🧠:<think>Okay, so I have this problem about a toroidal grid graph and then a related graph constructed from it. Let me try to break it down step by step.First, part 1: Determine the chromatic number of a toroidal grid graph G with dimensions 5x7. Hmm, I remember that a toroidal grid graph is like a grid where the top and bottom edges are connected, as well as the left and right edges, forming a doughnut shape, or a torus. So, it's a grid graph embedded on a torus.I also recall that the chromatic number of a graph is the smallest number of colors needed to color the vertices so that no two adjacent vertices share the same color. For planar graphs, the four-color theorem says that four colors suffice, but since this is a toroidal grid, it's not planar. So, the chromatic number might be higher.Wait, for grid graphs on a torus, the chromatic number depends on the dimensions. I think for a toroidal grid graph with dimensions m x n, the chromatic number is equal to the maximum of m and n if both are greater than 1. But I'm not entirely sure. Let me think.In a regular grid graph (not on a torus), the chromatic number is 2 if it's a bipartite graph, which it is if both m and n are even. But on a torus, things change because of the wrap-around edges. For example, a 2x2 toroidal grid is a complete graph K4, which has chromatic number 4. Similarly, a 3x3 toroidal grid would have a higher chromatic number.Wait, maybe it's related to the number of vertices. For a 5x7 grid, there are 35 vertices. But that doesn't directly help. Maybe it's related to the maximum degree. Each vertex in a toroidal grid has degree 4, right? So, the maximum degree Δ is 4. Then, according to Brooks' theorem, the chromatic number is at most Δ + 1, which would be 5. But Brooks' theorem says that any connected graph (except complete graphs and odd cycles) has chromatic number at most Δ. So, if the toroidal grid graph is not a complete graph or an odd cycle, its chromatic number should be at most 4.But wait, a toroidal grid graph is actually a 4-regular graph, and it's also bipartite? No, wait, a toroidal grid graph is not bipartite because it has cycles of even and odd lengths. For example, in a 5x7 grid, there are cycles of length 5 and 7, which are odd. So, it's not bipartite, meaning it can't be colored with just 2 colors.But then, maybe it's 3-colorable? Or 4-colorable? I think for toroidal grids, the chromatic number can be higher. Let me recall that for a torus, the Euler characteristic is 0, so the genus is 1. But how does that relate to the chromatic number?Wait, maybe I should think about the graph's properties. A toroidal grid graph is a Cayley graph for the group Zm x Zn, right? So, it's a torus grid, which is a 4-regular graph. Since it's 4-regular and not bipartite, it might require 3 or 4 colors.But I think for toroidal grids, the chromatic number is 2 if both m and n are even, 3 if one is even and the other is odd, and 4 if both are odd. Wait, is that correct?Let me test it with small grids. For example, a 2x2 toroidal grid is a complete graph K4, which requires 4 colors. A 2x3 toroidal grid: each vertex is connected to its neighbors, but since it's a torus, it's like a 2x3 grid with wrap-around. Is that 3-colorable?Yes, I think a 2x3 toroidal grid can be colored with 3 colors. Similarly, a 3x3 toroidal grid would require 4 colors because it contains odd-length cycles.So, in our case, m=5 and n=7, both odd. So, according to this logic, the chromatic number should be 4.Wait, but let me confirm. If both dimensions are odd, the toroidal grid graph is not bipartite and contains odd cycles, so it can't be 2-colored. It also might not be 3-colorable because of the odd cycles. For example, in a 3x3 grid, you can't color it with 3 colors without conflict because of the odd cycles. So, 4 colors are needed.Therefore, for a 5x7 toroidal grid graph, the chromatic number should be 4.Okay, moving on to part 2: Define a new graph H constructed from G by replacing each vertex with a complete graph K4, and connecting the new complete graphs according to the edges of G. Calculate the genus of H.Hmm, so H is a graph where each vertex in G is replaced by a K4, and edges between these K4s are added according to the edges in G. So, H is a kind of graph product, maybe a tensor product or a replacement product.Wait, more specifically, it's a graph formed by substituting each vertex with a K4 and connecting corresponding vertices if there was an edge in G. So, each edge in G corresponds to a complete bipartite graph K4,4 in H.But actually, no. If two vertices in G are connected, then in H, each vertex in the K4 corresponding to one vertex in G is connected to each vertex in the K4 corresponding to the other vertex in G. So, each edge in G becomes a complete bipartite graph K4,4 in H.Therefore, H is the tensor product of G and K4, or maybe the strong product. Wait, no, the tensor product of G and K4 would have edges only between corresponding edges, but in this case, it's replacing each vertex with K4 and connecting all corresponding vertices across edges.Actually, this is known as the "subdivision" or "expansion" of G by K4. Alternatively, it's the "composition" of G and K4, where each vertex of G is replaced by a copy of K4, and edges are added between corresponding vertices if the original vertices were adjacent.So, H is the composition of G and K4, denoted as G[K4].Now, to find the genus of H. The genus of a graph is the minimum genus of an orientable surface into which the graph can be embedded without crossings.I remember that the genus of a graph can be calculated using the formula involving the number of vertices, edges, and the Euler characteristic.The Euler characteristic χ is given by χ = V - E + F, where F is the number of faces. For an orientable surface of genus g, χ = 2 - 2g.So, if we can find the Euler characteristic of H, we can solve for g.First, let's find the number of vertices and edges in H.G is a 5x7 toroidal grid graph. So, G has 5*7=35 vertices. Each vertex is replaced by a K4, which has 4 vertices. So, the number of vertices in H is 35*4=140.Now, the number of edges in H: Each vertex in G is replaced by a K4, which has 6 edges (since K4 has 4 vertices, each connected to 3 others, so 4*3/2=6 edges). So, for each vertex in G, we have 6 edges in H. So, 35*6=210 edges from the K4s.Additionally, for each edge in G, we have a complete bipartite graph K4,4 connecting the two K4s. Each K4,4 has 4*4=16 edges. So, how many edges does G have?G is a 5x7 toroidal grid. Each vertex has degree 4, so the number of edges is (5*7*4)/2=70 edges.Therefore, the number of edges contributed by the connections between K4s is 70*16=1120 edges.So, total edges in H: 210 + 1120 = 1330 edges.So, V=140, E=1330.Now, to find the genus, we can use the Euler's formula for planar graphs, but since H is likely non-planar, we need to find its genus.The formula relating genus, vertices, and edges is:Euler's formula: V - E + F = 2 - 2gBut we need another relation to find F. For planar graphs, we have E ≤ 3V - 6, but since H is not planar, we can use the inequality for graphs of genus g: E ≤ 3V + 6g - 6.Wait, actually, the inequality is E ≤ 3V + 6g - 6 for orientable surfaces. So, rearranged, we can solve for g.But since we don't know F, maybe it's better to use the formula for the genus:g ≥ (E - 3V + 6)/6This is a lower bound on the genus. But sometimes, the genus can be higher, but for certain graphs, this gives the exact genus.So, let's compute (E - 3V + 6)/6.E=1330, V=140.So, 1330 - 3*140 +6 = 1330 - 420 +6 = 916.Then, 916 /6 ≈ 152.666...So, the genus is at least 153.But wait, that seems really high. Maybe I made a mistake in calculating E.Wait, let me double-check the number of edges in H.Each vertex in G is replaced by a K4, which has 6 edges. So, 35*6=210 edges.Each edge in G connects two K4s, and each such connection adds 16 edges. G has 70 edges, so 70*16=1120 edges.Total edges: 210 + 1120=1330. That seems correct.So, E=1330, V=140.So, plugging into the formula:g ≥ (1330 - 3*140 +6)/6 = (1330 - 420 +6)/6= (916)/6≈152.666, so g≥153.But is this the exact genus? Or is there a better way to compute it?Alternatively, since H is the composition of G and K4, maybe we can use some properties of graph compositions.I recall that the genus of a composition of two graphs can be bounded by the sum of their genera plus some function of their sizes.But I'm not sure about the exact formula. Alternatively, since G is embedded on a torus (genus 1), and K4 is planar (genus 0), maybe the composition's genus is related.But I think the composition can have a higher genus. Alternatively, perhaps we can use the fact that the composition of G and K4 is a graph that can be embedded on a surface of genus g, where g is related to the genus of G and the size of K4.Wait, another approach: the composition H = G[K4] can be seen as a graph where each vertex is a pair (v, w), where v is a vertex in G and w is a vertex in K4. So, each vertex in H corresponds to a vertex in G and a vertex in K4.The edges in H are of two types: those corresponding to edges within each K4, and those corresponding to edges between K4s when there's an edge in G.So, the number of vertices is |V(G)|*|V(K4)|=35*4=140.The number of edges is |E(G)|*|E(K4, K4)| + |V(G)|*|E(K4)|.Wait, no. Actually, for each edge in G, we have a complete bipartite graph K4,4, which has 16 edges. So, |E(G)|*16=70*16=1120.Plus, for each vertex in G, we have the edges within the K4, which is 6 per vertex, so 35*6=210.Total edges: 1120 + 210=1330. So, that's correct.Now, to find the genus, perhaps we can use the formula for the genus of a graph:g = (E - V + 1)/2 - h, where h is the number of handles, but I think that's not directly applicable.Wait, another formula: For a connected graph, the genus g satisfies:E ≤ 3V + 6g - 6.So, rearranged:g ≥ (E - 3V + 6)/6.As we calculated earlier, (1330 - 420 +6)/6=916/6≈152.666, so g≥153.But is this tight? Or is the genus actually higher?Wait, but in reality, the genus can't be more than that because the formula gives a lower bound. So, the genus is at least 153, but could it be higher?Wait, no, actually, the formula gives a lower bound, meaning the genus must be at least 153. But sometimes, the genus is exactly that. For example, for complete graphs, the genus is given by the formula, and it's tight.But in this case, H is a composition of G and K4. Since G has genus 1, and K4 is planar, maybe the composition's genus is related.Wait, but I think the composition can have a much higher genus because each edge in G is expanded into a K4,4, which is a complete bipartite graph. Each K4,4 has genus 1, because K3,3 is non-planar and has genus 1, and K4,4 is a subdivision of K3,3, so it also has genus 1.Wait, no, K4,4 is actually a bipartite graph, and its genus is 1 because it contains a K3,3 minor. So, each K4,4 has genus 1.But in H, we have 70 such K4,4 graphs connected together. So, each contributes to the genus.But I'm not sure how to sum the genera. Maybe the total genus is the sum of the genera of each component plus some interaction.Alternatively, perhaps the genus of H is equal to the number of K4,4 components times their genus, but that seems too simplistic.Wait, another approach: The graph H is a 4-regular graph? No, wait, each vertex in H is part of a K4, so each vertex has degree 3 within its K4, plus degree equal to the degree in G times 4, because each edge in G connects to 4 vertices in the adjacent K4.Wait, no, let me think. Each vertex in H is connected to 3 other vertices within its K4, and for each edge in G that the original vertex was part of, it connects to 4 vertices in the adjacent K4. Since each vertex in G has degree 4, each vertex in H has degree 3 + 4*4=3+16=19.Wait, that can't be right. Wait, no, each vertex in G has degree 4, so each vertex in H (which is part of a K4) is connected to 4 other K4s, each connection adding 4 edges (since each connection is to a K4, and each vertex connects to all 4 vertices in the other K4). So, for each edge in G, each vertex in the K4 is connected to 4 vertices in the adjacent K4, so each vertex in H has 4*4=16 edges from the connections, plus 3 edges within its own K4. So, total degree 19.So, H is a 19-regular graph with 140 vertices and 1330 edges.Now, to find the genus, we can use the formula:g ≥ (E - 3V + 6)/6.Plugging in E=1330, V=140:g ≥ (1330 - 3*140 +6)/6 = (1330 - 420 +6)/6= (916)/6≈152.666.So, g≥153.But is this the exact genus? Or is it higher?I think for graphs where E > 3V -6, the genus is at least (E - 3V +6)/6, but sometimes it's higher. However, for certain classes of graphs, this bound is tight.In this case, H is a 4-regular graph? Wait, no, it's 19-regular. But regardless, the formula gives a lower bound.But perhaps we can use another approach. Since H is the composition of G and K4, and G has genus 1, maybe the genus of H is related to the genus of G multiplied by something.Wait, I found a paper once that said the genus of a composition H = G[K_n] is at least n times the genus of G. But I'm not sure.Alternatively, since each edge in G is expanded into a K4,4, which has genus 1, and there are 70 edges, maybe the total genus is 70*1=70. But that seems too low compared to our earlier calculation.Wait, but each K4,4 is embedded on a torus, which has genus 1, but when you connect them all together, the overall genus might be additive in some way.But I'm not sure. Maybe the genus is actually 70, but our earlier calculation suggested it's at least 153.Wait, perhaps the formula I used earlier is more accurate because it's a general formula for any graph. So, if the lower bound is 153, then the genus must be at least 153.But I'm not sure if it's exactly 153 or higher. Maybe it's exactly 153 because the formula gives a tight bound for certain graphs.Alternatively, perhaps the genus is 153 because the formula gives the exact value when the graph is triangulated or something.Wait, but H is not necessarily triangulated. It's a composition of G and K4, so it's a more complex structure.Hmm, I'm a bit stuck here. Maybe I should look for another approach.Wait, another formula for the genus of a graph is:g = (E - V + 1)/2 - h,where h is the number of handles. But I don't know h.Alternatively, using the Euler characteristic:χ = V - E + F = 2 - 2g.But without knowing F, the number of faces, it's hard to compute.Wait, maybe we can use the average degree to estimate the genus. The average degree d = 2E/V = 2*1330/140 ≈ 19.For a graph with average degree d, the genus is at least (d - 2)V/6 - something. Wait, not sure.Alternatively, I think the formula g ≥ (E - 3V +6)/6 is the standard lower bound, and sometimes it's the exact genus.Given that, and since H is a composition of G and K4, which are both graphs with known properties, perhaps the genus is exactly 153.But I'm not entirely confident. Maybe I should check if 1330 ≤ 3*140 + 6g -6.So, 1330 ≤ 420 + 6g -6 → 1330 ≤ 414 + 6g → 6g ≥ 916 → g ≥ 152.666, so g=153.Therefore, the genus is 153.Wait, but let me think again. If H can be embedded on a surface of genus 153, then that's the minimal genus. But is there a way to confirm this?Alternatively, maybe the genus is higher because of the multiple K4,4 components. Each K4,4 has genus 1, and there are 70 of them, so maybe the total genus is 70. But that contradicts the earlier calculation.Wait, perhaps the genus is additive when combining multiple graphs, but I think that's only when the graphs are disconnected. Since H is connected, the genus isn't simply additive.I think the correct approach is to use the formula g ≥ (E - 3V +6)/6, which gives g≥153, and since H is a connected graph, this is the minimal genus.Therefore, the genus of H is 153.Wait, but I'm still a bit unsure because 153 seems very high. Maybe I made a mistake in calculating E or V.Let me double-check:G is a 5x7 toroidal grid, so 35 vertices, each with degree 4, so edges E(G)= (35*4)/2=70.H is constructed by replacing each vertex with K4, so V(H)=35*4=140.Each K4 has 6 edges, so 35*6=210 edges.Each edge in G connects two K4s, and each such connection adds 4*4=16 edges. So, 70 edges in G contribute 70*16=1120 edges.Total E(H)=210+1120=1330.Yes, that's correct.So, V=140, E=1330.Then, using the formula:g ≥ (1330 - 3*140 +6)/6 = (1330 - 420 +6)/6=916/6≈152.666, so g=153.Therefore, the genus is 153.I think that's the answer.</think>

🧠:<think>Alright, so I have two problems to solve here related to optimizing a multiplayer online game's network performance. Let me tackle them one by one.Starting with the first problem: I need to determine the maximum number of players ( n ) that can be supported without exceeding the server's bandwidth. The game can support up to ( n ) players, each sending a packet every ( t ) milliseconds. Each packet is ( s ) bytes. The server has a bandwidth ( B ) in bytes per millisecond. Additionally, the server uses ( k % ) of its bandwidth for other data like updates and maintenance. Each response packet from the server is ( r ) bytes.Okay, so I need to figure out how much bandwidth is used by the client-server communication and ensure it doesn't exceed the server's available bandwidth after accounting for the ( k % ) overhead.First, let's think about the data flow. Each player sends a packet every ( t ) milliseconds. So, the number of packets sent by each player per millisecond is ( frac{1}{t} ). Each packet is ( s ) bytes, so the data sent by one player per millisecond is ( frac{s}{t} ) bytes.Since there are ( n ) players, the total data sent from all clients to the server per millisecond is ( n times frac{s}{t} ).Now, the server also sends response packets back to each client. Each response is ( r ) bytes, and since the server sends a response every time it receives a packet, that's also every ( t ) milliseconds. So, similar to the client side, the server sends ( frac{r}{t} ) bytes per millisecond per player. Therefore, the total data sent from the server to all clients per millisecond is ( n times frac{r}{t} ).Adding both the incoming and outgoing data, the total bandwidth used by the game is ( n times left( frac{s}{t} + frac{r}{t} right) ).But the server also has ( k % ) of its bandwidth reserved for other purposes. So, the available bandwidth for the game is ( B times (1 - frac{k}{100}) ).Therefore, the total bandwidth used by the game must be less than or equal to the available bandwidth:[ n times left( frac{s}{t} + frac{r}{t} right) leq B times left(1 - frac{k}{100}right) ]I can factor out ( frac{1}{t} ):[ n times frac{s + r}{t} leq B times left(1 - frac{k}{100}right) ]To solve for ( n ), I can rearrange the inequality:[ n leq frac{B times left(1 - frac{k}{100}right) times t}{s + r} ]Since ( n ) must be an integer, I should take the floor of the right-hand side to get the maximum number of players.So, the formula is:[ n = leftlfloor frac{B times (1 - frac{k}{100}) times t}{s + r} rightrfloor ]Let me double-check this. Each player contributes ( frac{s + r}{t} ) bytes per millisecond, multiplied by ( n ) players, and this must be less than or equal to the available bandwidth after the ( k % ) deduction. Yes, that seems correct.Moving on to the second problem: The engineer implements a predictive algorithm that estimates the next move based on past ( m ) moves. The accuracy is given by ( A(m) = frac{1}{1 + e^{-0.1m}} ). We need to find the minimum ( m ) such that ( A(m) geq 0.8 ).So, setting up the inequality:[ frac{1}{1 + e^{-0.1m}} geq 0.8 ]I need to solve for ( m ).First, let's rewrite the inequality:[ 1 + e^{-0.1m} leq frac{1}{0.8} ]Calculating ( frac{1}{0.8} ) gives 1.25.So,[ 1 + e^{-0.1m} leq 1.25 ]Subtract 1 from both sides:[ e^{-0.1m} leq 0.25 ]Now, take the natural logarithm of both sides. Remember that ( ln(e^{x}) = x ), and since the exponential function is positive, the inequality direction remains the same.[ -0.1m leq ln(0.25) ]Calculating ( ln(0.25) ). I know that ( ln(1) = 0 ), ( ln(e) = 1 ), and ( ln(1/e) = -1 ). Since 0.25 is ( e^{-1.386} ) approximately, because ( e^{-1.386} approx 0.25 ). Let me verify:( e^{-1.386} approx e^{-1.386} approx 0.25 ). Yes, that's correct because ( ln(0.25) = -1.386294... ).So,[ -0.1m leq -1.386294 ]Multiply both sides by -1, which reverses the inequality:[ 0.1m geq 1.386294 ]Divide both sides by 0.1:[ m geq frac{1.386294}{0.1} ]Calculating that gives:[ m geq 13.86294 ]Since ( m ) must be an integer (number of past moves), we round up to the next whole number, which is 14.Let me verify this. Plugging ( m = 14 ) into ( A(m) ):[ A(14) = frac{1}{1 + e^{-0.1 times 14}} = frac{1}{1 + e^{-1.4}} ]Calculating ( e^{-1.4} approx 0.2466 ). So,[ A(14) approx frac{1}{1 + 0.2466} = frac{1}{1.2466} approx 0.802 ]Which is just above 0.8. If I try ( m = 13 ):[ A(13) = frac{1}{1 + e^{-1.3}} approx frac{1}{1 + 0.2725} approx frac{1}{1.2725} approx 0.785 ]Which is below 0.8. So, 14 is indeed the minimum number of past moves needed.I think that covers both problems. Let me recap:1. For the bandwidth calculation, the maximum number of players is determined by the available bandwidth after accounting for the ( k % ) overhead, divided by the per-player bandwidth usage (both incoming and outgoing packets). The formula gives ( n ) as the floor of that value.2. For the predictive algorithm, solving the inequality for ( A(m) geq 0.8 ) leads to ( m geq 13.86 ), so rounding up gives ( m = 14 ).I don't see any mistakes in my reasoning, so I think these are the correct answers.Final Answer1. The maximum number of players is boxed{leftlfloor dfrac{B times (1 - frac{k}{100}) times t}{s + r} rightrfloor}.2. The minimum number of past moves required is boxed{14}.</think>

🧠:<think>Okay, so I have this problem where a law student is analyzing crime data in a city with 10 districts. The crimes follow a Poisson distribution with an average of λ per district. The student wants to find the probability that at least one district has more than k crimes in a year. Then, there's another part about severity scores, which are normally distributed, and figuring out the threshold for the top 5% most severe crimes.Starting with the first part: Poisson distribution. I remember that the Poisson distribution is used for counting the number of events happening in a fixed interval of time or space. The probability mass function is P(X = x) = (λ^x * e^(-λ)) / x! where x is the number of occurrences.But the question is about the probability that at least one district has more than k crimes. So, since there are 10 districts, each with their own Poisson distribution, I think we can model this as 10 independent Poisson random variables. We need the probability that the maximum of these 10 variables is greater than k.Hmm, how do I calculate that? Well, the probability that at least one district has more than k crimes is equal to 1 minus the probability that all districts have k or fewer crimes. So, if I denote X_i as the number of crimes in district i, then:P(max(X_1, X_2, ..., X_10) > k) = 1 - P(X_1 ≤ k, X_2 ≤ k, ..., X_10 ≤ k)Since the districts are independent, this becomes:1 - [P(X ≤ k)]^10Where P(X ≤ k) is the cumulative distribution function (CDF) of the Poisson distribution evaluated at k.So, first, I need to compute the CDF for a Poisson distribution with parameter λ up to k. Then, raise that probability to the 10th power and subtract from 1.But wait, how do I compute the CDF for Poisson? It's the sum from x=0 to x=k of (λ^x * e^(-λ)) / x! That could get computationally intensive for larger k, but since we're just setting up the formula, maybe we can leave it in terms of the sum.Alternatively, if we have access to software or tables, we can compute it numerically. But for the sake of this problem, I think expressing it in terms of the CDF is sufficient.So, the formula is:Probability = 1 - [CDF_Poisson(λ, k)]^10Now, moving on to the second part: severity index. Each crime has a severity score from 1 to 10, normally distributed with mean μ and standard deviation σ. The student wants to find the threshold above which a crime is in the top 5% most severe.Since the severity scores are normally distributed, we can use the properties of the normal distribution to find this threshold. Specifically, we need the value such that 95% of the scores are below it, meaning it's the 95th percentile.In terms of z-scores, the 95th percentile corresponds to a z-score of approximately 1.645. So, we can calculate the threshold severity score (let's call it T) using the formula:T = μ + z * σWhere z is 1.645 for the 95th percentile.But wait, let me double-check that. The z-score for 95% confidence is indeed about 1.645, yes. So, plugging that in, we get the threshold.However, the severity index is from 1 to 10. If the normal distribution is being used, we have to make sure that the threshold T doesn't exceed 10 or go below 1. But since we're dealing with the top 5%, and assuming μ is somewhere in the middle, it's likely that T will be within the 1-10 range. If not, we might have to cap it, but I think the problem assumes that the normal distribution is a good fit, so T should be within the range.So, summarizing:1. For the Poisson part, the probability is 1 - [CDF_Poisson(λ, k)]^10.2. For the severity threshold, it's μ + 1.645σ.I think that's the approach. Let me just make sure I didn't miss anything.Wait, for the Poisson part, is it exactly 1 - [P(X ≤ k)]^10? Yes, because all districts must have ≤k for the max to be ≤k. So, the complement is at least one district having >k.And for the severity, since it's the top 5%, we're looking for the value where 95% are below, so yes, the 95th percentile.I think that's correct.</think>

🧠:<think>Okay, so I have this problem about calculating the waiting time for a digital lending service. Let me try to break it down step by step. First, the problem is divided into two parts. The first part is about calculating the expected waiting time without any premium features, and the second part involves calculating the waiting time after activating a premium feature which reduces the waiting time by a priority factor. Starting with part 1: The formula given for the waiting time ( W ) is ( W = frac{n}{lambda} + frac{D}{k} ). I need to figure out what each variable stands for. - ( n ) is the number of books currently borrowed by the user. In this case, the user has borrowed 5 books. So, ( n = 5 ).- ( lambda ) is the average rate at which the user returns books, modeled as a Poisson process. The user returns books at an average rate of 0.2 books per day. So, ( lambda = 0.2 ).- ( D ) is the current demand factor for the book, which is given as 15.- ( k ) is the user's preference coefficient, which is 2. It's a real number greater than 0, indicating how urgently the user needs the book.So, plugging these values into the formula:First, calculate ( frac{n}{lambda} ). That would be ( frac{5}{0.2} ). Let me compute that. 5 divided by 0.2 is the same as 5 multiplied by 5, which is 25. So, that part is 25 days.Next, calculate ( frac{D}{k} ). That's ( frac{15}{2} ). 15 divided by 2 is 7.5. So, that part is 7.5 days.Adding these two together, 25 + 7.5 equals 32.5 days. So, the expected waiting time without any premium features is 32.5 days.Wait, let me double-check that. ( n = 5 ), ( lambda = 0.2 ), so 5 / 0.2 is indeed 25. ( D = 15 ), ( k = 2 ), so 15 / 2 is 7.5. Adding them together gives 32.5. That seems correct.Moving on to part 2: The digital lending service offers a premium feature that reduces the waiting time by a priority factor ( P ). The formula for ( P ) is ( P = e^{-r cdot t} ).Given that ( r ) is the rate of decay, which is a constant value of 0.1, and ( t ) is the time in days since the user activated the premium feature. The user activates the premium feature today, so after 10 days, ( t = 10 ).First, I need to calculate ( P ). Plugging in the values, ( P = e^{-0.1 times 10} ). Let's compute that exponent first: 0.1 multiplied by 10 is 1. So, ( P = e^{-1} ).I remember that ( e^{-1} ) is approximately 0.3679. So, ( P approx 0.3679 ).Now, how does this priority factor affect the waiting time? The problem says the waiting time ( W ) is reduced by the priority factor ( P ). So, does that mean we subtract ( P ) from ( W ), or do we multiply ( W ) by ( P )?Looking back at the problem statement: "the waiting time ( W ) is reduced by a priority factor ( P )." The wording is a bit ambiguous, but in most cases, when something is reduced by a factor, it usually means multiplication. For example, if you reduce something by a factor of 2, it becomes half. So, I think it means ( W_{text{new}} = W times P ).Alternatively, if it meant subtracting, it would probably say "reduced by ( P )", but since it says "reduced by a priority factor ( P )", it's more likely a multiplicative factor.So, assuming that, the new waiting time after 10 days is ( W_{text{new}} = W times P ).From part 1, we have ( W = 32.5 ) days. So, ( W_{text{new}} = 32.5 times 0.3679 ).Let me compute that. 32.5 multiplied by 0.3679. First, 32 multiplied by 0.3679 is approximately 11.7728, and 0.5 multiplied by 0.3679 is approximately 0.18395. Adding them together, 11.7728 + 0.18395 ≈ 11.95675.So, approximately 11.95675 days. Rounding to a reasonable number of decimal places, maybe two, so 11.96 days.But let me verify the multiplication to be precise.32.5 * 0.3679:Break it down:32.5 * 0.3 = 9.7532.5 * 0.06 = 1.9532.5 * 0.0079 ≈ 0.25675Adding these together: 9.75 + 1.95 = 11.7; 11.7 + 0.25675 ≈ 11.95675. So, yes, that's correct.So, the waiting time after 10 days with the premium feature is approximately 11.96 days.Wait, but let me make sure I interpreted the reduction correctly. If the priority factor is ( P ), and it reduces the waiting time, is it ( W - P ) or ( W times P )?Looking back at the problem statement: "the waiting time ( W ) is reduced by a priority factor ( P )." The term "reduced by" can sometimes mean subtraction, but in the context of factors, it's more likely to mean multiplication. For example, if something is reduced by 50%, it's multiplied by 0.5.Moreover, the priority factor is given as ( P = e^{-r t} ), which is a value between 0 and 1 because ( r ) and ( t ) are positive. So, multiplying ( W ) by ( P ) would indeed reduce it. If it were subtraction, it might not make much sense because ( P ) is less than 1, and subtracting it from ( W ) might not have the intended effect of reducing the waiting time proportionally.Therefore, I think my initial interpretation is correct: ( W_{text{new}} = W times P ).So, with ( W = 32.5 ) and ( P approx 0.3679 ), the new waiting time is approximately 11.96 days.Let me just recap to ensure I didn't miss anything.In part 1, we calculated the waiting time using the given formula with the provided values, resulting in 32.5 days.In part 2, after activating the premium feature, the waiting time is reduced by the priority factor ( P ), which is calculated as ( e^{-0.1 times 10} approx 0.3679 ). Multiplying the original waiting time by this factor gives the new waiting time of approximately 11.96 days.I think that's all. I don't see any other steps or considerations needed here. The problem seems straightforward once you parse the formulas correctly.Final AnswerThe expected waiting time after 10 days with the premium feature is boxed{11.96} days.</think>