steppin stone method

8/7/2019 steppin stone method

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The Stepping-Stone Method

The stepping-stone method is an iterative technique for moving from aninitial feasible solution to an optimal solution.

It is used to evaluate the cost effectiveness of shipping goods via

transportation routes not currently in the solution.

Each unused cell ,or square, in the transportation table is tested by askingthe following question: What would happen to total shipping costs if one unit of

product (for example, one desk) were tentatively shipped on an unused route ?


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This testing of each unused square is conducted as follows:

Select an unused square to be evaluated.

Beginning at this square, trace a closed path back to the original squarevia squares that are currently being used (only horizontal and verticalmoves are allowed).

Beginning with a plus (+) sign at the unused square, place alternate

minus (-) signs and plus signs on each corner square of time closedpath just traced.Calculate an improvement index by adding together the unit costfigures found in each square containing a plus sign and then subtracting

the unit costs in each square containing a minus sign.

Repeat steps l 4 until an improvement index has been calculated forall unused squares. If all indices computed are greater than or equal tozero, an optimal solution lies been reached. If not, it is possible toimprove the current solution and decrease total shipping costs


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Consider the table given below


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` We can use the northwest corner rule to find an initial feasiblesolution to the Executive Furniture Corp. problem shown in the

Table.` It takes five steps in this example to make the initial

shopping assignments:

Assign 100 units from Des Moines to Albuquerque (exhaustingDes Moines supply).

Assign 200 units from Evansville to Albuquerque (exhaustingAlbuquerques demand).

Assign 100 units from Evansville to Boston (exhaustingEvansvilles supply).

Assign 100 units from Ft. Lauderdale to Boston (exhaustingBostons demand).

Assign 200 units from Ft. Lauderdale to Cleveland (exhaustingClevelands demand and Ft. Lauderdales supply).


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We can easily compute the cost of this shipping assignment


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` The solution given here is feasible since demand-and-supply constraints are all satisfied.

` It would be very lucky if this solution yielded theminimal transportation cost for the problem, however.

` It is more likely that one of the iterative proceduresdesigned to help reach an optimal solution shall have tobe employed.


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An improvement index for the Des Moines - Boston route is now

computed by adding unit costs in squares with plus signs and subtractingcosts in squares with minus signs. Hence

Des Moines Boston index = + $4 - $5 + $8 - $4 = +$3


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Des Moines Cleveland index = +$3 - $5 + $8 - $4 + $7 - $5 = $4Thus, opening this route will also not lower our total shipping costs. Theother two routes may be evaluated in a similar fashion.

Evansville Cleveland index = +$3 - $4 + $7 - $5 = +$1 (Closed path is

+EC EB + FB FC)Fort Lauderdale Albuquerque index = +$9 - $7 + $4 - $8 = -$2 (Closedpath is +FA FB + EB EA)Because the second index is negative, a cost savings may be attained by

making use of the (currently unused) Fort Lauderdale to Albuquerque

route.


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Each negative index represents the amount by which total transportationcosts could be decreased if one unit or product were shipped by that

sourcedestination combination.

The next step, then, is to choose that route (unused square) with the largestnegative improvement index.

W

e can then ship the maximum allowable number of units on that routeand reduce the total cost accordingly.

The maximum quantity that can be shipped on the new money-saving routecan be found by referring to the closed path of plus signs and minus signs

drawn for the route and selecting the smallest number found in thosesquares containing minus signs.


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To obtain a new solution, that number is added to all squares on the closedpath with plus signs and subtracted from all squares on the path assignedminus signs.

One iteration of the stepping-stone method is now complete. Again, we

must test to see if it is optimal or whether any further improvements can bemade.

That is done by evaluating each unused square as described in the stepslisted earlier.


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To improve Executive Furnitures solution we make use of theimprovement indices calculated and The largest (and only) negative index is

found on the Fort Lauderdale to Albuquerque route. We repeat thetransportation table for the problem below

The maximum quantity that may be shipped on the newly opened route(FA) is the smallest number found in squares containing minus signs-inthis case, 100 units.


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Hence, we add 100 units to the 0 now being shipped on route FA; thenproceed to subtract 100 from route FB, leaving zero in that square (but still

balancing the row total for F); then add 100 to route B yielding 200; andfinally, subtract 100 from route EA, leaving 100 units shipped. Note thatthe new members still produce the correct row and column totals as

required.

The new solution is shown in the following table:


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Total shipping cost has been reduced by (100 units) X ($2 saved/unit) =200,and is now $4,000, This cost figure can, of course, also be derived bymultiplying each unit shipping cost times the number of units transported

on its route, namely,

lOO($5) + l00( $8) + 100($9) + 200($5) = $4,000

Now again doing the same procedure to recheck if it is optimized or not


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` Again Calculating Improvementindex for each

unallocatedcell.

A B C Supply

D 5

100

4

$3

3

$2

100

E 8

100

4

200

3

$(-1)

300

F 9

100

7

$2

5

200

300

Demand 300 200 200 700


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A B C Supply

D 5

100

4

$3

3

$2

100

E 8

100

4

200

3

$(-1)

300

F 9

100

7

$2

5

200

300

Demand 300 200 200 700

Add

Add Less


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A B C Supply

D 5

100

4 3 100

E 8 4

200

3

100

300

F 9

200

7 5

100

300

Demand 300 200 200 700


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` Modified Transportation Cost

` TC=(100*$5)+(400*$2)+(100*$3)+(900*$2)+(100*$5)

` TC=$3900

Thank You

steppin stone method

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