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
5.6 Seventh
iteration……………………………………………………23
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
.
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, decisionmaking 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
poultrydirectional 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
Agroindustrial 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 
Stock 
Price 
Rules 

On 
On 

Silage, 
1250 
20 
2,1 
– 
Feed 
1000 
100 
1 
0,54 
Grain, 
132 
55 
– 
0,6 
Grazing 
580 
65 
1 
– 
Labor 
5300 
3,9 
10,5 
2,5 
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.
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).
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 C10D13
record the coefficients of the mathematical
model, G10G14
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 
40 
10 














Limitations 








Consumption 

Stock 


View 


Left 

Right 
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 
40 
10 














Limitations 








Consumption 

Stock 


View 


Left 

Right 
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.
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.
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 righthand 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 

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
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:
R_{CATTLE}=(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= (3076028191)/28191*100=9,1%
Thus, the economic
efficiency specialization
9.1%, or 2799.e.
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: 19063014512,5=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
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
Consider
agroindustrial 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 costeffective 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).