8 Pages

b_sample_Mid2_A

Course: MGF 405, Spring 2012
School: SUNY Buffalo
Rating:
 
 
 
 
 

Word Count: 2093

Document Preview

#:________________ SECTION:_____ FIRST STUDENT NAME:________________________ LAST NAME: ______________________ UNIVERSITY AT BUFFALO MGF405 ADVANCED CORPORATE FINANCE Professor: Sahn W. Huh SAMPLE MIDTERM II (TYPE A): Suggested Solution April SECTION I (1.5 point x 10 = 15 points): Choose only one of the 5 choices and clearly circle the letter in each question. No credit will be given based on scratch work in...

Register Now

Unformatted Document Excerpt

Coursehero >> New York >> SUNY Buffalo >> MGF 405

Course Hero has millions of student submitted documents similar to the one
below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.

Course Hero has millions of student submitted documents similar to the one below including study guides, practice problems, reference materials, practice exams, textbook help and tutor support.
#:________________ SECTION:_____ FIRST STUDENT NAME:________________________ LAST NAME: ______________________ UNIVERSITY AT BUFFALO MGF405 ADVANCED CORPORATE FINANCE Professor: Sahn W. Huh SAMPLE MIDTERM II (TYPE A): Suggested Solution April SECTION I (1.5 point x 10 = 15 points): Choose only one of the 5 choices and clearly circle the letter in each question. No credit will be given based on scratch work in or between lines without clear marking. 1. As the degree of sensitivity of a project to a single variable rises, the A. lower the forecasting risk of the project. B. smaller the range of possible outcomes given a pre-defined range of values for the input. C. more attention management should place on accurately forecasting the value of that variable. D. lower the maximum potential value of the project. E. lower the maximum potential loss of the project. Ans: C 2. Which of the following statements is/are correct concerning the accounting break-even point? I. The net income is equal to zero at the accounting break-even point. II. The net present value is equal to zero at the accounting break-even point. III. The quantity sold at the accounting break-even point is equal to the total fixed costs plus depreciation divided by the contribution margin. IV. The quantity sold at the accounting break-even point is equal to the total fixed costs divided by the contribution margin. A. I and III only B. I and IV only C. II and III only D. II and IV only E. I, II, and IV only Ans: A 3. Including the option to expand in your project analysis will tend to A. extend the duration of a project but not affect the project's net present value. B. increase the cash flows of a project but decrease the project's net present value. C. increase the net present value of a project. D. decrease the net present value of a project. E. have no effect on either a project's cash flows or its net present value. Ans: C 1 4. In the financial break-even analysis, the EAC is used to A. determine the opportunity cost of investment. B. allocate depreciation over the life of the project. C. allocate the initial investment over the life of the project. D. determine the contribution margin to fixed costs. E. None of the above. Ans: C 5. The relationship between nominal rates, real rates, and inflation is known as the A. Miller and Modigliani theorem. B. Fisher effect. C. Gordon growth model. D. term structure of interest rates. E. interest rate risk premium. Ans: B 6. Which of the following statements is/are true? (i) If the coupon rate of a bond is higher than the yield to maturity, the bond sells at a discount. (ii) When interest rates increase, the prices of bonds will also increase because investors buying bonds at that time will receive higher yield. A. (i) B. (ii) C. both (i) and (ii) D. none of the above E. insufficient information Ans: D 7. You find that yesterday the bid price of a 20-year Treasury bond was quoted as Bid 135:22 in the WSJ. Assume that the face value of the bond is $1,000. What was the actual bid price of the Treasury bond (in dollars) yesterday? A. B. C. D. E. $135.22 $1,356.88 $1,350.22 $1,352.20 insufficient information Ans: B 8. The total number of shares outstanding in UBCorp is 20,000 shares, 4 directors will be elected, and 7 candidates run for the BOD seats in the cumulative voting. How many shares will guarantee a seat in the board of UBCorp? A. 10,000 shares B. 10,001 shares C. 4,000 shares 2 D. 4,001 shares E. 2,501 shares Ans: D 9. A US firm plans to raise debt capital in Tokyo (Japan) by issuing bonds which will be denominated in the Japanese Yen. These bonds can be referred to as: A. B. C. D. E. Eurobonds Foreign bonds, specifically Samurai bonds Foreign bonds, specifically Yankee bonds Eurobonds, specifically Samurai bonds Eurobonds, specifically Yankee bonds Ans: B 10. The formula for calculating beta is given by the dividing the ___________ of the stock return with the market portfolio return by the ___________ of the market portfolio return. A. variance; covariance B. covariance; variance C. standard deviation; variance D. expected return; variance E. expected return; covariance Ans: B 3 SECTION II (85 points): Whenever necessary, show the equations or expressions leading to the final answer. Note that in some questions no partial credit (NPC) is available. In multiple choice questions, choose only one of the 5 choices and clearly circle the letter. In the final answers, show dollar amounts rounded up to a cent (e.g., $25.38) and other numbers rounded up to 4 decimal places (e.g., 2.3857). 1. (15 points) Your company has developed a new product called M. If M is successful, the PV of the payoff at the time M is brought to the market is $30,000. If M fails, the PV of the payoff is $10,000. If M goes directly to the market, there is a 50% chance of success. Alternatively, you can delay the launch by one year and spend $2,000 now to test-market the product. Testmarketing will improve M next year and thus increase the probability of success to 85%. The discount rate is 12%. You should decide which strategy is better. 1) (6 pts) Strategy A is to directly go to the market now without test-marketing. What is the NPV of this strategy? Sol 1) In the CF diagram, at t=0 you have $30,000 (with a prob. of 0.5) or $10,000 (with a prob. of 0.5), but nothing at t=1. Thus, NPV = 30,000 x (0.5) + 10,000 x (0.5) = $20,000 2) (6 pts) Strategy B is to test-market at t=0, and then go to the market next year. What is the NPV of Strategy B? Sol 2) In the CF diagram, at t=0 you have a cost of $2,000, and at t=1 you have $30,000 (with a prob. of 0.85) or $10,000 (with a prob. of 0.15) Thus, NPV = -2,000 + [30,000 x (0.85) + 10,000 x (0.15)]/(1.12) = $22,107.14 3) (3 pts) Based on the above information, which strategy should be chosen? Sol 3) Strategy B because its NPV is higher. 4 2. (19 points) Consider a coupon bond with a coupon rate of 10%. The par value is $1000 and the bond has 20 years to maturity. The yield to maturity is 8%. 1) (a) (5 pts: NPC) The coupon interest is paid annually. Show the equation for computing the price of the bond. Sol (a) m=1 (=k), CR=0.1, FV=1000, annual coupon=100, n=20, i=0.08. Draw a CF diagram. , or 100 , or . . . = 100 . . . , . . . . . 100 . . (b) (2 pts) What is the price of the bond (P0)?: A) 1,196.36 B) 1,916.36 C) 1,169.36 D) 1,619.36 E) 1,916.63 Sol: A. P0 = 981.81 + 214.55 = 1,196.36 2) The coupon interest is paid annually. The value of this bond at the end of year 1 (at t=1) consists two of components. (a) (1 pts) What is the accrued coupon interest at t=1? Sol 2) (a) At t=1, you will have a coupon interest of 100. (b) (5 pts: NPC) Show the equation to find the PV (evaluated at t=1) of the remaining CFs from the bond (i.e., P1). Sol (b) You will receive coupon interest from t=2 to t=20 as well as the FV=1000 at t=20. So you can evaluate all these CFs at t=1, i.e., get P1. Because one year has passed, note that at t=1 only 19 years are left over (n=19). Thus, use your another diagram, . . . = 100 or = 100 . . . or . . 100 . . . . . (c) (1 pts) What is the value of the bond at t=1 [i.e., V1 = (Coupon Interest at t=1) + P1]? A) 1,293.07 B) 1,292.07 C) 1,929.07 D) 1,392.07 E) 1,192.07 Sol: B. V1 = (coupon interest accrues at t=1) + P1 = 100 + (960.36 + 231.71) = 1,292.07 3) The coupon interest is paid semi-annually. (a) (3 pts: NPC) Show the equation for computing the price of the bond. Sol 1) (a) m=2 (=k), CR=0.1, FV=1000, annual coupon=100, semi-annual coupon = 100/2 = 50, n=20, i=0.08. Draw a CF diagram. , or =50 , or . . . . = 50 . . . . , . . . 50 . . (b) (2 pts) What is the price of the bond (P0)? A) 1,297.93 B) 1,719.93 C) 1,197.93 D) 1,917.93 E) 1,297.39 Sol (b): C. P0 = 989.64 + 208.29 = 1,197.93 5 3. (19 points) BBI has an outstanding 30-year maturity bond with a 9% coupon rate that can be called in one year. The bond makes annual coupon payments. The call premium is set at $130 (including the coupon interest of $90 accrued at t=1) over the par value. The face (par) value of the bond is $1,000. There is a 40% chance that the interest rate in one year will be 13%, and a 60% chance that the interest rate will be 8%. The current interest rate is 10%. To obtain the price of BBIs callable bond, answer the following questions. 1) Suppose that the interest rate increases to 13% in a year. (a) (5 pts: NPC) In this case, what is the probable value of the bond at the end of year 1 (at t=1)? (b) (2 pts) Will BBI call the bond at t=1? Why or why not? (c) (1 pts) Based on the available information, what is the actual value of the bond to the bondholder at t=1? Sol 1) (a) If i=0.1 => i=0.13 at t=1, the probable value of the bond at t=1 will be the coupon interest of 90, plus P1 (the present value of the coupon interest to be received from t=2 to t=30 (29 years left) plus FV, evaluated at t=1). So with CR=0.09 and i=0.13, = (coupon interest at t=1) + P1 = 90 + . = 90 + 90 . . . . . . . = 90 + 90 . . = 90 + 672.31 + 28.89 = $791.20 (b) This is lower than the exercise price of $1,130. So BBI will NOT call the bond. (c) So $791.20 is the actual value of the bond to bondholders at t=1. 2) Suppose that the interest rate decreases to 8% in a year. (a) (5 pts: NPC) In this case, what is the probable value of the bond to the bondholder at the end of year 1 (t=1)? (b) (2 pts) Will BBI call the bond at t=1? Why or why not? (c) (1 pts) Based on the available information, what is the actual value of the bond to the bondholder at t=1? Sol 2) (a) If i=0.11 => i=0.08 at t=1, the probable value at t=1 will be the coupon interest of 90, plus P1 (the present value of the coupon interest to be received from t=2 to t=30 (29 years left) plus FV, evaluated at t=1). So with CR=0.09 and i=0.08, = (coupon interest amount at t=1) + P1 = 90 + . = 90 + 90 . . . = 90 + 90 . . . . + . = 90 + 1,004.26 + 107.33 = $1,201.59 (b) This is higher than the exercise price of $1,130. So the company WILL CALL the bond at $1,130. (the firm will choose to give out the lower and get back the higher!). (c) So $1,130 is the actual value of the bond to bondholders at t=1 (not $1,201.59). 3) (a) (2 pts) Given the information above, what is the expected value of the bond to the bondholder at t=1? (b) (1 pts) Thus, what is the price of the callable bond? Sol 3) (a) V1 = 791.20 with a probability of 0.4 = 1,130 with a probability of 0.6 Thus, the expected value at t=1 is: E(V1) = (791.20)(0.4) + (1,130)(0.6) = 994.48 (b) Because i=0.1 for the first period, the price (P0) of the callable bond is: P0 = 994.48/(1.1) = $904.07 6 4. (15 points) UBC has 2 million shares of common shares outstanding. The stock sells at $25 per share. The firms debt is publicly traded and was recently quoted at 95% of the face value. It has a total face value of $8 million, and its yield to maturity is 11%. The risk-free rate is 6%, and the market risk premium is 7%. The firms beta is 1.8 and the corporate tax rate is 35%. 1) (5 pts) Calculate the weights for UBCs equity and debt based on their market values. Sol 1) E = 2M x $25 = $50M D = $8M x 0.95 = $7.6M So, xE = E/(E+D) = 50M/(50M+7.6M) = 0.8681 Thus, xD = D/(E+D) (or = 1 xE) = 7.6M/(50M+7.6M) = 0.1319 2) (5 pts) What is the cost of equity capital for UBC? Sol 2) RF=0.06, (RM RF) = 0.07, and =1.8. To get the cost of equity capital, use CAPM/SML: RE = RF + (RM RF) = 0.06 + 1.8(0.07) = 0.186 3) (5 pts) Given the information above, what is the weighted average cost of capital (RWACC) for UBC? Sol 3) RD=0.11, and tC=0.35. Lets get RWACC xE RE xD RD (1 tc ) E D RE RD (1 tc ) ED ED = (0.8681)(0.186) + (0.1319)(0.11)(1 0.35) = 0.1709 7 5. (17 points) AB Corp. (ABC) recently issued new securities to finance a project. The project cost $15 million and the firm paid a total of $850,000 in flotation costs. The equity issued had a flotation cost of 7%, whereas the debt issued has a flotation cost of 3%. ABC issued new securities in the same proportion as its target capital structure (i.e., target debt-equity ratio). 1) (4 pts) Based on the information given above, what is the weighted average flotation cost for the project? Sol 1) YNet = $15M, Flotation cost = $0.85M, so YGross = $15M + $0.85M = $15.85M. We can use the relation: YGross (1 fWAVG) = YNet. 15.85M(1 fWAVG) = 15M => (1 fWAVG) = 0.946372. Thus, fWAVG = 0.0536 .(a) 2) (7 pts) Now you like to find ABCs target debt-equity ratio. First, let D/E = u. Then fill out the following part using all the information available so that the right-hand side of the equation includes only one unknown (i.e., u). fWAVG = xEfE + xPfP + xDfD = Sol 2) Let D/E = u. Then D = uE. Also, fE = 0.07 and fD = 0.03 are given. Also xP = fP = 0. Thus, fWAVG = = = 0.07 0.03 0.07 0.03 .(b) 3) (6 pts) Given the information in 1) and 2) above, solve out for the target debt-equity ratio (i.e., u). 0.07 Sol 3) Let (a) = (b): 0.0536 = = . + . 0.03 = . . So u=0.6949 (or more precisely 0.6929 if we use more decimal places in (a) above). 8
Find millions of documents on Course Hero - Study Guides, Lecture Notes, Reference Materials, Practice Exams and more. Course Hero has millions of course specific materials providing students with the best way to expand their education.

Below is a small sample set of documents:

SUNY Buffalo - MGF - 405
Formula Sheet I:SummaryofCashFlowAnalysis:CF(A)CF(B)+CF(S)|OCF (= EBIT + Depr Curr. Tax)|NCS (= dNFA + Depr= FA Purch. FA Sold)|dNWC [= NWCt NWCt-1= (CAt CLt) - (CAt-1 CLt-1)]|Interest|New Borrowing(= LDt LDt-1= LD Issued LD Retired)Forad
SUNY Buffalo - MGF - 405
Formula Sheet II: up to Chap. 15SummaryofCashFlowAnalysis:CF(A)CF(B)+CF(S)|OCF (= EBIT + Depr Curr. Tax)|NCS (= dNFA + Depr= FA Purch. FA Sold)|dNWC [= NWCt NWCt-1= (CAt CLt) - (CAt-1 CLt-1)]|Interest|New Borrowing(= LDt LDt-1= LD Issued L
SUNY Buffalo - MGF - 405
Formula Sheet III: up to Chap. 22SummaryofCashFlowAnalysis:CF(A)CF(B)+CF(S)|OCF (= EBIT + Depr Curr. Tax)|NCS (= dNFA + Depr= FA Purch. FA Sold)|dNWC [= NWCt NWCt-1= (CAt CLt) - (CAt-1 CLt-1)]|Interest|New Borrowing(= LDt LDt-1= LD Issued
Chalmers University of Technology - ACC - 100
Five numbers you need to knowHere are five numbers every business owner should know.1.2.3.4.5.Reconciled cash balanceDays sales outstandingBreak-even pointMarginsYour special industry number1. Reconciled cash balanceYour reconciled cash balan
SUNY Buffalo - MGF - 405
HW1_chap0204: SolutionsChap 02:8.CF(B) = Interest Net new borrowing (=LDt LDt-1 )= $118,000 ($1,390,000 1,340,000)= $68,0009.CF(S) = Dividends Net new equity (= CS Issued incl. Cap Surp Repurch)= Dividends [(CSt + APISt) (CSt-1 + APISt-1)] or= Di
SUNY Buffalo - MGF - 405
HW2_chap06: Solutions1.Year12345Rev.9,500.009,500.009,500.009,500.009,500.00t=0.34Costs4,180.004,180.004,180.004,180.004,180.00i=0.14Depr2,400.002,400.002,400.002,400.002,400.00n=5EBT=Rev - Cost - DeprI/S02,920.002,920.
SUNY Buffalo - MGF - 405
HW3_chap0708: SolutionsChap07:1. Note that there is no salvage value and dNWC. So getting CFs for t=1 to t=8 is quite easy (thesame as OCF, right?). OCF is obtained from a tax-shield approach, although you canmake the usual (full) table to obtain CFs
SUNY Buffalo - MGF - 405
HW4_chap13: Solutions3.We have the information available to calculate the cost of equity using the CAPM and thedividend growth model. Using the CAPM, we find:ECAPM(R) = 0.05 + 0.85(0.08)= 0.1180 or 11.80%And using the dividend discount model, the co
SUNY Buffalo - MGF - 405
HW6_chap1617: SolutionsChap. 16:14. EBIT=$140,000, RD = 0.09, now all-equity, REU = 0.17, tC = 0.35 (with corporatetaxes)!a.VU =?From NI/REU = E (=VU), and NI = EBIT(1 tC)VU = EBIT(1 tC)/REU= $140,000(1 0.35)/0.17= $535,294.12b.If the firm issu
SUNY Buffalo - MGF - 405
HW7_chap1822: SolutionsChap. 18:1.Note that EBTD=140,000, tC=0.35, REU=0.13, and n=5.a.The maximum price that the company should be willing to pay for the fleet ofcars with all-equity funding is the price that makes the NPV of the transactionequal
SUNY Buffalo - MGF - 405
HW5_chap15: Solutions1.1) If the company uses straight voting, the board of directors is elected one at a time.You will need to own one-half of the shares, plus one share, in order to guaranteeenough votes to win the election. So, the number of shares
Vellore Institute of Technology - ECONOMICS - Y9020212
Singlylinkedlist void create() cfw_ int i; node *new,*temp; printf("\n enter the number of datas"); scanf("0",&n); head=(node*)malloc(sizeof(node); temp=head; printf("\n enter the datas:\n"); scanf("0",&head->data); for(i=2;i<=n;i+) cfw_ new=(node*)malloc
Vellore Institute of Technology - ECONOMICS - Y9020212
Binary Tree Declarationtypedef struct bin cfw_ int data; struct bin *left; struct bin *right; node;Creationnew=getnode(); printf("enter the element:"); scanf("0",&new->data); if(root=NULL) root=new; else insert(root,new);Get Nodenode *getnode()cfw_
Vellore Institute of Technology - ECONOMICS - Y9020212
#include<stdio.h> int heapSize; int hs; void print(int a[]) cfw_ int i; for (i = 0; i <hs; i+) cfw_ printf("0-",a[i]); printf("\n"); int left(int i) cfw_ return (2 * i) + 1; int right(int i) cfw_ return (2 * i) + 2; void heapify(int a[], int i) cfw_ i
Vellore Institute of Technology - ECONOMICS - Y9020212
Binary Search Tree Declaration: typedef struct node cfw_ int data; struct node *left, *right; tree; Insertion: tree *insert(int data, tree *temp) cfw_ if(temp=NULL) cfw_ temp=(tree*)malloc(sizeof(tree); temp->data=data; temp->left=NULL; temp->right=NULL;
Vellore Institute of Technology - ECONOMICS - Y9020212
LIST Implementation of data structure can be done with help of programs. To write any program we need an algorithm. Algorithm is nothing but collection of instruction which ahs to be executed in step by step manner. And data structure tells us the way to
Vellore Institute of Technology - ECONOMICS - Y9020212
Representation of stack: A stack is a special case of an ordered list, i.e. it is a ordered list with some restrictions on the way in which we perform various operations on a list. It is therefore quite natural to use sequential representation for impleme
Vellore Institute of Technology - ECONOMICS - Y9020212
Circular link list INSERTION: Beginning: Temp=head; While (temp->next!=head) Temp=temp->next; Temp->next=new; new->next=head; Head=new; Last: Temp=head; While (temp->next!=head) temp=temp->next Temp->next=new; New->next=head; Particular Position: Temp=hea
Vellore Institute of Technology - ECONOMICS - Y9020212
Applications of stack: The place where stacks are frequently used is in expression conversion. An arithmetic expression consist of operands and operators. The operands can be numeric values or numeric variables. The operators used in an arithmetic express
Vellore Institute of Technology - ECONOMICS - Y9020212
Doublylinkedlist INSERTION: AT FIRST: new->next=head; head->prev=new; head=new; AT LAST: last->next=new; new->prev=last; last=new; AT PARTICULAR POSITION: temp=head; for(i=1;i<pos;i+) cfw_ temp1=temp; temp=temp->next; temp1->next=new; new->prev=temp1; ne
Vellore Institute of Technology - ECONOMICS - Y9020212
Insertion Sortfor ( i = 1 ; i <n ; i+ ) cfw_ for ( j = 0 ; j < i ; j+ ) cfw_ if ( a[j] > a[i] ) cfw_ temp = a[j] ; a[j] = a[i] ; a[i]=temp ;
Vellore Institute of Technology - ECONOMICS - Y9020212
#include<stdio.h> int heapSize; int hs; void print(int a[]) cfw_ int i; for (i = 0; i <hs; i+) cfw_ printf("0-",a[i]); printf("\n"); int left(int i) cfw_ return (2 * i) + 1; int right(int i) cfw_ return (2 * i) + 2; void heapify(int a[], int i) cfw_ i
Vellore Institute of Technology - ECONOMICS - Y9020212
Stack using linkedlist PUSH OPERATION : if (top=NULL) top=new; else new->next=top; top=new;POP OPERATION: temp=top; top=top->next; free(temp);DISPLAY: temp=top; while(temp!=NULL) cfw_ printf("0",temp->data); temp=temp->next;
Vellore Institute of Technology - ECONOMICS - Y9020212
Bubble Sort for(i=0;i<n;i+) cfw_ scanf("0",&a[i]); for(i=0;i<n-1;i+) cfw_ for(j=0;j<n-1-i;j+) cfw_ if(a[j]>a[j+1]) cfw_ temp=a[j]; a[j]=a[j+1]; a[j+1]=temp; Selection sort for(g=0;g<=n;g+) cfw_ min_index=g; for(i=g+1;i<=n;i+) cfw_ if(a[i]<a[min_index]
Vellore Institute of Technology - ECONOMICS - Y9020212
Queue: What is queue? The queue can be formally defined as ordered collection of elements that has two ends named as front and rear. From the front one can delete the elements and from the rear end one can insert the elements. For example: The typical exa
Vellore Institute of Technology - ECONOMICS - Y9020212
Shell sort incr=abs(n/2); while(incr>0) cfw_ if(incr%2=0) incr+; for(i=0;i<n-incr;i+) cfw_ for(j=i;j<n;j=j+incr) cfw_ while(a[j+incr]<a[j])&(j+incr<n)&(j>=i) cfw_ temp=a[j+incr]; a[j+incr]=a[j]; a[j]=temp; j=j-incr; incr=abs(incr/2);
Vellore Institute of Technology - ECONOMICS - Y9020212
Types of Queue: Circular queue In case of linear queue the elements get deleted logically. Following fig can show this10 20 30 40 50FrontRearWe have deleted the elements 10,20 and 30 means simply the rear pointer is shifted ahead. We will consider a q
Vellore Institute of Technology - ECONOMICS - Y9020212
void qs(int b[],int LB,int UB) cfw_ int i,j,pivot,temp; int flag=1; if(LB<UB) cfw_ i=LB; j=UB; pivot=b[LB]; while(flag) cfw_ i=i+1; while(b[i]<pivot) cfw_ i+; while(b[j]>pivot) cfw_ j-; if(i<j) cfw_ temp=b[i]; b[i]=b[j]; b[j]=temp; else flag=0; temp=b
Vellore Institute of Technology - ECONOMICS - Y9020212
void qs(int b[],int LB,int UB) cfw_ int i,j,pivot,temp; int flag=1; if(LB<UB) cfw_ i=LB; j=UB; pivot=b[LB]; while(flag) cfw_ i=i+1; while(b[i]<pivot) cfw_ i+; while(b[j]>pivot) cfw_ j-; if(i<j) cfw_ temp=b[i]; b[i]=b[j]; b[j]=temp; else flag=0; temp=b
Vellore Institute of Technology - ECONOMICS - Y9020212
#include<stdio.h> void mergesort(int k[],int,int); void merge(int k[],int,int,int); void mergesort(int k[],int low,int high) cfw_ int mid; if(low<high) cfw_ mid=(low+high)/2; printf("\nCalling MS low & mid with low=0 mid=0",low,mid); mergesort(k,low,mid);
Vellore Institute of Technology - ECONOMICS - Y9020212
Radix sort: int bucket[10][10], buck_count[10]; int i,j,k,r,no_of_passes=0,divisor=1,largest,pass_no; largest=a[0]; for(i=1;i<n;i+) cfw_ if(a[i] > largest)/finding the largest no largest=a[i]; while(largest > 0) cfw_ no_of_passes+;/finding maximum no of
Vellore Institute of Technology - ECONOMICS - Y9020212
Deque The word Deque is a short form of double-ended queue and defines a data structure in which items can be added or deleted at either the front end or rear and, but no changes can be made elsewhere in the list. Thus a deque is a generalization of both
Vellore Institute of Technology - ECONOMICS - Y9020212
Applications of Queue Job scheduling In the operating system various programs are getting executed. We cal these programs as jobs. In this process, some programs are in executing state. The state of these programs is called as `running' state. Some progra
Vellore Institute of Technology - ECONOMICS - Y9020212
TreeDefinition Of tree A tree is a finite set of one or more nodes such that 1. There is a specially designated node called root 2. The remaining nodes are partitioned into n>=0 disjoint sets T1, T2, T3,. Tn Where T1, T2, T3,.Tn are called the sub trees
Vellore Institute of Technology - ECONOMICS - Y9020212
IT 203 Data Structures and Algorithms in C Model Question Paper Max.Marks:100 PART A 8*5=40 Answer all the Questions 1. Explain the necessity of algorithms in data structures and define asymptotic notations. 2. Write a Pseudo code to count no. of nodes in
Vellore Institute of Technology - ECONOMICS - Y9020212
IT 203 Prerequisite ObjectivesData Structures and Algorithms in C CSE 101 L 3T 1P 0C 4To understand pointer concept in C and related library functions To study linear and nonlinear data structures To master sorting and searching techniques and study
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT III Dynamic InitializationByK. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE3/12/2012Inline member functionclass sample cfw_ int x; public: inline void getdata( ); inline voi
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT I Object Oriented Programming - IntroductionBy K. John Singh M.Tech., (Ph.D)., Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE1/5/2012SoftwareSystem software 2. Application software1.Sys
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT IV InheritanceByK. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE3/22/2012Visibility Mode & Code reuse 1. Private 2. Public 3. Protected Software update Software reuse Ex. Micr
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT II Array and PointerBy K. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE2/29/2012Array Array is a collection of identical data objects which are stored inconsecutive memory lo
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT II Introduction to C+By K. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE2/29/2012 C+ is an expanded version of C. Bjarne Stroutstrup first invented the C+ in 1980 atBell Labor
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT III Class and ObjectsByK. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE3/12/2012Class User defined data typeclass class_name cfw_ private data and functions access-specifier
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT II Basic Elements of C+By K. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE2/29/2012Character Set Characters are used to form the words, numbers andexpressions. alphabets from
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT II FunctionBy K. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE2/29/2012Type Conversion1. Converting by assignment operator 2. Using casting methodSyntax: (cast_type) expressi
Vellore Institute of Technology - ECONOMICS - Y9020212
UNIT III Constructor and DestructorByK. John Singh Assistant Professor (Senior) School of Information Technology and Engineering VIT University1VIT - SITE3/12/2012ConstructorA constructor is a special member function for automatic initialization of
Vellore Institute of Technology - ECONOMICS - Y9020212
INTRODUCTION TO OBJECT ORIENTED PROGRAMMINGM. Lawanya Shri, AP SITEDynamic memory allocation New Delete The new operator is used to create a heap memory space for an object of a class. The new keyword calls upon the function operator new() to obtain
Vellore Institute of Technology - ECONOMICS - Y9020212
INTRODUCTION TO OBJECT ORIENTED PROGRAMMINGM. Lawanya Shri, AP SITEProgramming Paradigms Monolithic Programming Procedural Programming Structured Programming Object Oriented ProgrammingMonolithic ProgrammingMonolithic programming indicates the progr
Vellore Institute of Technology - ECONOMICS - Y9020212
INTRODUCTION TO OBJECT ORIENTED PROGRAMMINGM. Lawanya Shri, AP SITETokens A token is the smallest element of a C+ program that is meaningful to the compiler. The C+ parser recognizes these kinds of tokens: identifiers, keywords, literals, operators, p
Vellore Institute of Technology - ECONOMICS - Y9020212
Inheritance Conceptclass Rectanglecfw_PolygonRectangleTriangle;private: int numVertices; float *xCoord, *yCoord; public: void set(float *x, float *y, int nV); float area();class Polygoncfw_ private: int numVertices; float *xCoord, *yCoord; public:
Vellore Institute of Technology - ECONOMICS - Y9020212
Engineering H192 - Computer ProgrammingInheritance and OverloadingLecture 28Winter QuarterThe Ohio State University Gateway Engineering Education CoalitionLect 28Engineering H192 - Computer ProgrammingInheritance Objects are often defined in terms
Vellore Institute of Technology - ECONOMICS - Y9020212
In C+ Inheritance is a mechanism for building class types from other class types defining new class types to be a specialization augmentation of existing typesInheritanceInheritanceSubgroupings with respect to a parent are called Subclass Derived Clas
Vellore Institute of Technology - ECONOMICS - Y9020212
Functions in C+M. Lawanya Shri, AP SITEFunctions in C+ The function is a self contained block of statements which performs a coherent task of a same kind Experience has shown that the best way to develop and maintain large programs is to construct it
Vellore Institute of Technology - ECONOMICS - Y9020212
INHERITANCE Introduction:CPPReusability is important feature of OOP. Instead of creating the same which already exists, reuse something that is already exists. Reusability increases reliability and saves time as well as money. For instance, the reuse of
Vellore Institute of Technology - ECONOMICS - Y9020212
ProgramsFunction overloading#include<iostream.h> #include<stdlib.h> #include<conio.h> #define pi 3.14 class fn cfw_ public: void area(int); /circle void area(int,int); /rectangle void area(float ,int,int); /triangle ; void fn:area(int a) cfw_ cout<"Area
Vellore Institute of Technology - ECONOMICS - Y9020212
# include<iostream.h> # include<conio.h> class employee cfw_ protected : int eno; char name[20]; public: void getdata() cfw_ cin>eno; cin>name; ; class empacademic cfw_ protected: char designation[20]; float salary; public: void getdata1() cfw_ cin>desig
Vellore Institute of Technology - ECONOMICS - Y9020212
Gayatri infotech Introduction:OBJECTS AND CLASSESCPPA class is an extension of the idea of structure used in C, which provide a method for packing together data of different types. The data items are logically related. The standard C does not allow the
Vellore Institute of Technology - ECONOMICS - Y9020212
Constructors and DestructorsConstructorConstructor is a nonstatic special member function which that is automatically called when an instance(object ) of a class is created. The constructor has the same name as the class and it doesn't return any type .
Vellore Institute of Technology - ECONOMICS - Y9020212
CPP Introduction:CONSTRUCTOR AND DESTRUCTORWhile implementing classes, member functions are used to provide initial values to the I private member variables, (e.g. A. input ()All the 'function call' statements to member functions are used with appropri
Vellore Institute of Technology - ECONOMICS - Y9020212
Alderson-Broaddus College - PHYSICS - 9498
SOLUTIONS TO CONCEPTSCHAPTER 11.a) Linear momentumb) Frequency:: mv1= [MLT ]10 0 1= [M L T ]TForce [MLT 2 ]1 2= [ML T ]Area[L2 ]0 0 1a) Angular speed = /t = [M L T ]c) Pressure :2. M0L0 T 2 [M0L0T2]tT22 2c) Torque = F r = [MLT
Alderson-Broaddus College - PHYSICS - 9498
SOLUTIONS TO CONCEPTSCHAPTER 21. As shown in the figure, The angle between A and B = 110 20 = 90 | A | = 3 and | B | = 4m Resultant R = By R20 AA 2 B 2 2AB cos = 5 m Let be the angle between R and A 4 sin 90 1 = tan 1 = tan (4/3) = 53 3 4 cos 90