Friday, July 2, 2010

MGT602 Assignment 2 Solution 1 July 2010

Questions for analysis:



a. What could Tanveer have done to avoid this situation?


Q#1 Answers:

After review of his analysis, In my point of view, after developing and implementing the successful business plan, Tanveer have done to avoid some situations such as: tanveer must have to take the corrective actions. The very first step he must have to take is to register his partnership legally. As he is the financer of his firm, he must have to make ROI proportion to his investment to recover his losses. To monitor his activities regularly he must take daily statement of banking transaction prior to link his company’ Bank account with e-statement facility to get daily transactions alert on his email. By using this facility he can check his company’s inflows and outflows to avoid leakage. Properly depute his trustworthy person to keep the records through Gate Pass & Builty No’s. Also he must have to install the CCTV’s at the premises and regularly check the activities at every start of day.


b. What are his options after the fact?


Q#2 Answers:

After revealing the fact that partab singh has used company resources for his personal use and to avoid further embezzlement, tanveer decided to recover the losses through negotiation. If recovery does not exist then and negotiation fails to recovery the losses then he involve the mediator to recovery. But if this also fails then the last step is he takes the legal action against him.

CS101 Assignment 5 Solution 1 July 2010

Question # 01
  • XYZ is an organization wants to develop the network. Total numbers of computers are 7. Which topology is best suitable for the network? Justify your answer.

Solution:


If we talk about about globally so it comes internet network but in small organization - XYZ with 7 computers. So my point of view that System Bus is concerned to use.

1. A computer’s bus is an electronic pathway that carries information among devices in
a computer & due to this no any server required to pass the data from one computer to another.

2. if 1 computer’s network is down in all 7 so no affect in others computers because It depend the bus width & speed that how much information & how fast information can flow & travel respectively & further that computer receive the message at the same time & take action as well.

So, we can say that Computer communicates with its development cards and strategy through the system bus.



Question # 02

If computer A wants to transmit the data to computer F, then
  • What condition should A satisfy?
  • What protocol should be used?
  • And how the data will transmit?

Solution:


Definitely, if computer transmit the data to another computers so some fact must be take action.

In Condition, information may be communicate respond to the clockwise way.

In Protocol, System Bus build and handle the Computer’s networks based on proprietary protocols.

In Data Transmission, A computer process data Transmit from A to H & one by one data transmit to the next computer & process data take act

MGT201 Assignment # 2 Solution 1 July 2010



Question 01


Po
* = DIV1 / [ (rRF + (rM - rRF ) βA ) - g]
Where
Po*=80
DIV1= 5
g= 7%
rRF = 6%
rM = 10%
βA ) ?

Po* = DIV1 / [ (rRF + (rM - rRF ) βA ) - g]
80 = 5 / [ (6% + (10% - 6% ) BA ) -7%]
Now put the values and get answer which is = βA 1.8125


Question 02.


FORMULA
Bond Price = PV = C1/ (1+rD) + C2/ (1+rD) t2+ C3 / (1+rD) t3 + …….
+C/ (1+rD) n + PAR / (1+rD)n



A)
Bond Price = PV = C1/ (1+rD) + C2/ (1+rD) t2+ C3 / (1+rD) t3 + PAR / (1+rD) n3
Pv = bound value
C= coupon payment Pa = 2000*10.100= 200
rD= required rate of return = 14%=0.14
PAR = par value or face value= 2000
Maturity period
Bound A = t 3 ,n3
Bond Price = PV = C1/ (1+rD) + C2/ (1+rD) t2+ C3 / (1+rD) t3 + PAR / (1+rD) n3
= 200/ (1.14)+200/(1.14)t2 + 200/(1.14)t3+2000/(1.14)n3
= 175.43 + 153.89 +134.99 + 1349.98
= 1814.29 OR 1814
Bond Price = PV = C1/ (1+rD) + C2/ (1+rD) t2+ C3 / (1+rD)t3+C4(1+rD)t4+C5(1+rD)t5+PAR/(1+rD)n5
Pv = bound value
C= coupon payment Pa = 2000*10.100= 200
rD= required rate of return = 14%=0.14
PAR = par value or face value= 2000
Maturity period
Bound B = t 5, n5

B)
Bond Price = PV = C1/ (1+rD) + C2/ (1+rD) t2+ C3 / (1+rD)t3+C4(1+rD)t4+C5(1+rD)t5+PAR/(1+rD)n5
= 200/ (1+.14)+200/(1+.14)t2 + 200/(1+.14)t3+200(1+.14)t4+200/(1+.14)t5+ 2000/(1+.14)n5
= 200/(1.14) + 200/(1.14)t2+200/(1.14)t3+200(1.14)t4+200(1.14)t5(1.14)t5+2000/(1.14)n5
=175.44 + 154 + 135 + 118.42 + 104 + 1039
= 1725

(Part C)
Bond A is perfect decision having the shorter maturity period because of market interest risk diversify the portfolio as it can be spread and chose the bound with shorter duration