COURSE WORK

Question 1

A new highway has been proposed between Kampala and Entebbe at a construction cost of 700bn shillings. The highway has a 20 year design life. Construction work will require diversion of traffic from the existing motorway which will cost up to 80bn shillings. The project will also displace so many people whose cost of resettlement amounts to 6bn shillings. The new highway will lead to the permanent closure of the existing motorways which will otherwise require 5bn shillings for annual maintenance.

Recoverable and reusable resources from closed motorways and vacated properties are valued at 12bn which can be used on the new highway. The new highway is expected to yield taxes from vehicles using it at 5,000/= per ton per mile and the existing motorways will be closed and a total of 50 annual ton miles will be expected. It is also expected that the general tax revenue will decrease by 30m each year because of the new highway. On the other hand the new highway is estimated to cost 5bn per year to maintain.

Estimate the economic viability of the highway relative to the existing motorways assuming the project will be funded by 12% per year interest bearing bonds with interest compounded annually and the average annual inflation rate being 4%.

Solution to question 1

COSTS:

Construction costs, assume two year construction period and equal distribution of cost over the two years, i.e., Year 0 = 350bn and Year 1 = 350bn

Diversion of the motorway = 80bn in the first year (Year 0) Resettlement = 6bn in Year 0 Maintenance = 5bn per year but begins only after the completion of the construction, say Year 2 Loss of taxes = 0.03bn per year and this loss is realizes as a result of the new highway (possibly

due to traffic diversion from other motorways) assume it starts in Year 2

BENEFITS:

Recovery and reuse of resources = 12bn in the first year (Year 0) Revenue from road tolls = 50 ton miles per year x 5,000/= per ton mile = 0.00025bn per year,

beginning in Year 2 when the road is opened to public. Accruals from the maintenance of closed motorways at 5bn per year, which begins in the 1st year

(year 0)

INTEREST RATE:

Assume the interest rate given (12%) is nominal rate;

Nominal rate, i = 12%; Inflation rate, k = 4%; Real rate, r =?

From

r = [(1+i)/ (1+k)] – 1

r = [(1 + 0.12)/ (1 + 0.04)] – 1 = 0.0769 (or 7.69%)

This is the interest rate to be used for discounting the future values

From the discounting formula:

PV = FV¿¿

Where:

PV = Present Value FV = Future Value r = real interest rate (after adjusting for inflation) t = period in years

The table below shows the Economic Viability assessment using the Net Present Value (NPV) Method, derived from:

NPV =∑t=0

T (B¿¿ t−C t)

(1+r )t¿

Where Bt and Ct are benefits and cost in the t th year respectively.

USING NET PRESENT VALUE (NPV) METHOD

YEAR

COSTS IN BN SHS BENEFITS IN BN SHS

CON

STRU

CTIO

N

DIVE

RSIO

N

RESE

TTLE

MEN

T

MAI

NTE

NAN

CE

LOST

GEN

ERAL

TAX

ES

TOTA

L CO

STS

RECO

VERY

& R

EUSE

OF

RESO

URC

ES

REVE

NU

E FR

OM

TO

LLS

ACC

RUAL

S FR

OM

M

TCE

OF

CLO

SED

MO

TORW

AYS

TOTA

L BE

NEF

ITS

NET

BEN

EFIT

S

NET

PRE

SEN

T VA

LUE

0 350 80 6 0 0 436 12 5 17.00000 -419.00000 -4191 350 0 0 350 5 5.00000 -345.00000 -320.3642 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.025653 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.023824 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.022125 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.020546 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.019077 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.017718 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.016459 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.01527

10 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0141811 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0131712 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0122313 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0113614 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0105415 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0097916 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0090917 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0084418 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.0078419 5 0.03 5.03 0.00025 5 5.00025 -0.02975 -0.00728

NPV = -739.629

NPV < 0 THEREFORE PROJECT NOT VIABLE

The table below shows the Economic Viability assessment using the Benefit Cost Ratio (BCR) Method, derived from:

BCR=∑t=0

T Bt

(1+r)t

∑t=0

T C t

(1+r)t

Where Bt and Ct are benefits and cost in the t th year respectively.

USING BENEFIT COST RATIO (BCR) METHOD

YEAR

COSTS IN BN SHS BENEFITS IN BN SHS

CONS

TRUC

TIO

N

DIVE

RSIO

N

RESE

TTLE

MEN

T

MAI

NTEN

ANCE

LOST

GEN

ERAL

TAX

ES

TOTA

L CO

STS

RECO

VERY

& R

EUSE

OF

RESO

URCE

S

REVE

NUE

FRO

M T

OLL

S

ACC

RUAL

S FR

OM

M

TCE

OF

CLO

SED

MO

TORW

AYS

TOTA

L BE

NEFI

TS

Bt/(1

+r)^

t

Ct/(1

+r)^

t

0 350 80 6 0 0 436 12 5 17.00000 17.000000000 436

1 350 0 0 350 5 5.00000 4.642956635 325.007

2 5 0.03 5.03 0.00025 5 5.00025 4.311624833 4.337278

3 5 0.03 5.03 0.00025 5 5.00025 4.003737425 4.027558

4 5 0.03 5.03 0.00025 5 5.00025 3.717835848 3.739956

5 5 0.03 5.03 0.00025 5 5.00025 3.452350124 3.472891

6 5 0.03 5.03 0.00025 5 5.00025 3.205822383 3.224896

7 5 0.03 5.03 0.00025 5 5.00025 2.976898860 2.994611

8 5 0.03 5.03 0.00025 5 5.00025 2.764322463 2.780769

9 5 0.03 5.03 0.00025 5 5.00025 2.566925864 2.582198

10 5 0.03 5.03 0.00025 5 5.00025 2.383625094 2.397807

11 5 0.03 5.03 0.00025 5 5.00025 2.213413589 2.226583

12 5 0.03 5.03 0.00025 5 5.00025 2.055356662 2.067585

13 5 0.03 5.03 0.00025 5 5.00025 1.908586370 1.919942

14 5 0.03 5.03 0.00025 5 5.00025 1.772296750 1.782841

15 5 0.03 5.03 0.00025 5 5.00025 1.645739391 1.655531

16 5 0.03 5.03 0.00025 5 5.00025 1.528219325 1.537312

17 5 0.03 5.03 0.00025 5 5.00025 1.419091211 1.427534

18 5 0.03 5.03 0.00025 5 5.00025 1.317755790 1.325596

19 5 0.03 5.03 0.00025 5 5.00025 1.223656598 1.230937

66.110215213 805.7388BENEFIT-COST RATIO = 0.08204919 < 1BCR < 1 THEREFORE THE PROJECT IS NOT

VIABLEQuestion 2:

A contractor engaged by MoWT on a road Maintenance project presented the following certified works for payment over a period of 6 months:

Month 1 2 3 4 5 6Amount 20M 30M 43M 35M 35M 40M

If the cost of the work done for each month is 80% of the certified value and the presented request excluded 10% of money retained for liability to be paid 3 months after the 6 month; how much money does the contractor need to mobilize to do the project assuming the certified projects are paid within one month of presentation.

Solution 2:

MonthITEMS 1 2 3 4 5 6 7 8 9Certified works in Million Shs. 20 30 43 35 35 40

Total (including Retention) 22.2222222 33.333 47.7778 38.889 38.889 44.444 0 0 0

Retention 2.22222222 3.3333 4.7778 3.889 3.889 4.444 0 0 0Cost of work 16 24 34.4 28 28 32Presented Request 18 27 38.7 31.5 31.5 36 0 0 0Cash out 16 24 34.4 28 28 32 0 0 0Cash Outflow 16 24 34.4 28 28 32 0 0 0Cash in 18 18 27 38.7 31.5 31.5 0 0 0Retention paid 0 0 0 2.222 3.333 4.778 3.889 3.889 4.444Total Cash inflow 18 18 27 40.922 34.833 36.278 3.889 3.889 4.444