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AMAZING INF3708 EXAM PACK 4 recent EXAMS SOLUTIONS

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VERY DETAILED AND EASY TO UNDERSTAND EXPLANATIONS! 2019-06, 2018-10, 2018-06 and 2016-10. Contains exams Solutions. ALL IN ONE!

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  • October 13, 2020
  • 44
  • 2019/2020
  • Exam (elaborations)
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No answers provided for all question 1s since you can just look them up from
the textbook.

2016-10

, Invisibility: With software progress cannot immediately be visible which makes
things difficult.
 Complexity: Software projects contain more complexity than other engineered
artefacts
 Conformity: Software developers have to conform to the requirements of human
clients.
 Flexibility: Software must be flexible to accommodate other components which
makes development difficult. That Software is easy to change is a strength.

, +R60 000 +R70 000 +R165 000

Net Profit:
P1: +R60 000
P2: +R70 000
P3: +R165 000



P1:
AAP(Average annual Profit)
AAP = Net Profit/#years
= 60000/6
= R10 000
ROI = AAP/Total Investment*100
= (10 000) * 100
= 5.13%
P2:
AAP = Net Profit/#years
= 70000/6
= R11666.67
ROI = AAP/Total Investment*100
= (R11666.67/ 160 000) * 100
= 7.29%
P3:
AAP = Net Profit/#years

, = 165000/6
= R27500
ROI = AAP/Total Investment*100
= (27 000) * 100
= 9.32%




BreakEven Year(BEY) is when the investment (amount at year 0) will be equal to the income.
So to get BEY take amount at year 0 (with minus sign) then start adding up the amounts 1 at
a time until you get to a positive value. When you get to positive value then that will be the
year.
Payback period = BEY – (Profit in BEY/Income in BEY)
P1:
BEY: -R195000 + 15000
= -R180000 +30000
= -R150000 + 55000
= -R95000 +50000
= -R45 000 + 55000
= +R5000 Here we arrived at a positive amount when we added year’s 5’s income, thus
BEY= 5 years and the profit is R5000. And the income in year 5 is R55000 since it is the
amount listed.
Thus
BEY = 5
Profit = R5000
Income = R55000
Payback period = BEY – (Profit in BEY/Income in BEY)
= 5 – (5000/55000)
= 4.91 years
P2:
Investment is R160000. Thus go –R160000 on calculator then starting with year 1 add up all
the incomes until you get to positive value. Doing this we get:

,BEY = 6 (in year 6 we got to positive amount)
Profit = R70000 (The positive value we arrived at was R7000 when we finally reached a
positive value)
Income in BEY = R90000 (The amount at BEY)
Payback period = BEY – (Profit in BEY/Income in BEY)
= 6 – (70000/90000)
=5.22 years
P3:
Payback period = BEY – (Profit in BEY/Income in BEY)
= 5 – (50000/110000)
= 4.54 years




Chose one with highest Net Profit
Project 3 since it has the highest net profit.




Chose highest ROI
Project 3 since it has the highest ROI.




Chose the one with the least payback period.
Project 3 since it has the smallest payback period.

,Make table from table 1 and add a column for each project as well as a column for the
discount rate. Multiple the discounted rate with each value. Then add all the new values
which would be NPV. Note: Year 0 will have a discounted rate of 1.

Yea 12% P1 Discounted P2 Discounte P3 Discounte
r Discoun Cash flow d Cash d Cash
t flow flow
0 1 - 1*-200000= - 1*-160000 -295000 1*-295000
195000 --200000 160000 = -160000 =-295000
1 0.8929 +15000 0.8929*1500 +15000 0.8929* +30000 0.8929*
0 =13393.5 15000 30000
=13393.5 =+26787
2 0.7972 +30000 0.7972*3000 +15000 0.7972* +35000 0.7972*
0 =+23916 15000 35000
=+11958 =+27902
3 0.7118 +55000 +39149 +20000 +14236 +50000 +35590
4 0.6355 +50000 +31775 +35000 +22242.5 +12000 +76260
0
5 0.5674 +55000 +31207 +55000 +31207 +11000 +62414
0
6 0.5066 +50000 +25330 +90000 +45594 +11500 +58259
0
NPV -R35229 -R21369 -R7788




Chose project with the highest NPV.
I would select project 3 since it has the highest NPV.

,Steps to calculate:
1. Earliest date = Previous earliest date + Duration (DO this 1 st doing a forward pass
through diagram starting at the start going to the end)
2. Latest Date = Previous Latest Date – duration (DO backward pass starting from end
going to beginning)
3. SLACK = Latest date – Earliest date

, A->C->D->F->H->J
Duration= 117 days



Path 1:
A->C->D->G->I->J
Duration = 114
Path 2:
A->B->E->F->H->J
Duration = 109
Path 3:
A->B->E->G->I->J
Duration = 106

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