\u00a0In the original TCP protocol, MSL (maximum segment lifetime) is 120 seconds and sequence number space is 32-bit in size. (a) What is the maximum allowable window size in bytes so no wrap around occur? (b) Is the window size found in part (a) allowable if the bandwidth used is 45 Mbps {hint find how long it would take to transmit the window find in (a)?} (c) Repeat part (b) if BW=2.4Gbps. In this case is wraparound an issue?<\/p>\n\n\n\n
The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n (a) Maximum Allowable Window Size:<\/strong> Given that the sequence number space is 2322^{32}<\/strong> and MSL is 120 seconds<\/strong>, the transmission rate must be slow enough that data does not exceed the sequence number range within MSL<\/strong>.<\/p>\n\n\n\n The maximum sequence number space is: 232=4,294,967,296 bytes2^{32} = 4,294,967,296 \\text{ bytes}<\/p>\n\n\n\n Thus, the maximum allowable window size<\/strong>: W\u22644,294,967,296 bytes=4GBW \\leq 4,294,967,296 \\text{ bytes} = 4GB<\/p>\n\n\n\n (b) Check Allowability with Bandwidth = 45 Mbps:<\/strong> Since 127.33 minutes<\/strong> is far greater<\/strong> than MSL (120s)<\/strong>, the window size is too large<\/strong> to prevent wraparound. Hence, this window is not allowable<\/strong>.<\/p>\n\n\n\n (c) Check Allowability with Bandwidth = 2.4 Gbps:<\/strong> Time=4,294,967,296\u00d782.4\u00d7109=14.32 seconds\\text{Time} = \\frac{4,294,967,296 \\times 8}{2.4 \\times 10^9} = 14.32 \\text{ seconds}<\/p>\n\n\n\n Since 14.32s<\/strong> is less than<\/strong> MSL (120s), this window is allowable<\/strong>.<\/p>\n\n\n\n Is Wraparound an Issue?<\/strong> In TCP, the Maximum Segment Lifetime (MSL)<\/strong> defines the time a segment can exist in the network before being discarded. The sequence number space<\/strong> is finite (32-bit, meaning 2322^{32} unique values), requiring careful window size control to prevent wraparound.<\/p>\n\n\n\n For no wraparound<\/strong>, the sender must not send more data than the available sequence numbers within MSL. Thus, the maximum allowable window size<\/strong> is the entire sequence number space (4GB).<\/p>\n\n\n\n When the bandwidth is 45 Mbps<\/strong>, sending 4GB would take approximately 127.33 minutes<\/strong>, far exceeding MSL (120s). This causes wraparound<\/strong>, making the window too large<\/strong>.<\/p>\n\n\n\n With a higher bandwidth (2.4 Gbps)<\/strong>, transmission completes in 14.32s<\/strong>, well within MSL. Thus, the window size is valid<\/strong> and wraparound is not a concern<\/strong>.<\/p>\n\n\n\n Thus, TCP’s flow control and sequence space constraints<\/strong> demand adjusting window sizes based on bandwidth<\/strong> to avoid data corruption due to sequence number reuse<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" \u00a0In the original TCP protocol, MSL (maximum segment lifetime) is 120 seconds and sequence number space is 32-bit in size. (a) What is the maximum allowable window size in bytes so no wrap around occur? (b) Is the window size found in part (a) allowable if the bandwidth used is 45 Mbps {hint find how […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-199105","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199105","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=199105"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199105\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199105"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199105"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199105"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Answers:<\/strong><\/h3>\n\n\n\n
To prevent sequence number wraparound, the window size WW must satisfy: W\u2264MSL\u00d7Bandwidth8W \\leq \\frac{MSL \\times Bandwidth}{8}<\/p>\n\n\n\n
\n\n\n\n
First, calculate how long it takes to transmit 4GB<\/strong> with 45 Mbps<\/strong>: Time=4,294,967,296\u00d7845\u00d7106=7639.93 seconds\u2248127.33 minutes\\text{Time} = \\frac{4,294,967,296 \\times 8}{45 \\times 10^6} = 7639.93 \\text{ seconds} \\approx 127.33 \\text{ minutes}<\/p>\n\n\n\n
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No, because the data transmission is completed in 14.32s<\/strong>, much shorter than MSL (120s)<\/strong>, ensuring sequence numbers do not overlap.<\/p>\n\n\n\n
\n\n\n\nExplanation (300 words)<\/strong><\/h3>\n\n\n\n