solving a “transportation planning” problem through the programming language “c”

Solving a Transportation Planning Problem Through the Programming Language “C”

Presented by

Yousuf Mahid(0615012)

Shahadat Hossain Shakil(0615020)

Khadija Akhter(0615027)

Calculation of Inter-Zonal Trips (Gravity Model)

Problem:Calculation of inter-zonal trips in a town.(Gravity Model)

Example: A self-contained town consists of four residential areas A, B, C and D and two industrial areas X and Y. Generation equations show that for the design year the trips from home to work generated by each residential area per day are A=1000, B=2250, C=1750 and D=3200. There are 3700 jobs in X and 4500 jobs in Y. The frictional factor is inversely proportional to the travel time between zones as tabulated below and the value of exponent is 2.

Travel time between the zones in minutes

Zones X Y

A 15 20

B 15 10

C 10 10

D 15 20


We have to determine the correct value of the TAX, TAY, TBX, …………. TDY, until justifying both the column and the row total by continuous calculation through iteration.

X Y Total

A 1000

B 2250

C 1750

D 3200

Total 3700 4500 8200

DestinationOrigin

TAX TAY

TBX TBY

TCY

TDX

TCX

TDY


Solution:The most typical version of gravity model used in the transportation planning application is

Tij= (Oi × Djfij) ∕ ( ∑ Djfij)

Where

Tij = no. of trips from zone i to zone j.

Oi = total no. of trips produced in zone i.

Dj = total no. of trips attracted to zone j.

b = an exponential constant whose value is usually found between 1 and 2.

The frictional factor is an inverse function of the travel cost (time, distance, monetary cost).

fij = 1 / Cijb


TAX = 1000 × (3700/152) / (3700/152 + 4500/202) = 592

TAY = 1000 × (4500/202) / (3700/152 + 4500/202) = 407

………………………………………………………………

TDX = 3200 × (3700/152) / (3700/152 + 4500/202) = 1896

TDY = 3200 × (4500/152) / (3700/152 + 4500/202) = 1303

X Y Total

A 1000

B 2250

C 1750

D 3200

Calculated Attraction

Projected Attraction 3700 4500 8200

DestinationOrigin

3867 4329

592

590

789

1896 1303

1659

960

407

8196


The adjusted attraction = Projected attraction × (The attraction used in the previous iteration ∕ Calculated attraction in the previous iteration)

D1X = 3700 × (3700 ∕ 3867) = 3540

D1Y = 4500 × (4500 ∕ 4329) = 4677

TAX = 1000 × (3540/152) / (3540/152 + 4677/202) = 576

………………………………………………………….

TDY = 3200 × (4677/152) / (3540/152 + 4677/202) = 1353

X Y Total

A 1000

B 2250

C 1750

D 3200

Calculated Attraction

Projected Attraction 3700 4500 8200

Destination

Origin

3731 4465

576

553

756

1846

423

1696

993

1353

8196


Again the adjusted attractions after second iteration is

D2X = 3700 × (3540 ∕ 3731) = 3510

D2Y = 4500 × (4677 ∕ 4465) = 4713

Using this adjusted attraction value, 3rd iteration will be operated.

This process continues until the calculated attraction value and the

projected attraction value is being equalized or nearest most.

Using the “C” programming language we have to develop a program to do

this cumbersome mathematical calculation mechanically to save time and

energy and to get the accurate result in the shortest possible time.

Making Procedure of the Program in “C”

Divide the whole task among different functions.

Function declaration.

Declaring functions prototype before main function.

Then subdivide the claculation of a specific function into more smaller parts

to store the return value into different variables.

Variable Declaration-(Local and Global).

Global Array Declaration- (1D and 2D)-to store and manipulate same

kind of data.

Function calling - inside the main function and also inside the called

function.

Making Procedure of the Program in “C”

Library function used – main(); clrscr(); printf();scanf(); fflush(stdin);

getch(); (including related header files)

Type of variables used – int; long int; float;

(including ralated format specifier)

Operator used – arithmatic (addition, multiplication, division)

– assignment(=), relational(==),increment(++)

Control Statement Used – if statement;

– for loop;

A progarm largely dependent on the use of array and continuous iteration.

Limitations – deigned for 10 residential zone and 2 working zone.

Failure of converting the fraction trip value into the next higher integer number.

Thanks for your patience.

Q/A

solving a “transportation planning” problem through the programming language “c”

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