In this paper we shall give a new solution for the optimal assignation of workers on jobs from the point of view of execution total time minimization using the Simplex algorithm which can solve the problem using computers instead the known Little’s solution. In the second section, we shall give a new solution for the optimal assignation of workers on jobs from the point of view of maximize the number of allocates workers, using the Simplex algorithm which can solve the problem using computers instead the known graphical solution.Simplex, assignation, minimization

Catalin Angelo Ioan

English

Research Papers in Economics

49 
 
A SOLVING METHOD OF TWO ECONOMICAL PROBLEMS 
USING LINEAR PROGRAMMING IN INTEGER NUMBERS 
 
Catalin Angelo IOAN 
Danubius University, Galati, Romania 
 
 
Abstract: 
In this paper we shall give a new solution for the optimal assignation of workers on jobs from the point of 
view of execution total time minimization using the Simplex algorithm which can solve the problem using computers 
instead the known Little’s solution. In the second section, we shall give a new solution for the optimal assignation of 
workers on jobs from the point of view of maximize the number of allocates workers, using the Simplex algorithm 
which can solve the problem using computers instead the known graphical solution. 
 
Keywords Simplex, assignation, minimization 
 
Classification codes and keywords: C70 
 
I The Optimal Assignation Of Workers From The Point Of View Of Execution Total Time 
Minimization 
 
The problems of assignation appear usual in the process of targets allocation in an institution. 
Let consider A= A1,...,An  the set of workers in an institution and L= L1,...,Lm  the set of jobs which 
must be executed at a specific moment. In the execution of Lj, the worker Ai spends a time equal with tij 
units (hours, minutes, seconds etc.). Supposing that it exists workers which can execute a lot of jobs we 
put the problem of allocation on jobs such that the total time spending in the execution to be minimum. 
 
We shall assign an infinite value to tij if Ai is not able to execute the job Lj. Also, we shall understand that 
the number of workers is equal with those of jobs, in the opposite case introducing fictional workers or 
jobs with infinite times of execution to prevent the allocation of them. 
 
The method of Little suggests the following steps: 
 
Step 1 It is build the table of times (with workers on columns and jobs on rows) and after we shall 
compute the minimum on each row. After this we subtract these values from those of rows, compute the 
minimum on each column and after also, we shall subtract these from the values on the columns. After this 
step, on each row or column is at least one value equal with 0. 
 
Step 2 We shall compute the sum of all elements subtracted from rows and columns and noted with S1. 
 
Step  3  For  each  element  equal  with  0  in  the  last  table,  we  shall  compute  the  quantities  ij=min 
tik k j +min tpj p i   or,  in  other  words,  the  sum  of  the  elements  on  the  row  and  column 
corresponding to the null quantity. After this, we shall determine the maximum of that values and the 
appropriate allocation (s,r). We shall build a tree graph where the initial knot comes with the value S1. We 
shall build after a bend where we shall put the activities (s,r) and non(s,r) who will come with the values 
sr and  sr=S1+ sr respectively. 
 
Step 4 We shall erase the row s and the column r and we shall act like in the first step. 
 
Step 5 We shall compute S2 like sum of the elements of minimum of rows and columns and we shall 
modify the indicator  sr=S1+S2. 
 
Step 6 If the simplified table will has only one row and column the algorithm will close. If not it will be 
choose the minimum between  sr and  sr. If both values will be equal we shall choose the value  sr 
appropriate to an allocation and not to a reject of allocation. 50 
 
 
Step 7 If the choice value was  sr we shall return at the step 3. 
 
Step 8 If the choice value was  sr then we shall consider in the table previously of step 1: tsr=  and we 
shall compute the minimum of row s and column r, subtract these from the appropriate row and column 
and return at the third step. 
 
We can see that the algorithm is a little hard therefore we shall propose in what follows a new method 
based on the Simplex algorithm. 
 
II The New Method For Optimal Assignation Of Workers From The Point Of View Of 
Execution Total Time Minimization 
 
Let consider A‘= A1,...,An‘  the set of workers in an institution and L‘= L1,...,Lm‘  the set of jobs which 
must be executed at a specific moment. 
 
Let therefore f:A‘ P(L‘), f(Ai)=
k 1 i i L ,..., L
  i=1,...,n‘ the function who assign to Ai the jobs: 
k 1 i i L ,..., L
 
which he can realize if he has the necessary qualification for at least one job and f(Ai)=  in opposite 
cases. 
 
We shall restrict the set A' and we shall consider, from the beginning, the subset of those workers for 
which f(Ai)   Ai A. We shall note therefore A= A1,...,An  with n n‘ (after a possible renotation of 
workers). Let now (again after a possible renotation of workers):  
n
1 i
i) A ( f = L1,...,Lm  with m m‘. If 
m m‘  we  have  that  the  jobs  Lm+1,...,Lm‘  cannot  be  executed  from  any  workers,  therefore  will  be 
excludes. 
 
Finally, let consider: L= L1,...,Lm  and the new allocation function: f:A P(L). 
 
We shall define a matrix: 
M=
n
1
nm 1 n
m 1 11
m 1
A
...
A
a ... a
... ... ...
a ... a
L ... L   
 
where aij=1 if the worker Ai can execute the job Lj and 0 in the other cases. 
 
Let now consider the matrix A=( ij) where: 
ij=
j i
j i
L   of execution    in the   nominate not     will A  worker   the if   0
L   of execution    in the   nominate    will A  worker   the if   1
 
 
We shall, like in the previous section, build the matrix T=(tij) of execution times, assigning tij=  if Ai 
cannot execute Lj. 
 
In a distinction with Little‘s method we shall not enjoin restrictions to the number of workers or jobs. 
 
Let now the matrix B=( ijaij) who‘s elements belong to the set  0,1  and who has the following meaning: 
ijaij=1 if Ai will nominate to execute Lj and is also qualified for this thing and  ijaij=0 in the other 
cases. 
 51 
 
Because  no  one  can  execute  two  jobs  simultaneously,  we  have  therefore  the  condition: 
m
1 j
ij ij a 1 
i= n , 1 .. 
 
Also,  because  any  job  cannot  be  execute  simultaneously  by  two  different  workers  we  have  that: 
n
1 i
ij ij a 1  j= m , 1 . 
 
From the above conditions it follows that: aij ij 1  i= n , 1   j= m , 1 . 
 
The allocation problem will become: 
0
M a
1 a
1 a
) t min(
ij
n
1 i
m
1 j
ij ij
n
1 i
ij ij
m
1 j
ij ij
n
1 i
m
1 j
ij ij
 
where M is the number of workers proposed for the execution. 
 
Before solving the problem, let remark first that if it isn‘t a maximal allocation the problem will not have a 
solution and in other case if it has at the final we shall obtain effective the allocation. The value of 
minimum will be the searched total time. 
 
The problem will be solved in the following manner: we start with the value M=n. If it has not a solution 
we diminish M with a unit and we begin again to solve the new problem. Because M is a free term in the 
upper problem we shall reoptimize the older. 
 
The process is obviously finite because the problem has always a solution at least for M=0:  ij=0. 
 
III The Optimal Assignation Of Workers On Jobs 
 
Let consider now A‘= A1,...,An‘  the set of workers in an institution and L‘= L1,...,Lm‘  the set of jobs 
which must be executed at a specific moment. 
 
Because each worker can has a multiple qualification, but not all necessary for the entire set of jobs we put 
the problem of allocation on jobs such that they realize too much if it is possible of them. 
 
Let therefore f:A‘ P(L‘), f(Ai)=
k 1 i i L ,..., L
  i=1,...,n‘ the function who assign to Ai the jobs: 
k 1 i i L ,..., L
 
which he can realize if he has the necessary qualification for at least one job and f(Ai)=  in opposite 
cases. 
 
We shall restrict the set A' and we shall consider, from the beginning, the subset of those workers for 
which f(Ai)   Ai A. We shall note therefore A= A1,...,An  with n n‘ (after a possible renotation of 
workers). Let now (again after a possible renotation of workers):  
n
1 i
i) A ( f = L1,...,Lm   with m m‘. If 
m m‘  we  have  that  the  jobs  Lm+1,...,Lm‘  cannot  be  executed  from  any  workers,  therefore  will  be 
excludes. 
 
Finally, let consider: L= L1,...,Lm  and the new allocation function: f:A P(L). 52 
 
We shall define a matrix: 
M=
n
1
nm 1 n
m 1 11
m 1
A
...
A
a ... a
... ... ...
a ... a
L ... L   
 
where aij=1 if the worker Ai can execute the job Lj and 0 in the other cases. 
 
The graphical method presented in  2  proposes a construction of a simple graph (a decomposition of 
nodes  in  two  disjoint  subsets:  workers  and  jobs)  and  after  an  initial  allocation  a  succession  of 
improvements based on graphical observations. This method is good but cannot be easily implemented on 
computers. 
 
We shall propose in what follows a new method based on the Simplex algorithm. 
 
Let now, the matrix A=( ij) where: 
ij=
j i
j i
L   job    the execute not     will A  worker   the if   0
L   job    the execute    will A  worker   the if   1
 
 
and the matrix B=( ijaij) with elements in the set  0,1 . We have that  ijaij=1 if the worker Ai will 
execute the job Lj and if he is qualified for this thing and  ijaij=0 if the worker Ai will not execute the job 
Lj or he is not qualified to do this. How any worker cannot execute two jobs in the same time, we have the 
condition: 
m
1 j
ij ij a 1  i= n , 1 . 
Because  a  job  cannot  be  executed  in  the  same  time  by  two  workers  we  have  also  that: 
n
1 i
ij ij a 1 
j= m , 1 . From these conditions we have now that: aij ij 1  i= n , 1   j= m , 1 . 
 
The problem becomes now the following linear programming: 
0
1 a
1 a
) a max(
ij
n
1 i
ij ij
m
1 j
ij ij
n
1 i
m
1 j
ij ij
 
 
Because  ij=0 verify the restrictions we have that the problem has always a solution. One problem can 
appear after sloving: what is happened if the solutions will not be entire? It is possible, for example, on the 
i-th row to be a lot of elements equal with 1 (appropr iate to the fact that one worker can execute a few 
jobs), say k elements, and the optimal solution to contains the variables:  k
1
...
k 2 1 ij ij ij
. Because 
the objective function is 
n
1 i
m
1 j
ij ij a
 it follows that it will not modify if we replace all the cited values with, 
for example: 
1
p ij  for a 1 p k. 
 53 
 
Example 
 
Let the workers A1,A2,A3 and the jobs L1,L2,L3 which posibility of execution is in the following table: 
Worker  Jobs 
A1  L1,L3 
A2  L1,L2 
A3  L2 
Considering  the  matrix  M=
3
2
1
3 2 1
A
A
A
0 1 0
0 1 1
1 0 1
L L L
  and  A=
33 32 31
23 22 21
13 12 11
  we  have  the  following  linear 
programming problem: 
0 , , , ,
1
1
1
1
1
1
) max(
32 22 21 13 11
13
32 22
21 11
32
22 21
13 11
32 22 21 13 11
 
with the solution:  13=1,  32=1,  21=1. We have therefore that A1 will execute the job L3, A2 – L1 and 
A3 – L2. 
 
Bibliography 
 
Ioan C.A., 2008. Mathematics II, Ed. Sinteze, Galati 
 
Stancu-Minasian I.M., 1970. The study of the simulation problem using the graph theory‖, Studies and 
researches of economical calculus and cibernetics, nr.5. 
 
Teodorescu N. (coord.), 1972. Methods of operational researchs‖, Ed. tehnica, Bucharest. 
 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

The study of the simulation problem using the graph theory‖, Studies and researches of economical calculus and cibernetics,

A SOLVING METHOD OF TWO ECONOMICAL PROBLEMS USING LINEAR PROGRAMMING IN INTEGER NUMBERS

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A SOLVING METHOD OF TWO ECONOMICAL PROBLEMS USING LINEAR PROGRAMMING IN INTEGER NUMBERS

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