UNIVERSITY OF SOUTH AFRICA (UNISA)
College of Engineering, Science and Technology
⋄
Civil Engineering Financial Management
Assignment 02 — 2026
⋄
Module Code: FIM3701
Module Name: Civil Engineering Financial Management
Assignment No.: Assignment 02
Total Marks: 50
Semester: 2026
Submitted in partial fulfilment of the requirements for FIM3701
at the University of South Africa.
,UNISA | FIM3701 Assignment 02 — 2026
Question 1
An international textile company’s North America Division must decide which type of fabric
cutting machines it will use, straight knife or round knife. The estimates are summarised
below. Compare them on the basis of annual worth (AW) values at i = 10% per year using (a)
factors, and (b) single-cell spreadsheet functions.
Table 1: Machine Cost Estimates
Item Round Knife Straight Knife
First cost −$250,000 −$170,000
Annual operating cost −$31,000/year −$35,000/year
(AOC)
Overhaul in Year 2 $0 −$26,000
Salvage value $40,000 $10,000
Life 6 years 4 years
1.1 Round Knife: Annual Worth by Factors
The annual worth equation for the round knife is
AW = −P (A/P, i, n) − AOC + S(A/F, i, n)
with P = $250,000, AOC = $31,000 per year, S = $40,000, n = 6 years and i = 10%.
The capital recovery factor is
0.10(1.10)6
(A/P, 10%, 6) =
(1.10)6 − 1
Since (1.10)6 = 1.771561,
0.10(1.771561) 0.1771561
(A/P, 10%, 6) = = = 0.229607
1.771561 − 1 0.771561
The sinking fund factor is
0.10 0.10
(A/F, 10%, 6) = = = 0.129607
1.771561 − 1 0.771561
Page 1 of 10
,UNISA | FIM3701 Assignment 02 — 2026
The annual capital recovery amount is
250,000(0.229607) = 57,401.75
The annual salvage recovery amount is
40,000(0.129607) = 5,184.28
Substituting these values into the annual worth equation,
AW = −57,401.75 − 31,000 + 5,184.28
AWround knife = −$83,217.47 per year
1.2 Straight Knife: Annual Worth by Factors
The annual worth equation for the straight knife, which carries an additional overhaul cost in
Year 2, is
AW = −P (A/P, i, n) − AOC − Annual Overhaul + S(A/F, i, n)
with P = $170,000, AOC = $35,000 per year, overhaul = $26,000 in Year 2, S = $10,000, n = 4
years and i = 10%.
The capital recovery factor is
0.10(1.10)4
(A/P, 10%, 4) =
(1.10)4 − 1
Since (1.10)4 = 1.4641,
0.14641
(A/P, 10%, 4) = = 0.315471
0.4641
Page 2 of 10
,UNISA | FIM3701 Assignment 02 — 2026
The sinking fund factor is
0.10 0.10
(A/F, 10%, 4) = = = 0.215471
1.4641 − 1 0.4641
The Year 2 overhaul is first discounted to present worth,
26,000 26,000
Poverhaul = 2
= = 21,487.60
(1.10) 1.21
and then converted to an annual worth using the capital recovery factor,
AWoverhaul = 21,487.60(0.315471) = 6,779.03
The annual capital recovery amount is
170,000(0.315471) = 53,630.07
The annual salvage recovery amount is
10,000(0.215471) = 2,154.71
Substituting these values into the annual worth equation,
AW = −53,630.07 − 35,000 − 6,779.03 + 2,154.71
AWstraight knife = −$93,254.39 per year
1.3 Comparison and Selection
The round knife carries an annual worth of −$83,217.47 per year, and the straight knife carries
an annual worth of −$93,254.39 per year. The round knife’s annual worth is less negative,
meaning it has the lower equivalent annual cost.
Page 3 of 10
, UNISA | FIM3701 Assignment 02 — 2026
Select the Round Knife
1.4 Annual Worth by Single-Cell Spreadsheet Functions
Round Knife
The capital recovery and salvage recovery are combined in a single PMT function, with the
operating cost added separately.
=-PMT(10%,6,250000,40000)-31000
−$83,217.47 per year
Straight Knife
The Year 2 overhaul is nested inside a PV function to bring it to Year 0, and this present worth
is then nested inside a second PMT function to annualise it over the 4-year life.
=-PMT(10%,4,170000,10000)-35000-PMT(10%,4,PV(10%,2,0,-26000),0)
−$93,254.39 per year
Table 2: Summary of Annual Worth Results
Machine Annual Worth ($/year)
Round Knife −$83,217.47
Straight Knife −$93,254.39
The Round Knife is the preferred machine, since its annual worth is higher, meaning less
negative, than that of the Straight Knife.
Page 4 of 10
College of Engineering, Science and Technology
⋄
Civil Engineering Financial Management
Assignment 02 — 2026
⋄
Module Code: FIM3701
Module Name: Civil Engineering Financial Management
Assignment No.: Assignment 02
Total Marks: 50
Semester: 2026
Submitted in partial fulfilment of the requirements for FIM3701
at the University of South Africa.
,UNISA | FIM3701 Assignment 02 — 2026
Question 1
An international textile company’s North America Division must decide which type of fabric
cutting machines it will use, straight knife or round knife. The estimates are summarised
below. Compare them on the basis of annual worth (AW) values at i = 10% per year using (a)
factors, and (b) single-cell spreadsheet functions.
Table 1: Machine Cost Estimates
Item Round Knife Straight Knife
First cost −$250,000 −$170,000
Annual operating cost −$31,000/year −$35,000/year
(AOC)
Overhaul in Year 2 $0 −$26,000
Salvage value $40,000 $10,000
Life 6 years 4 years
1.1 Round Knife: Annual Worth by Factors
The annual worth equation for the round knife is
AW = −P (A/P, i, n) − AOC + S(A/F, i, n)
with P = $250,000, AOC = $31,000 per year, S = $40,000, n = 6 years and i = 10%.
The capital recovery factor is
0.10(1.10)6
(A/P, 10%, 6) =
(1.10)6 − 1
Since (1.10)6 = 1.771561,
0.10(1.771561) 0.1771561
(A/P, 10%, 6) = = = 0.229607
1.771561 − 1 0.771561
The sinking fund factor is
0.10 0.10
(A/F, 10%, 6) = = = 0.129607
1.771561 − 1 0.771561
Page 1 of 10
,UNISA | FIM3701 Assignment 02 — 2026
The annual capital recovery amount is
250,000(0.229607) = 57,401.75
The annual salvage recovery amount is
40,000(0.129607) = 5,184.28
Substituting these values into the annual worth equation,
AW = −57,401.75 − 31,000 + 5,184.28
AWround knife = −$83,217.47 per year
1.2 Straight Knife: Annual Worth by Factors
The annual worth equation for the straight knife, which carries an additional overhaul cost in
Year 2, is
AW = −P (A/P, i, n) − AOC − Annual Overhaul + S(A/F, i, n)
with P = $170,000, AOC = $35,000 per year, overhaul = $26,000 in Year 2, S = $10,000, n = 4
years and i = 10%.
The capital recovery factor is
0.10(1.10)4
(A/P, 10%, 4) =
(1.10)4 − 1
Since (1.10)4 = 1.4641,
0.14641
(A/P, 10%, 4) = = 0.315471
0.4641
Page 2 of 10
,UNISA | FIM3701 Assignment 02 — 2026
The sinking fund factor is
0.10 0.10
(A/F, 10%, 4) = = = 0.215471
1.4641 − 1 0.4641
The Year 2 overhaul is first discounted to present worth,
26,000 26,000
Poverhaul = 2
= = 21,487.60
(1.10) 1.21
and then converted to an annual worth using the capital recovery factor,
AWoverhaul = 21,487.60(0.315471) = 6,779.03
The annual capital recovery amount is
170,000(0.315471) = 53,630.07
The annual salvage recovery amount is
10,000(0.215471) = 2,154.71
Substituting these values into the annual worth equation,
AW = −53,630.07 − 35,000 − 6,779.03 + 2,154.71
AWstraight knife = −$93,254.39 per year
1.3 Comparison and Selection
The round knife carries an annual worth of −$83,217.47 per year, and the straight knife carries
an annual worth of −$93,254.39 per year. The round knife’s annual worth is less negative,
meaning it has the lower equivalent annual cost.
Page 3 of 10
, UNISA | FIM3701 Assignment 02 — 2026
Select the Round Knife
1.4 Annual Worth by Single-Cell Spreadsheet Functions
Round Knife
The capital recovery and salvage recovery are combined in a single PMT function, with the
operating cost added separately.
=-PMT(10%,6,250000,40000)-31000
−$83,217.47 per year
Straight Knife
The Year 2 overhaul is nested inside a PV function to bring it to Year 0, and this present worth
is then nested inside a second PMT function to annualise it over the 4-year life.
=-PMT(10%,4,170000,10000)-35000-PMT(10%,4,PV(10%,2,0,-26000),0)
−$93,254.39 per year
Table 2: Summary of Annual Worth Results
Machine Annual Worth ($/year)
Round Knife −$83,217.47
Straight Knife −$93,254.39
The Round Knife is the preferred machine, since its annual worth is higher, meaning less
negative, than that of the Straight Knife.
Page 4 of 10
