 Hello everyone. The topic for my today's lecture is Reliability Design Problem. At the end of the session, the students will be able to solve the reliability design problem using dynamic programming. We need to design a system which is composed of several devices that are connected in series. Now, this is the diagram for it. The device is D 1, D 2, D 3 up to D n that are connected in series. So, we have n number of devices. Let us assume that the reliability of the device D i is R i. So, therefore, the reliability of the entire system will be pi R i meaning the product of all such R i's that is the reliability of the devices. Now, reliability of individual devices is generally close to one, but the reliability of the entire system may not be good. For example, if we take n equal to 10 meaning 10 devices and R i equal to 0.99 meaning reliability of the devices 0.99 where we have i ranging from 1 to 10. So, pi R i would be equal to 0.904 that is the reliability of the entire system. Now, multiple copies of the same device type are connected in parallel through the use of switching circuits. This is how we have the devices connected in parallel in each stage and as we can see in every stage multiple copies of the devices are present at that stage. This is for D 1, D 2, D 3 and so on up to D n. The switching circuits determine which devices in any given group are functioning properly. If suppose the stage i contains m i number of copies of device D i, the probability that all the m i copies have a mal function is equal to 1 minus R i raise to m i where this R i would be the reliability of that device D i and m i would be the number of copies of the device D i. So, this is nothing, but the mal function probability. So, therefore, the reliability of that stage i will be equal to 1 minus that probability of mal functioning that is 1 minus into bracket 1 minus R i raise to m i. For example, if the reliability R i is equal to 0.99 and m i equal to 2 meaning 2 copies in that case the stage reliability will be equal to 1 minus into bracket 1 minus R i closing the bracket raise to m i. This will be equal to 1 minus as we know R i is 0.99. So, we substitute that value over here. So, this gives 1 minus into bracket 1 minus 0.99 bracket raise to 2 and it equals to 1 minus 0.01 square which equals to 0.9999. So, this is the stage reliability. The stage reliability in practical may not be equal to 1 minus into bracket 1 minus R i close the bracket raise to m i. Now, this is because of the reason that the switching circuits themselves are not fully reliable. The failures of the copies of the same device may not be fully independent if the failure is due to some design defect. So, the reliability of a stage i is generally given by a function that is phi i of m i where n is greater than 1. So, the reliability of the entire system of stages is equal to product of phi i m i where i goes from 1 to n. So, there is a need of device duplication so as to maximize the reliability. So, that tells us device duplication actually increases the reliability. The maximization needs to be carried out under some constraint of cost. So, let us assume that C i is a cost of each unit of the device D i and C be the maximum allowable cost of the system being designed. So, our maximization problem is maximize phi phi i m i where i goes from 1 to n subject to the constraints summation C i m i where i goes from 1 to n and it should be less than equal to c where m i is always greater than equal to 1 and integer i ranges between 1 to n. And as we know m i greater than equal to 1 indicates that at least one device copy must be present in every stage. Now, S i consists of the tuples of the form f comma x where f stands for reliability and x stands for cost and S i j represents a tuple which are obtainable from S i minus 1 by choosing m i value equal to j that is the number of copies equal to j. So, there is at most one tuple for each different x value that results from a sequence of decisions on m 1 m 2 up to m n meaning the number of copies m 1 m 2 up to m n. Now, the dominance rule holds that is f 1 x 1 dominates f 2 x 2 if and only if f 1 greater than equal to f 2 and x 1 less than equal to x 2. For example, there are two tuples 0.7 comma 30 and 0.65 comma 35. So, 0.7 comma 30 dominates 0.65 comma 35 if and only if the following condition holds true. So, since this condition is holding true 0.7 is greater than 0.65 and 30 is less than 35 due to that the first tuple dominates the second one and the dominated tuples can be discarded from S i. So, now we have the reliability design problem to be solved where n equal to 3, 3 stages the cost value C 1, C 2, C 3 equal to 30, 15, 20 and C the maximum cost available is equal to 105 dollars and the reliability is of the devices indicated as R 1, R 2, R 3 equal to 0.9, 0.8 and 0.5 respectively and U 1, U 2, U 3 equal to 233 that indicate the maximum copies available of devices D 1, D 2 and D 3 respectively. So, now S 0 will be equal to 1 comma 0 the set consisting of the tuple 1 comma 0 indicating reliability equal to 1 and cost involved in now is equal to 0. Now, the tuple that we are going to take into consideration is 0.9 comma 30 that is for the first device. So, S 11 is constructed as multiplying the reliability 0.9 to the reliability of S 0 that is 1. So, it gives value 0.9 and adding 32 the cost value of the tuple in S 0 that is it will be 30 plus 0 equal to 30. Now, this S 11 indicates that only 1 copy of device D 1 is being taken. Similarly, S 12 is then formed using 2 copies of device D 1 and so we take the tuple 0.9 comma 30 as it is and 0.99 comma 60 is the second tuple to be inserted. How is this 0.99 comma 60 forming a part of S 12? It is because of this formula 1 minus 1 minus 0.9 bracket square. So, this is nothing, but the reliability of the system where 0.9 is the reliability of device D 1 and 2 is the number of copies and 60 is nothing, but since we are taking 2 copies of device D 1. So, it is 2 into 30 equal to 60 now S 1 is formed by combining the tuples of S 11 and S 12. So, this is how we get S 1 moving ahead for S 21 we need to now consider the second object where f comma x equal to 0.8 and 15. So, when you take 1 copy of device D 2. So, its reliability is 0.8 and its cost is 15. So, this 0.8 multiplied by 0.9 gives 0.92 and 15 added to 30 gives 45. Similarly, 0.8 multiplied by 0.99 gives 0.792 and 15 added with 60 gives 75. So, this is S 21. Now for calculating tuples in S 22 we need to consider 2 copies of device D 2 and accordingly we get the tuples 0.864 comma 60. Now, the students are expected to think and write the answer to the following question is any tuple eliminated from S 22? Now pause the video and write your answer. Yes, the tuple 0.9504 comma 90 is eliminated. The next question is what is the reason for the tuple getting eliminated from S 22? Now once again pause the video and write your answer. So, the tuple is eliminated since when you see that 0.864 comma 60 is generated from this tuple and the tuple 0.9504 comma 90 is generated from this tuple and this is eliminated because if we consider that tuple the overall maximum cost available is 105 and the cost of the device D 3 is 20 dollars. So, if 20 is added to 90 it gives 110 dollars. So, this 110 since it is greater than 105 we cannot design the third stage. So, therefore, this tuple is eliminated. Now moving ahead for S 23 we now consider 3 copies of device D 2. So, we get the tuple 0.8928 comma 75 and this tuple is formed from the tuple 0.9 comma 30. Moving ahead so S 2 when we need to find S 2 we will combine or we will merge the tuples in S 21, S 22 and S 23 and we get these 3 tuples. So, 0.792 comma 75 is eliminated or purge since its reliability is less than this tuple and its cost is greater. Similarly, S 31 for S 31 now we go to the third stage the cost is 20 of the device DI and reliability is 0.5. So, this 0.36 comma 65, 0.432 comma 80 and 0.4464 comma 95 are generated from the tuples of S 2 by adding one copy of device D 3. Similarly, S 32 we consider 2 copies of device D 3 and generate these tuples and S 33 by considering 3 copies of device D 3 and generating this tuple. So, finally, when we take S 3 we are going to merge the tuples and we are going to get this final result. Now 0.648 comma 100 is the maximum reliability. So, it is generated from S 32. So, therefore, 2 copies of device D 3 and this tuple was generated from this S 2 tuple 0.864 comma 60 and it was coming from S 22. So, therefore, here 2 copies of device D 2 are taken. So, therefore, M 2 equal to 2 and this tuple 0.864 comma 60 was generated from 0.9 comma 30 and this 0.9 comma 30 was coming from S 11. So, it indicates one copy of device D 1. So, M 1 M 2 M 3 are having value 1, 2 and 2. The 3-stage system design has the reliability of 0.648 and the cost of 100 dollars where M 1 equal to 1, M 2 equal to 2 and M 3 equal to 2. This is the reference for the video lecture.