Does the Backlog Fit the Contact Window?
Does the Backlog Fit the Contact Window?
Ada re-derives this chapter’s own numbers step by step, at full precision
ADA · CALCULATION AUDIT
Does the Backlog Fit the Contact Window?
Two elephant collars brush past each other for just 30 seconds, and at 250 kbps that contact can move 937.5 KB — yet the backlog waiting to cross is only 250 KB, a collar’s own 50 KB five-day log plus about 200 KB of relayed data. The whole store-carry-forward promise rests on that one inequality. This audit re-derives the capacity, payload, and transfer time and asks whether the backlog really fits the contact window, and by how safe a margin.
Companion to the chapter Real-World DTN Deployments — every number here comes from that chapter.
See the relationship before changing it
The figure reads from left to right. The blue card is contact duration. The middle card applies this page's rule. The green card is transfer capacity. Walk the arrows once: set the input, apply the rule, then read the result with its unit.
The retained audit below checks several chapter fixtures. This model keeps those stated values fixed and changes only contact duration, so the numeric fixture does not switch without explanation.
Derive the baseline in four named moves
- 1
Name the input. The chapter baseline is 30 s.
- 2
Name the relationship. capacity = 250 kbit/s / 8 x contact seconds
- 3
Substitute with units. 250 / 8 x 30 = 937.5 KB
- 4
Read the result. Keep the unit beside the value. Use it only inside the technical boundary on this page.
Predict, then change contact duration
Try Predict the direction of capacity = 250 kbit/s / 8 x contact seconds. Test another contact duration, then compare transfer capacity.
Observe A longer contact opens a larger transfer budget, but protocol overhead still reduces usable payload. Reset contact duration to 30 and compare transfer capacity.
Explain A longer contact opens a larger transfer budget, but protocol overhead still reduces usable payload.
Check yourself
What should you do before trusting a moved-control result?
What does this small model leave out?
Ada: The whole store-carry-forward promise here rests on one inequality: the buffered payload must fit inside a brief encounter. This chapter’s contact-window rule is C = R x t, and it claims a 30-second collar meeting at 250 kbps moves 937.5 KB while the payload needs only 250 KB. Let me check both, using the chapter’s decimal convention throughout (1 kbps = 1000 bit/s, 1 KB = 1000 bytes).
- Contact capacity:
250,000 bit/s x 30 s = 7,500,000 bits = 937,500 bytes = 937.5 KB. - Payload to move: own five-day backlog
10 KB/day x 5 = 50 KB, plus about 200 KB of relayed data, so50 + 200 = 250 KB. - Transfer time:
250 KB x 8 bit/byte / 250,000 bit/s = 2,000,000 / 250,000 = 8.0 s. - Contact margin:
30 - 8 = 22 sleft for setup, handshake, and retries. - Window used:
250 / 937.5 = 0.2667 = 26.7%of the capacity a full 30-second contact offers. - Short-contact stress test: even a 10-second brush carries
250,000 x 10 / 8 = 312,500 bytes = 312.5 KB, still clearing the 250 KB payload.
The audit confirms the design claim and shows why it is robust rather than lucky: the backlog consumes about a quarter of the window, so a contact cut to a third of its planned length still delivers everything. In DTN sizing the margin, not the headline capacity, is what turns a chance encounter into a dependable delivery opportunity.
Every number above is taken from the chapter’s own material and re-derived step by step.
Audit result
The payload uses 26.7% of ideal contact capacity; 22 s is available for all omitted overhead and impairment.