🧠:<think>Okay, so I have this problem about a professor analyzing a DJ's educational efforts on an audience's appreciation for country music. The appreciation level is modeled by the function ( A(t) ), which is defined as the integral from 0 to t of ( 5 + 2cos(pi x) ) dx. There are two sub-problems here. Let me tackle them one by one.Starting with Sub-problem 1: I need to calculate the total appreciation level ( A(3) ) at the end of the 3-hour performance. Also, I have to find the maximum appreciation level reached during the performance and the time at which it occurs.Alright, so first, ( A(t) ) is given as an integral. To find ( A(3) ), I just need to compute the definite integral from 0 to 3 of ( 5 + 2cos(pi x) ) dx. Let me recall how to integrate such functions.The integral of 5 with respect to x is straightforward. It's just 5x. Then, the integral of ( 2cos(pi x) ) with respect to x. I remember that the integral of ( cos(kx) ) is ( frac{1}{k}sin(kx) ). So, applying that here, the integral of ( 2cos(pi x) ) should be ( 2 times frac{1}{pi}sin(pi x) ), which simplifies to ( frac{2}{pi}sin(pi x) ).Putting it all together, the antiderivative of ( 5 + 2cos(pi x) ) is ( 5x + frac{2}{pi}sin(pi x) ). Therefore, ( A(t) ) is equal to this evaluated from 0 to t. So, ( A(t) = [5t + frac{2}{pi}sin(pi t)] - [5(0) + frac{2}{pi}sin(0)] ). Since ( sin(0) ) is 0, this simplifies to ( A(t) = 5t + frac{2}{pi}sin(pi t) ).Now, to find ( A(3) ), I substitute t = 3 into this expression. So, ( A(3) = 5(3) + frac{2}{pi}sin(3pi) ). Calculating each term, 5 times 3 is 15. Then, ( sin(3pi) ) is equal to 0 because sine of any integer multiple of pi is 0. So, the second term is 0. Therefore, ( A(3) = 15 + 0 = 15 ). So, the total appreciation at the end of the performance is 15.Next, I need to find the maximum appreciation level during the performance and the time at which it occurs. Since ( A(t) ) is a continuous function on the interval [0,3], its maximum must occur either at a critical point within the interval or at one of the endpoints.To find critical points, I need to take the derivative of ( A(t) ) with respect to t and set it equal to zero. The derivative of ( A(t) ) is the integrand itself, right? Because ( A(t) ) is defined as the integral from 0 to t of that function, so by the Fundamental Theorem of Calculus, ( A'(t) = 5 + 2cos(pi t) ).So, ( A'(t) = 5 + 2cos(pi t) ). To find critical points, set this equal to zero:( 5 + 2cos(pi t) = 0 )Solving for ( cos(pi t) ):( 2cos(pi t) = -5 )( cos(pi t) = -frac{5}{2} )Wait a minute, cosine of any real number can only be between -1 and 1. So, ( cos(pi t) = -frac{5}{2} ) is impossible because -5/2 is less than -1. That means there are no real solutions for t where the derivative is zero. Therefore, there are no critical points where the derivative is zero.Hmm, so if there are no critical points, the maximum must occur at one of the endpoints, which are t=0 and t=3. Let's compute ( A(0) ) and ( A(3) ).We already found ( A(3) = 15 ). For ( A(0) ), substituting t=0 into ( A(t) ), we get ( 5(0) + frac{2}{pi}sin(0) = 0 ). So, ( A(0) = 0 ).Therefore, the maximum appreciation level is 15, occurring at t=3. But wait, is that the case? Because sometimes, even if the derivative doesn't cross zero, the function might have a maximum somewhere in between. But in this case, since the derivative is always positive or always negative?Let me check the derivative ( A'(t) = 5 + 2cos(pi t) ). The cosine function oscillates between -1 and 1, so ( 2cos(pi t) ) oscillates between -2 and 2. Therefore, ( A'(t) ) oscillates between 5 - 2 = 3 and 5 + 2 = 7. So, the derivative is always positive because the minimum value is 3, which is greater than 0. Therefore, the function ( A(t) ) is strictly increasing on the interval [0,3]. That means the maximum appreciation occurs at t=3, which is 15, and the minimum at t=0, which is 0.So, for Sub-problem 1, the total appreciation at the end is 15, and the maximum appreciation is also 15 at t=3.Moving on to Sub-problem 2: The professor hypothesizes that the educational impact is maximized when the rate of change of appreciation is greatest. So, we need to determine the time t at which this rate is maximized, and find the corresponding rate of change.The rate of change of appreciation is given by ( A'(t) = 5 + 2cos(pi t) ). So, we need to find the maximum value of ( A'(t) ) over the interval [0,3], and the time t where this maximum occurs.Since ( A'(t) = 5 + 2cos(pi t) ), and cosine oscillates between -1 and 1, the maximum value of ( A'(t) ) occurs when ( cos(pi t) ) is maximized, which is 1. So, the maximum rate of change is ( 5 + 2(1) = 7 ).Now, we need to find the time t in [0,3] where ( cos(pi t) = 1 ). The cosine function equals 1 at integer multiples of 2π. So, ( pi t = 2pi k ), where k is an integer. Solving for t, we get ( t = 2k ). Now, within the interval [0,3], the possible values of k are 0 and 1 because 2*1=2 is within [0,3], but 2*2=4 is outside.Therefore, t=0 and t=2 are the points where ( cos(pi t) = 1 ). So, the maximum rate of change occurs at t=0 and t=2.But wait, let's think about this. At t=0, the rate of change is 7, but at t=2, it's also 7. However, we need to check if these are within the interval [0,3]. Yes, both t=0 and t=2 are within [0,3]. So, does the maximum occur at both t=0 and t=2?But wait, let me verify the values. At t=0, ( A'(0) = 5 + 2cos(0) = 5 + 2(1) = 7 ). At t=2, ( A'(2) = 5 + 2cos(2π) = 5 + 2(1) = 7 ). So, yes, both times have the same maximum rate of change.However, the problem says "the time t at which this rate is maximized". So, if there are multiple times, we need to report all of them. But perhaps the question expects the first occurrence or all occurrences. Let me check the problem statement again.It says: "Determine the time t at which this rate is maximized during the performance, and find the corresponding rate of change of appreciation at that time."Hmm, it says "the time t", which might imply a single time, but in reality, it's maximized at multiple times. So, maybe I should report all times where the maximum occurs.Alternatively, perhaps I made a mistake because cosine can also reach 1 at other points? Wait, no, cosine reaches 1 only at multiples of 2π, so in terms of t, it's t=0, t=2, t=4, etc. But since our interval is [0,3], t=0 and t=2 are the points where ( cos(pi t) = 1 ).Wait, hold on, actually, ( cos(pi t) = 1 ) when ( pi t = 2pi k ), so t = 2k. So, t=0, 2, 4,... but in [0,3], only t=0 and t=2.But let me think again. The function ( A'(t) = 5 + 2cos(pi t) ). The maximum value is 7, which occurs when ( cos(pi t) = 1 ). So, ( pi t = 2pi k ), so t=2k. So, in [0,3], k=0 gives t=0, k=1 gives t=2, k=2 gives t=4 which is outside. So, yes, only t=0 and t=2.But wait, is t=0 considered part of the performance? The performance is from t=0 to t=3, so t=0 is the start, and t=2 is within the performance. So, both are valid.But the problem says "during the performance", so t=0 is the beginning, but maybe it's included. So, perhaps both t=0 and t=2 are times where the rate is maximized.However, let me think about whether the rate of change is actually maximum at both points. Since the function ( A'(t) ) is periodic, it reaches its maximum at t=0, then decreases, reaches a minimum, then increases again, and reaches maximum again at t=2, and then decreases again towards t=3.Wait, let me plot or think about the behavior of ( A'(t) ). Since ( A'(t) = 5 + 2cos(pi t) ), which is a cosine wave with amplitude 2, shifted up by 5. The period of this cosine function is ( frac{2pi}{pi} = 2 ). So, every 2 hours, it completes a full cycle.So, starting at t=0, ( A'(0) = 7 ). Then, as t increases, ( cos(pi t) ) decreases, reaching -1 at t=1, so ( A'(1) = 5 + 2(-1) = 3 ). Then, at t=2, ( cos(2pi) = 1 ), so ( A'(2) = 7 ). Then, at t=3, ( cos(3pi) = -1 ), so ( A'(3) = 3 ). So, the rate of change oscillates between 3 and 7 every 2 hours.Therefore, the maximum rate of 7 occurs at t=0, t=2, t=4, etc. But within [0,3], it occurs at t=0 and t=2.So, the times when the rate of change is maximized are t=0 and t=2. The corresponding rate of change is 7.But the question says "the time t at which this rate is maximized". So, it's possible that both times are acceptable answers. Alternatively, maybe the professor is considering the first occurrence, but the problem doesn't specify. So, perhaps I should mention both times.Alternatively, maybe I made a mistake in interpreting the function. Let me double-check.Wait, the function is ( A(t) = int_0^t (5 + 2cos(pi x)) dx ). So, ( A'(t) = 5 + 2cos(pi t) ). So, yes, that's correct. So, the derivative is 5 + 2cos(πt), which oscillates between 3 and 7 with period 2.So, the maximum occurs at t=0, t=2, t=4, etc. So, within [0,3], t=0 and t=2.But t=0 is the starting point, so maybe the professor is more interested in t=2 as the time during the performance when the rate is maximized again. But the question doesn't specify, so I think both are correct.But let me check the problem statement again: "Determine the time t at which this rate is maximized during the performance, and find the corresponding rate of change of appreciation at that time."It says "the time t", which is singular, but perhaps it's expecting all times. Alternatively, maybe it's expecting the first time after t=0 when the rate is maximized, which would be t=2.Alternatively, perhaps the maximum rate is achieved at t=0 and t=2, so both are correct. So, maybe I should report both times.But let me think about the physical meaning. At t=0, the DJ starts, and the rate of appreciation is 7, which is the maximum. Then, as the performance goes on, the rate decreases to 3 at t=1, then increases again to 7 at t=2, then decreases again to 3 at t=3.So, the maximum rate occurs at the start and then again at t=2. So, both are valid.Therefore, the times are t=0 and t=2, with the rate of change being 7 at both points.But let me check if t=0 is considered part of the performance. The performance is from t=0 to t=3, so t=0 is included. So, yes, both times are during the performance.Therefore, the answer is that the rate of change is maximized at t=0 and t=2, with the maximum rate being 7.Wait, but the problem says "the time t", so maybe it expects a single answer. Hmm, perhaps I should consider that t=0 is the starting point, and the next maximum is at t=2. So, maybe the answer is t=2, because t=0 is the beginning, and the question is about during the performance, so maybe t=2 is the time when the rate is maximized again.But the problem doesn't specify, so perhaps I should mention both times.Alternatively, maybe the function is only considered for t>0, so t=0 is excluded. But the interval is [0,3], so t=0 is included.Alternatively, perhaps the maximum rate occurs at t=0 and t=2, so both are correct.I think the safest answer is to state both times, t=0 and t=2, with the rate of change being 7 at both points.But let me think again. The problem says "during the performance", which is from t=0 to t=3. So, t=0 is the start, and t=2 is within the performance. So, both are valid.Therefore, I think the answer is that the rate of change is maximized at t=0 and t=2, with the maximum rate being 7.But let me check the derivative again. At t=0, ( A'(0) = 7 ). At t=2, ( A'(2) = 7 ). So, yes, both are correct.Therefore, the time t is 0 and 2, with the rate of change being 7.But the problem says "the time t", so maybe it expects all times. So, perhaps I should write both.Alternatively, maybe the problem expects only t=2, as t=0 is the starting point, and the maximum rate is achieved again at t=2.But since the problem doesn't specify, I think it's better to mention both times.Wait, but in the context of the problem, the DJ is performing for 3 hours, so t=0 is the beginning, and t=2 is during the performance. So, both are valid.Therefore, I think the answer is that the rate of change is maximized at t=0 and t=2, with the maximum rate being 7.But let me think again. The function ( A'(t) = 5 + 2cos(pi t) ) has a period of 2, so it reaches maximum every 2 hours. So, in the interval [0,3], it reaches maximum at t=0, t=2, and would reach again at t=4, which is outside. So, within [0,3], t=0 and t=2.Therefore, the times are t=0 and t=2.But the problem says "during the performance", so t=0 is the start, and t=2 is during. So, perhaps both are acceptable.Alternatively, maybe the problem expects the first time after t=0 when the rate is maximized, which is t=2.But since the problem doesn't specify, I think it's safer to mention both times.Therefore, the time t is 0 and 2, with the rate of change being 7.But let me check the problem statement again: "Determine the time t at which this rate is maximized during the performance, and find the corresponding rate of change of appreciation at that time."It says "the time t", which is singular, but in reality, there are two times. So, perhaps the answer expects both times.Alternatively, maybe the problem expects the first time after t=0, so t=2.But I think the correct answer is that the maximum rate occurs at t=0 and t=2, with the rate being 7.Therefore, I think I should state both times.But let me think again. The function ( A'(t) ) is 5 + 2cos(πt). The maximum value is 7, which occurs when cos(πt)=1, which is at t=0,2,4,... So, in [0,3], t=0 and t=2.Therefore, the times are t=0 and t=2, with the rate being 7.So, I think that's the answer.Final AnswerSub-problem 1: The total appreciation at the end is (boxed{15}), and the maximum appreciation level is (boxed{15}) at (t = boxed{3}) hours.Sub-problem 2: The rate of change is maximized at times (t = boxed{0}) and (t = boxed{2}) hours, with the maximum rate being (boxed{7}).</think>

🧠:<think>Alright, so I'm trying to tackle this problem inspired by the movie "Cowboys & Aliens." It's about creating a defense system using graph theory concepts. Let me break it down into the two parts and think through each step carefully.Part 1: Finding the Minimum Spanning Tree (MST) and Proving UniquenessFirst, the scenario describes a connected, undirected graph ( G ) with ( n ) nodes and ( m ) edges. Each edge has a weight ( w(e_i) ). The goal is to find an MST, which is a subset of the edges that connects all the nodes together without any cycles and with the minimum possible total edge weight.I remember that an MST is a spanning tree with the smallest possible sum of edge weights. Since the graph is connected, an MST exists. Now, the problem states that all edge weights are distinct. I need to prove that under this condition, the MST is unique.Let me recall some properties of MSTs. If all edge weights are unique, then there is only one possible MST. This is because when you use algorithms like Krusky's or Prim's, the choices at each step are forced due to the distinct weights. For example, in Krusky's algorithm, you sort all edges by weight and add them one by one, avoiding cycles. Since all weights are different, there's no ambiguity in the order, so the MST must be unique.To formalize this, suppose there are two different MSTs, say ( T_1 ) and ( T_2 ). Since they are different, there must be at least one edge in ( T_1 ) that is not in ( T_2 ). Let ( e ) be the edge with the smallest weight that is in ( T_1 ) but not in ( T_2 ). Adding ( e ) to ( T_2 ) would create a cycle. The cycle must have another edge ( f ) with the same or higher weight than ( e ) because ( e ) was the smallest. But since all weights are distinct, ( f ) must have a higher weight. Removing ( f ) from ( T_2 ) and adding ( e ) would result in a spanning tree with a smaller total weight, contradicting the assumption that ( T_2 ) is an MST. Therefore, no such ( T_2 ) can exist, proving the uniqueness.Part 2: Making the MST ( k )-Edge-ConnectedAfter establishing the MST, the defense system needs to be resilient against up to ( k ) edge disruptions. This means the graph must remain connected even if any ( k ) edges are removed. Such a graph is called ( k )-edge-connected.I need to determine the minimum number of edges ( m' ) to add to the MST to achieve this. Let me recall that a ( k )-edge-connected graph remains connected whenever fewer than ( k ) edges are removed. For a tree, which is minimally connected, the edge connectivity is 1 because removing any single edge disconnects it. So, starting from an MST (which is a tree), we need to increase its edge connectivity to ( k ).What's the minimum number of edges to add to a tree to make it ( k )-edge-connected? I remember that for a tree, the edge connectivity is 1, and to make it ( k )-edge-connected, we need to ensure that between any pair of nodes, there are at least ( k ) edge-disjoint paths.One approach is to consider the concept of making the tree ( k )-edge-connected by adding edges. A tree has ( n-1 ) edges. To make it ( k )-edge-connected, we need to add edges such that the resulting graph is ( k )-edge-connected.I think the formula for the minimum number of edges to add is related to the number of leaves in the tree. Wait, no, perhaps it's more straightforward. For a tree, the edge connectivity is 1, so to make it ( k )-edge-connected, we need to add edges such that the minimum degree of the graph is at least ( k ). But wait, that's vertex connectivity, not edge connectivity.Actually, for edge connectivity, a ( k )-edge-connected graph requires that the minimum degree is at least ( k ). But since we're starting from a tree, which has nodes of degree 1 (leaves), we need to increase the degrees of these leaves.But I think there's a more precise way. Let me recall that the edge connectivity ( lambda ) of a graph is the minimum number of edges that need to be removed to disconnect the graph. For a tree, ( lambda = 1 ). To make ( lambda = k ), we need to ensure that every node has degree at least ( k ), because if a node has degree less than ( k ), removing all its incident edges would disconnect it, which would make ( lambda ) less than ( k ).But wait, actually, the edge connectivity is the minimum number of edges to remove to disconnect the graph, which can be less than the minimum degree. For example, in a cycle, the edge connectivity is 2, and the minimum degree is 2. But in a graph where a node has degree 3 but is connected to a part of the graph with only two edges, the edge connectivity might still be 2.Hmm, maybe I need a different approach. Let me think about the concept of making a tree ( k )-edge-connected. A tree is minimally connected, so to make it ( k )-edge-connected, we need to add edges such that between any two nodes, there are at least ( k ) edge-disjoint paths.One way to achieve this is to make the tree ( k )-edge-connected by adding edges in such a way that the resulting graph is ( k )-edge-connected. The minimum number of edges to add can be determined by considering the structure of the tree.I remember that for a tree, the number of edges to add to make it ( k )-edge-connected is ( lceil frac{k}{2} rceil (n - 1) ). Wait, no, that doesn't sound right.Alternatively, perhaps the number of edges to add is ( (k - 1)(n - 1) ). Let me think. If we have a tree, which has ( n - 1 ) edges, to make it ( k )-edge-connected, we need to add edges such that the total number of edges is at least ( kn/2 ), but that might not be precise.Wait, another approach: For a graph to be ( k )-edge-connected, it must be connected and have at least ( kn/2 ) edges. But since we're starting from a tree with ( n - 1 ) edges, the number of edges to add is ( kn/2 - (n - 1) ). However, this might not always be an integer, and it's a lower bound.But actually, the exact formula for the minimum number of edges to add to a tree to make it ( k )-edge-connected is ( lceil frac{k}{2} rceil (n - 1) ). Wait, that can't be right because for ( k = 1 ), it would give ( lceil 1/2 rceil (n - 1) = n - 1 ), but a tree is already 1-edge-connected, so we don't need to add any edges.I think I'm getting confused. Let me look for a formula or theorem related to this.I recall that for a tree, the minimum number of edges to add to make it ( k )-edge-connected is ( lceil frac{k}{2} rceil (n - 1) - (n - 1) ). Wait, that simplifies to ( (lceil frac{k}{2} rceil - 1)(n - 1) ). But I'm not sure.Alternatively, perhaps the formula is ( (k - 1)(n - 1) ). Let me test this for small values.For ( k = 1 ), we don't need to add any edges, so ( m' = 0 ). The formula would give ( 0 ), which is correct.For ( k = 2 ), we need to make the tree 2-edge-connected, which means it should be a block graph where each block is a 2-edge-connected component. For a tree, which is a minimally connected graph, making it 2-edge-connected requires adding edges to make it a 2-edge-connected graph. The minimum number of edges to add is ( n - 1 ). Wait, no, that can't be right because adding ( n - 1 ) edges to a tree with ( n - 1 ) edges would give ( 2n - 2 ) edges, which is more than necessary.Wait, actually, to make a tree 2-edge-connected, you need to add edges such that every edge is part of a cycle. The minimum number of edges to add is ( n - 1 ). For example, turning a tree into a cycle requires adding one edge, but for a general tree, you might need more.Wait, no, for a tree with ( n ) nodes, the number of edges to add to make it 2-edge-connected is ( n - 1 ). Because each edge in the tree can be part of a cycle by adding one edge, but actually, you need to add edges in such a way that the tree becomes a 2-edge-connected graph. The minimum number of edges to add is ( n - 1 ). Wait, that seems high.Wait, let's think about a simple tree, like a straight line of 3 nodes: A connected to B connected to C. To make it 2-edge-connected, we need to add an edge between A and C. So we added 1 edge. The formula ( n - 1 ) would give 2 edges, which is incorrect. So my previous thought was wrong.So for ( n = 3 ), ( k = 2 ), we need to add 1 edge. For ( n = 4 ), a tree is a straight line A-B-C-D. To make it 2-edge-connected, we need to add edges to form cycles. Adding A-C and B-D would make it 2-edge-connected, but actually, adding just A-C and B-D might not be necessary. Alternatively, adding A-C and C-D? Wait, no, that might not cover all edges.Wait, actually, for a tree with ( n ) nodes, the minimum number of edges to add to make it 2-edge-connected is ( n - 2 ). For ( n = 3 ), that gives 1 edge, which is correct. For ( n = 4 ), it would give 2 edges, which seems correct because adding two edges can create the necessary cycles.But I'm not entirely sure. Let me think differently. The edge connectivity ( lambda ) of a graph is the minimum number of edges that need to be removed to disconnect the graph. For a tree, ( lambda = 1 ). To make ( lambda = k ), we need to ensure that the graph remains connected after removing any ( k - 1 ) edges.One way to achieve this is to make the graph ( k )-edge-connected by adding edges. The minimum number of edges to add can be determined by considering the structure of the tree and ensuring that between any two nodes, there are at least ( k ) edge-disjoint paths.I think the formula for the minimum number of edges to add to a tree to make it ( k )-edge-connected is ( (k - 1)(n - 1) ). Wait, let's test this.For ( k = 1 ), ( m' = 0 ), which is correct.For ( k = 2 ), ( m' = (2 - 1)(n - 1) = n - 1 ). But earlier, for ( n = 3 ), we only needed to add 1 edge, which is ( 3 - 1 = 2 ), which is incorrect. So this formula is wrong.Wait, perhaps it's ( lceil frac{k}{2} rceil (n - 1) ). For ( k = 2 ), it would be ( 1 times (n - 1) = n - 1 ), which again is incorrect for ( n = 3 ).I think I need to find a better approach. Let me recall that for a tree, the number of edges to add to make it ( k )-edge-connected is ( (k - 1)(n - 1) ). Wait, no, that can't be right because for ( k = 2 ), it would require adding ( n - 1 ) edges, which is too many.Wait, perhaps the formula is ( lfloor frac{k}{2} rfloor (n - 1) ). For ( k = 2 ), that would be ( 1 times (n - 1) ), which again is too high.I'm getting stuck here. Let me think about the concept of making a tree ( k )-edge-connected. A tree has ( n - 1 ) edges and is minimally connected. To make it ( k )-edge-connected, we need to add edges such that the graph remains connected after removing any ( k - 1 ) edges.One way to achieve this is to make the graph ( k )-edge-connected by ensuring that every node has degree at least ( k ). However, this is a sufficient condition but not necessary. For example, in a cycle, every node has degree 2, and the edge connectivity is 2.But for a tree, the leaves have degree 1. To make the edge connectivity ( k ), we need to increase the degrees of these leaves. Each leaf needs to have at least ( k ) edges. Since each edge added can connect two leaves, the number of edges needed is roughly ( lceil frac{k}{2} rceil times text{number of leaves} ). But the number of leaves in a tree can vary, so this approach might not give a general formula.Wait, perhaps a better way is to consider that for a tree, the number of edges to add to make it ( k )-edge-connected is ( (k - 1)(n - 1) ). Let me test this for small values.For ( n = 3 ), ( k = 2 ): ( (2 - 1)(3 - 1) = 2 ). But we only need to add 1 edge, so this is incorrect.Wait, maybe it's ( lfloor frac{k}{2} rfloor (n - 1) ). For ( k = 2 ), ( 1 times (n - 1) ). For ( n = 3 ), that's 2 edges, which is still incorrect.I think I'm overcomplicating this. Let me recall that the minimum number of edges to add to a tree to make it ( k )-edge-connected is ( (k - 1)(n - 1) ). Wait, no, that can't be right because for ( k = 2 ), it would require adding ( n - 1 ) edges, which is too many.Wait, perhaps the formula is ( lceil frac{k}{2} rceil (n - 1) ). For ( k = 2 ), ( 1 times (n - 1) ), which is still too high.I think I need to look for a different approach. Let me consider that a ( k )-edge-connected graph must have at least ( kn/2 ) edges. Since the tree has ( n - 1 ) edges, the number of edges to add is ( kn/2 - (n - 1) ). However, this might not always be an integer, and it's a lower bound.But for example, for ( n = 3 ), ( k = 2 ): ( 2*3/2 - 2 = 3 - 2 = 1 ), which is correct. For ( n = 4 ), ( k = 2 ): ( 4 - 3 = 1 ). But wait, for ( n = 4 ), a tree has 3 edges. To make it 2-edge-connected, we need to add at least 1 edge, which is correct because adding one edge between two non-adjacent nodes creates a cycle, making it 2-edge-connected.Wait, so the formula ( m' = lceil frac{kn}{2} rceil - (n - 1) ) might work. Let's test it.For ( n = 3 ), ( k = 2 ): ( lceil 3 rceil - 2 = 3 - 2 = 1 ). Correct.For ( n = 4 ), ( k = 2 ): ( lceil 4 rceil - 3 = 4 - 3 = 1 ). Correct.For ( n = 4 ), ( k = 3 ): ( lceil 6 rceil - 3 = 6 - 3 = 3 ). So adding 3 edges to a tree with 3 edges gives 6 edges, which is a complete graph ( K_4 ), which is 3-edge-connected. Correct.For ( n = 5 ), ( k = 2 ): ( lceil 5 rceil - 4 = 5 - 4 = 1 ). Adding 1 edge to a tree with 4 edges gives 5 edges. Is 5 edges sufficient for 2-edge-connectedness? Let's see. A tree with 5 nodes has 4 edges. Adding one edge creates a single cycle. The edge connectivity is now 2 because you can't disconnect the graph by removing one edge. So yes, correct.Wait, but for ( n = 5 ), ( k = 3 ): ( lceil 7.5 rceil - 4 = 8 - 4 = 4 ). So adding 4 edges to a tree with 4 edges gives 8 edges. The complete graph ( K_5 ) has 10 edges, so 8 edges might not be sufficient for 3-edge-connectedness. Wait, no, 3-edge-connectedness requires that the graph remains connected after removing any 2 edges. A graph with 8 edges on 5 nodes might have edge connectivity 3, but I'm not sure.Wait, actually, the formula ( m' = lceil frac{kn}{2} rceil - (n - 1) ) gives the minimum number of edges to add to a tree to make it ( k )-edge-connected. Let me verify for ( n = 5 ), ( k = 3 ):( lceil frac{3*5}{2} rceil = lceil 7.5 rceil = 8 ). So ( m' = 8 - 4 = 4 ). Adding 4 edges to a tree with 4 edges gives 8 edges. Now, does a graph with 5 nodes and 8 edges have edge connectivity at least 3?The maximum edge connectivity for a graph with 5 nodes is 4 (since each node can have degree 4). To have edge connectivity 3, the graph must remain connected after removing any 2 edges. A graph with 8 edges on 5 nodes is quite dense. Let's see: the complete graph ( K_5 ) has 10 edges, so 8 edges is 2 edges less than complete. It's likely that such a graph is 3-edge-connected because it's missing only 2 edges, which can't disconnect the graph by removing 2 edges.Wait, but actually, if you remove 2 edges, the graph might still be connected. So yes, 8 edges would suffice for 3-edge-connectedness.Therefore, the general formula for the minimum number of edges ( m' ) to add to the MST to achieve ( k )-edge-connectedness is:( m' = lceil frac{kn}{2} rceil - (n - 1) )Simplifying this:( m' = lceil frac{kn}{2} rceil - n + 1 )But since ( lceil frac{kn}{2} rceil ) can be written as ( lfloor frac{kn + 1}{2} rfloor ), but perhaps it's better to leave it as is.Alternatively, since ( kn ) might be even or odd, we can express it as:( m' = leftlceil frac{kn}{2} rightrceil - n + 1 )But let's test this formula for ( n = 3 ), ( k = 2 ):( lceil 3 rceil - 3 + 1 = 3 - 3 + 1 = 1 ). Correct.For ( n = 4 ), ( k = 2 ):( lceil 4 rceil - 4 + 1 = 4 - 4 + 1 = 1 ). Correct.For ( n = 5 ), ( k = 3 ):( lceil 7.5 rceil - 5 + 1 = 8 - 5 + 1 = 4 ). Correct.Another test: ( n = 2 ), ( k = 1 ):( lceil 1 rceil - 2 + 1 = 1 - 2 + 1 = 0 ). Correct, since a single edge is already 1-edge-connected.Wait, but for ( n = 2 ), the tree is just one edge, and to make it 1-edge-connected, we don't need to add any edges. So correct.Another test: ( n = 4 ), ( k = 3 ):( lceil 6 rceil - 4 + 1 = 6 - 4 + 1 = 3 ). So adding 3 edges to a tree with 3 edges gives 6 edges. Is a graph with 4 nodes and 6 edges 3-edge-connected? Yes, because ( K_4 ) has 6 edges and is 3-edge-connected. So correct.Therefore, the formula seems to hold.But wait, let me think again. The formula ( m' = lceil frac{kn}{2} rceil - (n - 1) ) gives the number of edges to add. But is this always the case? For example, for ( n = 6 ), ( k = 2 ):( lceil 6 rceil - 5 = 6 - 5 = 1 ). So adding 1 edge to a tree with 5 edges gives 6 edges. Is a graph with 6 edges on 6 nodes 2-edge-connected? Not necessarily. For example, if the added edge creates a single cycle, the graph might still have edge connectivity 2. Wait, no, actually, a tree with 5 edges plus one edge makes 6 edges, which is a connected graph with cyclomatic number 1. But edge connectivity is 2 only if every edge is part of a cycle. Wait, no, edge connectivity is the minimum number of edges to remove to disconnect the graph. If the graph has a single cycle, removing one edge from the cycle won't disconnect the graph, but removing another edge might. Wait, no, if you have a tree plus one edge, you have exactly one cycle. The edge connectivity is 2 because you need to remove at least two edges to disconnect the graph. Wait, no, actually, if you have a cycle, the edge connectivity is at least 2. Because to disconnect the graph, you need to remove at least two edges from the cycle. So yes, adding one edge to a tree with 5 edges makes it 2-edge-connected. So the formula holds.Wait, but for ( n = 6 ), ( k = 3 ):( lceil 9 rceil - 5 = 9 - 5 = 4 ). So adding 4 edges to a tree with 5 edges gives 9 edges. Is a graph with 6 nodes and 9 edges 3-edge-connected? The complete graph ( K_6 ) has 15 edges, so 9 edges is less than that. But 9 edges is more than ( 3*6/2 = 9 ), so it's exactly the threshold. So a graph with 9 edges on 6 nodes is 3-edge-connected if it meets the necessary conditions. I think it is, because the edge connectivity is at least 3.Wait, actually, the formula ( m' = lceil frac{kn}{2} rceil - (n - 1) ) gives the minimum number of edges to add to a tree to make it ( k )-edge-connected. This is because the total number of edges needed for a ( k )-edge-connected graph is at least ( frac{kn}{2} ), and since the tree already has ( n - 1 ) edges, the number of edges to add is ( lceil frac{kn}{2} rceil - (n - 1) ).Therefore, the general formula for ( m' ) is:( m' = leftlceil frac{kn}{2} rightrceil - (n - 1) )Simplifying this:( m' = leftlceil frac{kn}{2} rightrceil - n + 1 )But we can also write this as:( m' = leftlceil frac{kn - 2(n - 1)}{2} rightrceil )Which simplifies to:( m' = leftlceil frac{(k - 2)n + 2}{2} rightrceil )But this might not be necessary. The original expression is sufficient.Wait, let me check for ( n = 5 ), ( k = 2 ):( lceil 5 rceil - 4 = 5 - 4 = 1 ). Correct.For ( n = 6 ), ( k = 2 ):( lceil 6 rceil - 5 = 6 - 5 = 1 ). Correct.For ( n = 6 ), ( k = 3 ):( lceil 9 rceil - 5 = 9 - 5 = 4 ). Correct.Therefore, the formula seems consistent.So, to summarize:1. The MST is unique if all edge weights are distinct because any two MSTs would lead to a contradiction when considering the smallest differing edge.2. The minimum number of edges ( m' ) to add to the MST to make it ( k )-edge-connected is ( leftlceil frac{kn}{2} rightrceil - (n - 1) ).</think>

🧠:<think>Okay, so I have this problem about a retired teacher who's analyzing voter turnout data. There are two questions here, both related to probability models. Let me try to tackle them one by one.Starting with the first question: It says that voter arrivals follow a Poisson process with an average rate of 250 voters per hour. We need to find the probability that exactly 20 voters will arrive between 9 AM and 10 AM.Hmm, Poisson process, right? So, I remember that a Poisson process is a counting process where the number of events (in this case, voter arrivals) in a given interval follows a Poisson distribution. The key parameters are the rate parameter λ, which is the average number of events per interval.Given that the average rate is 250 voters per hour, and we're looking at a specific one-hour window from 9 AM to 10 AM. So, the interval length t is 1 hour. Therefore, the expected number of voters in this interval, λ, should be 250 * 1 = 250.Wait, but the question is asking for the probability of exactly 20 voters arriving in that hour. That seems low because the average is 250. Maybe I misread something? Let me check again.No, it says 250 voters per hour, so over one hour, the expected number is indeed 250. So, the probability of exactly 20 voters arriving would be extremely low, but let's compute it anyway.The formula for the Poisson probability mass function is:P(k) = (λ^k * e^(-λ)) / k!Where k is the number of occurrences, which is 20 here.So plugging in the numbers:P(20) = (250^20 * e^(-250)) / 20!But wait, 250^20 is a huge number, and e^(-250) is a very small number. This might result in an underflow if I try to compute it directly, but maybe I can compute it using logarithms or recognize that it's practically zero.Alternatively, maybe I made a mistake in interpreting the rate. Let me think again. The rate is 250 voters per hour, so over one hour, λ is 250. So, yes, 20 is way below the mean, which is 250. So, the probability is going to be extremely small.But perhaps the question is correct, and I just need to compute it as is. Let me proceed.Calculating P(20):First, compute 250^20. That's 250 multiplied by itself 20 times. That's an astronomically large number. Then, e^(-250) is approximately... Well, e^(-250) is like 1 divided by e^250, which is a number so small that it's effectively zero for most practical purposes.But let's see if we can compute this using logarithms to avoid underflow.Taking the natural logarithm of P(20):ln(P(20)) = 20 * ln(250) - 250 - ln(20!)Compute each term:ln(250) ≈ 5.52126So, 20 * 5.52126 ≈ 110.4252Then, subtract 250: 110.4252 - 250 = -139.5748Now, compute ln(20!). 20! is 2432902008176640000. The natural log of that is approximately ln(2.43290200817664 × 10^18) = ln(2.43290200817664) + ln(10^18) ≈ 0.889 + 41.558 ≈ 42.447So, ln(P(20)) ≈ -139.5748 - 42.447 ≈ -182.0218Therefore, P(20) ≈ e^(-182.0218). Let me compute that.e^(-182) is like 1 / e^182. e^182 is approximately... Well, e^100 is about 2.688117 × 10^43, so e^182 is e^100 * e^82. e^82 is about e^80 * e^2 ≈ (1.9073486 × 10^34) * 7.389056 ≈ 1.409 × 10^35. So, e^182 ≈ 2.688 × 10^43 * 1.409 × 10^35 ≈ 3.787 × 10^78. Therefore, e^(-182) ≈ 2.64 × 10^(-79). But wait, that's just e^(-182). But our exponent is -182.0218, which is slightly less, so maybe around 2.6 × 10^(-79). So, P(20) is approximately 2.6 × 10^(-79). That's an incredibly small probability, practically zero.But let me double-check if I did the calculations correctly. Maybe I made a mistake in the logarithms.Wait, ln(20!) is actually known. Let me recall that ln(20!) is approximately 42.3356. Let me verify that.Yes, using Stirling's approximation: ln(n!) ≈ n ln n - n + (ln(2πn))/2. For n=20:ln(20!) ≈ 20 ln 20 - 20 + (ln(40π))/2Compute 20 ln 20: 20 * 2.9957 ≈ 59.914Subtract 20: 59.914 - 20 = 39.914Compute (ln(40π))/2: ln(40 * 3.1416) ≈ ln(125.664) ≈ 4.834, divided by 2 is 2.417So, total approximation: 39.914 + 2.417 ≈ 42.331, which is close to the actual value of 42.3356. So, that seems correct.So, ln(P(20)) ≈ -182.0218, so P(20) ≈ e^(-182.0218) ≈ 2.6 × 10^(-79). So, yes, that's correct.But wait, is this the right approach? Because in reality, when λ is large, the Poisson distribution can be approximated by a normal distribution. But in this case, we're looking for a very small k compared to λ, so maybe the Poisson formula is still the way to go.Alternatively, maybe the rate is given per hour, but the interval is one hour, so λ is 250. So, yes, the calculation is correct.So, the probability is approximately 2.6 × 10^(-79), which is effectively zero. So, the answer is practically zero, but in exact terms, it's (250^20 * e^(-250)) / 20!.But maybe the question expects the answer in terms of the formula, not the numerical value, because the number is too small to compute practically.Wait, let me check the question again: "what is the probability that exactly 20 voters will arrive between 9 AM and 10 AM?" It doesn't specify whether to compute it numerically or just write the formula. Since it's a probability, maybe they expect the formula.But in the initial problem statement, it's a teacher analyzing data, so perhaps they expect a numerical answer, but given the magnitude, it's effectively zero. Alternatively, maybe I misread the rate.Wait, 250 voters per hour over 12 hours (from 7 AM to 7 PM) would be 3000 voters, but the registered voters are 5000. So, maybe the rate is 250 per hour, but the total expected voters would be 250 * 12 = 3000, which is less than 5000, so that makes sense.But in any case, for the one-hour interval, λ is 250, so the calculation is correct.So, the answer is (250^20 * e^(-250)) / 20! which is approximately 2.6 × 10^(-79).But maybe I should express it in terms of the Poisson PMF.Alternatively, perhaps the rate is given per hour, but the interval is one hour, so λ is 250, so the formula is correct.Okay, moving on to the second question: The teacher collects data on the time gaps between consecutive voter arrivals. If the time gaps are exponentially distributed, what is the expected time gap between two consecutive voters arriving at the polling station?Hmm, exponential distribution is the inter-arrival time in a Poisson process. The expected value of an exponential distribution is 1/λ, where λ is the rate parameter.But wait, in the Poisson process, the rate λ is the average number of events per unit time. So, if the average rate is 250 voters per hour, then the expected time between arrivals is 1/λ, which would be 1/250 hours per voter.But let me think carefully. The rate λ is 250 per hour, so the expected number of arrivals per hour is 250. Therefore, the expected time between arrivals is 1/250 hours. To convert that into minutes, since 1 hour is 60 minutes, 1/250 hours is 60/250 minutes, which is 0.24 minutes, or 14.4 seconds.Wait, that seems very fast. 250 voters per hour is about 4.1667 voters per minute, so the expected time between arrivals is 1/4.1667 minutes, which is 0.24 minutes, which is 14.4 seconds. That seems correct.But let me make sure. In a Poisson process, the inter-arrival times are exponentially distributed with parameter λ, where λ is the rate per unit time. So, the expected value is 1/λ.Given that λ is 250 per hour, the expected time between arrivals is 1/250 hours, which is 0.004 hours. To convert to minutes, multiply by 60: 0.004 * 60 = 0.24 minutes, which is 14.4 seconds.Yes, that seems correct.Alternatively, if we consider the rate in per minute terms, 250 per hour is 250/60 ≈ 4.1667 per minute. So, the expected time between arrivals is 1/4.1667 ≈ 0.24 minutes, which is consistent.So, the expected time gap is 1/250 hours, or 0.24 minutes, or 14.4 seconds.But the question asks for the expected time gap, so probably in hours or minutes. Since the voting hours are given in hours, but the first question was about a one-hour interval, maybe the answer is expected in hours.But let me see. The question says "the time gaps between consecutive voter arrivals." It doesn't specify the unit, but since the rate is given per hour, maybe the expected time gap is in hours.But 1/250 hours is 0.004 hours, which is 0.24 minutes, which is 14.4 seconds. So, depending on the unit, but perhaps the answer is 1/250 hours.Alternatively, maybe the question expects the answer in minutes, so 0.24 minutes or 14.4 seconds.But let me check the units. The voting hours are from 7 AM to 7 PM, which is 12 hours, but the rate is given per hour, so the unit is per hour.Therefore, the expected time gap is 1/250 hours, which is 0.004 hours. Alternatively, in minutes, it's 0.24 minutes, or 14.4 seconds.But the question doesn't specify the unit, so maybe we can leave it in hours or specify both.But in probability questions, unless specified, it's safer to use the same unit as given in the problem. Since the rate is given per hour, the expected time gap would be in hours.So, 1/250 hours is the expected time gap.Alternatively, if we convert it to minutes, it's 0.24 minutes, which is 14.4 seconds. But since the question is about time gaps, maybe minutes or seconds are more intuitive.But let me think again. In the Poisson process, the inter-arrival times are exponential with parameter λ, where λ is per unit time. So, if λ is 250 per hour, then the expected inter-arrival time is 1/250 hours.So, the answer is 1/250 hours, which is 0.004 hours, or 0.24 minutes, or 14.4 seconds.But since the question doesn't specify the unit, perhaps it's best to express it in hours, as that's the unit used in the rate.So, the expected time gap is 1/250 hours.But let me make sure. The exponential distribution's expected value is 1/λ, where λ is the rate parameter. So, if λ is 250 per hour, then the expected time between arrivals is 1/250 hours.Yes, that's correct.So, summarizing:1. The probability is (250^20 * e^(-250)) / 20!, which is approximately 2.6 × 10^(-79).2. The expected time gap is 1/250 hours, which is 0.004 hours, or 0.24 minutes, or 14.4 seconds.But for the first question, since the numerical value is extremely small, maybe it's better to leave it in the Poisson PMF form, unless a numerical approximation is specifically requested.But the question says "what is the probability," so perhaps they expect the formula.Alternatively, if they expect a numerical answer, it's approximately 2.6 × 10^(-79), which is practically zero.But let me check if I can write it in terms of factorials and exponents.Yes, the exact probability is (250^20 * e^(-250)) / 20!.So, that's the answer.For the second question, the expected time gap is 1/250 hours, which is 0.004 hours.Alternatively, in minutes, it's 0.24 minutes, or 14.4 seconds.But since the rate was given per hour, maybe the answer is expected in hours.So, to answer:1. The probability is (250^20 * e^(-250)) / 20!.2. The expected time gap is 1/250 hours.But let me double-check if I interpreted the rate correctly. The rate is 250 voters per hour, so λ = 250 per hour. Therefore, the expected inter-arrival time is 1/λ = 1/250 hours.Yes, that's correct.Alternatively, sometimes the exponential distribution is parameterized with β = 1/λ, so the expected value is β. So, in that case, β = 1/250 hours.Yes, that's consistent.So, I think that's the answer.Final Answer1. The probability is boxed{dfrac{250^{20} e^{-250}}{20!}}.2. The expected time gap is boxed{dfrac{1}{250}} hours.</think>

🧠:<think>Alright, so I have this problem where a construction company owner is trying to decide between two suppliers, A and B, for heavy machinery. The decision is based on total costs over five years, including initial, maintenance, and downtime costs. Hmm, okay, let me break this down step by step.First, I need to figure out the total cost for each supplier over five years. That means I have to consider three components: initial cost, maintenance costs, and downtime costs. Each of these components has different factors, so I need to calculate each one separately and then sum them up for each supplier.Starting with Supplier A:1. Initial Cost: This is straightforward. It's a one-time cost of 500,000. So, that's easy.2. Maintenance Costs: Supplier A charges 30,000 per year, but this increases by 5% each subsequent year. So, I need to calculate the maintenance cost for each of the five years, taking into account the 5% increase each year. That sounds like a geometric progression where each year's cost is 1.05 times the previous year's.3. Downtime Costs: Supplier A's machines have a 2% downtime per year, and each 1% downtime costs 20,000. So, 2% downtime would cost 2 * 20,000 = 40,000 per year. Is this cost fixed each year, or does it increase? The problem doesn't specify an increase, so I think it's a flat 40,000 per year for downtime.Similarly, for Supplier B:1. Initial Cost: 450,000, which is less than Supplier A.2. Maintenance Costs: 35,000 per year with a 3% increase each year. Again, this is a geometric progression, but with a 3% increase each year.3. Downtime Costs: 3% downtime per year, with each 1% costing 25,000. So, 3% downtime would cost 3 * 25,000 = 75,000 per year. Again, I think this is a flat cost unless specified otherwise.Okay, so to summarize, for each supplier, I need to calculate:- Initial Cost: one-time- Maintenance Costs: annual, increasing at a certain rate- Downtime Costs: annual, fixedThen, sum all these up over five years.Let me structure this into tables for clarity.For Supplier A:- Year 0 (Initial Cost): 500,000- Year 1: Maintenance = 30,000, Downtime = 40,000- Year 2: Maintenance = 30,000 * 1.05, Downtime = 40,000- Year 3: Maintenance = 30,000 * (1.05)^2, Downtime = 40,000- Year 4: Maintenance = 30,000 * (1.05)^3, Downtime = 40,000- Year 5: Maintenance = 30,000 * (1.05)^4, Downtime = 40,000Similarly, for Supplier B:- Year 0 (Initial Cost): 450,000- Year 1: Maintenance = 35,000, Downtime = 75,000- Year 2: Maintenance = 35,000 * 1.03, Downtime = 75,000- Year 3: Maintenance = 35,000 * (1.03)^2, Downtime = 75,000- Year 4: Maintenance = 35,000 * (1.03)^3, Downtime = 75,000- Year 5: Maintenance = 35,000 * (1.03)^4, Downtime = 75,000Wait, but the problem says "over a 5-year period," so does that include year 0? I think so, because year 0 is the initial cost, and then years 1 through 5 are the subsequent years. So, total of 5 years of operation, each with their respective maintenance and downtime costs.So, for each supplier, the total cost is:Total Cost = Initial Cost + Sum of Maintenance Costs over 5 years + Sum of Downtime Costs over 5 yearsLet me calculate each component step by step.Calculating for Supplier A:1. Initial Cost: 500,0002. Maintenance Costs: - Year 1: 30,000 - Year 2: 30,000 * 1.05 = 31,500 - Year 3: 31,500 * 1.05 = 33,075 - Year 4: 33,075 * 1.05 = 34,728.75 - Year 5: 34,728.75 * 1.05 ≈ 36,465.19 Let me verify these calculations: - Year 1: 30,000 - Year 2: 30,000 * 1.05 = 31,500 - Year 3: 31,500 * 1.05 = 33,075 - Year 4: 33,075 * 1.05 = 34,728.75 - Year 5: 34,728.75 * 1.05 = let's compute 34,728.75 * 1.05 34,728.75 * 1.05: 34,728.75 + (34,728.75 * 0.05) = 34,728.75 + 1,736.4375 = 36,465.1875, which is approximately 36,465.19 So, the maintenance costs over five years are: 30,000 + 31,500 + 33,075 + 34,728.75 + 36,465.19 Let's add these up: 30,000 + 31,500 = 61,500 61,500 + 33,075 = 94,575 94,575 + 34,728.75 = 129,303.75 129,303.75 + 36,465.19 ≈ 165,768.94 So, total maintenance cost for Supplier A is approximately 165,768.943. Downtime Costs: Each year, it's 40,000. So, over five years, that's 5 * 40,000 = 200,000 So, total downtime cost is 200,0004. Total Cost for Supplier A: Initial + Maintenance + Downtime = 500,000 + 165,768.94 + 200,000 Let's compute that: 500,000 + 165,768.94 = 665,768.94 665,768.94 + 200,000 = 865,768.94 So, approximately 865,768.94Calculating for Supplier B:1. Initial Cost: 450,0002. Maintenance Costs: - Year 1: 35,000 - Year 2: 35,000 * 1.03 = 36,050 - Year 3: 36,050 * 1.03 ≈ 37,131.50 - Year 4: 37,131.50 * 1.03 ≈ 38,245.45 - Year 5: 38,245.45 * 1.03 ≈ 39,392.81 Let me verify these calculations: - Year 1: 35,000 - Year 2: 35,000 * 1.03 = 36,050 - Year 3: 36,050 * 1.03 = 36,050 + (36,050 * 0.03) = 36,050 + 1,081.50 = 37,131.50 - Year 4: 37,131.50 * 1.03 = 37,131.50 + (37,131.50 * 0.03) = 37,131.50 + 1,113.945 ≈ 38,245.45 - Year 5: 38,245.45 * 1.03 ≈ 38,245.45 + (38,245.45 * 0.03) ≈ 38,245.45 + 1,147.36 ≈ 39,392.81 So, maintenance costs over five years: 35,000 + 36,050 + 37,131.50 + 38,245.45 + 39,392.81 Let's add these up: 35,000 + 36,050 = 71,050 71,050 + 37,131.50 = 108,181.50 108,181.50 + 38,245.45 = 146,426.95 146,426.95 + 39,392.81 ≈ 185,819.76 So, total maintenance cost for Supplier B is approximately 185,819.763. Downtime Costs: Each year, it's 75,000. So, over five years, that's 5 * 75,000 = 375,000 So, total downtime cost is 375,0004. Total Cost for Supplier B: Initial + Maintenance + Downtime = 450,000 + 185,819.76 + 375,000 Let's compute that: 450,000 + 185,819.76 = 635,819.76 635,819.76 + 375,000 = 1,010,819.76 So, approximately 1,010,819.76Comparing the Total Costs:- Supplier A: ~865,768.94- Supplier B: ~1,010,819.76So, clearly, Supplier A is more cost-effective over the five-year period.Wait, but let me double-check my calculations to make sure I didn't make any errors.For Supplier A:- Maintenance: 30k, 31.5k, 33.075k, 34.72875k, 36.4651875kAdding them up:30 + 31.5 = 61.561.5 + 33.075 = 94.57594.575 + 34.72875 = 129.30375129.30375 + 36.4651875 ≈ 165.7689375kYes, that's correct.Downtime: 5 * 40k = 200kTotal: 500k + 165.7689k + 200k ≈ 865.7689kFor Supplier B:- Maintenance: 35k, 36.05k, 37.1315k, 38.24545k, 39.39281kAdding them up:35 + 36.05 = 71.0571.05 + 37.1315 ≈ 108.1815108.1815 + 38.24545 ≈ 146.42695146.42695 + 39.39281 ≈ 185.81976kDowntime: 5 * 75k = 375kTotal: 450k + 185.81976k + 375k ≈ 1,010.81976kSo, the calculations seem correct.Therefore, over five years, Supplier A is cheaper by approximately 1,010,819.76 - 865,768.94 = 145,050.82That's a significant difference. So, the owner should choose Supplier A.But wait, just to be thorough, let me consider if the downtime costs are indeed fixed each year or if they also increase with the maintenance costs. The problem states:"the effects of annual increases in maintenance costs and annual downtime costs based on the given percentages."Wait, hold on. Let me re-read the problem statement."2. Maintenance Costs: ... annual increases ... 3. Downtime Costs: ... annual downtime costs based on the given percentages."Wait, does that mean that downtime costs also increase annually? Or is it just maintenance costs that increase?Looking back:"2. Maintenance Costs: ... annual maintenance cost ... with a 5% increase each subsequent year. ... 3. Downtime Costs: ... downtime costs ... based on the given percentages."Hmm, the problem says "based on the given percentages," but it's unclear whether downtime costs also increase annually. Let me check the exact wording:"3. Downtime Costs: The reliability of the machines impacts the downtime costs. Supplier A's machines have an expected downtime of 2% per year, which translates to a loss of 20,000 per 1% downtime annually. Supplier B's machines have an expected downtime of 3% per year, translating to a loss of 25,000 per 1% downtime annually."So, it says "annually" for downtime costs as well. So, does that mean the downtime cost is fixed each year, or does it increase? The problem doesn't specify an increase rate for downtime costs, only for maintenance costs. So, I think downtime costs are fixed each year.But wait, the problem says "based on the given percentages." The given percentages are 2% and 3% for downtime, and 20k and 25k per 1%. So, perhaps the downtime cost is calculated each year based on the downtime percentage, which is fixed, but maybe the cost per 1% downtime changes? But the problem doesn't specify that. It just gives a flat rate per 1% downtime.Therefore, I think downtime costs are fixed each year. So, my initial calculations are correct.Alternatively, if downtime costs also increased annually, but the problem doesn't specify a rate, so I think it's safe to assume they are fixed.Therefore, my conclusion remains that Supplier A is more cost-effective.Final AnswerThe total cost for Supplier A is boxed{865769} dollars and for Supplier B is boxed{1010820} dollars. Therefore, Supplier A is more cost-effective.Wait, but the problem asks to present the final answer within boxes, but it's two separate answers. Hmm, the instructions say "put your final answer within boxed{}." Maybe I should present both in separate boxes or indicate which is more cost-effective.But looking back, the problem has two sub-problems: 1) calculate total costs, 2) determine which is more cost-effective.So, perhaps I need to present both total costs and then state which is better.But the user instruction says "put your final answer within boxed{}." Maybe they expect just the conclusion, but the initial problem has two parts. Hmm.Alternatively, perhaps I should present both total costs boxed and then the conclusion.But the initial problem says "solve the following sub-problems: 1. Calculate the total cost... 2. Determine which supplier is more cost-effective."So, perhaps I need to answer both.But the user instruction says "put your final answer within boxed{}." Maybe they expect the numerical answers for each sub-problem.But in the initial problem, the first sub-problem is to calculate the total cost for each, and the second is to determine which is more cost-effective.So, perhaps I should present both total costs boxed, and then a separate box for the conclusion.But the system might expect one box. Hmm.Alternatively, perhaps I can write both total costs and then the conclusion in words.But the user instruction says "put your final answer within boxed{}." So, maybe just the numerical total costs for each, and then the conclusion outside the box.But the initial problem is in Chinese, and the user translated it, but the instructions are in English.Wait, looking back, the user wrote:"Please reason step by step, and put your final answer within boxed{}."So, perhaps the final answer is just the conclusion, i.e., which supplier is more cost-effective, boxed.But the first sub-problem is to calculate the total costs, which are numerical, and the second is to determine which is better.But the user instruction says "put your final answer within boxed{}." So, maybe the final answer is the conclusion, i.e., Supplier A.Alternatively, perhaps both total costs and the conclusion.But in the initial problem, it's two sub-problems, so perhaps both answers are needed.But the user instruction says "put your final answer within boxed{}." So, maybe just the conclusion.But I think, to be safe, I'll present both total costs and then the conclusion.But since the user instruction is a bit ambiguous, I'll follow the initial problem's sub-problems.So, first, the total costs for each supplier, then the conclusion.But in terms of the final answer, perhaps just the conclusion is needed.But the initial problem says "solve the following sub-problems: 1. Calculate... 2. Determine...". So, the final answer should address both.But the user instruction says "put your final answer within boxed{}." So, perhaps the conclusion is the final answer, which is that Supplier A is more cost-effective.But to be thorough, I think I should present both total costs and the conclusion.But since the user instruction says "put your final answer within boxed{}", perhaps just the conclusion.But in the initial problem, the first sub-problem is to calculate the total cost, which is numerical, and the second is to determine which is better, which is a choice.But the user instruction says "put your final answer within boxed{}." So, maybe the numerical total costs for each supplier, and then the conclusion.But the system might expect one box. Hmm.Alternatively, perhaps I can write the total costs for each supplier in separate boxes and then the conclusion in words.But the user instruction says "put your final answer within boxed{}." So, perhaps the conclusion is the final answer, which is a single box.But in the initial problem, the first sub-problem is to calculate the total cost, which is numerical, and the second is to determine which is better, which is a choice.So, perhaps the final answer is just the conclusion, i.e., Supplier A.But to be safe, I think I should present both total costs and then the conclusion.But given the user instruction, I think the expected answer is the conclusion, which is Supplier A.But to be thorough, I think I should present both total costs and the conclusion.But since the user instruction says "put your final answer within boxed{}", perhaps I need to box the conclusion.Alternatively, perhaps the user expects both total costs boxed and the conclusion.But given the ambiguity, I think I'll present both total costs and the conclusion, with the conclusion boxed.So, the total cost for Supplier A is approximately 865,769, and for Supplier B, approximately 1,010,820. Therefore, Supplier A is more cost-effective.But to present it as per the user instruction, perhaps:The total costs are Supplier A: boxed{865769} and Supplier B: boxed{1010820}. Therefore, Supplier A is more cost-effective.But the user instruction says "put your final answer within boxed{}", so perhaps just the conclusion.But given the initial problem's sub-problems, I think both are needed.But to comply with the user instruction, perhaps just the conclusion.But I'm not sure. Maybe I should follow the initial problem's structure.Wait, in the initial problem, the user wrote:"Please reason step by step, and put your final answer within boxed{}."So, perhaps the final answer is just the conclusion, i.e., which supplier is more cost-effective, boxed.But the first sub-problem is to calculate the total costs, which are numerical, but the second is to determine which is better.But the user instruction says "put your final answer within boxed{}." So, perhaps the conclusion is the final answer, which is a single boxed answer.But in the initial problem, the user wrote:"Given these factors, solve the following sub-problems:1. Calculate the total cost...2. Based on the total cost calculations, determine which supplier is more cost-effective..."So, the final answer is the conclusion, which is the answer to the second sub-problem.Therefore, the final answer is that Supplier A is more cost-effective, which can be boxed.But in the initial problem, the user wrote:"Please reason step by step, and put your final answer within boxed{}."So, perhaps the final answer is just the conclusion, boxed.But to be thorough, I think I should present both total costs and the conclusion.But given the user instruction, I think the expected answer is the conclusion, which is a single box.But to be safe, I'll present both total costs and the conclusion, with the conclusion boxed.So, the total cost for Supplier A is approximately 865,769, and for Supplier B, approximately 1,010,820. Therefore, the more cost-effective supplier is boxed{A}.But the user instruction says "put your final answer within boxed{}", so maybe just the conclusion.Alternatively, perhaps I should present both total costs in boxes and then the conclusion.But given the ambiguity, I think I'll present both total costs and the conclusion, with the conclusion boxed.So, final answer:The total cost for Supplier A is boxed{865769} dollars and for Supplier B is boxed{1010820} dollars. Therefore, the more cost-effective supplier is boxed{A}.But the user instruction says "put your final answer within boxed{}", so perhaps just the conclusion.But I think the problem expects both parts answered, so I'll proceed accordingly.Final AnswerThe total cost for Supplier A is boxed{865769} dollars and for Supplier B is boxed{1010820} dollars. Therefore, the more cost-effective supplier is boxed{A}.</think>