Hey guys so ive been working on these Radix for quite sometime but I keep hitting roadblocks on my todos. Ive implement the first TODO but I cant get the stringRadixSort to work properly as well nor the ChainOfNodes mergeTwoChainsIntoThird() function. Please feel free to rewrite anything you see wrong with it. Ive also provided the Lab Description down below as well for what the output should be and what to do. Thanks again for your guys help! 

1) Implement a radix sort of String objects in a chain of linked nodes, where strings consist only of upper-case characters. This sort will need 27 buckets: 26 for letters A to Z and one bucket for a space.  Each bucket holds a ChainOfNodes object, which will hold appropriate Strings during the sorting. 

2) The ChainOfNodes is a private class inside the SortChain class, its add method adds new elements to the end on the chain.

3) SortChain constructor is already defined and it calls a private method generateList that randomly generates the given number of String objects with number of characters no longer than the given maxStringLength and adds them to the given list. Utilize StringBuffer to build each String.

4) You will create two lists: this.linkedDataOne and this.linkedDataTwo. Each list will be sorted with the stringRadixSort method which you need to implement following the radix sort algorithm.

5) Finally, you will create a third list with mergeTwoChainsIntoThird method. The new list will contain the same data the this.linkedDataOne and this.linkedDataTwo contain. The new list will also be sorted in the ascending order and the original list will remain unchanged. 

See sample runs (with seed 101):
RUN #1
How many nodes in the first chain?
 It should be an integer value greater than or equal to 0.
5
How many characters in the longest string in the first chain?
 It should be an integer value greater than or equal to 1.
6

How many nodes in the second chain?
 It should be an integer value greater than or equal to 0.
7
How many characters in the longest string in the second chain?
 It should be an integer value greater than or equal to 1.
3

The original listOne is: 
CLYUD RWUMOX OVW A CEZ 
The are 5 elements in the list
  The longest string has 6 characters, so the number of needed passes is 6
The original listOne sorted with RadixSort: 
A CEZ CLYUD OVW RWUMOX 
The are 5 elements in the list


The original listTwo is: 
AY NCQ UBT D TF A UAR 
The are 7 elements in the list
  The longest string has 3 characters, so the number of needed passes is 3
The original listTwo sorted with RadixSort: 
A AY D NCQ TF UAR UBT 
The are 7 elements in the list

The mergeChain is:
A A AY CEZ CLYUD D NCQ OVW RWUMOX TF UAR UBT 
The are 12 elements in the list
The listOne sorted should be unchanged: 
A CEZ CLYUD OVW RWUMOX 
The are 5 elements in the list
The listTwo sorted should be unchanged: 
A AY D NCQ TF UAR UBT 
The are 7 elements in the list

Process finished with exit code 0


RUN #2
How many nodes in the first chain?
 It should be an integer value greater than or equal to 0.
0
How many characters in the longest string in the first chain?
 It should be an integer value greater than or equal to 1.
5

How many nodes in the second chain?
 It should be an integer value greater than or equal to 0.
5
How many characters in the longest string in the second chain?
 It should be an integer value greater than or equal to 1.
3

The original listOne is: 
The list is empty
The original listOne sorted with RadixSort: 
The list is empty


The original listTwo is: 
CL UD RWU O S 
The are 5 elements in the list
  The longest string has 3 characters, so the number of needed passes is 3
The original listTwo sorted with RadixSort: 
CL O RWU S UD 
The are 5 elements in the list

The mergeChain is:
CL O RWU S UD 
The are 5 elements in the list
The listOne sorted should be unchanged: 
The list is empty
The listTwo sorted should be unchanged: 
CL O RWU S UD 
The are 5 elements in the list

Process finished with exit code 0

RUN #3
How many nodes in the first chain?
 It should be an integer value greater than or equal to 0.
25
How many characters in the longest string in the first chain?
 It should be an integer value greater than or equal to 1.
4

How many nodes in the second chain?
 It should be an integer value greater than or equal to 0.
4
How many characters in the longest string in the second chain?
 It should be an integer value greater than or equal to 1.
25

The original listOne is: 
CLY D RWU OX OVWM MCE O Y NCQD BTQ WT GA UAR BNOZ KK H U JBAO DF DIN RYBA BMZG GFLU SJX X 
The are 25 elements in the list
  The longest string has 4 characters, so the number of needed passes is 4
The original listOne sorted with RadixSort: 
BMZG BNOZ BTQ CLY D DF DIN GA GFLU H JBAO KK MCE NCQD O OVWM OX RWU RYBA SJX U UAR WT X Y 
The are 25 elements in the list


The original listTwo is: 
OAQTRHPCWYKQJLDVISONZN ORLBRYUSSXFNFQ YSLFLIASALCGMYBNXISCIJQ OPEMSYQVZUMCCVDS 
The are 4 elements in the list
  The longest string has 23 characters, so the number of needed passes is 23
The original listTwo sorted with RadixSort: 
OAQTRHPCWYKQJLDVISONZN OPEMSYQVZUMCCVDS ORLBRYUSSXFNFQ YSLFLIASALCGMYBNXISCIJQ 
The are 4 elements in the list

The mergeChain is:
BMZG BNOZ BTQ CLY D DF DIN GA GFLU H JBAO KK MCE NCQD O OAQTRHPCWYKQJLDVISONZN OPEMSYQVZUMCCVDS ORLBRYUSSXFNFQ OVWM OX RWU RYBA SJX U UAR WT X Y YSLFLIASALCGMYBNXISCIJQ 
The are 29 elements in the list
The listOne sorted should be unchanged: 
BMZG BNOZ BTQ CLY D DF DIN GA GFLU H JBAO KK MCE NCQD O OVWM OX RWU RYBA SJX U UAR WT X Y 
The are 25 elements in the list
The listTwo sorted should be unchanged: 
OAQTRHPCWYKQJLDVISONZN OPEMSYQVZUMCCVDS ORLBRYUSSXFNFQ YSLFLIASALCGMYBNXISCIJQ 
The are 4 elements in the list

Process finished with exit code 0


   

import java.util.InputMismatchException;
import java.util.Random;
import java.util.Scanner;

/**
 * A class Radix Sort tester
 *
 * @author YOUR NAME
 * @version 10/6/2020
 */
public class SortChain>
{
    private ChainOfNodes linkedDataOne;
    private ChainOfNodes linkedDataTwo;
    private Random generator;

    public SortChain(int listSizeOne, int maxStringLengthOne, int listSizeTwo, int maxStringLengthTwo)
    {
        this.generator = new Random(101);
        this.linkedDataOne = generateList(listSizeOne, maxStringLengthOne);
        this.linkedDataTwo = generateList(listSizeTwo, maxStringLengthTwo);
    }

    /**
     * Creates a ChainOfNodes object called list, randomly generates String objects and adds them to the list
     *
     * @param listSize        - number of STring objects to generate
     * @param maxStringLength - the maximum number of characters in the generated String object
     * @return created list
     */
    private ChainOfNodes generateList(int listSize, int maxStringLength)
    {
        final int ASCII_Z = 90;
        final int ASCII_A = 65;
        ChainOfNodes list = new ChainOfNodes<>();
        //TODO Project 4 #1

        for (int i = 0; i < listSize; i++)
        {
            StringBuilder builder = new StringBuilder();
            int length = generator.nextInt (maxStringLength) + 1;

            for (int j = 0; j < length; j++)
            {
                char c = (char) (generator.nextInt ((ASCII_Z - ASCII_A) + 1) + ASCII_A) ;
                builder.append (c);
            }
            String word = builder.toString();
            list.add (word);
        }

        return list;
    }

    public void sortLinkedDataAndDisplayResults()
    {
        System.out.println("\nThe original listOne is: ");
        displayChain(this.linkedDataOne.firstNode);
        stringRadixSort(this.linkedDataOne);
        System.out.println("The original listOne sorted with RadixSort: ");
        displayChain(this.linkedDataOne.firstNode);

        System.out.println("\n\nThe original listTwo is: ");
        displayChain(this.linkedDataTwo.firstNode);
        stringRadixSort(this.linkedDataTwo);
        System.out.println("The original listTwo sorted with RadixSort: ");
        displayChain(this.linkedDataTwo.firstNode);
    }

    /**
     * RADIX SORT
     * sorts String objects in linkedData chain of nodes
     */
    private void stringRadixSort(ChainOfNodes linkedData)
    {
        // TODO Project 4 #2
        Node currentNode = linkedData.firstNode;
        final int NUMBER_OF_UPPER_CASE_LETTERS = 26;

        // do nothing if no more than one node

        // otherwise
        // create buckets

        // traverse the chain to find the longest string

        if (currentNode.next == null)
        {
            return;
        }

        Node[] buckets = new Node[NUMBER_OF_UPPER_CASE_LETTERS];


        String longestString = currentNode.data;
        while (currentNode != null)
        {
            if (currentNode.data.length () > longestString.length ())
            {
                longestString = currentNode.data;
            }
            currentNode = currentNode.next;
        }




        int max = longestString.length (); // THIS IS A STUB
        System.out.println("\u001B[35m\u001B[1m  The longest string has " + max + " characters so the number of needed passes is " + max + "\u001B[0m\n");


        // start soring

        // for each pass:
        // distribute nodes to appropriate buckets
        // clear the original chain and add the nodes back in "buckets" order
        // clear buckets


        for (int i = 0; i < max; i++)
        {

            currentNode = linkedData.firstNode;
            int length = 1;

            while (currentNode.data != null)
            {
                if (currentNode.data.length () > i)
                {
                    int charValue = (int) (currentNode.data.charAt (i)) - 'A';

                }
                currentNode = currentNode.next;
                length++;
            }
            linkedData.clear ();
        }

    }

    public void createSortedChainFromTwoAndDisplayResults()
    {
        ChainOfNodes mergeChain = mergeTwoChainsIntoThird();
        System.out.println("\n\u001B[35m\u001B[1mThe mergeChain is:\u001B[0m");
        displayChain(mergeChain.firstNode);

        System.out.println("The listOne sorted should be unchanged: ");
        displayChain(this.linkedDataOne.firstNode);

        System.out.println("The listTwo sorted should be unchanged: ");
        displayChain(this.linkedDataTwo.firstNode);
    }

    /**
     * Merges two sorted chains: this.linkedDataOne and this.linkedDataTwo into a new one
     * and returns a reference to the new chain.
     * Each chain is in ascending order.
     * The resulting chain should be in ascending order
     * The original chains remain unchanged
     */
    private ChainOfNodes mergeTwoChainsIntoThird()
    {
        ChainOfNodes newChain = new ChainOfNodes<>();
        Node firstCurrent = this.linkedDataOne.firstNode;
        Node secondCurrent = this.linkedDataTwo.firstNode;

        while ((firstCurrent != null) && (secondCurrent != null))
        {
            // TODO Project 4 #3
            // Create new node for merged chain and decide what data belongs in it


        } // end while


        // Exhausted one or both chains; add any remaining data


        return newChain;
    }


    /**
     * Displays all entries in the given chain of nodes.
     */
    private void displayChain(Node currentNode)
    {
        if (currentNode == null)
            System.out.println("The list is empty");
        else
        {
            int count = 0;
            while (currentNode != null)
            {
                System.out.print(currentNode.data + " ");
                count++;
                currentNode = currentNode.next;
            }
            System.out.println("\nThe are " + count + " elements in the list");
        }
    }

    /**
     * Chain of nodes class with add method that adds elements to the end of the list
     */
    private class ChainOfNodes>
    {
        private Node firstNode;       // reference to first node
        private Node tailNode;        // reference to the last node in the chain

        public ChainOfNodes()
        {
            this.firstNode = null;
            this.tailNode = null;
        } // end default constructor

        /**
         * Adds a new entry to the end of the chain.
         *
         * @param newEntry the object to be added as a new entry
         * @return true
         */
        public boolean add(T newEntry) // OutOfMemoryError possible
        {
            // add to the end of the chain:
            Node newNode = new Node<>(newEntry);
            if (this.firstNode == null)
            {
                this.firstNode = newNode;
                this.tailNode = newNode;
            }
            else
            {
                this.tailNode.next = newNode;
                this.tailNode = newNode;
            }
            return true;
        } // end add

        /**
         * Removes all entries from this bag.
         */
        public void clear()
        {
            this.firstNode = null;
            this.tailNode = null;
        } // end clear
    } // end ChainOfNodes

    private class Node
    {
        private S data; // entry in bag
        private Node next; // link to next node

        private Node(S dataPortion)
        {
            this(dataPortion, null);
        } // end constructor

        private Node(S dataPortion, Node nextNode)
        {
            this.data = dataPortion;
            this.next = nextNode;
        }
    } // end Node

    public static void main(String args[])
    {
        int listOneSize = 0;
        int maxStringLengthListOne = 0;
        int listTwoSize = 0;
        int maxStringLengthListTwo = 0;
        boolean invalidInput;

        // get input
        do
        {
            try
            {
                invalidInput = false;
                Scanner keyboard = new Scanner(System.in);
                System.out.println("How many nodes in the first chain?" + "\n It should be an integer value greater than or equal to 0.");
                listOneSize = keyboard.nextInt();

                System.out.println("How many characters in the longest string in the first chain?" + "\n It should be an integer value greater than or equal to 1.");
                maxStringLengthListOne = keyboard.nextInt();

                System.out.println("\nHow many nodes in the second chain?" + "\n It should be an integer value greater than or equal to 0.");
                listTwoSize = keyboard.nextInt();

                System.out.println("How many characters in the longest string in the second chain?" + "\n It should be an integer value greater than or equal to 1.");
                maxStringLengthListTwo = keyboard.nextInt();
            } catch (InputMismatchException ime)
            {
                System.out.println("Could not convert input to an integer");
                invalidInput = true;
            } catch (Exception e)
            {
                System.out.println("There was an error with System.in");
                System.out.println(e.getMessage());
                invalidInput = true;
            }
        } while (invalidInput);

        SortChain tester = new SortChain(listOneSize, maxStringLengthListOne, listTwoSize, maxStringLengthListTwo);
        tester.sortLinkedDataAndDisplayResults();
        tester.createSortedChainFromTwoAndDisplayResults();
    }
} // end RadixSort

Financial analysis of new products at Bay City Electronics had always been rather informal. Bill Roberts, who founded the firm in 1970, knew residential electronics because he had worked for almost seven years for another firm specializing in home security systems. But, he had never been trained in financial analysis. In fact, all he knew was what the bank had asked for every time he went to discuss his line of credit.

Bay City had about 45 full?time employees (plus a seasonal factory work force) and did in the neighborhood of $18 million in sales. His products all related to home security and were sold by his sales manager, who worked with a group of manufacturers' reps, who in turn called on wholesalers, hardware and department store chains, and other large retailer. He did some consumer advertising, but not much.

Bill was inventive, however, and had built the business primarily by coming up with new techniques. His latest device was a remote­-controlled electronic closure for any door in the home. The closure was effected by a special ringing of the telephone: for example, if a user wanted to leave a back door open until 9:00 p.m. it was simple to call the house at 9:00 and wait for 10 rings, after which the electronic device would switch the door to a locked position. A similar call would reopen the door.

The bank liked the idea but wanted Bill to do a better job of financial analysis. Based on his understanding of this market, Bill filled out the FINANCIAL worksheet as appears at the end of the exercise. To date, Bay City had spent $85,000 in expense money for supplies and labor developing the closure and had invested $15,000 in a machine (asset). If the company decided to go ahead, it would have to invest $50,000 more in a new facility, continue R&D to validate and improve the product, and--if things went according to expectations--invest another $45,000 in year 3 to expand production capability.

1) Use the given data to calculate the NPV for the electronic closure product. Do the numbers look good?

2) How is NPV affected if the following contingencies occur? (Assess each of these separately.)

a) Direct manufacturing cost estimate may be overly optimistic, and may never get below the original $16.

b) Competition may force higher marketing costs – what if starting in year 2 the level that must be spent is exactly twice what was forecasted above?

Assumptions:

Question

Calculate Pay Back Period, Net Present Value, Internal Rate of Return and Profitability Index (Benefit Cost Ration) of each of these projects and decide and provide your analysis which project is better. Critically evaluate your decision.

On January 11, 2005, the finance committee of Harding Plastic Molding Company (HPMC) met to consider eight capital budgeting projects. Present at the meeting were Robert L. Harding, President and founder, Susan Jorgensen, comptroller, and Chris Woelk, head of research & development. Over the past five years this committee has met every month to consider and make final judgment on all proposed capital outlays brought up for review during the period.

Harding Plastic Molding Company was founded in 1954 by Robert L. Harding to produce plastic parts and molding for the Detroit automakers. For the first 10 years of operations, HPMC worked solely as a subcontractor for the automakers, but since then has made strong efforts to diversify in an attempt to avoid the cyclical problems faced by the auto industry. By 1970 this diversification attempt had led HPMC into the production of over 1000 different items, including kitchen utensils, camera housings, phonographic and recording equipment. It also led to an increase in sales of 500 percent during 1964 to 1974 prod. As this dramatic increase in sales was paralleled by a corresponding increase in production volume, HPMC was forced, in late 1973, to expand production facilities. This plant and equipment expansion involved capital expenditure of approximately $ 10.5 million and resulted in an increase of production capacity of about 40 percent. Because of this increased production capacity, HPMC has made a concerted effort to attract new business, and consequently, has recently entered into contracts with a large toy firm and a major discount department store chain. While non-automotive related business has grown significantly, it still only represents 32 percent of HPMC’s overall business. Thus, HPMC has continued to solicit non-automotive business, and as a result of this effort and its internal research and development, the firm has four sets of mutually exclusive projects to consider at this month’s finance committee meeting.

Over the past 10 years, HPMC’s capital budgeting approach has evolved into a somewhat elaborate procedure in which new proposals are categorized into three areas – profit, research and development, and safety. Projects falling into the profit or research and development area are evaluated by using present value techniques. Assuming a 10% opportunity cost, those falling into the safety classification are evaluated in a more subjective framework. Although research and development projects have to receive favorable results from the present value criteria, there is also a total dollar limit assigned to projects of this category, typically running about $ 750,000 per year. This limitation was imposed by Harding primarily because of the limited availability of quality researchers in the plastics industry. Harding felt that if more funds than this were allocated, “We simply couldn’t find the manpower to administer them properly”. The benefits derived from safety projects, on the other hand, are not in terms of cash flows; hence, present value methods are not used at all in the evaluation. The subjective approach used to evaluate safety projects is a result of the pragmatically difficult task of quantifying the benefits from these projects into dollar terms. Thus, these projects are subjectively evaluated by a management worker committee with a limited budget. All eight projects to be evaluated in January are classified as profit projects.

The first set of projects listed on the meeting’s agenda for examination involves utilization of HPMC’s precision equipment. Project A calls for production of vacuum containers for thermos bottles produced for a large discount hardware chain. The containers would be manufactured in five different size and colour combinations. This project would be carried out over a three-year period. Project B involves manufacture of inexpensive photographic equipment for a national photography outlet. Although HPMC currently has excess plant capacity, both of these projects would utilize precision equipment of which the excess capacity is limited. Thus adopting either project would tie up all precision facilities. In addition, the purchase of new equipment would be both prohibitively expensive and involve a time delay of about two years, thus making these projects mutually exclusive. (The cash flows associated with these two projects are given in exhibit-1)

CASE STUDY: Excavation Buckets Design and Manufacture

Peter Border is a qualified mechanical engineer who graduated from the QTech University two years ago. Peter works for Trueblood, a small mechanical design and manufacturing company. Owner and founder of the company is William Trueblood.

William qualified as a mechanical tradesman and saw the opportunity to build a business based on designing and manufacturing complex parts for large earthmoving equipment. The business was founded 35 years ago and today employs 55 people. Trueblood Enterprises currently has three professional engineers, Rohan Petronis (25 years of experience), Claude Weatherly (15 years of experience), and Peter. Claude is in charge of the manufacturing area while Rohan and Peter comprise the design and analysis division.

Two months ago, Trueblood Enterprises were contracted by Cranbrook Excavators to design and manufacture an excavation buckets for a range of large excavators and draglines that the company manufactures. Cranbrook Excavators is a large company with total worldwide sales of about $2 billion (Australian). Trueblood Enterprises was elated to gain the contract as they had been trying for several years to secure a contract with Cranbrook Excavators. It is hoped that this initial contract will lead to further large contracts between the two companies.

Design of the excavation buckets was undertaken by Rohan and Peter. The designed part was extremely difficult to analyse and eventually they adopted a design which they considered was adequate and safe, but with which they were not entirely happy. The design was done manually without modern 3D modelling and simulation tools. They would have liked to have had more time to carry out further analysis work, but the production area needed to get the parts into production in order to meet the timelines associated with the contract. The first batch of parts (10) has now been manufactured and delivered and Cranbrook Enterprises has expressed their pleasure at the way in which the contract has been fulfilled to date. The contract calls for the manufacture of a further 100 parts over the next 18 months.

The contract price for the parts is $22 000 each, and Trueblood Enterprises currently estimates that the total cost of design and manufacture will be $18 500 each.

Although busy with other work since the finalisation of the design for the excavation buckets, Peter has continued to ponder how the analysis of the part could be improved. Last night he had a sudden flash of inspiration and two hours’ calculation this morning has provided a much improved understanding of the stress distribution which is likely to occur in the bucket design. On reviewing the new analysis, Peter becomes concerned that the existing design may create the possibility of fatigue failure in the longer term. Further analysis leads him to the conclusion that the premature failure of the existing units is a distinct possibility, although failure is unlikely to occur until 15,000 hours, though this needs to be further validated. The original contract specification asked for a minimum fatigue life of 20,000 hours. Peter also does a quick estimate of the likely cost of using an improved design in manufacturing and estimates that the cost per part will rise to $20 500.

Peter discusses his findings with Rohan. Initially Rohan is reluctant to take any action whatsoever, as he considers it would reflect poorly on the design and analysis division, and particularly on his inherent leadership of that area based on his extended years of experience. When Peter presses the issue and threatens to go directly to William Trueblood, Rohan agrees to set up a meeting between William, Peter and himself.

At the meeting, Peter presents his findings and recommends that the new design be adopted for production, and that the parts already manufactured and supplied be recalled from Cranbrook Excavators. Predictably, William Trueblood gets very upset and irate. He asks if the parts that have already been supplied are in danger of imminent failure and Rohan says no. William Trueblood states that his decision is that the current parts will not be recalled and the production process will continue to manufacture the existing design and not the new design. He says that the existing part is "safe enough" and the company cannot afford to increase the cost of production. He also says that he is extremely disappointed with the performance of Rohan and Peter, and that the design and analysis division needs to "get its act together or the company will have to consider closing this division and outsourcing its design work". He also says that if Rohan or Peter so much as blink an eyelid out of place in the future they will be sacked from the company!

- Identify and discuss the management, contractual and ethical issues involved in this case. What courses of action would be appropriate for Peter to follow (starting immediately)?

- The answer should be no more than 3000 words. This is merely a guide and there is no penalty associated with this word count. The final section of the main body of the report should clearly identify the courses of action that Peter should follow. This section will be a major section of the report on which technical content will be judged. The conclusions reached and action recommended, however, will need to be supported by the arguments presented in the previous sections of the report. This final section should be between 200 and 250 words in length.

- Your report should have a formal format with title page, executive summary, contents page and references. The report should be word processed

A very large study showed that aspirin reduced the rate of first heart attacks by 35% .A pharmaceutical company thinks it has a drug that will be more effective than​ aspirin, and plans to do a randomized clinical trial to test the new drug. Complete parts a through d below.

​a) What is the null hypothesis the company will​ use?

A.The new drug is not more effective than aspirin.

B.The new drug is more effective than aspirin.

C.The new drug is as effective as aspirin.

D.The new drug is less effective than aspirin.

​b) What is their alternative​ hypothesis?

A.The new drug is more effective than aspirin.

B.The new drug is less effective than aspirin.

C.The new drug is as effective as aspirin.

D.The new drug is not more effective than aspirin.

​c) The company conducted the study and found that the group using the new drug had somewhat fewer heart attacks than those in the aspirin group. The​ P-value from the hypothesis test was 0.46.

What do you​ conclude?

A.There is not sufficientis not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

B.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

C.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

D.There is not sufficientis not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

​d) What would you have concluded if the​ P-value had been 0.004​?

A.There is not sufficientis not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

B.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

C.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

D.There is not sufficientis not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

7. Capstone, Inc. (Chapter 12)

Capstone, Inc. puts much emphasis on cash flow when it plans for capital investments. The company chose its discount rate of 8% based on the rate of return it must pay its owners and creditors. Using that rate, Capstone, Inc. then uses different methods to determine the best decisions for making capital outlays.

In 2020 Capstone, Inc. is considering buying five new backhoes to replace the backhoes it now has. The new backhoes are faster, cost less to run, provide for more accurate trench digging, have comfort features for the operators, and have 1-year maintenance agreements to go with them. The old backhoes are working just fine, but they do require considerable maintenance. The backhoe operators are very familiar with the old backhoes and would need to learn some new skills to use the new backhoes.

The following information is available to use in deciding whether to purchase the new backhoes.

Old Backhoes          New Backhoes

Purchase cost when new                                             $90,000                     $225,000

Salvage value now $42,000

Investment in major overhaul needed in next year    $55,000

Salvage value in 8 years                                             $15,000                       $65,000

Remaining life                                                              8 years                      8 years

Net cash flow generated each year                            $30,425                       $43,900

Instructions

1. Evaluate in the following ways whether to purchase the new equipment or overhaul the old equipment. (Hint: For the old machine, the initial investment is the cost of the overhaul. For the new machine, subtract the salvage value of the old machine to determine the initial cost of the investment.)
   1. Using the net present value method for buying new or keeping the old.
   2. Using the payback method for each choice. (Hint: For the old machine, evaluate the payback of an overhaul.)
   3. Comparing the profitability index for each choice.
   4. Comparing the internal rate of return for each choice to the required 8% discount rate.
2. Are there any intangible benefits or negatives that would influence this decision?
3. What decision would you make and why?

Philadelphia Fastener Corporation manufactures nails, screws, bolts, and other fasteners. Management is considering a proposal to acquire new material-handling equipment. The new equipment has the same capacity as the current equipment but will provide operating efficiencies in labor and power usage. The savings in operating costs are estimated at $150,000 annually.

The new equipment will cost $300,000 and will be purchased at the beginning of the year when the project is started. The equipment dealer is certain that the equipment will be operational during the second quarter of the year it is installed. Therefore, 60 percent of the estimated annual savings can be obtained in the first year. The company will incur a one-time expense of $30,000 to transfer production activities from the old equipment to the new equipment. No loss of sales will occur, however, because the processing facility is large enough to install the new equipment without interfering with the operations of the current equipment. The equipment is in the MACRS 7-year property class. The firm would depreciate the machinery in accordance with the MACRS depreciation schedule.

The current equipment has been fully depreciated. Management has reviewed its condition and has concluded that it can be used an additional eight years. The company would receive $10,000, net of removal costs, if it elected to buy the new equipment and dispose of its current equipment at this time. The new equipment will have no salvage value at the end of its life. The company is subject to a 40 percent income-tax rate and requires an after-tax return of at least 12 percent on any investment.

Use Appendix A and Exhibit 16-9 for your reference. (Use appropriate factor(s) from the tables provided.)

Required:

1. 1.Calculate the annual incremental after-tax cash flows for Philadelphia Fastener Corporation’s proposal to acquire the new equipment.

2. 2-a. Calculate the net present value of the proposal to acquire the new equipment using the cash flows calculated in requirement (1), Assume all cash flows take place at the end of the year.

3. 2-b. Should management purchase the new equipment?

The Allied Group intends to expand the company's operation by making significant investments in several opportunities available to the group. Accordingly, the group has identified a need for additional financing in preferred and new common stock and new bond issues. The (Krf) risk-free rate for the company is 7%, and the appropriate tax rate is 40%. Also, the beta coefficient for the company is 1.3 and the market risk premium (Km) is 12%. New Debt (Kd) The company has been advised that new bonds can be sold on the market at par ($1000) with an annual coupon of 8%, for 30 years. New Common Stock Market analysis has determined that given the positive history of the firm, new common stock can be sold at $29 per share, with the last dividend being paid of $2.25 per share. The growth rate on any new delete the words highlighted in yellow common stock has been estimated at a constant rate of 15% per year for the next 3 years. Preferred Stock New Preferred Stock can be issued with an annual dividend of 10% of par and is paid annually and currently would sell for $90 per share. Tasks: Using the Capital Asset Pricing Model (CAPM), discuss and calculate the cost of new common stock (Ks). What would the dividend yield as a percentage (i.e., per dividend payment divided by the book value of a share of stock) today and a year from now if the dividend growth rate is 12%? What is the after-tax cost as a percentage (e.g., interest rate) of new debt today? What are your recommendations for raising capital based on your answers to the above questions plus considering other factors (e.g., current and potential changes in the economy locally, regionally, nationally and worldwide, changes in the demand and/or supply plus cost of materials, skilled labor, management and/or leadership, changes in interest, tax, inflation and/or supply of investment capital)?

A very large study showed that aspirin reduced the rate of first heart attacks by 45%. A pharmaceutical company thinks it has a drug that will be more effective than aspirin, and plans to do a randomized clinical trial to test the new drug. Complete parts a through d below.

a) What is the null hypothesis the company will use?

A. The new drug is not more effective than aspirin.

B. The new drug is more effective than aspirin.

C. The new drug is less effective than aspirin.

D. The new drug is as effective as aspirin.

b) What is their alternative hypothesis?

A. The new drug is as effective as aspirin.

B. The new drug is not more effective than aspirin.

C. The new drug is less effective than aspirin.

D. The new drug is more effective than aspirin.

c) The company conducted the study and found that the group using the new drug had somewhat fewer heart attacks than those in the aspirin group. The P-value from the hypothesis test was 0.002. What do you conclude?

A. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

B. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

C.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

D. There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

d) What would you have concluded if the P-value had been 0.42?

A.There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so small.

B. There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

C. There is evidence to conclude that the alternative hypothesis is true because the​ P-value was so large.

D.There is not sufficient evidence to conclude that the alternative hypothesis is true because the​ P-value was so

small.