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000-858 - i5 iSeries WebSphere Technical Solutions V5R3 - Dump Information
| Vendor | : | IBM |
| Exam Code | : | 000-858 |
| Exam Name | : | i5 iSeries WebSphere Technical Solutions V5R3 |
| Questions and Answers | : | 94 Q & A |
| Updated On | : | October 30, 2017 |
| PDF Download Mirror | : | 000-858 Brain Dump |
| Get Full Version | : | Pass4sure 000-858 Full Version |
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QUESTION: 82
An
iSeries development group with no Java programming skills has been
asked to provide a web-based application for querying information from
DB2 UDB for iSeries. There will be several categories of users, each
with varying levels of access to the application. Which of the following
minimizes the development effort?
- Create iSeries Access for Web Database requests.
- Generate a web application with the WDSc for iSeries SQL wizard.
- Build a 5250 application and web-enable it via the IBM WebFacing Tool.
- Use IBM Query for iSeries and web-enable it with the IBM WebFacing Tool.
Answer: A
QUESTION: 83
A
customer wants to utilize caching to minimize the overhead of serving
common servlet output from WAS. How do they implement this?
- Using cachespec.xml files in the Web application
- Using the Fast Cache Response Accelerator
- Using Java Servlet Version 2.3 filter code
- Using the WDSc deployment descriptor editor
Answer: A
QUESTION: 84
A
new J2EE WAS-based application is being developed, which will require a
high volume of DB2 UDB for iSeries reads on the same machine. Which of
the following techniques should be utilized?
- JDBC datasource using the IBM Toolbox for Java
- JDBC datasource using the iSeries Native JDBC provider
- JDBC datasource using the Enterprise JavaBean query language
- JDBC datasource using the IBM Toolbox for Java record level access provider
Answer: B
QUESTION: 85
A
J2EE application developer would like to determine the overall
application response time contribution by individual Java components and
methods. Which of the following tools should be used?
- iDoctor for iSeries
- Tivoli Performance Viewer
- OS/400 Collection Services
- WDSc Profiling Perspective
Answer: D
QUESTION: 86
An
i825 customer requires separate WAS production, test, and development
environments on their system. Their applications require JavaServer
Pages and servlets plus read/write access to DB2 UDB for iSeries. Which
of the following provides these capabilities at the lowest cost?
- WAS (Base)
- WAS - Express for iSeries
- WAS Single Server Edition
- WAS Network Deployment Edition
Answer: B
QUESTION: 87
An
independent software vendor (ISV) is developing a new WAS V5 ERP
application. Which of the following WAS utility options should be
investigated to streamline the installation and configuration tasks?
- WSCP scripting
- WSADMIN scripting
- CONFIGWASSVR shell script
- Administrative console command macros
Answer: B
QUESTION: 88
A
customer is running a new WAS-based application on their development
iSeries. They are now ready to move this application into a production
environment. Which tool should be used to determine what iSeries server
resources are required?
- PM/400
- iSeries Job Watcher
- Patrol for iSeries - Predict
- iSeries Performance Explorer
Answer: C
QUESTION: 89
An
existing customer's RPG development group is reluctant to move away
from host tools such as PDM, SEU, and SDA. Which of the following
benefits of the workstation based CODE tools should be emphasized to the
customer?
- CODE Program Generator produces more efficient compiled code.
- CODE Designer and Editor can work disconnected from the host.
- CODE Designer is the only tool that can specify screen colors.
- CODE Editor has a text find and replace feature that complements SEU.
Answer: B
QUESTION: 90
Which of the following communication protocols allows WebSphere Application Server
version 4.0 and later releases to communicate with a Web server?
- IIOP
- HTTP and HTTPS
- Remote OSE Transport
- SOAP
Answer: B
QUESTION: 91
An
i825 customer wants to build a new order entry Web application that
avoids interactive workload and traditional 5250 screen-to-screen
navigation. However, the customer would like to reuse the existing
RPG-based business logic for calculating discounts, sales tax, and
shipping costs. What is an appropriate strategy?
- Rewrite the RPG business logic as callable ILE Service Programs.
- Utilize the IBM WebFacing Tool.
- Import the RPG business logic into VisualAge RPG and rewrite the application.
- Use the XML Toolkit for iSeries product to Web-enable the RPG business logic.
Answer: A
QUESTION: 92
A
customer has an i820 FC #2437/1526 (two-way, 2350 CPW/1050 ICPW).
Measurements show that after they have used the IBM WebFacing Tool to
convert all their 5250 applications, they are using 1600 CPW. The
customer is planning to double the capacity being used. Which of the
following upgrades will meet their requirements and minimize the total
upgrade cost?
- eServer i5 520 Standard Edition
- eServer i5 520 Enterprise Edition
- eServer i5 520 Value Edition
- eServer i5 550 Solution Edition
Answer: A
QUESTION: 93
WAS-based applications running on OS/400 differ from traditional RPG and COBOL
based applications in which of the following ways?
- Traditional applications use more memory.
- WAS applications use less operating system services.
- WAS applications require more system input/output services.
- Traditional applications always provide the maximum performance.
Answer: B
QUESTION: 94
A
customer is looking at options for web enabling their DB2 UDB for
iSeries resources. Which of the following is a key advantage of using a
WAS-based application over a Net.Data-based application?
- WAS-based applications can utilize standard SQL syntax.
- WAS-based applications can call iSeries program objects.
- WAS-based applications can use database connection pooling.
- WAS-based applications can access DB2 UDB on a local or remote iSeries.
Answer: C
IBM 000-858 Exam (i5 iSeries WebSphere Technical Solutions V5R3) Detailed Information
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ADMS 3330 task four solution.pdf - ADMS 3330 task...
Unformatted text
preview: ADMS 3330 assignment four answer issue 1 (60 marks) a) [15 pts]
The choice variables: X: # of comedy advertisements Y: # of football
commercials The purpose feature is: minimize 50 X + one hundred Y [They
are actually 50K and 100K but does not make a difference.] The
constraints are: 7X + 2Y >= 28 (attain to excessive-profits girls –
advertising and marketing constraint) 2X + 12Y >= 24 (reach to
high-profits men – marketing constrdaint) The system will also be
written as: reduce 50 X + 100 Y field to 7X + 2Y >= 28 2X + 12Y >=
24 X, Y >= 0 As will also be from the graphical solution there are
three severe points: (0,14), (three.6,1.4), and (12,0). The most
excellent answer is (X*,Y*) = (three.6,1.four) for (three.6)(50,000)
+(1.four)(one hundred,000) = 320.000. Y (0, 14) possible vicinity (3.6,
1.four) (12, 0) X 7X + 2Y = 28 Obj. func: 50X + 100 Y = 320 2X + 12Y =
24 b and c) [15 points] To compute the range of goal feature
coefficients, where the existing basis continue to be top-rated, need to
make sure the slope of the goal characteristic continues to be within
the boundaries described with the aid of the. therefore, for part b)
-7/2 <= - c1/100 <= -2/12, which ends up in 3.5 >= c1/a hundred
>= 1/6, and then 350 > = c1 > = sixteen.67. in a similar
fashion for part c) -7/2 < = -50/c2 <= -2/12, which then results
in 300 >= c2 >= 14.28. d and e) [15 pts] These components are not
coated in all sections, and for this reason usually are not graded. each
person may be awarded 15 facets. besides the fact that children, for
the sake of completeness are offered. To compute these stages we need to
examine how the objective solution basis adjustments when the
appropriate-hand-aspect (r.h.s.) of the constraints adjustments. The
subsequent determine indicates it for the constraint 7X + 2Y >= 28,
which suggests that after the r.h.s. of the constraint can change
between 7*0 + 2*2 = 4 t0 7*12 +2*0 = eighty four and the most efficient
basis does not trade. The identical category of analysis finds the 2nd
constraint’s r.h.s. range as 8 and 168. Y (0, 14) possible area (3.6,
1.four) (12, 0) X 7X + 2Y = 28 Obj. func: 50X + one hundred Y = 320 2X +
12Y = 24 f) [ 15 points] To compute the dual costs we should alternate
the r.h.s of constraints for a unit (or any quantity provided that the
greatest answer foundation doesn't change) and the compute the change
within the objective function. because it can be viewed from the next
graphic, if the r.h.s of the constraint changes from 24 to 25, the ideal
foundation continue to be the same, but now the highest quality
solution is (X, Y ) = (3.575, 1.4875), which results in an objective
feature value of 327.5. therefore the twin fee is – 320 – 327.5 = 7.5.
[Shown in the next figure.] If the same is repeated for the different
constraint (i.e. 7X + 2Y = 29), the most fulfilling solution turns into,
(X,Y) = (three.seventy five,1.375), for most efficient goal 325. Then
the dual expense of the primary constraint is 320 – 325 = -5 Y (0, 14)
possible place (three.575, 1.4875) (3.6, 1.four) (12, 0) X 7X + 2Y = 28
Obj. func: 50X + a hundred Y = 320 2X + 12Y = 24 issue 2 (40 marks) a)
[20 points] The primary choice variables are how a lot long term
borrowing to do initially and how a lot brief time period borrowing is
performed in each and every length. [Note that it does not make sense to
make borrowing in any other month other than July. Also it does not
make sense to do short term borrowing in December.] Let, Y: volume of
future borrowing Xi: volume of short time period borrowing performed in
period i, i=1,2,..5 [July…Nov] furthermore, let Ii denote the money
accessible on the conclusion of length I, where I0 = 1,000. The revenues
and the bills in every length are denoted by Ri and Bi. The main goal
here is to pay the entire expenses and debts the usage of revenues and
other borrowing, whereas minimizing the can charge of borrowing. The
leading constraint is that there ought to now not be terrible money
accessible [that is the firm does not go bankrupt]. The objective
function is: minimize Z = 0.06Y + 0.02X1 + 0.02X2 + 0.02X3 + 0.02X4 +
0.02X5 + 0.02X6 There are cash stability constraints for each and every
length. they're: I1 = I0 + Y + X1 + R1 – B1 [Month 1] [Months 2-5] I2 =
I1 + X2 –(1.02)X1 + R2 – B2 I3 = I2 + X3 –(1.02)X2 + R3 – B3 I4 = I3 +
X4 –(1.02)X3 + R4 – B4 I5 = I4 + X5 –(1.02)X4 + R5 – B5 I6 = I5
–(1.02)X5 – (1.06)Y + R6 – B6 [Month 6] The constraints can even be
written by using substituting parameters: I1 = Y + X1 -3,000 I2 = I1 +
X2 – (1.02)X1 -3,000 I3 = I2 + X3 – (1.02)X2 – 4,000 I4 = I3 + X4 –
(1.02)X3 +1,000 I5 = I4 + X5 – (1.02)X4 +four,000 I6 = I5 –(1.02)X5 –
(1.06)Y +15,000 The formula will also be written as: cut Z = 0.06Y +
0.02X1 + 0.02X2 + 0.02X3 + 0.02X4 + 0.02X5 field to: I1 - Y - X1 =
-three,000 I2 - I1 - X2 + (1.02)X1 = - three,000 I3 - I2 - X3 + (1.02)X2
= – four,000 I4- I3 - X4 + (1.02)X3 = 1,000 I5 - I4 - X5 + (1.02)X4 =
four,000 I6 - I5 + (1.02)X5 + (1.06)Y = 15,000 Ii, Xi, Y >=0 b) [20
points] The critical a part of the administration Scientist is given
under. therefore, the top-rated answer is to get a long-time period debt
of 6,000 in July and brief term bills of four,000 and 3,080 within the
months of October and November. the overall financing cost is $501.60.
most efficient solution purpose characteristic cost = 501.600 Variable
-------------I1 I2 I3 I4 I5 I6 Y X1 X2 X3 X4 X5 cost
--------------3000.000 0.000 0.000 0.000 858.400 9498.400 6000.000 0.000
0.000 4000.000 3080.000 0.000 reduced costs -----------------0.000
0.020 0.020 0.020 0.000 0.000 0.000 0.021 0.001 0.000 0.000 0.020
Constraint -------------1 2 three four 5 6 Slack/Surplus
--------------0.000 0.000 0.000 0.000 0.000 0.000 dual fees
------------------0.060 -0.060 -0.040 0.020 0.000 0.000 ...View Full
document
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