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Solving a Certainty Equivalent (Decision Analysis) problem

Solving ATSP problem for large-scale problemHow to formulate this scheduling problem efficiently?Online Education for OR and Developing Decision Support SystemsApproaches for choosing a “risk” factor in an Inventory Optimization problem?Scheduling Optimization ProblemCan an integer optimization problem be convex?How could we simplify solving the large scale MIPs without using any advanced methods like decompositions?

3

$begingroup$

I am solving a Certainty Equivalent (Decision Analysis) problem.

The problem is a Risk-Averse Case - a deal of $60%$ chance to win $$100,!000$ and $40%$ chance to lose $$10,!000$.

Suppose the decision-maker is risk-averse with a risk tolerance of $$20,!000$ and his utility function is:

$$u(x)=1.0067837 (1-e^-x/20,000).$$

The answer shows:
beginalignu(rm CE)&= 0.6 u(100,000) + 0.4 u(-10,000)\&= 0.4(1.00) + 0.4(-0.65312)\&= 0.338751\impliesrm CE&=u^-1(0.338751)=$8,!203.59.endalign

Why does $0.6 u(100,000)$ equal to $0.4(1.00)$, and likewise $0.4 u(-10,000)$ equals to $0.4(-0.65312)$?

Also, with $u^-1(0.338751)$, how does it arrive at $$8,!203.59$?

optimization

share|improve this question

edited 7 hours ago

TheSimpliFire♦

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asked 8 hours ago

Mark KMark K

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New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome to OR.SE, the calculation you mentioned in your question is not correct. In the second line there is a typo, instead of $0.4(1.00)$ it should be $0.6(1.00)$.
  $endgroup$
  – Oguz Toragay
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OguzToragay Thank you for the comment. Can you please post an answer?
  $endgroup$
  – Mark K
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

3

$begingroup$

I am solving a Certainty Equivalent (Decision Analysis) problem.

The problem is a Risk-Averse Case - a deal of $60%$ chance to win $$100,!000$ and $40%$ chance to lose $$10,!000$.

Suppose the decision-maker is risk-averse with a risk tolerance of $$20,!000$ and his utility function is:

$$u(x)=1.0067837 (1-e^-x/20,000).$$

The answer shows:
beginalignu(rm CE)&= 0.6 u(100,000) + 0.4 u(-10,000)\&= 0.4(1.00) + 0.4(-0.65312)\&= 0.338751\impliesrm CE&=u^-1(0.338751)=$8,!203.59.endalign

Why does $0.6 u(100,000)$ equal to $0.4(1.00)$, and likewise $0.4 u(-10,000)$ equals to $0.4(-0.65312)$?

Also, with $u^-1(0.338751)$, how does it arrive at $$8,!203.59$?

optimization

share|improve this question

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

asked 8 hours ago

Mark KMark K

11611 bronze badge

New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Hi, welcome to OR.SE, the calculation you mentioned in your question is not correct. In the second line there is a typo, instead of $0.4(1.00)$ it should be $0.6(1.00)$.
  $endgroup$
  – Oguz Toragay
  8 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OguzToragay Thank you for the comment. Can you please post an answer?
  $endgroup$
  – Mark K
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

I am solving a Certainty Equivalent (Decision Analysis) problem.

The problem is a Risk-Averse Case - a deal of $60%$ chance to win $$100,!000$ and $40%$ chance to lose $$10,!000$.

Suppose the decision-maker is risk-averse with a risk tolerance of $$20,!000$ and his utility function is:

$$u(x)=1.0067837 (1-e^-x/20,000).$$

The answer shows:
beginalignu(rm CE)&= 0.6 u(100,000) + 0.4 u(-10,000)\&= 0.4(1.00) + 0.4(-0.65312)\&= 0.338751\impliesrm CE&=u^-1(0.338751)=$8,!203.59.endalign

Why does $0.6 u(100,000)$ equal to $0.4(1.00)$, and likewise $0.4 u(-10,000)$ equals to $0.4(-0.65312)$?

Also, with $u^-1(0.338751)$, how does it arrive at $$8,!203.59$?

optimization

share|improve this question

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

asked 8 hours ago

Mark KMark K

11611 bronze badge

New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

asked 8 hours ago

Mark KMark K

11611 bronze badge

New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 7 hours ago

TheSimpliFire♦

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asked 8 hours ago

Mark KMark K

11611 bronze badge

New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

asked 8 hours ago

Mark KMark K

11611 bronze badge

New contributor

Mark K is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Mark KMark K

11611 bronze badge

2 Answers
2

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oldest

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2

$begingroup$

There is a typo in the calculation that you mentioned.
$$u(rm CE) = 0.6 u(100,000)+0.4 u(-10,000)=0.6(1.0000)+0.4(-0.65312)=0.338751$$

For your second question, if $y=f(x) text then x=f^-1(y).$

For the calculations:

beginalignu(100,000)&=1.0067837(0.993262053000)=1.000000044789
\u(-10,000)&=1.0067837(-0.64872127070)=-0.65312200118endalign

share|improve this answer

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Why $0.6u(100,000) = 0.6(1.0000)$, and $0.4u(−10,000) = 0.4(−0.65312)$?
  $endgroup$
  – Mark K
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkK Just replace $x$ with the value $100,000$ in the $u(x)$ function.
  $endgroup$
  – Oguz Toragay
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

If $u(d)=c$ then $d=u^-1(c)$ since $ucirc u^-1$ forms the identity. Thus in general, under suitable constraints for $a,b,c$,beginaligna(1-e^-d/b)=c&implies1-e^-d/b=frac ca\&implies e^-d/b=1-frac ca=fraca-ca\&implies-frac db=lnfraca-ca&&\&implies d=-blnfraca-ca=blnfrac aa-c.endalign Now substitute the values of $a=1.0067837$, $b=20,000$ and $c=0.338751$ to obtain $d=u^-1(c)$.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

$endgroup$

add a comment
 | 

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2

$begingroup$

There is a typo in the calculation that you mentioned.
$$u(rm CE) = 0.6 u(100,000)+0.4 u(-10,000)=0.6(1.0000)+0.4(-0.65312)=0.338751$$

For your second question, if $y=f(x) text then x=f^-1(y).$

For the calculations:

beginalignu(100,000)&=1.0067837(0.993262053000)=1.000000044789
\u(-10,000)&=1.0067837(-0.64872127070)=-0.65312200118endalign

share|improve this answer

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Why $0.6u(100,000) = 0.6(1.0000)$, and $0.4u(−10,000) = 0.4(−0.65312)$?
  $endgroup$
  – Mark K
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkK Just replace $x$ with the value $100,000$ in the $u(x)$ function.
  $endgroup$
  – Oguz Toragay
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

2

$begingroup$

There is a typo in the calculation that you mentioned.
$$u(rm CE) = 0.6 u(100,000)+0.4 u(-10,000)=0.6(1.0000)+0.4(-0.65312)=0.338751$$

For your second question, if $y=f(x) text then x=f^-1(y).$

For the calculations:

beginalignu(100,000)&=1.0067837(0.993262053000)=1.000000044789
\u(-10,000)&=1.0067837(-0.64872127070)=-0.65312200118endalign

share|improve this answer

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you. Why $0.6u(100,000) = 0.6(1.0000)$, and $0.4u(−10,000) = 0.4(−0.65312)$?
  $endgroup$
  – Mark K
  7 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarkK Just replace $x$ with the value $100,000$ in the $u(x)$ function.
  $endgroup$
  – Oguz Toragay
  7 hours ago
  
  

  
  
  
  

add a comment
 | 

$begingroup$

There is a typo in the calculation that you mentioned.
$$u(rm CE) = 0.6 u(100,000)+0.4 u(-10,000)=0.6(1.0000)+0.4(-0.65312)=0.338751$$

For your second question, if $y=f(x) text then x=f^-1(y).$

For the calculations:

beginalignu(100,000)&=1.0067837(0.993262053000)=1.000000044789
\u(-10,000)&=1.0067837(-0.64872127070)=-0.65312200118endalign

share|improve this answer

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

$endgroup$

share|improve this answer

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

edited 7 hours ago

TheSimpliFire♦

2,62977 silver badges3939 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

answered 7 hours ago

Oguz ToragayOguz Toragay

3,88211 gold badge33 silver badges3030 bronze badges

2

$begingroup$

If $u(d)=c$ then $d=u^-1(c)$ since $ucirc u^-1$ forms the identity. Thus in general, under suitable constraints for $a,b,c$,beginaligna(1-e^-d/b)=c&implies1-e^-d/b=frac ca\&implies e^-d/b=1-frac ca=fraca-ca\&implies-frac db=lnfraca-ca&&\&implies d=-blnfraca-ca=blnfrac aa-c.endalign Now substitute the values of $a=1.0067837$, $b=20,000$ and $c=0.338751$ to obtain $d=u^-1(c)$.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

$endgroup$

add a comment
 | 

2

$begingroup$

If $u(d)=c$ then $d=u^-1(c)$ since $ucirc u^-1$ forms the identity. Thus in general, under suitable constraints for $a,b,c$,beginaligna(1-e^-d/b)=c&implies1-e^-d/b=frac ca\&implies e^-d/b=1-frac ca=fraca-ca\&implies-frac db=lnfraca-ca&&\&implies d=-blnfraca-ca=blnfrac aa-c.endalign Now substitute the values of $a=1.0067837$, $b=20,000$ and $c=0.338751$ to obtain $d=u^-1(c)$.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

$endgroup$

add a comment
 | 

$begingroup$

If $u(d)=c$ then $d=u^-1(c)$ since $ucirc u^-1$ forms the identity. Thus in general, under suitable constraints for $a,b,c$,beginaligna(1-e^-d/b)=c&implies1-e^-d/b=frac ca\&implies e^-d/b=1-frac ca=fraca-ca\&implies-frac db=lnfraca-ca&&\&implies d=-blnfraca-ca=blnfrac aa-c.endalign Now substitute the values of $a=1.0067837$, $b=20,000$ and $c=0.338751$ to obtain $d=u^-1(c)$.

share|improve this answer

edited 6 hours ago

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

$endgroup$

share|improve this answer

edited 6 hours ago

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges

answered 7 hours ago

TheSimpliFire♦TheSimpliFire

2,62977 silver badges3939 bronze badges