Um, mic test, mic test. Oh, we got somebody, we're in a game. Hold on, I'm just still setting up. Looks like we're live on YouTube, I think. Either YouTube or Facebook, I can't tell which one. Okay, I am testing out. This is my second time to game with this new mouse. Let me just turn off the music, because the music gets kind of weird. Oh, there we go. Perfect, good timing. I was able to switch it beforehand. So I am testing out the Razer Viper Ultimate. If you look at my title, it's actually an old mouse. This was released in... Eliminate all enemies or catch the overtime flash and win. Okay, it's one-on-one. Okay, so it was released in, I think, around September 2019. So this is an old mouse. However, there was some controversy over the buttons of the mouse. And then later on, like in... I think it was around 2021, they made a version 2 of this. Is this... It's really... Okay, this guy has some skill. This guy has some skill. All right, let's go. All right. This is serious. Okay, all right. All right, all right, okay. Let's go, let's go. So let's see if this is really still a... If this can improve a average first-person shooter player with the beating A. The above-average gamer. Oh! What the hell, dude? Look at that. Okay. All right, all right. Okay, okay. All right. Okay. Night machine gun. Okay, this is a spray and pill. Let's see where he's at. Let's see where he's at. There are two of them. Oh, man. I thought it was still one-on-two. Oh, my gosh. Let's go. Oh, my gosh. Okay, okay, okay. Come on, guys. It's a horrible start. Okay. Teammates talking to me. Oh, I need a light machine gun. Okay, I think I can do better with this one. Come on. Come on. Come on, Razor Viper ult. Oh, okay. At least I got one killed before I died. Let's see.  Oh. Come on, teammates. We lose this. I don't think we lose this. Yeah, it's game over for this one, for this round, for this matchup. Oh. I think he's behind you. Oh, you have no cover. Oh, I saw him. Oh, is this another round? I thought it was first hit. I had played a lot through the end. Let's lower the volume. Okay, slowly. All right, all right. Hey, Lee Magni, what's going on? What are you doing on a nice cloudy Sunday morning? I hope everyone is okay, by the way, from the storm yesterday. Grabbing storm? That was a close call. That was a close call. Oh, hey, my ally 6A. Let's win. Well, if you gotta win, you're gonna bring in some kills. They're all match points. We can't take another loss. Yeah, boys. Where's he at? Not bad. My first time playing Call of Duty. Maybe close to five months. I played PUBG for the first time. They're all match points. We can't take another loss. Okay, guys, I don't know if I'm gonna do this. I'm not really good with snipes. Oh, I really walked into that one. Oh, it's okay. It's up to your team. It's up to you, teammate. Nice. Nice kill. Nice. Nice. Nice. Let's keep going. Okay, let's rush it. Let's rush it. Oh, I walked into that again. I didn't think... I was hoping it was gonna panic. GG. All right, lads. Clean yourselves up. At least we didn't get wiped. Ugh. Not bad. Not bad. Well, actually, it wasn't fair. I don't know. Whatever. It's my first Call of Duty game in a while. I'm actually planning... I was planning to get the latest Modern Warfare, the Modern Warfare 2 game. But the problem is that the campaign does sucks, which is sad, because I really like the campaign of Modern Warfare. In general, I like the campaign of Modern Warfare. But then the latest campaign, according to the IG interviews, it's just satisfactory. And it's just pre-ordered for the campaign. So if it's just okay, I'll just wait for it to go on discount. So actually, so some... If you guys didn't see me on the stream last PUBG, I did some research. You know, if you're in the market for a new mouse, the Razer Viper Ultimate is still a good contender, mostly because it's cheaper than the latest V2 Pro. And second, they updated the mechanisms of the mouse post-2021, I think. So this should be a longer-lasting mouse, which is cheaper than the latest one. This is not by any means a budget mouse, even though it's three years older. This is still a very expensive mouse. But, well, as I said, I'm willing to give Razer a shot because I do like that it's ambidextrous. So what that means is that you get the same buttons. It's essentially the same shape, regardless if you're right-handed or left-handed. And you have the same buttons on both sides. And even I'm right-handed, and I find the other side useful for like opening other windows or other forms of macro. Okay. So yeah, let's try it. Let's see, let's do a Warzone. Did it hang? Ah, there we go. Okay. And then also what I wanted to do, I also wanted to game now because I wanted to test out my new hard drive. So as a content creator, I have a lot of B-roll, which ate up my hard drive, and then finally moving it into an external hard drive. And now I have actually finally have space. I have space to actually do the updates because I had Warzone apparently on the hard drive, which was full. And so even if it was fully installed, I couldn't play it because the latest updates were just so big. Hi Gino, T-Zone. Actually, I think there is a setting for you to turn off. But for one thing, I believe Counter Strike is capped at, I would say 60 frames, but no, that's not true. There's a motion blur option in COD, I think. Let me see. Mouse sensitivity, Invert, mouse acceleration, mouse smoothing. What's this do? And smooths aim. Now this has nothing to do with graphics. It should be in graphics. I know my motion blur data, which you can turn off. I also turned it off because it's disorienting. Dismemberment, bullet impacts, particle texture, stream quality, render resolution, film grain. Motion blur. To answer your question, Gino, you can turn off motion blur. I will now turn it off for you. See, there we go, as it says. This is without, this is with motion blur. I think motion blur is to create that appearance now. It's like an action movie kind of thing. Wow. There's no one playing this game anymore. Okay, that's... Does Warzone really exist? I don't know what's Uso now in the first-person shooter world, like the multiplayer. But every time I play Warzone, I think about the pandemic because that's when I actually sort of got into Warzone. I keep playing it beforehand. Wow. Oh, it's surreal. I haven't played it since ancient times. Oh, okay. There we go. There we go. There we go. Okay. Oh! All right, all right. So this is just try practice. I'm good with practice. Ah, yeah, I thought that was a top building. Whoa! Man, I haven't changed this gun. It's so long. Enough training. Now you fight a real enemy. All right, here we go. Teams, bro. Teams, bro. Yeah, teams, bro. Since I haven't played in so long, I'll just let someone else take over. But if they're not going to see anything... Let's head this way. Surgeons! Go, go, go! Let's go. That's what the machine guns for, guys. Just to waste them. Okay, we're still in the safe zone. Let me get back into this game. How do I... Does anyone remember how to... Oh, he's on top. Oh!  How could they all have just died like that? I have to get a gun first. I have to stay alive and rebirth island, guys. Does everyone remember that? Searching for supplies. Let's get some new toys. Finish scanning. Exiting airspace. Bounty target is up. Take him down. How come my teammates left me, guys? Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! Oh! What I did there. Oh! Okay, let's go. Let's go. At least stay alive. I don't know why my teammates just totally left me. They parachuted somewhere else. Many three people. It's all relatively close to finishing up now. Oh, okay. At least they're complete. Let's see if we can get a gun. Alright. I do would like my care package. Oh, before anything else. You've got a second chance. I can count. I need windows. But where can I get windows? There's a lot of options. There's a lot of prices. Oh, here it is. Good to know. I have CD keyovers.com. It's easy to order. Search for the software you need. Add to cart. Go to their payment options. There's no more five minutes. Finished. There's a legit working CD key for your windows. Use our code to get discounts. That's why if you're looking for legit, cheap and original software, check out CD keyovers.com. Okay, you guys missed it. I, a guy ate my grenade launcher. An enemy team is hunting you. Don't become prey. Okay, sniper rifle. Okay. This is moving in. Use safe zone highlighted. This station costs are adjusted. Set the climb off. Shit, I already forgotten how to use the... How do you use the special monitor? Enemy UAV active. Be active. I'm still alive. Still a lot of people. A vigilant. Resurgence window is about to close. Enemy UAV active. That wasn't in. That wasn't intense. The level three? I used to be higher than that. Get in the safe zone. All right, all right, all right. Hi, Boris. Hi. So recently switched to an Ergo chair. Thanks. Thanks to you. My back is now happy. I got the Ergo Prime foam from Furniture. I'm glad I was able to help. I'm glad I was able to help Boris. Well, I recognize your profile photo from before. I'm not familiar with that specific chair. However, I am familiar with Furniture. My familiarity with that chair model is that people were asking us to review it before. I haven't gotten around to reviewing that specific model, but I'm glad that you made the switch to an ergonomic chair instead of sticking with the gaming chairs. Yeah, I mean, actually, I'm planning. I'm still trying. I'm working on a video right now on the top chairs to avoid. The top chair types, not so much chair models that you should avoid, but the chair types. One of them that's super basic, which you should always avoid is the chairs that don't have armrests, which go up and down. You should be very weary of chairs that have permanently fixed arms because you're going to be married to that chair's height. And then, depending on your table, the only time that I would say that it would be okay to get a chair type with armrests that are permanent is if you can promise to yourself that if your table is adjustable. Because if it's not weird, you're going to be like this. If you're not aligned with your armrests table. And actually, lots of ergonomic chairs times to the pandemic are very affordable now.  Probably cheaper.  Playtime's over. Stand by for deployment to the wars. Ergonomic chairs over game chairs. Opposition suggests you start running. All right, there we go. Survive to redeploy fallen to return them faster. Oh, okay. We're getting the gun. We need a gun. Where is he at? It's the way to go. Tanger, see you at 10K, guys. Tano, decent. Thanks for, I'm glad I was able to help you out.  They're all in the hard brush of her stores in case you guys want to check them out. Well, be shy. Feel free to ask our team if you could sit on them for a while. Oh, shoot. Oh, shoot. Oh, shoot. Oh, shoot. Oh, my gosh. He's behind me. I know he's behind me. Oh, gosh. Stress. I'm glad my PC is only being together.  Last one. There's someone here. It takes so long to fire the RPG. Are you OR? Is there a new chair? I don't have any new chairs lined up for review because I'll be traveling. However, what I have lined up is other content. However, I'm always open. Actually, I've had to decline people actually wanting to send me chairs with a review. But thanks to all of you guys, we're kind of popular now, locally, because of my chair reviews. Something I didn't know that people were interested in, but apparently, my appeal... I think anyone's going to notice me here.  I like how, you know, to be honest, I'm the type of person who would watch reviews of products that I already own. I thought I was the only one, but apparently, there are a lot of people. Like, for instance, if you bought this mouse, even though it's been a day, fine, it's been a day, I still watch videos reviewing this mouse, even though there's no way I can return it. But I still watch it anyway. Same with my phone. I'm sure when I bought my S21 Ultra last year, I was like, oh yeah, I want to see if my purchase is justified. What are other people saying about my phone? Like, I feel good when someone tells me that they're enjoying the same product that they already paid for. Like, it doesn't make any sense to some people. But apparently, a lot of people get a high from that. Oh yes, validation. Ah, my purchase is okay. Right now, I'm sporting the Herman Miller Aeron chair, mostly because I really bought this chair. Let me show it. I really bought this chair to last. I'm really thankful that it became a really worthwhile investment because it was retailing locally for $80,000. And then I got it for $60,000 because I joined in with a bunch of other friends to buy the chairs. I think we were a total of four. So I was able to bring in $60,000.  And the strange thing is that it's really, it's still the same chair. What's the inflation? I don't think, you know, it's a weird thing. The pricing for luxury furniture. The pricing for luxury furniture is strange. Like, I am literally paying more. Like, people who buy this chair now will be paying a lot more money for the same product that's been being sold since 2015, I think. Because they only made two versions of this Herman Miller Aeron, the original one, and then this one. I forget what they call this latest one. But the latest one, Herman Miller Aeron, they call it the remastered. They made this chair, I think, like maybe six, seven years ago. I forget, just check it out. But essentially, it's the same chair. I mean, it's not like electronics, the value of the electronics of the gear or the hardware goes down immediately after purchase. And then after that, there's pressure on the company to release a updated version. Like, you know, there's always pressure on top of Razer, for instance, to create... They won't make the same mouse. Like, for instance, Razer, if it feels successful, like with the Death Adder, they will make newer versions of the Death Adder. But it's not the same Death Adder. Like, the Death Adder that's being sold now by Razer is not the same one that was released back in 2010. I don't know when the Death Adder was released. But definitely, it's not the same model type. It is similar in shape and functionality, but it's not the same one. There's Herman Miller Aeron, though. It's the same one, and I expect that it's gonna be the same one for the next, like, at least five more years. They don't seem like they're gonna be changing it. So yeah, it's weird. I think that if I wanted to buy the same chair now, it's an additional, like, 40-50,000 pesos. It's crazy. I think that it's the Herman Miller Aeron worth it for its price. You know, if you remember my video, I originally found it very uncomfortable. For me, it's worth it. It's worth it because for one thing, I was able to avoid the inflation price. And I like buying things which are for life. This is going to be the for-life chair. So stick around, watch the YouTube channel to see if I'm gonna be still using the same chair for the next 20 years. But yeah, I really meant... I really meant I didn't want to buy anymore a new chair after this. Like, this is supposed to be the chair to... Oh! I ran out of bullets, man. I was the last bullet. I like how the last bullet got it, man. I really recommend that if you're gonna invest that much money in a chair, you really need to try it out first in the showroom. And even then... And even then, there's the risk that you won't be able to notice if it's a good fit for you. Like, for instance, for me, it took me four days for my body to adjust to the chair. I mean, it wasn't love at first seating. But what are you paying for with the Aeron? What you're paying for with the Aeron is... Give him a proper English welcome. ...is, number one, the design. It looks really nice. If you're a fan of this design, pretty unique. Second, ergonomics. Got him. So the first one is design. It looks aesthetically, I like it. Second, it's comfortable. There are several levels of comfort. One is, it's breathable, as you know, with a mesh chair. And it's durable, really good. Like, this has... Like, lots of people have said that it's really quite durable. That's why lots of people are still selling chairs which are like, super old.  Give him a proper English welcome. I love playing pistols, man. I love playing pistols. Got him. Let's stick together. That was close. See that jump shot? Oh, I shot his knee. What are windows? But where do you get these windows? Lots of options, lots of prices. Oh, here. Good to know, there's a CD key of RISDAT.com. It's easy to order. Search for the software you need. Add to cart. Send it to their payment option. There's no more five minutes. Finished. There's a legit working CD key for your windows. Use my code to get discounts. So if you're looking for legit, cheap and original software. Check it out, CD key of RISDAT.com. Left me, guys. That's your hardware sugar. Okay, it's one on two. I think I can do this. Let's go. I think I should have slammed my new house. It's okay. This is what they were meant for. It's meant to be thrown around. I really am liking this white color. Thought I could get away with that one. Oh, come on. Man, I left me. Come on. I realized after using... Wait, I still have my Corsair iron claw here. My golf. My gun. Okay, there we go. I realized after being away from a... After using a really bulky mouse like the iron claw. I could see why this isn't the pinnacle of first person shooter. It's really heavy. I'm looking at the... You know, it's important to have weight. At first I was like, ah, no. It doesn't really matter to me the weight. It's not like I'm lifting it like weights and stuff. It's really nice when your mouse is just gliding across the surface. And yeah, if you are just tuning in now, the rubber grip of my iron claw has come undone. But it's okay. It's all right. Actually, no. It's not okay that the iron claw scroll wheel broke. But I don't feel like I wasted money a month. Because it's always handy to have a backup mouse. So just in case this one breaks, or I have to bring it in for warranty, my backup of an iron claw. If that... But if this mouse breaks down, I'll just get a deep cool MC310. I love that mouse. The RGB of the deep cool MC310 is out of this world. That's the goal I made here. You can't tell. You can't tell, bro. Who am I fighting against? Hi, it's Adik, the MC310 from my review. Or the Viper Ultimate. Let's get this done. It's not glorious enough. I like it. Pretty witty.  Totally walked into that one. Yeah, that's true. They were trying to go for the honeycomb thing. Ah! However, what set that mouse apart? The MC310. Oh, come on, come on, come on! Oh!   Oh, yikes! He got headshot with a melee. Gotta go slow with this gun. I like gunfight as the mode, because at least you get the practice. Nice. That was a good kill. That was a good kill. We strafed him. Sorry, flanked his problem in the carrier room. What's this, SMG? That was a quick kill. I think I head shot at him.  Went up here. Who got that kill? Good teamwork. Glorious, oh, way down at number five. What's this? Sniper rifle shotgun? Oh, did you see that? Whoa! Nice! I just hit his spleen. That jump shot was pretty good. Go out in a month, though. Okay, well, voilà. I want the win. I don't mind the win. Flank this guy. Come on, get him! Nice! Hi, Christian Espinosa. Good afternoon. How is your afternoon doing? GG, GG. I'm really glad that I was able to fix my camera. I'm so much clearer now. I was having a hard time fixing up the Fujifilm. I just kept dying on me. To be honest, what I just didn't know was I factored the Fujifilm. You know, unrelated to hardware and technology, my favorite new gear, which I hope we get sponsored by, is eco-vessel. This water tumbler is my new favorite something. It's a fast... It's a quick... Quick removal top? What's the better term for that? What's that? You can remove the top quicker with just one spin. What I really don't like about most tumblers is that you have to unscrew the whole thing. That takes a lot of time, especially if you want to drink water. You end up not drinking water.  This is an easy twist. I think that's an easy twist. It has this strainer. I feel like ice water like me. I totally love ice water. It has this strainer here. If you have small ice cubes, it gets strained. You're never drinking. You don't need to worry about swallowing crushed ice anymore. It's a small thing, really. I haven't seen that in even the other premium canteens. They should really work on the physics of that. Mind your own weirdness. His body distorts in the midair. Okay, I think... What is this? This is some sort of assault rifle? I got him! I'm supposed to deagle. Damn, look at this. It's so big, you can't even see the guy anymore when you're shooting with it. I think I can do this. I walked right into that one. I think it's Addick. That's the Razer Viper Pro. This is the Razer Viper Ultimate. There's a V2 Pro, which is the latest one. The problem with that, it's good though, for the other people. It's more lightweight than this, but it's like 9000 bucks. And it's a downgrade. No charging dock that's good. Yeah, Viper Ultimate. It's the mouse that was released back in late 2019. You get RGB. If you get the latest one because there's no RGB, which I know isn't the most important thing, but it's still a good thing to have, aesthetics-wise. Why are you gonna pay more for less? Find the only major good thing that you get... The only thing that you get out of getting the V2 Pro is that it's lighter, period. Oh, and USB-C. This one is to micro USB or USB-A, but I don't really care because it has a cool charging dock. But it's important to note that a lot of reviewers say the Viper Ultimate's charging dock breaks down. Even my friend RJ, his Viper Ultimate's dock broke down. What happened? He was aiming down sight and he headshot. Oh, that was a nice shot. I was jumping. I'm good with mid-range. What's your mid-range? I don't even know if it's a mouse that costs $50,000. It's wild. You know, but one thing that I'm really happy with is my stream deck. This is aged very well. The Elgato stream deck... I have a hard time using my computer without it. But I got the 3D stream line so many things together. I'll start in the area. Counting on you. Focus on the next round. The best for sound effects. That's what happens when you have poor morale. Just have to run. Come on, let's finish this game. I've lost gunfights before. Oh, man. This is serious. Let's win this and go home.  This is match point. The goal is to keep using the same hardware for at least two years. I'm already impressed with myself. GG, guys. $20. Viper wired. With my two-year old... Hey, Liam. How's the rubber grip there? Where's your Viper? Is it still attached? Is it already peeled off? Or what? How's that? That's what I'm really concerned about. I told myself because I wouldn't buy a... I wouldn't buy a razor ever again, but... Well, I've decided... Well, I avoided them for five years. I think it's about time that I gave them a second chance. One thing you guys should always look out for also is the feet. The rubber feet. They also get really fudged up. They also get fudged up. There we go. This is what it's like new. I'll show you what happened to my iron claw. There we go. You can see that deformity there. There, that deformity. It's like it melted, man. It's like it melted. It's also very rough to the point that not only is the iron claw heavy, but it's no longer gliding on the surface of the mouse. This really has super melted down. I don't know what happened here. I guess overused, but I wasn't even overusing it that much. I play FPS, but not intense. Anyway, I wish there were more videos about how to replace... how to replace... mouse feet. That's what they call it. I would totally just like to keep on customizing. You know, but it was the ROG. They have replaceable switches. Which comes with the mouse that they give. And by the way, I was actually planning to get the ROG Chakram. Chakram, the wireless version. But then our supplier was sold out. Maybe it's a mouse pad. I used to have the Razer Heavy Drutey mouse pad. The... Ballistic Blade Starcraft 2. Any... I'm not... I don't think so. It's in terms of shape. I don't think Logitech has a similar one to that. I didn't even aim down sight. Look at this. It's skull just took my whole shell. Okay. Okay. Everywhere else. Logitech and Razer have a similar mouse. But it's not the Death Adder one. It's the one for right-handed people. That's the one thing that... You know what to be honest. As much as I hated Razer over the years, I really do find their mice quite comfortable. My only problem with Razer was that they just hit it fast. I'll show you guys my onion, my script, my mouse. Actually, I don't think Anton knows that I bought this one. He's also an anti-Razer. But my keyboard, by the way, is still the Corsair K70. So far, the K70 is aged very well. Nothing is melting. The RGB is cool. I don't really play around with the RGB all that much. Okay, two more. Two more. Hey! With a P50 SMG. P90, sorry. What happened? Did you just commit suicide? Oh, nice throw! Oh, nice throw! Look, my ally threw a bolt of cocktail. Okay, nice. I like to P90. It's really a cool-looking SMG. It has a lot of ammo.  It's also really satisfying. Get two kills, but with the same clip. Okay, though. I totally dig pistol fights, man. I love pistol fights.  Cocktail. Okay, who are we rooting for here? My ally is really good. Okay, he's gonna fight the guy who's asked 3-6. My ally is 6-5. I don't know who's worse. Reload, man. Michael Espinosa, I thought you were testing out a new rig for a second there. Nope! Same rig. My rig, by the way, is August 2019. So... Hi, guys, sir! 17XD. What's going on? Nice seeing you again in the chat. Actually, I started this stream a little bit early. Usually, I mean, I game at night, because I actually have to take my dog out to walk, because she's getting a bit obese. Not a lot of obese, but she's a little bit overweight. And she's already 15 years old, and so she has arthritis. So I think I'll... But the important thing is she needs to also get exercise, otherwise she'll get really fat. So I want to allocate some time for my dog, Vega, and I to go to the park first, so she can exercise. When you have an old dog who has arthritis, and then the paws are quivering, she's having a hard time standing up, but then she doesn't make noise yet. It's so depressing. So I told myself that I would spend the night with Vega tonight. So thanks for joining our stream here. And I'm dropping a video... Actually, I dropped the video already. I dropped a top 7 budget tech accessories, which I use, which are below 200 pesos. And I use them every... I've been using them for the past 2 years, and actually if you've been watching my videos, you know these accessories already, because I've shown them before, but I've never actually made a compilation video, which is what I did. However, I mistakenly released it a day earlier than I should have. So it'll be dropping down tonight at midnight. Check it out tomorrow. I think you guys would really like this. And I highly recommend that you give them to your tech friends. So it's below 200 bucks, all these items. Some of them are 65 pesos long. But they're super handy. Super super duper handy. And you know, if you're in for Christmas shopping, these are really really practical things to give people. So what's your... What to get your tech Tito? Yeah. Yeah, as a tech Tito, totally. Okay guys, I'll see you in the next one. And as I was watching that video, I just wanted to... Ah, you watched it. Yeah, it's in the members. So for all you guys, thank you for supporting the channel guys, the members. Thanks for watching the video. Yes, so you've already seen. You've already seen what's available. I have the links there in the description. Enjoy and let me know. Let me know if people you're gifting it to enjoy it as much as I do. I'll stay safe everyone. And I hope everyone... I hope everyone fulfills everything they need to do for whatever got broken during the storm. Even here in the household, there's a lot happening. But you know, we have it better off than most. So my prayers go out to everybody who was affected by the storm. Also a lot. I think more people died in Seoul in that stampede than in the entire Philippines during this latest typhoon. Maybe wrong, but like a hundred something died just yesterday partying. Man, that's just... Gabien in E-T1. I was really depressing to hear that. That's something that could have been completely avoided. GG guys, stay safe.


Hey everyone, welcome to another video lesson from navigationtrading.com. In this video, I wanna show you how to calculate your profit or loss if you get assigned on a position. So last week on 418, we had a short call vertical spread in DIA, we held the trade all the way up till the last trading day before expiration and it was well in the money and so we ended up getting assigned. Now this doesn't happen very often. In fact, this was the first assignment we've had in our trade alerts portfolio. So it's been over two years, but assignment does happen and so if you ever do get assigned, I want you to be able to refer back to this video to kinda help walk you through so you can calculate the profit or loss.  So that's exactly what we're gonna do here. There are just a few little things that you need to be aware of because it's a little bit different when you get assigned.  But the first thing that we did on 3-8, we rolled a short call vertical when we simply just rolled our strikes down, same expiration cycle and so don't pay attention to this overall credit of $1.64 here because we already counted for that on our trade tracker. All we're gonna focus on is the position that we entered on that date. It's just the 255-260 call spread, okay? So the ones we sold, we collected 415, the ones we were long, net of two, so the net between those two is $2.15, okay? So on our initial credit, we got 215 times three contracts equals 6.45 and then times 100 because each contract controls 100 shares of stock, so that equals $645 because we want to convert all of these into a specific dollar amount because we were actually assigned stock or in this case, short stock, okay? So our initial credit is $645 and then we were assigned on the 255 calls, okay?  That's the actual dollar amount that we got. You can see that right here, okay? So 76,500 for that assignment of a short 300 DIA calls. And so we were short 300 DIA calls, okay? So that's a credit of 76,500. Then we also had the remaining 260 long calls, okay? Which we ended up selling for $5.55. So we take the 260, 260 long calls and we actually, so we sold them for $5.55, okay? Times three equals 1,000, make sure I got my math in my head correct. Let's just do it here, 5.55 times three equals 1665 times 100, so that equals $1,665, right? So these are all the credits that we received, okay? So let's add these together. So we've got $645 plus 76,500 plus 1,665 equals, okay? So our total credits, now that we've taken in equal 78,810. Now, what we did is we bought back our one debit in this transaction is we bought back those 300 short shares. So if we take 300, it's actually minus 300, just to be correct, shares, that equals, and you can see right here on the platform, we bought those back and my order actually got filled in two different orders. So we bought 100 and we bought 200, got filled at the exact same price, it just filled us separately. And so for a total of, so we gotta add those together. So we've got 26,560.9 plus, 53,121.8 equals, okay? So that's 79,000, 79,682.7, that's a negative. Okay, so that's what we bought those back for.  So let's do this, 78,810 minus 79,682.70, and that equals 872.70. So our total loss on that trade was $872.70. Okay, so that's how you do it. It's just like anything else, you gotta take all the credits, you get to subtract the debit, and so we took a loss on that trade. We ended up just closing it out because we didn't want to carry short shares in our account. And so that is how you calculate it. I hope that's helpful. If you have any questions, you wanna see what we're doing here at Navigation Trading, just go to navigationtrading.com. See you there.


So thank you all for coming, and I'd like to thank the organizers for giving me a chance to speak here. So this is joint work with Joav Len who spoke yesterday, and in the sense my talk is a continuation of his talk, but I don't want to assume that you're there for his talk yesterday. So I just want to begin with a few basic definitions. So tropical geometry is a kind of combinatorial version of algebraic geometry or more accurately a polyhedral version. And I don't want to get too much into the details, I just want to talk about a few examples today. So the analogs of algebraic curves in tropical geometry are graphs. So with or without a metric. So let's first talk about ordinary graphs. So a graph can have loops, it can have multiple edges between a pair of vertices. So combinatorials would call this multigraph. And the genus of a graph, it's simply its first betting number, so it's edges minus vertices plus one. So if it's a planar graph, it's the number of faces. And on a divisor on a graph is just an element of the free of viewing group generated by the vertices of the graph. And so the degree of the divisor is the sum of the coefficients. And so one of the starting points for tropical algebraic geometry was the realization that came in the first in the combinatorics community and then was taken up by algebraic geometers that divisor theory on graphs behaves in a way which is remarkably similar to divisor theory on a Riemann surface. So there is a Riemann rock theorem and so on, but let's first talk about what linear equivalence is. So linear equivalence for divisors on a graph is generated by what's called chip firing operation. So a chip firing operation is kind of an elementary principle divisor. So what you do is you fix the vertex on your graph. So this I'm fixing the red one here.  So just kind of a positive integer from this vertex to all neighboring vertices. And if there are multiple edges between a pair of vertices, then I move that many chips. So the divisor obtained by firing this vertex V is going to be one chip at this neighboring vertex here, two chips at this vertex here because there are two edges connecting. And then however many chips I give away, I put the negative of that number at the red vertex. So clearly the degree is equal to zero. And so the subgroup of the divisor group spanned by these chip firing devices is what's called the group of principle divisors. And then you do the same thing as normally in algebraic geometry. The Picard group is the group of divisors module of the principle divisors. Then degree is well-defined on the Picard group. So the Jacobian group is the degree zero Picard group. And there is the standard exact sequence. OK. And so a theorem which is really quite kind of impossible to attribute because it came together as kind of grew as an understanding. Not wasn't just someone proving this is that the number of that first of all, the Jacobian group of a graph is a finite Abelian group. And the number of elements in the group is the number of spanning trees of your graph. So for example, this graph you can count has five spanning trees. And in fact, the Jacobian is a cyclic group of four to five. And so, for example, the Jacobian of a graph is trivial, if and only if the graph is a tree itself, that is to say if and only if it is genus zero. So that's the tropical version of the statement that the Jacobian of a Riemann surface is trivial, if and only if it's a P1, if it has genus zero. OK. So now let's talk a little bit about metric graphs. So in a metric graph, what I'm going to do is I'm going to take a certain underlying combinatorial graph, but now I'm going to put edge lengths along all of its edges. And when I look at divisors on a metric graph, I'm now allowed to put, I treat the graph as a metric space in an obvious way, and I'm allowed to put points on the interiors of edges. So there's now going to be infinitely many points on the graph. And so a divisor on such a graph is simply the an element of the free abelian group generated by its points. So what is the chip firing operation on a metric graph? Well, the idea is the same. You fix a point so it can be a vertex, but it can also be a point in the interior of an edge. And then in every direction from that point, you move a chip and you move them all a certain equal distance, which you can also select. So, for example, if you put a vertex, if this is your chosen vertex, you can find some small distance epsilon and then just move one chip a distance of epsilon in this direction, this direction and this direction. And then you get some divisor, which also has degree zero. And so you take the subgroup of the divisor group, which is spanned by divisors of this kind. So that's going to be the group of principal divisors. And then you do the same thing. You define the precar group as the set of divisors module of principal divisors. You might want to look at the divisors of degree zero. So that gives the Jacobian group. And what's the difference now? Well, so the way to think about this is that the Jacobian of a graph is a compact group over z. So that's a finite group. The Jacobian of a metric graph is a compact group over r. So that is to say it's a real force. And in fact, it can be identified with h1 coefficients in r mod h1 coefficients in z. So it's a real torus of dimension g. And in fact, it has much more structure than I'm able to describe today. It's what's called a tropical principally polarized abelian variety. It has something called an integral structure. It has a dual torus to which it is isomorphic in some sense. So there are quite a few details that unfortunately I'll have to state. So now let's talk about our Kirchhoff's theorem for graphs. So Kirchhoff's theorem for an ordinary graph says that the number of elements in the Jacobian is the number of spanning trees, which is to say it's the number of subsets of the edge set with a property that the complement is a tree. Now it's easy to see that if you have a graph of genus g, then the number of edges you need to remove to obtain a tree is exactly g. So here we're summing over certain g element subsets of the edge set of the graph. Now, the Jacobian is a torus. And the fact that it's a tropical PPV in particular means that it has a certain product on it. So it's a Riemannian manifold. So you can ask what is the volume of this Riemannian manifold? And so the answer is given by this paper on Baker-Couperbergen-Schokere, which is that here's what you do. You just take this sum where you sum over all g element subsets of your graph, such that the complement is a tree. But instead of taking a one for each such subset, what you do is you just take the product of the edge length that you're removing. So for every g edges that form such that removing them forms give you a spanning tree, you take a product of those g edges and then you add up all of these volumes that you get. So it would seem for dimension reasons that the volume of the Jacobian should be the sum. But there's kind of a quirk here that when you translate the metric geometry of the graph into Riemannian geometry on your Jacobian, the dimensions all double. So in some sense, the unit of length on a graph is meters squared and not meters. So, for example, if you have a graph, which is a circle of length L, then the volume of that circle viewed as its own Jacobian is actually going to be square root L and not L. That's just kind of a slightly unusual circumstance. OK, so now I'm telling you that the volume squared of the Jacobian is a sum of certain sets, right? And so now think about it this way. If I tell you that a certain polygon has a perimeter A plus B plus C, you may be tempted to ask, well, maybe this polygon is a triangle and A, B and C are the lengths of its sides. So the question that naturally arises is, is there some kind of way of representing the Jacobian of a metric graph, cutting it up into pieces such that this decomposition of its volume as a sum represents just simply this decomposition of the Jacobian into pieces. And so this is what's called the A, B, K, S decomposition of the tropical Jacobian. So here's how it works. So what we're going to do is consider the tropical apple Jacobi map. So you're mapping from the g-th symmetric power of a curve to the g-th symmetric to the pic g. And it's exactly the analog of the algebraic one where you're just taking g points and you look at the line bundle of the generate, which I'm just identifying with a corresponding divisor. Now, in the algebraic setting, this map has degree one, right? It's onto and if you want to construct the pre-image map, so the locus of indeterminacy is precisely the theta divisor, so it's the set of line bundles which have a non-trivial section. OK, so what we want to do is we want to understand the structure of the tropical apple Jacobi map. So here's how the structure works. So if you have a metric graph, then the symmetric product carries a natural cellular decomposition in the following way. So if you have a set of points, a set of g-points on your tropical curve, you just look at the edges on which those points lie, right? And so each point is going to lie in a certain cell. Some of them might lie in several cells, which are indexed by g-tuples of edges of your graph. So if you have a graph and you're looking at SIM 3 of it, well, you can just say, OK, well, let me pick any three edges on the graph. I could pick the same edge twice. And then I'm just going to allow points to move along those specific edges. And that gives me a cell in the symmetric product. OK, and so now the main idea is that the cells in the symmetric product come in two flavors. So I'm going to say that C of f is a break cell. If the edges, along which I am moving the points of my divisor, have the property that their complement is a spanning tree of my original graph. So that's going to be called a break cell. And all other cells I'm really not going to care about. OK, and so what the ABKSD composition looks like before I state the theorem. Let me just show you the picture. What ABKSD proved is that the apple Jacobium apple looks kind of like this. So this is SIMG of your curve. This is PICG. So PICG is a torus. So locally, you can just think of it as Euclidean space. It's just this nice manifold. SIMG is what's called a polyhedral complex. So it's something that's put together from from these blocks. And then the break cells evenly cover the in SIMG evenly cover PICG. Whereas the other cells that are not break cells are kind of sticking out and are contracted by this map. So you can think of the other cells as being kind of the tropical version of all of those line bundles, that all of those g tuples of points, which have a non trivial section. So this is the this is kind of the the locus that you're contracting. And the break cells evenly cover PICG. OK, so in algebraic geometry, if you have a map of degree one, which is, you know, the blow up of something that, of course, that map does not have a section, right? Because if you try to construct the inverse map, there's an indeterminacy locus and tropical geometry that actually doesn't take place. Whenever you have kind of a map of degree one, you can just take due to simple continuity reasons, you can take a section of this map. And so this is the this is the exact statement of the results of the ABKS, which, in fact, were actually proved earlier by Mikhail Kanan-Jakov in 2008, is that the tropical Able-Jakovian map has a unique continuous section. So if you have an element of PICG on on a graph, every element has a unique preferred representative as an actual sum of points, which is certainly not the case in algebraic geometry. And so what they also proved is that this decomposition of, so this map has a section and the image of the section is precisely the union of the break cells. And so the volume formula has a geometric interpretation in the sense that you can try to find the volume of the Jacobian, which is, of course, the volume of PICG because they're the same thing. So just one of them is a translate of the other by summing together the volumes of these cells. And so what they proved is that if you take a break cell here, then the volume of its image here is just the product of the edge length divided by a factor, which happens to be the volume of the entire Jacobian. Whereas if you have a cell which is not a break cell, then it gets contracted by the map, so the volume of its image is zero. And so then you can find the volume of the Jacobian by just adding together the volumes of all the break cells. And so what you get is this weighted sum. And then it comes with a coefficient, which happens to be the reciprocal of the volume that you're trying to find. And so then if you multiply out by this volume, you get volume squared, isn't this sum? Okay, so these are the results of ABKS. And so what I wanna talk about today is a version of all these results for the tropical print array. Okay, so first of all, let's talk about double covers. So you all have mentioned them in his talk yesterday. So a double cover of a graphs, for me would just be the naive notion you just treat the graph as a topological space and you look at a covering space of degree two. So any such cover is yellow, of course. And if I'm looking at double covers of metric graphs, I'm also going to require that my map be a local isometry. So all of the edge lengths in this graph are induced from edge lengths in this graph. Okay, and so it's an easy exercise to see is that if the target graph has genus G, then the source graph has genus two G minus one. So the numbers work out exactly as they do in the algebraic setting. Okay, so before we talk about metric graphs, let's just try to talk about ordinary graphs. If you have a double cover of finite graphs, then these two graphs have Jacobian groups, which are finite groups. And there's a natural way to define the push forward or the norm map from the Jacobians, which is essentially just take any point and map it downstairs. So take any divisor and map it downstairs. Okay, and so a theorem that Joab and I discovered last year, but which turns out to have been known in other language by combinatorialists, is that, well, we want to say in the algebraic setting, the prime variety has a, the kernel of the push forward map when you have a tall level cover of curves has two connected components. And then the connected component of the identity is the prime variety. Now these are finite groups. So it doesn't really make sense to talk about connected component of the identity. So instead, what you have to do is introduce a certain parity on this kernel. And then the prim group is just the even subgroup of the kernel corresponding to this parity. And then you could say, okay, well, I've defined a certain prim group for a double cover of finite graphs. So what is the order of this prim group? Okay, and so this is the formula that we found that as I wanna emphasize was actually known before. By the way, if you ever prove a, if you ever write a paper with a combinatorics result that you think is new, when you post that paper, make sure to cross-list it on math.co. That will save you some possible embarrassment later. Anyway, so literally the next day after we posted it, the criner from, I think he's from University of Minnesota pointed out that this result is not. Okay, so what we want to do is we want some kind of version of what it means to count spanning trees, except that now instead of having a single graph and we're trying to find the order of its Jacobian, we're trying to find the order of the prim groups of the kernel of this F of a double cover. Okay, and so here's what you need to do. You need to count the number of odd genus one decompositions of your graph G. These decompositions are stratified by rank. And so each, a odd genus one decomposition of rank K comes with a coefficient of four to the K minus one. So let me explain what these odd genus one decompositions are. Okay, so first of all, how am I going to describe a double cover? Well, so this, if you just kind of give this as an extended exercise in your algebraic topology class, I'm sure the students would come up with something like this. So if you have a double cover of graphs, what you first do is you choose a spanning tree for your target graph. So this will be in thick black. And then it's pre-image is a pair of trees. It's a spanning course for the source graph. Okay, and so then all of the complimentary edges come in two flavors. There is, there are edges with the property that their pre-images are somehow trivial in the sense that, for example, this black edge here, its pre-images connect the same vertices on the same tree as they do down here. And then there are the blue edges whose pre-images are flipped. So for example, this vertex is connected to this vertex via this blue edge on this graph here. This vertex is connected to the corresponding vertex on the other tree via the blue edge, right? So essentially to specify your graph, you just have to specify your spanning tree in thick black and then say which of the remaining edges are you going to color blue and which of them are going to be thin black edges? Okay, and so there's a natural way of identifying the set of choices you could make this way with H upper one of your graph with coefficients in Z mod two. Okay, so now here's what I'm going to do. I'm going to say, okay, now I have a graph of genus G and genus one decomposition of this graph is a way of removing G minus one edges from the graph. And so here's how this is going to work. So if you have your graph here, this graph has genus three, let's try to think what can happen if I remove two edges on the graph. And if you play around with it a little bit, you'll see that either one of two things happen. If you remove two edges from the graph, either you have one or possibly several graphs of genus one. So either your graph falls apart into a bunch of cycles or it doesn't or there's going to be a component of genus two and maybe a component of genus zero. And so a genus one decomposition of the graph is a way of removing G minus one edges from the graph such that what remains is a collection of cycles. And the rank of this decomposition is simply the number of cycles that you get this way. So removing this edge here and removing this loop is a genus one rank one decomposition. Removing this edge, this bridge and this edge here, genus one rank two, this is not a genus one decomposition. Okay, so that's what a genus one decomposition is. And I'm going to say that a genus one decomposition is odd and only after we came up with this terminology we realized that that shortens to O God. So an O God is a genus one decomposition with the following properties. So remember that all of we have a curve with some marked edges in it and that curve has a double cover, right? So now imagine removing a pair of edges from this graph and let's also remove their pre-images here. And so now your double cover is restricted to this collection of cycles that remains. Now it's a topological exercise that if you have a cycle and you have a degree two cover of it well either that cover is trivial or it's not, right? So the A because there are two representations of I of Z into Z mod two. And so I'm going to say that a genus one decomposition is odd if the restriction of this cover to each of these genus one components that you get is a non-trivial cover. And when is it a non-trivial cover? Well, the answer is that if there's precisely an odd number of edges in each cycle that remains. So for example, here, there's an odd number of edges in this cycle. Here there's an odd number of edges here and odd number of blue, sorry, blue edges here that we're counting. This is not odd because there's an even number of blue edges along this cycle. And so if you look at the covers that you get from these three decompositions, the way you can tell is that if your number, if you have an O God, if you have an odd number of edges in your cycle, then the pre-image of the corresponding component, each of these components is going to be connected. Whereas if you have a genus one decomposition that's not odd, then this cycle here, for example, is going to have two pre-images. So this is the trivial cover. Okay, so an odd genus one decomposition is a way of removing two cycles, G minus, sorry, is a way of removing G minus one edges such that the restricted cover is non-trivial on each connected component. Okay, and so we can count the genus one decompositions for this cover that we have here. So it turns out that there are eight of them that are odd of genus of rank one and four that are odd of rank two. So C one is going to be eight, C two is four. And so I predict that the order of the perm group is eight plus four times four, so that's 24. And so you can go back and try to count the spanning trees. This one we counted has five spanning trees. This one, if you want to count, it has 240 spanning trees. And so you take the ratio of these and you divide by two because remember the perm group is only half of the kernel. And so you get the predicted result, the number of elements in the perm group is 24. Okay, so it's so far so simple. And like I said, this is a result that was already known in not in the modern language, but it's already understood about 40 years ago. And so now what I want to do is I want to say, okay, so let's look at the tropical prim variety. So now if I have a double cover of metric graphs, then you can try to construct the prim variety of this in the same way that you do an algebraic geometry, which is that you look at the push forward map, the norm map on the Jacobians, you look at its kernel and you try to understand what that kernel looks like. And so there are two papers about this by Jensen and Lennon-Olerge that, so together they proved the following that essentially the same holds true as, the same statements hold true as in the algebraic setting, which is that the kernel of this map has two connected components. The connected component of the identity is a tropical principally polarized ability variety called the tropical prim variety. And in fact, the principal polarization on it is exactly half the principal polarization that's induced from the ambient Jacobian. And everything tropicalizes nicely, so I'm not really gonna have time to talk about the relationship to algebraic geometry, but essentially if you have a family of curves, so a curve over a non-archimedean field, then you can look at what's called the Berkovich analytification of your curve, look at the Berkovich analytification of the Jacobian. And then if you look at the skeleton of all that, then what you get is the tropical prim variety. So there is a direct relationship to algebraic geometry, which unfortunately I don't have time to explore. Okay, and so what I want to do now in my remaining few minutes is to just talk about a version of the ABKSD composition for the tropical prim variety. So first of all, is there a formula for the volume of the tropical prim variety? And so the answer is yes, and it's very similar to the ABKS formula. So the order of the discrete prim group is a sum over all o-gods of your graph, of your target graph, where each o-god is weighted by a factor which is four to the power of rank minus one. And so what we proved is that if you want to find the volume of the tropical prim variety of a double cover of metric graphs, then you just take the same formula except that you weigh it by a term which is volume F, which is just the product of the edge length that you're removing every time you have an o-god. And so again, there is this strange situation that the units don't really match up in some sense that what you get is a volume squared, not the volume. Okay, and so again, the question is, so I have proven a formula for the volume of the tropical prim variety. Is there some kind of way of taking your tropical prim variety and cutting it up into pieces so that the volumes of the individual pieces correspond to the terms in this sum? So the answer is yes, but the way to do this is more complicated than what you do with the Jacobian, as you would expect. And the idea is that instead of looking at the ABL Jacobian map, I'm gonna look at what's called the ABL prim map. So I'm going to look at the G minus first symmetric power of the top curve. And then I'm going to map it to the prim variety by just taking some divisor and mapping it to D minus the action of the involution on this divisor. And so to be extra careful here, you don't exactly map to the prim. You map to the connected component of the identity of the prim, of the kernel of the norm map, which has the same parity as G minus one, right? So for example, if you take a point minus the involution of that point that always lands you in the odd component of the of the prim, not the unit component. Okay, so there was a four-chlor result, which is that in the algebraic setting, the ABL prim map has degree two to the G minus one. And so we're very thankful to Sebastian Castellana-Martin who took the time to write up an 11 page appendix to our paper where he proves this and a number of other useful statements about the prim variety. So I encourage you to read our paper if only to look at his appendix because I think it's marvelously well-written. Okay, and so our main result, which I'm completely out of time now is that the tropical prim map is what's called a harmonic morphism of polyhedral complexes of degree two to the G minus one. And so perhaps instead of trying to explain what this is, I'm just going to show you a picture. So remember in the four Jacobians, the picture of what the ABL Jacobian maps look like is this. So you have cells corresponding to break divisors and you have all the other cells and these cells evenly cover the car group and all of the other cells are contracted. So in our case, the picture is a little bit more complicated but it's also kind of reasonable to understand is that there are cells corresponding to ogods and those cells cover your prim except that now you might have several cells that are ogods in the symmetric product over any particular cell of the prim variety. But if you count the number of cells that you have over any particular point, you're going to always get two to the G minus one. So there's a way of defining degree for these cells that are not contracted and the degrees are always powers of two in the same way in the way that this degree is locally constant in some sense. So if you have a cell, if you have two cells of degree two to the K meaning along a co-dimension one cell, then if there's another one here then the total degree here is the total degree here. And so the sum of all of the degrees and the pre-image is two to the G minus one. So well, unfortunately, there's no short way of saying this but just say I have completely out of time that if you have a element of the prim group then it has exactly two to the G minus one representatives of the form D minus Yoda of D where you count those representatives with certain specified weights. And so this is the tropical version of the statement that the ABLA prim map has degree two to the G minus one. So I think I'm going to stop here. So thank you very much. Thank you very much. So is there any question, Mark? I have a question which is maybe a bit philosophical. So do you expect that there is a way to connect the two results so that it's not a coincidence that the both degrees are the same but that in principle one could be used to prove the other. So that would of course be the ideal way to prove this result. You would say that first you prove that this degree, this map has degree two to the G minus one in the algebraic setting then degree is preserved by tropicalization in some sense. And therefore this result also holds in the tropical setting. We tried to do this and the machinery has not yet been developed for it. So we think that this can be done. In some sense, this is above my pay grade. So I think somebody else, not me, but yeah. So this is certainly something that's plausible. I've talked to a few people experts and they said that yeah, this is something that sounds feasible but our proof is entirely combinatorial. You essentially just count how many pre-images you have and then as you move around you make sure that everything glues together nicely. So yes, there should be some general theorem from which this follows but the techniques haven't been developed yet. Thank you. Other questions? I have only one curiosity. So do you think that it would make sense to consider also ramified double covers? Like not different? Yeah, so first of all, let me just say a few things. Before you consider ramified double covers you want to consider all un-ramified double covers. And in tropical geometry, un-ramified is a more general thing than being a free cover. So there's a collection of covers where you can have what you naively would think would be called ramification where an edge might have one pre-image or a vertex might have one pre-image instead of two but some of those covers are un-ramified in the tropical setting because essentially they are the tropicalizations of a tall double covers of curves. And so this theorem is actually true. I even erased a comment because I thought I would extra this. This theorem is also true for these un-ramified double covers except that the kernel that is now has a single connected component and the principle polarization. There is a principle polarization on it but it's not the restriction. It's not half of the restriction of the ambient one. And so we believe that all of these results also hold for the prim variety of un-ramified double covers of curves which just haven't worked that out yet. And then if you have ramification, yeah, the natural question is, let's say you have a cover of tropical curves which is ramified at exactly two points. Is that still a tropical prim variety? We don't know. I mean, I didn't make a list but there are exactly three papers about the tropical prim variety at this point. These two and the ones that I'm talking about. So there's plenty of low hanging fruit and a lot of work that remains to be done. But yeah, so looking at that's certainly a nice, I don't know, master's thesis program. Look at ramified covers and see if you can prove something about their tropical prim variety. Okay, thank you very much. Other questions? Marks, observations. Okay, so if not, we thank again Professor Zaharov and we go further on. So I ask the next speaker to prepare.