Optimization of the production program the AMS and the management of stocks of resources – Summary

The Ministry
agriculture


Russian
Federation


Krasnoyarsk
state agrarian University


Department
management and


administrative
management


Optimization
the production program of APP


and
inventory management resources


Calculation and graphic
work on logistics

Fulfilled:


student
4th course Iiab Gritsuk N.


Checked:


Professor
Eldestein, J. M.



Krasnoyarsk 2010


The contents



Introduction……………………………………………………………3



1.
Logistic arrangements for the production
system agricultural
enterprise………………………………………………………………4


2.
Statement of the problem……………………………………………………..6


3.
Analytical method of solving
tasks………………………………7


3.1.
The solution of the problem by an analytical method

with
using Excel………………………………………………….7


4.
The graphical method for solving
tasks………………………………..10


4.1.
The decision problem graphically

with
using Excel………………………………………………………..10


5.
The solution of the problem……………………………………………………….13


5.1
The second iteration…………………………………………………….13


5.2
The third iteration…………………………………………………….15


5.3
The fourth iteration………………………………………………….17


5.4
The fifth iteration………………………………………………………19


5.5
Sixth iteration……………………………………………………21


5.6 Seventh
iteration……………………………………………………23


6.
The solution of the problem in terms of narrow
specialization…………………25


6.1.
Optimization of resources in
the lower limit (required
production)……………………………………………………………27


6.2.
Optimization of resources in
the upper limit (limit
implementation)……………………………………………………………29


Conclusion………………………………………………………………..31


Introduction



The object of study
discipline “logistics” are
material and related
the flow of information. Logistics – the science
about optimization of material and
information flows with the aim of increasing
control over the production processes
and minimize the related financial
costs.



The main objective
agriculture – achieving
sustainable production growth
agricultural products, reliable
ensuring the nation’s food supply
and agricultural raw materials. However
in the conditions of market relations, a sharp
competition one
of the most important becomes the task of minimizing
the cost of production. One of the
the main components of the cost
are the costs of creating and storing
stock resources. In addition, distraction
money for the establishment and storage unnecessarily
large stocks of some resources through
the other directly affects the value of
reverse means and output of
of products and, consequently, on the income
.



1. Logistics
diagram of the production system

agricultural enterprises



One of the main links
in the supply chain is the
guide. From the General leadership
all the other links in the chain aimed
information flows: the flow of documents
reports, queries, decision-making and
etc General management
full management and control on
the enterprise and oversees the functioning
main units.



From the General leadership
to the crop, livestock and
poultry-directional flow of information
in the form of documents that contain
information about production plans for
month, year, etc., and various
directives, orders, and a number of other
documentation. With the main production
workshops to the General user
sent information about
the implementation of plans about the
the process of production,
production equipment,
the presence or lack of production
capacity etc.



General guide
makes a request to universities and
provides space to work
specialists with higher education
(information flow). Experts from
universities are sent to work
either the top management or placed
by key departments (material
stream). It all happens through the Department of
frames. Of the production workshops
(livestock and poultry)
directed information flow in
the requests for feed crop production.
Back sent material
flows in the form of supply feed,
silage and other feedstuffs.


So
as every enterprise is open
system, it is included in the system
procured resources and leaving the
system end products.



Logistics diagram of the structure of
agricultural enterprises are given
figure 1.




Department of animal husbandry

Department of poultry

The separation of cooking
silage

Department of plant




Office feed

grain

General guide









Krasgau

College

The personnel Department






Accounting








Fleet


catering







Warehouse workwear























2. Statement of the problem



Agro-industrial enterprise
(AMS) has an opportunity for production
cattle (cattle) and poultry.
There is appropriate
resources, norms which reserves are presented
in the table.


Table 1. The source data






















































Name
resource



Stock
resource



Price
resource for the unit.e.



Rules
resource issues



On
1 cow



On
a dozen birds



Silage,
t



1250



20



2,1





Feed
t



1000



100



1



0,54



Grain,
t



132



55





0,6



Grazing
ha



580



65



1





Labor
resources, people



5300



3,9
.e. on 1 slave. A year



10,5
person/days



2,5
person/days



The task is to:
iterative managing resources,
to get their full use
(the remains should not exceed 1% of
stocks).



Let’s make a mathematical model
tasks:



Let X1 – number of cattle, which
it is possible to produce in these reserves
resources



X2 – the number of tens
turkeys, which can be produced at
these stocks of resources.



2,1X1 ≤ 1250;



X1 + 0,54Х2 ≤ 1000;



0. 6X2 ≤ 132;



X1 ≤ 580;



X1 ≥ 0, X2 ≥ 0;



F
= 40Х1 + 10X2 
max.



3. Analytical
method of solving the problem



Analytical methods
absolutely accurate. They provide an opportunity
for accurate quantitative estimates of
excess of available resources.



Bring the problem to
canonical form:



2,1X1 + X3 = 1250;



X1 + 0,54Х2 + X4 = 1000;



0. 6X2 + X5 = 132;



X1 + X6 = 580.



Additional variables
X3, X4and X5
equal to the difference between the left and right
parts of the constraints and characterize
the shortfall of this restriction (in
this case, the excessive stock).



3.1.
The solution of the problem by an analytical method

with
using Excel



Let there be math
model:



2,1X1 ≤ 1250;



X1 + 0,54Х2 ≤ 1000;



0. 6X2 ≤ 132;



X1 ≤ 580;



X1 ≥ 0, X2 ≥ 0;



F
= 40Х1 + 10X2 
max



To solve this problem
need to cells C10-D13
record the coefficients of the mathematical
model, G10-G14
the right parts of restrictions, but C3
and D3
– the objective function coefficients (Fig.
1).

































































































































 



Cattle



Bird



Profit



max



 



 



 



The value



505



0



20190,47619



 



 



 



 



 



 



 



 



 



 



 



 



The coefficients
Fit



40



10



 



 



 



 



 



 



 



 



 



 



 



 



 



 



Limitations



 



 



 



 



 



 



 



Consumption



 



Stock



 



 



View
resource



 



 



Left
part



 



Right
part



The rest



Balance,%



Silage



2,1



0



1060



<=



1250



190



15,2



feed



1



0,54



504,7619048



<=



1000



495,238



49,5



Labor



10,5



2,5



5300



<=



5300



0



0,0



Grain



0



0,6



0



<=



132



132



100,0



Grazing



1



0



504,7619048



<=



580



75,2381



13,0




Figure 1. An example of the formation
source data



Next, you should click
click on Service,
and then Finding solutions.
On the screen appears a window
shown in figure 2.


Figure 2. Sample screen Search solutions



After clicking on Execute
a window appears, as shown in Fig.
3 where you can see the results and
to explore sustainability and limits.


Figure 3. Results
finding a solution



After the OKappears on the screen
the results of the decision, an example of which
shown in Fig. 4.












































































































































 



Cattle



Bird



Profit



max



 



 



 



The value



505



0



20190,47619



 



 



 



 



 



 



 



 



 



 



 



 



The coefficients
Fit



40



10



 



 



 



 



 



 



 



 



 



 



 



 



 



 



Limitations



 



 



 



 



 



 



 



Consumption



 



Stock



 



 



View
resource



 



 



Left
part



 



Right
part



The rest



Balance,%



Silage



2,1



0



1060



<=



1250



190



15,2



feed



1



0,54



504,7619048



<=



1000



495,238



49,5



Labor



10,5



2,5



5300



<=



5300



0



0,0



Grain



0



0,6



0



<=



132



132



100,0



Grazing



1



0



504,7619048



<=



580



75,2381



13,0




Figure 4. An example of the final solution



The results of the calculations of X1 and x2
are in cells C3 and D3.
The objective function is in cell E3.
The column “Left part” means
the actual use of resources.
Balance” is determined by the difference
between existing reserves (Right
part
) and their actual use
(The left part).



After each iteration of the calculation
the results should be retained
under a new name.



To complete each subsequent
iteration need only in column Fto change the number characterizing
resources.



Each iteration should be completed
the financial test, which is
that money invested in resources,
should remain unchanged. In addition
is also a need to ensure that
the objective function after each iteration
increased. If it does happen
so, purchased not a scarce resource
or excess resource is sold too
large quantities and became scarce.



4. The graphical method
solution of the problem



The graphical method
characterised by simplicity and clarity,
however, he is not precise and is applicable
only for tasks with no more than three
variables. For each
the analytical method of solving the problem
there is a corresponding
a graphical method.



4.1.
The decision problem graphically

with
using Excel



Microsoft
Excel
-2000 is designed to work with electronic
tables that allows you to collect,
analyze and report quantitative
information in automatic mode.
File created in Excel
is called a workbook.


For
image lines characterizing the
limitations, the coordinates of the corresponding
points should be written as
shown in Fig. 5.


In
cells M7
M11
– the right-hand sides of constraints – inventory
resources. In the column In
values of X1.
The Coordinates X2
calculated by equation characterizing
limitations.


































































































Resource



 



X2



X1



Silage



Combo



Grain



Grazing



Work.RES.



F



gradF



Silage



595,2381



0



 



595,2381



2900



 



Feed



0



1851,852



 



800



370,3704



 



Grain



0



220



 



700



220



 



Grazing



580



0



 



580



2900



 



Work.RES.



0



2120



 



700



-820



 



F



0



0



 



-25



200



 



gradF



0



0



200



 



 



 



 



 



 



25




Figure 5. An example entry
source data



As the cells M7
M11
are the values
characterize the resources,
each time you change resources,
we introduce in these cells is updated
data and obtain the solution graphic
method.











































































































Resource



 



X2



X1



Silage



Combo



Grain



Grazing



Work.RES.



F



gradF



Silage



595,2381



0



 



595,2381



2900



 



Feed



0



1851,852



 



800



370,3704



 



Stocks
resources



Grain



0



220



 



Silage



1250



700



220



 



Feed



1000



Grazing



580



0



 



Grain



132



580



2900



 



Grazing



580



Work.RES.



0



2120



 



Work.RES.



5300



700



-820



 



F



0



0



 



-25



200



 



gradF



0



0



200



 



 



 



 



 



 



25








Figure
6. An example of the final solution
graphically



5. The solution of the problem



5.1 Second
iteration







5.2 Third
iteration







The fourth iteration







The fifth iteration







Sixth iteration






Seventh
iteration















6. The solution of this problem
conditions

specialization



Specialization allows
to improve the quality and increase
performance by making it easier
the typical schemes of objects of the labor movement.



We define the profitability
production of cattle
and birds as the quotient of net profit
the costs of production units
products. As appropriate
to produce a more cost effective product.





In this problem, the profitability of
production of cattle
is:



RCATTLE=(40/(2,1*20+1*100+1*65+10,5*3,9)*100=16%.



Profitability
bird is:



RP=(10/(0,54*100+0,6*55+2,5*3,9)*100=5,4%.



Thus, in this case
decisions on the production only
cattle.



Available cash
the funds invested in the resources are
190630.e.



The money should be invested
only in raising cows.



We determine the amount of
cows:



X1=190630/(2,1*20+1*100+1*65+10,5*3,9)=768,82



Then resources can be
found from the original model with x2=0:



Silage
2,1*769=1614,9 t;



Feed
1*769=769 t;



Grazing
1*769=769т.;



Labor costs
10,5*769=8074,5



Then rename Unmet in
Narrow specialization (CS) and input
estimated data in the inventory.



Next, you should click
click on Service,
and then Finding solutions.
Then press to Execute.



You
the profit of F=40*769=30760
.e.



In this case, the mathematical
the model will be:



2,1X1 ≤ 769



X1 + 0,54Х2 ≤ 769



0. 6X2 ≤ 0



X1 ≤ 769



X1≥0, X2≥0



F
= 40Х1 + 10X2 
max



Analytical solution
this task is shown in Fig. 7, graphical
the solution in Fig. 8





Figure 7. Analytical
the solution of the problem in terms of narrow
specialization of production





Figure 8. Graphic solution
tasks in a narrow

specialization
production



Evaluation of economic efficiency
the solution of the optimization problem of the inventory when
specialization of production



Economic efficiency
is:



E= (30760-28191)/28191*100=9,1%



Thus, the economic
efficiency specialization
9.1%, or 2799.e.



6.1. Optimization
stock resources by limiting
bottom

(required manufacturing)


Accept
the decision, for their own consumption
need to make 1500 of the birds, i.e., 150 dozen.


For
the necessary resources:



Silage
0



Feed
0,54*150=81 t



Grazing 0



Labor costs
2,5*150=375т.



Grain
0.6 x 150=90T.


For
the necessary funds:



D = 0 + 81*100 + 375*3,9 + 90*55 + 0 =
14512,5



Then on the production
Cattle remains: 190630-14512,5=176117,5



The number of cattle will be
equal to:



X1 = 176117,5/(2,1*20+1*100+1*65+10,5*3,9)
= 710,3



Then the stocks of resources is equal to:



Silage
710,3*2,1=1491,63 t



Feed
710,3*1 + 150*0,54 = 791,3 t



Grain
150*0,6 = 90 t



Grazing
710,3*1=710,3 ha



Workforce
710,3*10,5 + 150*2,5 = 7833,15 person*days



Analytical solution of this
tasks are shown in figure 9, a graphical
the solution in Fig. 10.





Figure 9. Analytical solution
when the lower limit of





Figure 10. Graphic solution
when the lower limit of



6.2.
Optimization of resources in
the upper limit is

(restrictions on
implementation)



Market analysis showed that it is possible
implement cattle not more than
400 heads.



For this you will need
resources :



Silage 400 * 2,1 =
840 T.



Feed
400 * 1 = 400 T.



Grain
0 t



Grazing
400 * 1 = 400 ha



Workforce
400 * 10,5 = 4200 persons*days



Determine the money needed
for the production of a given amount
Cattle:



D = 400 * (2,1*20 + 1*100 + 65*1 + 10,5*3,9) = 99180



Production remains: 74873



X=74873 / (0,54*100+0,6*55+2,5*3,9) =302
birds



Analytical solution of this
tasks are shown in figure 11, graphical
the solution in Fig. 12.





Figure 11. Analytical solution
when the upper limit is





Figure 12. Graphic solution
when the upper limit is


Conclusion


Consider
agro-industrial enterprise has
logistics system, consisting of –
General management, OU
for crop production,
unit production
livestock and poultry production, workshops
processing: meat, mill flour,
dairy.


When
the solution of a linear programming problem
traditional iterative method
managed to achieve almost full
the use of resources. Their
residues amounted to less than one percent
from reserves. In this case, can be
made of 600 head of cattle and 343 of the tens
turkeys. This may be obtained
25713.e. profit.



Narrow
the specialization of work is very effective
since it allows to organize the model
the pattern of movement of objects of labor, sharply
to reduce the number of process
routes to improve performance.
It was found that beef production
the most cost-effective than the production
birds.



The task then
was solved with additional
restrictions “below” (mandatory
poultry production for own
consumption) and from “above” (restriction
the possibilities of realization of cattle).

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