EXTRA PROBLEM 6: SOLVING DECISION TREES p being …

1 EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp1 EXTRA PROBLEM 6: SOLVING DECISION TREESRead the following DECISION PROBLEM and answer the questions manufacturer produces items that have a probability of .pbeing defectiveThese items are formed into . Past experience indicates thatbatches of 150some are of and others are of (batches)good quality ( p= )bad quality( p= ). Furthermore, of the batches produced are of 80%good qualityand These items are then used in an20% of the batches are of bad , and ultimately their quality is determined before the final assemblyleaves the plant.

2 The manufacturer can either in a batch andscreen each itemreplace defective items at a total average cost of $10 per item or use the itemsdirectly without latter action If is chosen, the cost of rework isultimately $100 per defective item. For these data, the costs per batch can becalculated as follows:p = p = $1500$1500Do not Screen $750$3750 Because screening requires scheduling of inspectors and equipment, thedecision to screen or not screen must be made 2 days before the potentialEMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr.

3 J. R. van Dorp2screening takes place. However, the manufactures may take one item takenfrom a batch and sent it to a laboratory, and the test results (defective or non-defective) can be reported must bebefore the screen/no-screen decisionmade. After the laboratory test, Thethe tested item is returned to its of this initial inspection is $125. Also note that the probability that a randomsample item is defective * + * = ,and the probability that an item in a lot is of good quality given a randomlysampled item is defective is and the probability that an item in a lot isof good quality given a randomly sampled item is not defective is manufactur wants to minimize his or her Derive the cost figures in the table above.

4 Clearly show your = = *$10 = $1500150*$10 = $1500Do not Screen 150* *$100 = $750 150* *$100 = $3750B. What Law of Probability was used in deriving:EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp3Pr(Randomly Sampled Item is Defective)Pr(Sample Item is Defective|Batch of Good Quality) = (Sample Item is Defective|Batch of Bad Quality) = (Batch of Good Quality)= Pr(Batch of Bad Quality)= (Randomly Sampled Item is Defective) =Pr(Sample Item is Defective|Batch of Good Quality)Pr(Batch of Good Quality)+Pr(Sample Item is Defective|Batch of Bad Quality)Pr(Batch of Bad Quality)= * + * , THE LAW OF TOTAL PROBABILITY was used.

5 EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp4C. Show that: Pr(Batch of Good Quality|Sampled Item is Defective) = Pr(Batch of Good Quality|Sampled Item is Not Defective) = WMH "Batch of Good Quality" "Sampled Item is Defective"; = "Sampled Item is Not Defective"WMHT< FKUWMH |T < WMHlFKU T < FKU T< WMH ! !& ! )!! !* ! %%%T< FKUWMH |T < WMHlFKU T < FKU T< WMH ! )$& " ! !& ! )! " ! !* EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp5D.

6 Calculate and show your calculations: FKU WMH "Batch of Bad Quality""Sampled Item is Not Defective" Pr(Bad Quality|Sampled Item is Defective) = ? Pr(Bad Quality|Sampled Item is Not Defective) = ?T< FKUWMH " T< FKUWMH ||" ! %%% ! &&'T< FKUWMH " T< FKUWMH ||" ! )$& ! "'&EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp6E. Model the DECISION PROBLEM in a DECISION tree and fill in ALL the detailsTest Item?Screen?Passed TestFailed TestScreen?Screen?No TestTestNo ScreenBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityNo ScreenNo ScreenScreenScreenScreen( )( )( )( )( )( )( )( )$125$1500$1500$1500$750$750$750$3750$37 50$3750$875$875$750$3750$1500$1625$1625$ 3875$3875 COSTEMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr.

7 J. R. van Dorp7F. How many cumulative risk profiles can be drawn for the DECISION tree underE? Provide an Item?Screen?Passed TestFailed TestScreen?Screen?No TestTestNo ScreenBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityNo ScreenNo ScreenScreenScreenScreen( )( )( )( )( )( )( )( )$125$1500$1500$1500$750$750$750$3750$37 50$3750$875$875$750$3750$1500$1625$1625$ 3875$3875 COSTU pper Part: 4 , , , (T,{NS,NS})(T,{S,S})(T,{NS,S})(T,{S,NS}) Lower Part: 2 , Total: 6 Strategies(NT,NS)(NT, S)EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr.

8 J. R. van Dorp8G. Solve the tree under E using EMV and clearly show your calculations in thetree under ETest Item?Screen?Passed TestFailed TestScreen?Screen?No TestTestNo ScreenBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityNo ScreenNo ScreenScreenScreenScreen( )( )( )( )( )( )( )( )$125$1500$1500$1500$750$750$750$3750$37 50$3750$875$875$750$3750$1500$1625$1625$ 3875$3875 COST$ $1625$ $ $ $1350$1350$1350 EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr.

9 J. R. van Dorp9H. Describe in words the optimal DECISION not Sample an Item Randomly from a Batch for Testing,and Do Not Screen the entire Batch(No Test, No Screen in the DECISION Tree)EMSE 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp10I. Draw the Cumulative Risk Profile for each alternative of the immediatedecision, taking optimal decisions from thereon. What can you conclude withrespect to dominance considerations? (Hint: You should be drawing 2 Cumulative Risk Profiles).Test Item?Screen?Passed TestFailed TestScreen?

10 Screen?No TestTestNo ScreenBatch of Good QualityBatch of Bad QualityBatch of Good QualityBatch of Bad QualityNo ScreenScreen( )( )( )( )( )( )$125$1500$750$750$3750$3750$875$750$375 0$1625$3875 COST$ 269 - Elements of PROBLEM SOLVING and DECISION MakingInstructor: Dr. J. R. van Dorp11 Strategy: Test Random Item and If Item Passed Test then do not Screen theentire Batch and if Item Failed Test then Screen the entire BatchT < G9=> )(& ! *" ! )$& ! ('!() T < G9=> "'#& ! !*() T < G9=> $)(& ! *" ! "'& ! "&!()T < G9=> )(& !