UNIVERSITY OF SOUTH AFRICA
College of Engineering, Science and Technology
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FIM3701: Civil Engineering Financial Management
Assignment 02 | 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FIM3701
Module Code:
Civil Engineering Financial Management
Module Name:
Annual Worth, Present Worth and Future
Assignment Topic:
Worth Analysis
02
Assignment Number:
2026
Due Date:
50
Total Marks:
Submitted in partial fulfilment of the requirements for
Civil Engineering Financial Management, UNISA 2026
,UNISA | FIM3701 Financial Management Assignment 02
Question 1
An international textile company’s North America Division must decide which type of fabric
cutting machine 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. (20 marks)
Table 1: Cost and Life Estimates for the Two Machines
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 first cost is P = $250,000, the annual operating cost is $31,000 per year, the salvage value
is $40,000, the life is n = 6 years, and the interest rate is i = 10%.
The annual worth equation is
AW = −P (A/P, i, n) − AOC + S(A/F, i, n)
The capital recovery factor is
0.10(1.10)6
(A/P, 10%, 6) =
(1.10)6 − 1
Since (1.10)6 = 1.771561, this gives
0.1771561
(A/P, 10%, 6) = = 0.229607
0.771561
The sinking fund factor is
0.10 0.10
(A/F, 10%, 6) = = = 0.129607
1.771561 − 1 0.771561
Page 2 of 11
,UNISA | FIM3701 Financial Management Assignment 02
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 into the annual worth equation gives
AW = −57,401.75 − 31,000 + 5,184.28 = −$83,217.47peryear
1.2 Straight Knife: Annual Worth by Factors
The first cost is P = $170,000, the annual operating cost is $35,000 per year, the overhaul cost
in Year 2 is $26,000, the salvage value is $10,000, the life is n = 4 years, and the interest rate is
i = 10%.
The annual worth equation is
AW = −P (A/P ) − AOC − annualoverhaul + S(A/F )
The capital recovery factor is
0.10(1.10)4
(A/P, 10%, 4) =
(1.10)4 − 1
Since (1.10)4 = 1.4641, this gives
0.14641
(A/P, 10%, 4) = = 0.315471
0.4641
The sinking fund factor is
0.10 0.10
(A/F, 10%, 4) = = = 0.215471
1.4641 − 1 0.4641
Page 3 of 11
,UNISA | FIM3701 Financial Management Assignment 02
The overhaul cost is first converted to a present worth at Year 0
26,000 26,000
P = = = 21,487.60
(1.10)2 1.21
and then to an equivalent annual worth
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 all amounts into the annual worth equation gives
AW = −53,630.07 − 35,000 − 6,779.03 + 2,154.71 = −$93,254.39peryear
1.3 Comparison and Selection
Table 2: Annual Worth Comparison
Machine Annual Worth ($/year)
Round Knife −83,217.47
Straight Knife −93,254.39
The Round Knife carries the less negative annual worth, meaning it has the lower equivalent
annual cost of the two alternatives. The North America Division should select the Round
Knife.
1.4 Annual Worth by Single-Cell Spreadsheet Functions
Each alternative can be evaluated directly in a single spreadsheet cell, avoiding the intermedi-
ate factor lookups used above.
Page 4 of 11
, UNISA | FIM3701 Financial Management Assignment 02
Round Knife
=-PMT(10%,6,250000,40000)-31000
This cell returns −$83,217.47 per year, matching the factor solution.
Straight Knife
=-PMT(10%,4,170000,10000)-35000-PMT(10%,4,PV(10%,2,0,-26000),0)
The nested PV(10%,2,0,-26000) term first converts the Year 2 overhaul to a Year 0 value, and
the outer PMT then annualises it over the 4-year life. This cell returns −$93,254.39 per year,
matching the factor solution.
Both methods confirm the Round Knife as the lower-cost alternative.
Page 5 of 11
College of Engineering, Science and Technology
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FIM3701: Civil Engineering Financial Management
Assignment 02 | 2026
⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄ ⋄⋄
FIM3701
Module Code:
Civil Engineering Financial Management
Module Name:
Annual Worth, Present Worth and Future
Assignment Topic:
Worth Analysis
02
Assignment Number:
2026
Due Date:
50
Total Marks:
Submitted in partial fulfilment of the requirements for
Civil Engineering Financial Management, UNISA 2026
,UNISA | FIM3701 Financial Management Assignment 02
Question 1
An international textile company’s North America Division must decide which type of fabric
cutting machine 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. (20 marks)
Table 1: Cost and Life Estimates for the Two Machines
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 first cost is P = $250,000, the annual operating cost is $31,000 per year, the salvage value
is $40,000, the life is n = 6 years, and the interest rate is i = 10%.
The annual worth equation is
AW = −P (A/P, i, n) − AOC + S(A/F, i, n)
The capital recovery factor is
0.10(1.10)6
(A/P, 10%, 6) =
(1.10)6 − 1
Since (1.10)6 = 1.771561, this gives
0.1771561
(A/P, 10%, 6) = = 0.229607
0.771561
The sinking fund factor is
0.10 0.10
(A/F, 10%, 6) = = = 0.129607
1.771561 − 1 0.771561
Page 2 of 11
,UNISA | FIM3701 Financial Management Assignment 02
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 into the annual worth equation gives
AW = −57,401.75 − 31,000 + 5,184.28 = −$83,217.47peryear
1.2 Straight Knife: Annual Worth by Factors
The first cost is P = $170,000, the annual operating cost is $35,000 per year, the overhaul cost
in Year 2 is $26,000, the salvage value is $10,000, the life is n = 4 years, and the interest rate is
i = 10%.
The annual worth equation is
AW = −P (A/P ) − AOC − annualoverhaul + S(A/F )
The capital recovery factor is
0.10(1.10)4
(A/P, 10%, 4) =
(1.10)4 − 1
Since (1.10)4 = 1.4641, this gives
0.14641
(A/P, 10%, 4) = = 0.315471
0.4641
The sinking fund factor is
0.10 0.10
(A/F, 10%, 4) = = = 0.215471
1.4641 − 1 0.4641
Page 3 of 11
,UNISA | FIM3701 Financial Management Assignment 02
The overhaul cost is first converted to a present worth at Year 0
26,000 26,000
P = = = 21,487.60
(1.10)2 1.21
and then to an equivalent annual worth
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 all amounts into the annual worth equation gives
AW = −53,630.07 − 35,000 − 6,779.03 + 2,154.71 = −$93,254.39peryear
1.3 Comparison and Selection
Table 2: Annual Worth Comparison
Machine Annual Worth ($/year)
Round Knife −83,217.47
Straight Knife −93,254.39
The Round Knife carries the less negative annual worth, meaning it has the lower equivalent
annual cost of the two alternatives. The North America Division should select the Round
Knife.
1.4 Annual Worth by Single-Cell Spreadsheet Functions
Each alternative can be evaluated directly in a single spreadsheet cell, avoiding the intermedi-
ate factor lookups used above.
Page 4 of 11
, UNISA | FIM3701 Financial Management Assignment 02
Round Knife
=-PMT(10%,6,250000,40000)-31000
This cell returns −$83,217.47 per year, matching the factor solution.
Straight Knife
=-PMT(10%,4,170000,10000)-35000-PMT(10%,4,PV(10%,2,0,-26000),0)
The nested PV(10%,2,0,-26000) term first converts the Year 2 overhaul to a Year 0 value, and
the outer PMT then annualises it over the 4-year life. This cell returns −$93,254.39 per year,
matching the factor solution.
Both methods confirm the Round Knife as the lower-cost alternative.
Page 5 of 11
