei0j0, the optimal solution of the assignment problem
G(ei0j0,C)is the optimal solution of the time limit assignment problem (2.1).
Proof. By (i) of the Lemma , we have that
minx∈Sg(eij,x)≥minx∈Sg(e0,x)>0
for every eij <e
0. By (ii) of the Lemma, g(F(x),x)=0, and then there exists some eij such
that minx∈Sg(eij,x)=0. Let
ei0j0= min{eij |minx∈Sg(eij,x)=0,1≤i, j ≤n}.
It is clear that g(eij ,x)>0for all eij <e
i0j0. This is (i).
(ii) Let ei0j0= min{eij |eij >e
0,1≤i, j ≤n}.
If minx∈Sg(ei0j0,x)=0, then eij ≤e0∀eij <e
i0j0, and hence minx∈Sg(eij,x)≥
minx∈Sg(e0,x)>0by (i) of the Lemma. Then ei0j0is the critical time efficiency we want.
Otherwise, replacing e0by ei0j0, we repeat the above process. We can obtain a critical time
efficiency ei0j0within at most n2−n+1steps.
(iii) Suppose that ei0j0is a critical time efficiency and x∗is an optimal solution of the assign-
ment problem G(ei0j0,C), combining (i) of this theorem above and (ii) of the Lemma, we have
that F(x∗)≥ei0j0.
Assume that ei1j1=F(x∗)>e
i0j0, then (i1,j
1)/∈B(ei0j0)and we have
G(ei0j0,x∗)=
(i,j)∈B(ei0j0)
cijx∗
ij +M
(i,j)/∈B(ei0j0)
x∗
ij ≥M.
On the other side, let x0be an optimal solution of the assignment problem g(ei0j0,E), then
g(ei0j0,x0)=0. Therefore, eij ≤ei0j0if x0
ij =0, and then (i, j)∈B(ei0j0), and hence
(i,j)/∈B(ei0j0)x0
ij =0. Then
G(ei0j0,x0)=
(i,j)∈B(ei0j0)
cijx0
ij +M
(i,j)/∈B(ei0j0)
x0
ij =
n
i=1
n
j=1
cijx0
ij M≤G(ei0j0,x∗).
It contradicts that x∗is an optimal solution of the assignment problem G(ei0j0,C), and hence
F(x∗)=ei0j0.
(iv) Since minx∈Sg(e0,x)>0, (i) of the Lemma yields that
F(x) = max{eij |xij =0,1≤i, j ≤n}>e
0
for every x∈S.
Suppose that ei0j0is a critical time efficiency and x∗is an optimal solution of the assignment
problem G(ei0j0,C). By the definition of ei0j0, we have that ei0j0≤F(x)∀x∈S. By (iii)
above, we have that F(x)≥F(x∗)>e
0for every x∈S, and hence p1(x)≥p1(x∗)>0.
36 International Journal of Applied Mathematics & Statistics
If there exists some x1∈Ssuch that p1(x1)=p1(x∗), that is, F(x1)−e0=F(x0)−e0,
then F(x1)=ei0j0, and hence (i,j )/∈B(ei0j0)x1
ij =0. Since x∗is an optimal solution of the
assignment problem G(ei0j0,C),wehave
n
i=1
n
j=1
cijx∗
ij ≤
(i,j)∈B(ei0j0)
cijx∗
ij +M
(i,j)/∈B(ei0j0)
x∗
ij ≤
≤
(i,j)∈B(ei0j0)
cijx1
ij +M
(i,j)/∈B(ei0j0)
x1
ij =
n
i=1
n
j=1
cijx1
ij.
Thus, x∗is an optimal solution of the assignment problem (2.1).
In a time limit assignment problem, the key point is to have all tasks be accomplished before
the warning time. We therefore introduce the following definition.
Definition 3. Let xbe an implementable solution to the time limit assignment problem with
the warning time eand write D−(e, x) = max{0,−(F(x)−e)}and D+(e, x) = max{0,F(x)−
e}. Then D−(e, x)is called the negative deviation of the implementable solution xto the
warning time eand D+(e, x)is called the positive deviation of the implementable solution x
to the warning time e.
It is obvious that D−(e, x)·D+(e, x)=0 in every time limit assignment problem. If the
negative deviation D−(e, x)is positive, then all tasks can be accomplished before the warning
time e, and then there exists a surplus time D−(e, x). If the positive variation D+(e, x)is
positive, then it is impossible to accomplish all jobs before the warning time e, and the time of
completion is at least be delayed by D+(e, x).
Hungarian Approach is a well known tool to solve the traditional assignment problems. We
provide the following computational steps to solve time limit assignment problem (2.1) by The-
orem 1 and Theorem 2.
Step 0. Transfer the time limit assignment problem to balancing problem if it is necessary.
Step 1. Let eij =0 if eij ≤e0for each eij in the time efficiency matrix E.
Step 2. Let e1=e0if there exists some eij =0in each low and in each column, then go to
Step 5. Otherwise, go to Step 3.
Step 3. Compute ri= min{eij |j=1,2,··· ,n},1≤i≤nand cj= min{eij |i=
1,2,···,n},1≤j≤n, and then let
e1= max{ri,c
j|1≤i, j ≤n}.
Step 4. In the efficency matrix E, let eij =0 if eij ≤e1.
Step 5. Search for nindependent zero elements using the same searching method in Hun-
garian Approach of solving the traditional assignment problem. If the nindependent zero
elements have been found, go to Step 7. Otherwise, go to Step 6.
Step 6. Compute
e2= min{eij >e
1|1≤i, j ≤n},
Int. J. Appl. Math. Stat.; Vol. 13, No. S08, September 2008 37
replacing e1by e2, and then go to Step 5.
Step 7. Based on the given cost efficiency matrix C=(cij ), we establish a new matrix C
by letting c
ij =Mif eij >e
1where Mis a sufficiently large number, and letting c
ij =cij
otherwise. Then, Using Hungarian Algorithm (or other approaches) to obtain an optimal
solution x0.x0is an optimal solution for the time limit assignment problem too. Go to Step
8.
Step 8. Compute the deviations D−(e0,x0)and D+(e0,x0).
Clearly, if the given problem is a balancing problem, this algorithm converges to the optimal
solution of the time limit assignment problem.
4 An Example
The example in Section 1 is a typical time limit assignment problem. By the data provided in
that example, the time matrix and the cost matrix are
E=⎡
⎢
⎢
⎢
⎢
⎣
M25 20 21
20 21 24 25
19 24 18 23
17 18 20 19
⎤
⎥
⎥
⎥
⎥
⎦
,C=⎡
⎢
⎢
⎢
⎢
⎣
M115 111 104
110 105 114 115
110 115 108 113
108 109 115 109
⎤
⎥
⎥
⎥
⎥
⎦
,
and the warning time is e0=20. We solve the problem (2.1) with e0,E and Cin the
following steps.
1). Based on the matrix E, let e
ij =0 if eij ≤e0=20 and e
ij =eij otherwise, we have
the new efficiency matrix Efor the g(20,E)assignment problem where
E=⎡
⎢
⎢
⎢
⎢
⎣
M25 0 21
0 212425
024023
0000
⎤
⎥
⎥
⎥
⎥
⎦
.
2). Since Edoes not have 4 (n=4) independent zero elements, we have
e
1=e1= min{eij >20 |i, j =1,2,3,4}=21.
3). Based on E, let e
ij =0 if e
ij ≤e
1=21 and e
ij =e
ij otherwise, we have the time
efficiency matrix E for the g(21,E)assignment problem where
E =⎡
⎢
⎢
⎢
⎢
⎣
M25 0 0
0 0 24 25
024023
0000
⎤
⎥
⎥
⎥
⎥
⎦
.
4). There are 4 independent zero elements in E because e
13 =e
22 =e
31 =e
44 =0.
38 International Journal of Applied Mathematics & Statistics
5). Based on the cost matrix C, let c
ij =Mif eij >e
1=21 where Mis a sufficiently
large number, and let c
ij =cij otherwise, we have the efficiency matrix Cfor the G(21,C)
problem where
C=⎡
⎢
⎢
⎢
⎢
⎣
MM111 104
110 105 MM
110 M108 M
108 109 115 109
⎤
⎥
⎥
⎥
⎥
⎦
.
6). Using the Hungarian Algorithm, we find the optimal solution x∗for the G(21,C)assign-
ment problem where x∗
14 =x∗
22 =x∗
33 =x∗
41 =1, and x∗
ij =0 otherwise. x∗is also an
optimal solution for the time limit assignment problem, and
min F(x∗)=21,min f(x∗) = 425.
7). For the warning time e0=20, the deviations of the optimal solution to the warning time
e0=20 are
D−(20,x∗)=0,D
+(20,x∗)=1.
Because D+(20,x∗)=1>0, it is impossible to accomplish the project within 20 months. The
optimal solution tells us that the best possible result is that the project may be accomplished
within e0+D+(e0,x∗)=21 months with the expense of 425 million dollars.
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