MPV_APPE 2018, block 6
Deriving price from the demand curves – problem for practice
1. Prague wants to construct a new tunnel Bianca, which will improve the traffic connection
between two districts. Ne tunnel will decrease the costs per trip between the districts by 50
CZK for saved fuel, by 150 CZK for saved time, and moreover drivers will be much less
stressed (without appraisal). On the other hand the drivers would pay a 20 CZK toll per each
trip. Assume that the fee would become revenue in the municipal budget and the drivers
would only use the tunnel. Currently there are 2 million trips between the districts per year,
and with the tunnel it is estimated to increase up to 2.6 million. Calculate the total benefit of
building this new tunnel for the city and for the drivers?
(Solution: annual benefit for the drivers is 414 mil. CZK, revenue for Prague is 52 mil. CZK)
WTP – problem for practice
1. Small community in town is deciding between keeping the current playground for children,
and building new parking lots. Help them decide if social benefits of the playground have
been evaluated by experts as 7.5 million CZK (a big one). In case of new parking lot, based on
the questionnaire the demand for the first 30 parking lots is 𝑑1 = 1000 ∗ (200 − 4𝑞), for
the rest 𝑑2 = 1000 ∗ (110 − 𝑞). Calculate social benefits of the new parking lot using WTP
method (do not consider investment costs).
(Solution: Area under the pseudo-demand curve for parking lots is only 7.4 million CZK)
Hedonic method – problem for practice
1. You are supposed to calculate (using hedonic method) the value of negative impact of
building a new road on the houses along it. There are 50 houses in total. Price of the house is
estimated as:
𝑝𝑖 = 𝛼 + 𝛽(𝑟𝑜𝑜𝑚) + 𝛾(𝑛𝑜𝑖𝑠𝑒) + 𝛿(𝑥𝑖) + 𝜀𝑖
where 𝑝𝑖 means estimates price of the house, 𝑟𝑜𝑜𝑚 amount of rooms in the house, 𝑛𝑜𝑖𝑠𝑒
level of noise from the road, and 𝑥𝑖 some other house characteristics. Coefficients have
values of 𝛼 = 80, 𝛽 = 30, 𝛾 = −90 a 𝛿 = 1. New road will increase noise level from
originally 0.1 to 0.6.
(Solution: Value of the houses will decrease in total by 50 ∗ (−45) = −2250)
TCM – problem for practice
1. Somewhere in deep woods there is an old ruin of the fortress. Local authority is deciding,
whether to level it down and build there a wind power plant that would bring revenue of 10
million CZK per year. Based on the data below, using TCM estimate social benefits of the
fortress and decide whether to keep the ruin or build a plant. Assume discontinuous changes
in demand for visits and no visit from additional zones. Also assume the visit are one purpose
only, there is nothing of interest anywhere around.
MPV_APPE 2018, block 6
Zone Distance Population
Probability of visit
per year
Total costs per visit
1 10 10 000 15% 50 CZK
2 30 100 000 7% 170 CZK
3 70 300 000 3% 400 CZK
4 150 1 000 000 0.5% 1 000 CZK
(Solution: Economic benefits of the ruin are 9 865 000 CZK per year, what is less than
revenues from wind power plant – authority should thus build a new power plant)
Dominated and non-dominated variants – problems for practice
1. Choose which variants are non-dominated and dominated (and by which ones).
k1 (min) k2 (max) k3 (max)
Variant 1 50 54 24
Variant 2 28 72 39
Variant 3 21 77 51
(Solution: Non-dominated variant is 3; var 1 is dominated by 2 and 3; var 2 is dominated by 3)
2. Choose which variants are non-dominated and dominated (and by which ones). Determine
ideal and basal variant and full solution.
*hint: you can try to use Excel add-in SANNA from prof. Jablonský from PSE
k1 (min) k2 (max) k3 (min) k4 (max) k5 (min)
Variant 1 48 64 84 64 18
Variant 2 24 82 6 105 15
Variant 3 26 88 146 101 7
Variant 4 33 67 22 56 20
Variant 5 47 60 126 70 18
Variant 6 28 88 166 75 19
(Solution: Non-dominated variants are 2 and 3 (full solution); variants 1, 4, 5 are domin by 2,
variant 6 is domin by 3; basal variant has values 48, 60, 166, 56, 20; ideal 24, 88, 6, 105, 7)
Transformation of minimizing criteria to maximizing, normalizing – problems for practice
1. Transform the following criteria to maximizing.
k1 (min) k2 (max) k3 (min) k4 (max) k5 (min)
Variant 1 48 64 84 64 18
Variant 2 24 82 6 105 15
Variant 3 26 88 146 101 7
Variant 4 33 67 22 56 20
Variant 5 47 60 126 70 18
Variant 6 28 88 166 75 19
MPV_APPE 2018, block 6
(Solution: for transforming min criterion values to max values use y(max) = B(min) – y(min))
T-k1 (min) k2 (max) T-k3 (min) k4 (max) T-k5 (min)
Variant 1 0 64 82 64 2
Variant 2 24 82 160 105 5
Variant 3 22 88 20 101 13
Variant 4 15 67 144 56 0
Variant 5 1 60 40 70 2
Variant 6 20 88 0 75 1
2. Transform the following matrix of parameters to the normalized values.
k1 (min) k2 (min) k3 (max)
Variant 1 50 54 24
Variant 2 28 72 39
Variant 3 21 77 51
(Solution: we transform max criterion values using (y-B)/(I-B), and min using (B-y)/(B-I))
T-k1 (min) T-k2 (min) T-k3 (max)
Variant 1 0 1 0
Variant 2 0.76 0.22 0.56
Variant 3 1 0 1
WSA – problems for practice
1. We have 5 evaluation criteria, that were assigned points based on their importance: k1) 3, k2)
6, k3) 7, k4) 1, k5) 5. Calculate weights of these criteria (for possible further calculations).
(Solution: individual weights are calculated as the ratio of kn from Σ(k1 ...kn), thus
k1) 0.136; k2) 0.273; k3) 0.318; k4) 0.045; k5) 0.227; what sums up as 1.000)
2. An investor has decided to build a factory and chooses between 4 alternatives. Individual
parameters and weights are in the table. Use WSA for evaluation of the variants.
Investment
costs
Running
costs
Production
of item 1
Production
of item 2
Production
of item 3
Variant 1 58 9.7 58 58 67
Variant 2 55 5.4 59 69 121
Variant 3 54 9.2 63 50 31
Variant 4 69 11.8 43 90 190
weights 8% 12% 15% 22% 43%
(Solution: 1) 35.2%; 2) 66.3%; 3) 27.9%; 4) 65.0%)
3. An investor has again decided to build a factory and chooses between 3 alternatives.
Individual parameters and weights are in the table. Use WSA for evaluation of the variants.
Investment
costs
Running
costs
Production
of item 1
Production
of item 2
Production
of item 3
Negatíva
voči okoliu
Variant 1 46 5.9 68 57 122 5
Variant 2 31 8.4 61 92 81 6
Variant 3 55 10.3 88 111 144 14
weights 5 4 8 11 14 12
(Solution: 1) 53.8%; 2) 45.4%; 3) 61.1%)
MPV_APPE 2018, block 6
Lexicographic method – problem for practice
1. A family wants to buy a new car. They have preliminary chosen 5 models. Use lexicographic
method, which models should be considered for further decision.
Price Trunk size Power Fuel cons. Safety Looks
A 488 430 105 6.9 10 6
B 416 401 102 5.1 8 8
C 694 555 108 7.2 10 10
D 449 439 93 6.2 6 7
E 580 445 108 5.6 10 5
Criterion
preference
1. 3. 5. 4. 2. 6.
Limit ≤600 ≥400 ≥100 ≤7,0 ≥8 ≥6
(Solution: A family would consider models A, B for further decision making)
Public procurement evaluation – problems for practice
1. Municipality wants to purchase a new lift as a public procurement. Rank the bids.
Price
(mil. CZK)
Delivery
(weeks)
Guarantee
(months)
Response to
problem (hours)
Aesthetics
(0-5 points scale)
OTIS 1.95 9 59 4 4
KONE 1.51 12 55 7 1
SCHINDLER 1.66 11 65 5 4
Weights 67 15 20 29 18
(Solution: SCHINDLER 87.8%; OTIS 86.2%; KONE 77.4%)
2. Rank bind in a public procurement evaluation. How can you change k2 level to make the last
bid become the best? And how can you change k3 level in case of the second best bid?
k1 (min) k2 (max) k3 (min)
Bid 1 22 17 22
Bid 2 8 39 25
Bid 3 22 40 20
Bid 4 14 29 20
Weights 50% 15% 35%
(Solution: ranks of bids – 1) 56.4%; 2) 92.6%; 3) 68.2%; 74.4%;
changing parameter k2 cannot make the worst offer becoming the best;
to make 4) better than 2) by changing k3 the 4) must lower the acquired 28% of 2) in k3 to less
than 9.82% – as it is a minimizing criterion, we need to decrease the value of parameter k3 in
case of bid 4) to 20/(28/9.82) = 7.014 or less in order to make 4)get better score than 2))