RISK EVALUATION AND DECISION TREES Business Decisions involve risk What is risk and how to measure it Use of decision trees to represent real life investment, return and risk An example of a product launch to illustrate the use of decision trees Determination of optimal strategies

PRODUCT LAUNCH WITH Decision Trees

A new consumer product has been developed by a company and preliminary marketing potential has been estimated as a chance of 0.4 for successful launch which would result in an annual profit with the present worth of Rs.2,000,000.Failure of product would mean a loss of Rs.300,000. The company can either drop the idea , or launch the product immediately , or try in a test market ( at a cost of Rs.40,000) before deciding to drop or launch.

The test market would classify the sample response in one of the following three categories:-(a) Less than 10% of the public try the new product ,

(b) More than 10% try the product , but less than 25% of those who try buy it on a second or subsequent occasion,

(c) more 10% try the product and the repeat purchase rate is 25% or more.

PRODUCT LAUNCH WITH Decision Trees

Based on the subjective estimates of experts the following joint probability table is available.

Test Market response(a)(b)(c)

Success(S)0.050.100.25

Failure(F)0.450.150.00

Use a decision tree to determine what is best for the company to do . Also obtain the probability distribution of outcomes for each choice , and comment on the risk in each choice.

DECISION TREE

OPTIMAL DECISION TREE

PROBABILITY DATAP(S,a) = 0.05 P(S,b) = 0.10 P(S,c) = 0.25

P(F,a) = 0.45 P(F,b) = 0.15 P(F,c) = 0.00

P(S) = (0.05+0.10+0.25) = 0.40

P(F) = (0.45+0.15+0.00) = 0.60

P(a) = (0.05+0.45) = 0.50

P(b) = (0.10+0.15) = 0.25

P(c) = (0.25+0.00) = 0.25

TREE EVALUATION

Let annual profit be (P) = 2,000,000Let loss due to failure be (F)=300,000Let test market cost be (T) = 40,000

If product is launched direct then,At node 2, value (E2) = P(S)*P + P(F)*F = 0.4*2,000,000 + 0.6*(-300,000)(E2)= 620,000

If test market of product is done then,At node 9, value = [P(S,c)*P+P(F,c)*F]/P(c) =(0.25*2,000,000 + 0.00*(-300,000)) /0.25 (E9) = 2,000,000

TREE EVALUATION

At node 8, value = [P(S,b)*P+P(F,b)*F]/P(b) =(0.10*2,000,000 + 0.15*(-300,000)) /0.25 (E8) = 620,000

At node 7, value = [P(S,a)*P+P(F,a)*F]/P(a) =(0.05*2,000,000 + 0.45*(-300,000)) /0.50 (E7) = -70,000Since nodal value @ node 7 is negative indicating loss thereby it is better to drop the product @ node 4 but launch the product @ node 5 & node 6 since nodal value @ node 8 & 9 are positive indicating profit .

The value @ node 6 (E6) = value @ node 9 =2,000,000The value @ node 5 (E5) = value @ node 8 =620,000The value @ node 4 (E4) = 0Thereby P(F)=P(F,b)+P(F,c)=0.15+0=0.15

TREE EVALUATION

At node 3,value(E3)=P(a)*E4+P(b)*E5+P(c)*E6 = 0.5*0 + 0.25*620,000 + 0.25*2,000,000 = 655,000

If the company drop the idea then expected return is Rs 0If the company directly launches the product then the expected return is Rs 620,000If company tests the products potential by market survey then expected return is Rs (655,000-40,000)=615,000

Thereby to get some return company should not drop the idea, instead it should go for test market since expected return in direct launch and test market is near about same but the chance of failure in direct launch =0.6 while in test market it is only 0.15

ENUMERATION OF ALL STRATEGIES

The various possible strategies can be

(1) Drop or Status quo.

(2) Direct Launch

(3) Test Market

TEST MARKET STRATEGIES

There are three possible cases(a),(b)&(c) each with two possibilities drop(D) or launch(L), leading to totally.They are:-

CasesPossibilities

(i)(ii)(iii)(iv)(v)(vi)(vii)(viii)

(a)DDDDLLLL

(b)DDLLLLDD

(c)DLLDLDDL

RISK EVALUATION

STRATEGIES

(1) DROP OR STATUS QUOE(return) = E(1) = 0P(S)= 0 P(F)= 0 P(0)= 1

(2) DIRECT LAUNCHP(S)= 0.4 P(F)= 0.6 P(0)= 0

RISK EVALUATION

3.TEST MARKET

(i)D D D

E(3(i)) =0*0.5+0*0.25+0*0.25-40000=Rs -40,000P(S)= 0 ,P(F)= 0P(-0.04x106 )= 1

(ii)D D L

P(S)=P(S,c)=0.25 ,P(F)=P(F,c)=0, P(0)=0.75

(iii)D L L

P(S) = P(S,b)+P(S,c)=0.10+0.25=0.35P(F) = P(F,b)+P(F,c)=0.15+0.00=0.15P(0) = 1-0.35+0.15= 0.50

(iv)D L DP(S)=P(S,b)=0.10P(F)=P(F,b)=0.15P(0)=0.75

(v)LLL

P(S) = P(S,a)+P(S,b)+P(S,c) = 0.05+0.10+0.25=0.4P(F) = P(F,a)+P(F,b)+P(F,c) = 0.45+0.15+0.00=0.6P(0) = 0

(vi)LLDP(S) = P(S,a)+ P(S,b)=0.05+0.10=0.15P(F) = P(F,a)+P(F,b) =0.45+0.15=0.6P(0) =1-0.15-0.6=0.25

(vii)L D D

P(S)=P(S,a)=0.05,P(F)= P(F,a) =0.45, P(0)= 0.5

(viii)L D LP(S)= P(S,a) +P(S,c) =0.05+0.25=0.3P(F)= P(F,a) +P(F,c) =0.45+0.00=0.45P(0)= 0.25

STRATEGY 1

DROP OR STATUS QUO.

Expected return (E1)=0.Therefore probability of success or failure = 0 while probability of zero return = 1

STRATEGY 2

DIRECT LAUNCH

Expected return (E1) = (E2) = 0.62 millions.Therefore ,the probability of success = 0.4 and of failure = 0.60 while probability of zero return = 0

STRATEGY 3(i)

TEST MARKET(DDD)

Expected return (E1) = (E3(i)) =- 0.04millions.Therefore ,the probability of success and of failure = 0 while probability of zero return = 1

STRATEGY 3(ii)

TEST MARKET(DDL)

Expected return (E3(ii)) = 0.046 xmillions.Therefore ,the probability of success = 0.25 and of failure = 0 while probability of zero return = 0.75

STRATEGY 3(iii)

TEST MARKET(DLL)

Expected return (E3(iii)) = 0.615 xmillions.Therefore ,the probability of success = 0.35 and of failure = 0.15 while probability of zero return = 0.5

STRATEGY 3(iv)

TEST MARKET(DLD)

Expected return (E3(iv)) = 0.115 xmillions.Therefore ,the probability of success = 0.1 and of failure = 0.15 while probability of zero return = 0.75

STRATEGY 3(v)

TEST MARKET(LLL)

Expected return (E3(v)) = 0.58 xmillions.Therefore ,the probability of success = 0.4 and of failure = 0.6 while probability of zero return = 0.75

STRATEGY 3(vi)

TEST MARKET(LLD)

Expected return (E3(vi)) = 0.08 xmillions.Therefore ,the probability of success = 0.15 and of failure = 0.6 while probability of zero return = 0.25

STRATEGY 3(vii)

TEST MARKET(LDD)

Expected return (E3(vii)) = -0.075 xmillions.Therefore ,the probability of success = 0.05 and of failure = 0.45 while probability of zero return = 0.5

STRATEGY 3(viii)

TEST MARKET(LDL)

Expected return (E3(viii)) = 0.425 xmillions.Therefore ,the probability of success = 0.3 and of failure = 0.45 while probability of zero return = 0.25

Summary of all ten strategies

S.NoStrategyProfit (Prob)Loss (Prob)Expected GainRisk

1D0(1)0(1)00

2L2(0.4)0.3(0.6)0.620.6

3(i)D0(1)0(1)-0.040

3(ii)DDL2(0.25)0(0.75)0.460.75

3(iii)DLL2(0.35)0.3(0.15)0.6150.15

3(iv)DLD2(0.10)0.3(0.15)0.1150.15

3(v)LLL2(0.4)0.6(0.6)0.580.6

3(vi)LLD2(0.15)0.3(0.6)0.080.6

3(vii)LDD2(0.05)0.3(0.45)-0.0750.45

3(viii)LDL2(0.25)0.3(0.45)0.4250.45

It is seen that the srategy DLL(3(iii)) is the best both on account of the expected gain and reasonable risk.

CONCLUSIONS

Risk defined as the probability of a negative payoff (loss) Business situations abound in capital inflows and risky revenue realization Decision trees capture such situations with finite number of decision alternatives Example of Product launch to assess optimal strategy.