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000-858 - i5 iSeries WebSphere Technical Solutions V5R3 - Dump Information

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Exam Code : 000-858
Exam Name : i5 iSeries WebSphere Technical Solutions V5R3
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Updated On : February 19, 2018
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000-858 Questions and Answers



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?

  1. Create iSeries Access for Web Database requests.

  2. Generate a web application with the WDSc for iSeries SQL wizard.

  3. Build a 5250 application and web-enable it via the IBM WebFacing Tool.

  4. Use IBM Query for iSeries and web-enable it with the IBM WebFacing Tool.

Answer: A


A customer wants to utilize caching to minimize the overhead of serving common servlet output from WAS. How do they implement this?

  1. Using cachespec.xml files in the Web application

  2. Using the Fast Cache Response Accelerator

  3. Using Java Servlet Version 2.3 filter code

  4. Using the WDSc deployment descriptor editor

Answer: A


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?

  1. JDBC datasource using the IBM Toolbox for Java

  2. JDBC datasource using the iSeries Native JDBC provider

  3. JDBC datasource using the Enterprise JavaBean query language

  4. JDBC datasource using the IBM Toolbox for Java record level access provider

Answer: B


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?

  1. iDoctor for iSeries

  2. Tivoli Performance Viewer

  3. OS/400 Collection Services

  4. WDSc Profiling Perspective

Answer: D


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?

  1. WAS (Base)

  2. WAS - Express for iSeries

  3. WAS Single Server Edition

  4. WAS Network Deployment Edition

Answer: B


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?

  1. WSCP scripting

  2. WSADMIN scripting

  3. CONFIGWASSVR shell script

  4. Administrative console command macros

Answer: B


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?

  1. PM/400

  2. iSeries Job Watcher

  3. Patrol for iSeries - Predict

  4. iSeries Performance Explorer

Answer: C


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?

  1. CODE Program Generator produces more efficient compiled code.

  2. CODE Designer and Editor can work disconnected from the host.

  3. CODE Designer is the only tool that can specify screen colors.

  4. CODE Editor has a text find and replace feature that complements SEU.

Answer: B


Which of the following communication protocols allows WebSphere Application Server

version 4.0 and later releases to communicate with a Web server?

  1. IIOP

  2. HTTP and HTTPS

  3. Remote OSE Transport

  4. SOAP

Answer: B


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?

  1. Rewrite the RPG business logic as callable ILE Service Programs.

  2. Utilize the IBM WebFacing Tool.

  3. Import the RPG business logic into VisualAge RPG and rewrite the application.

  4. Use the XML Toolkit for iSeries product to Web-enable the RPG business logic.

Answer: A


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?

  1. eServer i5 520 Standard Edition

  2. eServer i5 520 Enterprise Edition

  3. eServer i5 520 Value Edition

  4. eServer i5 550 Solution Edition

Answer: A


WAS-based applications running on OS/400 differ from traditional RPG and COBOL

based applications in which of the following ways?

  1. Traditional applications use more memory.

  2. WAS applications use less operating system services.

  3. WAS applications require more system input/output services.

  4. Traditional applications always provide the maximum performance.

Answer: B


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?

  1. WAS-based applications can utilize standard SQL syntax.

  2. WAS-based applications can call iSeries program objects.

  3. WAS-based applications can use database connection pooling.

  4. 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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    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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