Slide 5.1

Video 23: Part 2 Explosion of Customer Journeys Across Channels

IBMS Semester 6a Fall 2014

Mr. George Szanto

Topics & Learning Goals• Customer Journey mathematics• Calculate total number of possible customer journey

paths for non ordered and ordered customer online interactions

----------------------------------------------------------• Be able to make simple decisions about content

deployment across channels based on journey mathematics

2

3

Journey Mathematics – how many paths are possible?

Consider 5 different websites (#1- #5) and 3 different content types (A, B, C)

CH # 1

A

B

C

CH # 2

A

B

C

CH # 3

A

B

C

CH # 4

A

B

C

CH # 5

A

B

D

4

Introduce Integer Decompositions

• Decompose N different content types.

Example, let’s start with N = 3

Voettekst

5

Non Ordered Journeys (5 channels x 3 content types) CH #

1A

B

C

CH # 2A

B

C

CH # 3A

B

C

CH # 4A

B

C

CH # 5A

B

D

Integer Start 2nd CH 3rd CH Total

decomp. 3:

6

Non Ordered Journeys – what happens if content types is greater than channels?(3 channels x 5 content types)

Integer Start 2nd CH 3rd CH Total

decomp. 5:

CH # 1A

B

C

D

E

CH # 2A

B

C

D

E

CH # 3A

B

C

D

E

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Ordered Journeys

Journey A1B1C2D3 ≠ C2D3B1A1;

Exact order does matter!

8

Review Permutations

• Permutation of “a” content types taken “b” at a time

• Factorials: a!

• Example:

• Note: 1! = 1 and 0! = 1

9

Ordered Journey 5 channels x 3 content types

CH # 1A

B

C

CH # 2A

B

C

CH # 3A

B

C

CH # 4A

B

C

CH # 5A

B

D

(5 channels x 3 content types)

Integer Start 2nd CH 3rd CH Total

decomp. 3: [ 5 x P ] x [4 x P ] x [3 x P ]

10

4 channels x 4 content types - ordered

• Scenario 6: Journey (1,2,1), or look at 1 pieces of content on 1st website, then 2 pieces at 2nd website, and then last piece at 3rd website

Website 1 = 4 x 4Permutation1 = 4x [ ] = 4x [] = 16

Website 2 = 3 x 3Permutation2 = 3x [ ] = 3x [] = 18

Website 3 = 2 x 1Permutation1 = 2 x [ ] = 2 x [] = 2

Scenario 6 Journeys = 16 x 18 x 4 = 576

Number of Possible Customer Journeys

Scenario Example

# of possibleChannelStart pts.

# of possible permutations  

1stChannel

(4x)

2ndChannel

(3x)

3rdChannel

(2x)

4thChannel

(1x)Total # ofJourneys

1 A1B1C1D1 4 24 0 0 0 962 A1B2C2D2 4 4 18 0 0 2883 A1B1C1D2 4 24 3 0 0 2884 A1B1C2D2 4 12 6 0 0 2885 A1B2C3D3 4 4 9 4 0 5766 A1B2C2D3 4 4 18 4 0 5767 A1B1C2D3 4 12 6 2 0 5768 A1B2C3D4 4 4 9 4 1 576

Total Number of Journeys Across All Scenarios 3264

1 42 1 33 3 14 2 25 1 1 26 1 2 17 2 1 18 1 1 1 1

(Video, Data Sheet, Testimonial, Demo)

Possible Channel Trips to Get All Information

Scenario

Content Types Collected Per Channel

11

So What’s The Best Journey Strategy?

• Option 1: Lots of content on a few sites– For non ordered content this makes sense;

reduces management costs

• Option 2: Split limited content across several sites– For ordered content this is a good option, it will limit

# of possible journeys by limiting paths and keeping content contextual

• Option 3: Place lots of content on many sites– Expensive; hard to track journeys.

End of Video - Thank you

• It is strongly suggested that you practice a few examples using the homework assignment posted on the class website next to this video recording

• See: Customer Journey Practice HW (PDF)